introduction to modern physics-r.b.singh

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Copyright © 2009, 2002, New Age International (P) Ltd., PublishersPublished by New Age International (P) Ltd., Publishers

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PREFACE TO THE SECOND EDITION

The standard undergraduate programme in physics of all Indian Universities includes courses onSpecial Theory of Relativity, Quantum Mechanics, Statistical Mechanics, Atomic and MolecularSpectroscopy, Solid State Physics, Semiconductor Physics and Nuclear Physics. To provide study materialon such diverse topics is obviously a difficult task partly because of the huge amount of material andpartly because of the different nature of concepts used in these branches of physics. This book comprisesof self-contained study materials on Special Theory of Relativity, Quantum Mechanics, StatisticalMechanics, Atomic and Molecular Spectroscopy. In this book the author has made a modest attempt toprovide standard material to undergraduate students at one place. The author realizes that the way hehas presented and explained the subject matter is not the only way; possibilities of better presentationand the way of better explanation of intrigue concepts are always there. The author has been verycareful in selecting the topics, laying their sequence and the style of presentation so that student maynot be afraid of learning new concepts. Realizing the mental state of undergraduate students, everyattempt has been made to present the material in most elementary and digestible form. The author feelsthat he cannot guess as to how far he has come up in his endeavour and to the expectations ofesteemed readers. They have to judge his work critically and pass their constructive criticism either tohim or to the publishers so that they can be incorporated in further editions. To err is human. Theauthor will be glad to receive comments on conceptual mistakes and misinterpretation if any that haveescaped his attention.

A sufficiently large number of solved examples have been added at appropriate places to make thereaders feel confident in applying the basic principles.

I wish to express my thanks to Mr. Saumya Gupta (Managing Director), New Age International(P) Limited, Publishers, as well as the editorial department for their untiring effort to complete thisproject within a very short period.

In the end I await the response this book draws from students and learned teachers.

R.B. Singh


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PREFACE TO THE FIRST EDITION

This book is designed to meet the requirements of undergraduate students preparing for bachelor'sdegree in physical sciences of Indian universities. A decisive role in the development of the presentwork was played by constant active contact with students at lectures, exercises, consultations andexaminations. The author is of the view that it is impossible to write a book without being in contactwith whom it is intended for. The book presents in elementary form some of the most exciting conceptsof modern physics that has been developed during the twentieth century. To emphasize the enormoussignificance of these concepts, we have first pointed out the shortcomings and insufficiencies ofclassical concepts derived from our everyday experience with macroscopic system and then indicatedthe situations that led to make drastic changes in our conceptions of how a microscopic system is to bedescribed. The concepts of modern physics are quite foreign to general experience and hence for theirbetter understanding, they have been presented against the background of classical physics.

The author does not claim originality of the subject matter of the text. Books of Indian andforeign authors have been freely consulted during the preparation of the manuscript. The author isthankful to all authors and publishers whose books have been used.

Although I have made my best effort while planning the lay-out of the text and the subject matter,I cannot guess as to how far I have come up to the expectations of esteemed readers. I request themto judge my work critically and pass their constructive criticisms to me so that any conceptual mistakesand typographical errors, which might have escaped my attention, may be eliminated in the next edition.

I am thankful to my colleagues, family members and the publishers for their cooperation duringthe preparation of the text.

In the end, I await the response, which this book draws from the learned scholars and students.

R.B. Singh


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CONTENTS

UNIT I

SPECIAL THEORY OF RELATIVITY

CHAPTER 1 The Special Theory of Relativity .............................................................................. 3–46

1.1 Introduction ............................................................................................................................... 31.2 Classical Principle of Relativity: Galilean Transformation Equations ..................................... 41.3 Michelson-Morley Experiment (1881) ..................................................................................... 71.4 Einstein’s Special Theory of Relativity ..................................................................................... 91.5 Lorentz Transformations ........................................................................................................ 101.6 Velocity Transformation .......................................................................................................... 131.7 Simultaneity ............................................................................................................................. 151.8 Lorentz Contraction................................................................................................................. 151.9 Time Dilation ........................................................................................................................... 16

1.10 Experimental Verification of Length Contraction and Time Dilation ..................................... 171.11 Interval ..................................................................................................................................... 181.12 Doppler’s Effect ...................................................................................................................... 191.13 Relativistic Mechanics ............................................................................................................. 221.14 Relativistic Expression for Momentum: Variation of Mass with Velocity ............................. 221.15 The Fundamental Law of Relativistic Dynamics ................................................................... 241.16 Mass-energy Equivalence ........................................................................................................ 261.17 Relationship Between Energy and Momentum ....................................................................... 271.18 Momentum of Photon ............................................................................................................. 281.19 Transformation of Momentum and Energy ........................................................................... 281.20 Verification of Mass-energy Equivalence Formula ................................................................ 301.21 Nuclear Binding Energy .......................................................................................................... 31

Solved Examples ..................................................................................................................... 31Questions.................................................................................................................................. 44Problems .................................................................................................................................. 45

x Contents

UNIT II

QUANTUM MECHANICS

CHAPTER 1 Origin of Quantum Concepts ................................................................................. 49–77

1.1 Introduction .......................................................................................................................... 491.2 Black Body Radiation ............................................................................................................ 501.3 Spectral Distribution of Energy in Thermal Radiation ........................................................ 511.4 Classical Theories of Black Body Radiation ........................................................................ 521.5 Planck’s Radiation Law ........................................................................................................ 541.6 Deduction of Stefan’s Law from Planck’s Law ................................................................. 561.7 Deduction of Wien’s Displacement Law ............................................................................. 57

Solved Examples ................................................................................................................... 581.8 Photoelectric Effect .............................................................................................................. 60

Solved Examples ................................................................................................................... 631.9 Compton’s Effect ................................................................................................................. 65

Solved Examples ................................................................................................................... 681.10 Bremsstrahlung ..................................................................................................................... 701.11 Raman Effect ........................................................................................................................ 72

Solved Examples ................................................................................................................... 741.12 The Dual Nature of Radiation .............................................................................................. 75

Questions and Problems ....................................................................................................... 76

CHAPTER 2 Wave Nature of Material Particles ........................................................................ 78–96

2.1 Introduction .......................................................................................................................... 782.2 de Broglie Hypothesis ........................................................................................................... 782.3 Experimental Verification of de Broglie Hypothesis ............................................................. 802.4 Wave Behavior of Macroscopic Particles ............................................................................ 822.5 Historical Perspective ........................................................................................................... 822.6 The Wave Packet .................................................................................................................. 832.7 Particle Velocity and Group Velocity .................................................................................... 862.8 Heisenberg’s Uncertainty Principle or the Principle of Indeterminacy ............................. 87

Solved Examples ................................................................................................................... 89Questions and Problems ....................................................................................................... 96

CHAPTER 3 Schrödinger Equation ............................................................................................. 97–146

3.1 Introduction .......................................................................................................................... 973.2 Schrödinger Equation ........................................................................................................... 983.3 Physical Significance of Wave Function ....................................................................... 1023.4 Interpretation of Wave Function in terms of Probability Current Density ................... 1033.5 Schrödinger Equation in Spherical Polar Coordinates ....................................................... 1053.6 Operators in Quantum Mechanics ..................................................................................... 106

Contents xi

3.7 Eigen Value Equation ............................................................................................................1123.8 Orthogonality of Eigen Functions ....................................................................................... 1133.9 Compatible and Incompatible Observables .........................................................................115

3.10 Commutator .........................................................................................................................1163.11 Commutation Relations for Ladder Operators ................................................................... 1203.12 Expectation Value ................................................................................................................ 1213.13 Ehrenfest Theorem ............................................................................................................. 1223.14 Superposition of States (Expansion Theorem) .................................................................. 1253.15 Adjoint of an Operator ........................................................................................................ 1273.16 Self-adjoint or Hermitian Operator ..................................................................................... 1283.17 Eigen Functions of Hermitian Operator Belonging to Different Eigen

Values are Mutually Orthogonal ........................................................................................ 1283.18 Eigen Value of a Self-adjoint (Hermitian Operator) is Real .............................................. 129

Solved Examples ................................................................................................................. 129Questions and Problems ..................................................................................................... 144

CHAPTER 4 Potential Barrier Problems ................................................................................. 147–168

4.1 Potential Step or Step Barrier ............................................................................................. 1474.2 Potential Barrier (Tunnel Effect) ........................................................................................ 1514.3 Particle in a One-dimensional Potential Well of Finite Depth ........................................... 1594.4 Theory of Alpha Decay ...................................................................................................... 163

Questions ............................................................................................................................. 167

CHAPTER 5 Eigen Values of ˆ2L and ˆzL Axiomatic: Formulation of

Quantum Mechanics ............................................................................................... 169–188

5.1 Eigen Values and Eigen Functions of 2L And zL ............................................................. 169

5.2 Axiomatic Formulation of Quantum Mechanics ............................................................... 1765.3 Dirac Formalism of Quantum Mechanics ......................................................................... 1785.4 General Definition of Angular Momentum ........................................................................ 1795.5 Parity ................................................................................................................................... 186

Questions and Problems ..................................................................................................... 187

CHAPTER 6 Particle in a Box .................................................................................................... 189–204

6.1 Particle in an Infinitely Deep Potential Well (Box) ............................................................ 1896.2 Particle in a Two Dimensional Potential Well .................................................................... 1926.3 Particle in a Three Dimensional Potential Well .................................................................. 1956.4 Degeneracy ......................................................................................................................... 1976.5 Density of States ................................................................................................................. 1986.6 Spherically Symmetric Potential Well ................................................................................ 200


xii Contents

CHAPTER 7 Harmonic Oscillator ............................................................................................. 205–217

7.1 Introduction ........................................................................................................................ 205Questions and Problems ..................................................................................................... 215

CHAPTER 8 Rigid Rotator ......................................................................................................... 218–224

8.1 Introduction ........................................................................................................................ 218Questions and Problems ..................................................................................................... 224

CHAPTER 9 Particle in a Central Force Field ........................................................................ 225–248

9.1 Reduction of Two-body Problem in Two Equivalent One-body Problem in aCentral Force ...................................................................................................................... 225

9.2 Hydrogen Atom ................................................................................................................... 2289.3 Most Probable Distance of Electron from Nucleus .......................................................... 2389.4 Degeneracy of Hydrogen Energy Levels ........................................................................... 2419.5 Properties of Hydrogen Atom Wave Functions ................................................................. 241


UNIT III

STATISTICAL MECHANICS

CHAPTER 1 Preliminary Concepts .......................................................................................... 251–265

1.1 Introduction ........................................................................................................................ 2511.2 Maxwell-Boltzmann (M-B) Statistics ................................................................................. 2511.3 Bose-Einstein (B-E) Statistics ............................................................................................ 2521.4 Fermi-Dirac (F-D) Statistics .............................................................................................. 2521.5 Specification of the State of a System ............................................................................. 2521.6 Density of States ................................................................................................................. 2541.7 N-particle System ............................................................................................................... 2561.8 Macroscopic (Macro) State ............................................................................................... 2561.9 Microscopic (Micro) State ................................................................................................. 257

Solved Examples ................................................................................................................. 258

CHAPTER 2 Phase Space ........................................................................................................... 266–270

2.1 Introduction ........................................................................................................................ 2662.2 Density of States in Phase Space ....................................................................................... 2682.3 Number of Quantum States of an N-particle System ....................................................... 270

CHAPTER 3 Ensemble Formulation of Statistical Mechanics ............................................. 271–291

3.1 Ensemble ............................................................................................................................. 271

Contents xiii

3.2 Density of Distribution (Phase Points) in -space ........................................................... 2723.3 Principle of Equal a Priori Probability ................................................................................ 2723.4 Ergodic Hypothesis ............................................................................................................. 2733.5 Liouville’s Theorem ............................................................................................................ 2733.6 Statistical Equilibrium ......................................................................................................... 277

Thermodynamic Functions

3.7 Entropy ................................................................................................................................ 2783.8 Free Energy ......................................................................................................................... 2793.9 Ensemble Formulation of Statistical Mechanics ................................................................ 280

3.10 Microcanonical Ensemble ................................................................................................... 2813.11 Classical Ideal Gas in Microcanonical Ensemble Formulation .......................................... 2823.12 Canonical Ensemble and Canonical Distribution ............................................................... 2843.13 The Equipartition Theorem ................................................................................................. 2883.14 Entropy in Terms of Probability ......................................................................................... 2903.15 Entropy in Terms of Single Particle Partition Function Z1 ............................................... 291

CHAPTER 4 Distribution Functions ......................................................................................... 292–308

4.1 Maxwell-Boltzmann Distribution ........................................................................................ 2924.2 Heat Capacity of an Ideal Gas ............................................................................................ 2974.3 Maxwell’s Speed Distribution Function ............................................................................. 2984.4 Fermi-Dirac Statistics ......................................................................................................... 3024.5 Bose-Einstein Statistics ....................................................................................................... 305

CHAPTER 5 Applications of Quantum Statistics ................................................................... 309–333

Fermi-Dirac Statistics

5.1 Sommerfeld’s Free Electron Theory of Metals ................................................................. 3095.2 Electronic Heat Capacity .................................................................................................... 3175.3 Thermionic Emission (Richardson-Dushmann Equation) ................................................ 3185.4 An Ideal Bose Gas ............................................................................................................... 3215.5 Degeneration of Ideal Bose Gas ......................................................................................... 3245.6 Black Body Radiation: Planck’s Radiation Law ................................................................. 3285.7 Validity Criterion for Classical Regime ............................................................................... 3295.8 Comparison of M-B, B-E and F-D Statistics ..................................................................... 331

CHAPTER 6 Partition Function ................................................................................................ 334–358

6.1 Canonical Partition Function .............................................................................................. 3346.2 Classical Partition Function of a System Containing N Distinguishable Particles ........... 3356.3 Thermodynamic Functions of Monoatomic Gas .............................................................. 3376.4 Gibbs Paradox ..................................................................................................................... 338

xiv Contents

6.5 Indistinguishability of Particles and Symmetry of Wave Functions ................................. 3416.6 Partition Function for Indistinguishable Particles ............................................................. 3426.7 Molecular Partition Function .............................................................................................. 3446.8 Partition Function and Thermodynamic Properties of Monoatomic Ideal Gas ............... 3446.9 Thermodynamic Functions in Terms of Partition Function ............................................. 346

6.10 Rotational Partition Function .............................................................................................. 3476.11 Vibrational Partition Function ............................................................................................. 3496.12 Grand Canonical Ensemble and Grand Partition Function ................................................ 3516.13 Statistical Properties of a Thermodynamic System in Terms of Grand

Partition Function ............................................................................................................... 3546.14 Grand Potential ............................................................................................................... 3546.15 Ideal Gas from Grand Partition Function .......................................................................... 3556.16 Occupation Number of an Energy State from Grand Partition Function:

Fermi-Dirac and Bose-Einstein Distribution ...................................................................... 356

CHAPTER 7 Application of Partition Function ...................................................................... 359–376

7.1 Specific Heat of Solids ....................................................................................................... 3597.1.1 Einstein Model .......................................................................................................... 3597.1.2 Debye Model ............................................................................................................ 362

7.2 Phonon Concept ................................................................................................................. 3657.3 Planck’s Radiation Law: Partition Function Method ......................................................... 367

Questions and Problems ..................................................................................................... 369Appendix–A ......................................................................................................................... 370

UNIT IV

ATOMIC SPECTRA

CHAPTER 1 Atomic Spectra–I .................................................................................................. 379–411

1.1 Introduction ........................................................................................................................ 3791.2 Thomson’s Model ............................................................................................................... 3791.3 Rutherford Atomic Model .................................................................................................. 3811.4 Atomic (Line) Spectrum ..................................................................................................... 3821.5 Bohr’s Theory of Hydrogenic Atoms (H, He+, Li++) ........................................................ 3851.6 Origin of Spectral Series .................................................................................................... 3891.7 Correction for Nuclear Motion .......................................................................................... 3911.8 Determination of Electron-Proton Mass Ratio (m/MH)..................................................... 3941.9 Isotopic Shift: Discovery of Deuterium ............................................................................ 394

1.10 Atomic Excitation ............................................................................................................... 3951.11 Franck-Hertz Experiment ................................................................................................... 3961.12 Bohr’s Correspondence Principle ...................................................................................... 397

Contents xv

1.13 Sommerfeld Theory of Hydrogen Atom............................................................................ 3981.14 Sommerfeld’s Relativistic Theory of Hydrogen Atom ...................................................... 403


CHAPTER 2 Atomic Spectra–II ................................................................................................. 412–470

2.1 Electron Spin ....................................................................................................................... 4122.2 Quantum Numbers and the State of an Electron in an Atom ........................................... 4122.3 Electronic Configuration of Atoms .................................................................................... 4152.4 Magnetic Moment of Atom ................................................................................................ 4162.5 Larmor Theorem................................................................................................................. 4172.6 The Magnetic Moment and Lande g-factor for One Valence Electron Atom .................. 4182.7 Vector Model of Atom ........................................................................................................ 4202.8 Atomic State or Spectral Term Symbol ............................................................................. 4262.9 Ground State of Atoms with One Valence Electron (Hydrogen and Alkali Atoms) ......... 426

2.10 Spectral Terms of Two Valence Electrons Systems (Helium and Alkaline-Earths) ......... 4272.11 Hund’s Rule for Determining the Ground State of an Atom ............................................ 4342.12 Lande g-factor in L-S Coupling ......................................................................................... 4352.13 Lande g-factor in J-J Coupling ......................................................................................... 4392.14 Energy of an Atom in Magnetic Field ................................................................................ 4402.15 Stern and Gerlach Experiment (Space Quantization): Experimental Confirmation for

Electron Spin Concept ........................................................................................................ 4412.16 Spin Orbit Interaction Energy ............................................................................................ 4432.17 Fine Structure of Energy Levels in Hydrogen Atom......................................................... 4462.18 Fine Structure of H Line ................................................................................................... 4492.19 Fine Structure of Sodium D Lines ..................................................................................... 4502.20 Interaction Energy in L-S Coupling in Atom with Two Valence Electrons ...................... 4512.21 Interaction Energy In J-J Coupling in Atom with Two Valence Electrons ...................... 4552.22 Lande Interval Rule ............................................................................................................. 458


CHAPTER 3 Atomic Spectra-III ............................................................................................... 471–498

3.1 Spectra of Alkali Metals ...................................................................................................... 4713.2 Energy Levels of Alkali Metals ........................................................................................... 4713.3 Spectral Series of Alkali Atoms ......................................................................................... 4743.4 Salient Features of Spectra of Alkali Atoms ...................................................................... 4773.5 Electron Spin and Fine Structure of Spectral Lines .......................................................... 4773.6 Intensity of Spectral Lines.................................................................................................. 481

Solved Examples ................................................................................................................. 484

xvi Contents

3.7 Spectra of Alkaline Earths .................................................................................................. 4873.8 Transitions Between Triplet Energy States ........................................................................ 4933.9 Intensity Rules .................................................................................................................... 493

3.10 The Great Calcium Triads .................................................................................................. 4933.11 Spectrum of Helium Atom.................................................................................................. 494

Questions and Problems ..................................................................................................... 497

CHAPTER 4 Magneto-optic and Electro-optic Phenomena ................................................... 499–519

4.1 Zeeman Effect ..................................................................................................................... 4994.2 Anomalous Zeeman Effect ................................................................................................. 5034.3 Paschen-back Effect .......................................................................................................... 5064.4 Stark Effect ......................................................................................................................... 512


CHAPTER 5 X-Rays and X-Ray Spectra ................................................................................. 520–538

5.1 Introduction ........................................................................................................................ 5205.2 Laue Photograph ................................................................................................................. 5205.3 Continuous and Characteristic X-rays ............................................................................... 5215.4 X-ray Energy Levels and Characteristic X-rays ............................................................... 5235.5 Moseley’s Law .................................................................................................................... 5265.6 Spin-relativity Doublet or Regular Doublet ........................................................................ 5275.7 Screening (Irregular) Doublet ............................................................................................ 5285.8 Absorption of X-rays .......................................................................................................... 5295.9 Bragg’s Law ........................................................................................................................ 532


UNIT V

MOLECULAR SPECTRA OF DIATOMIC MOLECULES

CHAPTER 1 Rotational Spectra of Diatomic Molecules ....................................................... 541–548

1.1 Introduction ........................................................................................................................ 5411.2 Rotational Spectra—Molecule as Rigid Rotator ................................................................ 5431.3 Isotopic Shift ...................................................................................................................... 5471.4 Intensities of Spectral Lines ............................................................................................... 548

CHAPTER 2 Vibrational Spectra of Diatomic Molecules ...................................................... 549–554

2.1 Vibrational Spectra—Molecule as Harmonic Oscillator .................................................... 549

Contents xvii

2.2 Anharmonic Oscillator ........................................................................................................ 5502.3 Isotopic Shift of Vibrational Levels .................................................................................... 553

CHAPTER 3 Vibration-Rotation Spectra of Diatomic Molecules ........................................ 555–561

3.1 Energy Levels of a Diatomic Molecule and Vibration-rotation Spectra ........................... 5553.2 Effect of Interaction (Coupling) of Vibrational and Rotational Energy on

Vibration-rotation Spectra ................................................................................................... 559

CHAPTER 4 Electronic Spectra of Diatomic Molecules ........................................................ 562–581

4.1 Electronic Spectra of Diatomic Molecules ........................................................................ 5624.2 Franck-Condon Principle: Absorption ............................................................................... 5734.3 Molecular States ................................................................................................................. 579

Examples ............................................................................................................................. 581

CHAPTER 5 Raman Spectra ...................................................................................................... 582–602

5.1 Introduction ........................................................................................................................ 5825.2 Classical Theory of Raman Effect ..................................................................................... 5845.3 Quantum Theory of Raman Effect .................................................................................... 586


CHAPTER 6 Lasers and Masers ................................................................................................ 603–612

6.1 Introduction ........................................................................................................................ 6036.2 Stimulated Emission ............................................................................................................ 6036.3 Population Inversion ........................................................................................................... 6066.4 Three Level Laser ............................................................................................................... 6086.5 The Ruby Laser .................................................................................................................. 6096.6 Helium-Neon Laser ............................................................................................................. 6106.7 Ammonia Maser ...................................................................................................................6116.8 Characteristics of Laser .......................................................................................................611

Questions and Problems ..................................................................................................... 612Index ........................................................................................................................... 613–618


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SPECIAL THEORY OFRELATIVITY

UNIT


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CHAPTER

THE SPECIAL THEORY OF RELATIVITY

1.1 INTRODUCTION

All natural phenomena take place in the arena of space and time. A natural phenomenon consists ofa sequence of events. By event we mean something that happens at some point of space and at somemoment of time. Obviously the description of a phenomenon involves the space coordinates andtime. The oldest and the most celebrated branch of science –mechanics- was developed on the conceptsspace and time that emerged from the observations of bodies moving with speeds very small comparedwith the speed of light in vacuum. Guided by intuitions and everyday experience Newton wrote aboutspace and time: Absolute space, in its own nature, without relation to anything external, remains alwayssimilar and immovable. Absolute, true and mathematical time, of itself, and from its own nature,flows equably without relation to anything external and is otherwise called duration.

In Newtonian (classical) mechanics, it assumed that the space has three dimensions and obeysEuclidean geometry. Unit of length is defined as the distance between two fixed points. Other distancesare measured in terms of this standard length. To measure time, any periodic process may be used toconstruct a clock. Space and time are supposed to be independent of each other. This implies thatthe space interval between two points and the time interval between two specified events do not dependon the state of motion of the observers. Two events, which are simultaneous in one frame, are alsosimultaneous in all other frames. Thus the simultaneity is an absolute concept. In addition to this,the space and time are assumed to be homogeneous and isotropic. Homogeneity means that all pointsin space and all moments of time are identical. The space and time intervals between two givenevents do not depend on where and when these intervals are measured. Because of these properties ofspace and time, we are free to select the origin of coordinate system at any convenient point andconduct experiment at any moment of time. Isotropy of space means that all the directions of spaceare equivalent and this property allows us to orient the axes of coordinate system in any convenientdirection.

The description of a natural phenomenon requires a suitable frame of reference with respect towhich the space and time coordinates are to be measured. Among all conceivable frames of reference,the most convenient ones are those in which the laws of physics appear simple. Inertial frames havethis property. An inertial frame of reference is one in which Newton’s first law (the law of inertia)holds. In other words, an inertial frame is one in which a body moves uniformly and rectilinearly in

4 Introduction to Modern Physics

absence of any forces. All frames of reference moving with constant velocity relative to an inertialframe are also inertial frames. A frame possessing acceleration relative to an inertial frame is callednon-inertial frame. Newton’s first law is not valid in non-inertial frame. Reference frame with itsorigin fixed at the center of the sun and the three axes directed towards the stationary stars was supposedto be the fundamental inertial frame. In this frame, the motion of planets appear simple. Newton’slaws are valid this heliocentric frame. Let us see whether the earth is an inertial frame or not. Themagnitude of acceleration associated with the orbital motion of earth around the sun is 0.006 m/s2

and that with the spin motion of earth at equator is 0.034 m/s2. For all practical purposes theseaccelerations are negligibly small and the earth may be regarded as an inertial frame but for precisework its acceleration must be taken into consideration. The entire classical mechanics was developedon these notions of space and time it worked efficiently. No deviations between the theoretical andexperimental results were noticed till the end of the 19th century. By the end of 19th century particles(electrons) moving with speed comparable with the speed of light c were available; and the departuresfrom classical mechanics were observed. For example, classical mechanics predicts that the radius rof the orbit of electron moving in a magnetic field of strength B is given by r = mv/qB, where m, vand q denote mass, velocity and charge of electron. The experiments carried out to measure the orbitradius of electron moving at low velocity give the predicted result; but the observed radius of electronmoving at very high speed does not agree with the classical result. Many other experimentalobservations indicated that the laws of classical mechanics were no longer adequate for the descriptionof motion of particles moving at high speeds.

In 1905 Albert Einstein gave new ideas of space and time and laid the foundation of specialtheory of relativity. This new theory does not discard the classical mechanics as completely wrong butincludes the results of old theory as a special case in the limit (v/c) 0. i.e., all the results of specialtheory of relativity reduce to the corresponding classical expressions in the limit of low speed.

1.2 CLASSICAL PRINCIPLE OF RELATIVITY: GALILEANTRANSFORMATION EQUATIONS

The Galilean transformation equations are a set of equations connecting the space-time coordinatesof an event observed in two inertial frames, which are in relative motion. Consider two inertial framesS (unprimed) and S' (primed) with their corresponding axes parallel; the frame S' is moving alongthe common x-x' direction with velocity v relative to the frame S. Each frame has its own observerequipped with identical and compared measuring stick and clock. Assume that when the origin O ofthe frame S' passes over the origin O of frame S, both observers set their clocks at zero i.e., t = t' = 0.The event to be observed is the motion of a particle. At certain moment, the S-observer registers thespace-time coordinates of the particle as (x, y, z, t) and S'- observer as (x', y', z', t'). It is evident thatthe primed coordinates are related to unprimed coordinates through the relationship

x' = x – vt, y' = y, z' = z, t' = t ...(1.2.1)

The last equation t' = t has been written on the basis of the assumption that time flows at thesame rate in all inertial frames. This notion of time comes from our everyday experiences with slowlymoving objects and is confirmed in analyzing the motion of such objects. Equations (1.2.1) are calledGalilean transformation equations. Relative to S', the frame S is moving with velocity v in negative

The Special Theory of Relativity 5

direction of x-axis and therefore inverse transformation equations are obtained by interchanging theprimed and unprimed coordinates and replacing v with –v. Thus

x = x' + vt', y = y', z = z', t = t' ...(1.2.2)

Fig. 1.2.1 Galilean transformation

Transformation of Length

Let us see how the length of an object transforms on transition from S to S'. Consider a rod placedin frame S along its x-axis. The length of rod is equal to the difference of its end coordinates: l = x2

– x1. In frame S', the length of rod is defined by the difference of its end coordinates measuredsimultaneously. Thus:

l' = ′ ′−2 1x x

Making use of Galilean transformation equations we havel' = (x2 – vt) – (x1 – vt) = x2 – x1 = l

Thus the distance between two points is invariant under Galilean transformation.

Transformation of Velocity

Differentiating the first equation of Galilean transformation, we have

′= −

′′ = −x x

dx dxv

dt dtu u v

...(1.2.3)

where ux and u'x are the x-components of velocity of the particle measured in frame S and S'respectively. Eqn. (1.2.3) is known as the classical law of velocity transformation. The inverse law is

ux = u' + v ...(1.2.4)

These equations show that velocity is not invariant; it has different values in different inertialframes depending on their relative velocities.

Transformation of Acceleration

Differentiating equation (1.2.3) with respect to time, we have

′′= ⇒ =x xx x

du dua a

dt dt ...(1.2.5)


where ax and a'x are the accelerations of the particle in S and S'. Thus we see that the acceleration isinvariant with respect to Galilean transformation.

Transformation of the Fundamental Law of Dynamics (Newton’s Law)

The fundamental law of mechanics, which relates the force acting on a particle to its acceleration, is

ma = F ...(1.2.6)

In classical mechanics, the mass of a particle is assumed to be independent of velocity of themoving particle. The well known position dependent forces–gravitational, electrostatic and elasticforces and velocity dependent forces- friction and viscous forces are also invariant with respect toGalilean transformation because of the invariance of length, relative velocity and time. Hence thefundamental law of mechanics is also invariant under Galilean transformation. Thus

m a = F in frame S

m' a' = F' in frame S'

The invariance of the basic laws of mechanics ensures that all mechanical phenomena proceedidentically in all inertial frames of reference consequently no mechanical experiment performed whollywithin an inertial frame can tell us whether the given frame is at rest or moving uniformly in a straightline. In other words all inertial frames are absolutely equivalent, and none of them can be preferredto others. This statement is called the classical (Galilean) principle of relativity.

The Galilean principle of relativity was successfully applied to the mechanical phenomena onlybecause in Galileo’s time mechanics represented the whole physics. The classical notions of space,time and matter were regarded so fundamental that nobody ever felt necessity to raise any doubtsabout their truth. The Galilean principle of relativity did not worry physicists too much by the middleof the 19th century. By the middle of 19th century other branches of physics—electrodynamics, opticsand thermodynamics—were developing and each of them required its own basic laws. A natural questionarose: does the Galilean principle of relativity cover all physics as well? If the principle of relativitydoes not apply to other branches of physics then non-mechanical phenomena can be used to distinguishinertial frames thereby choosing a preferred frame. The basic laws of electrodynamics—Maxwell’sfield equations—predicted that light was an electromagnetic phenomenon. The light propagates invacuum with speed c = (00) –½ = 3 × 108 m/s. The wave nature of light compelled the then physiciststo assume a medium for the propagation of light and hypothetical medium luminiferous ether waspostulated to meet this requirement. Ether was regarded absolutely at rest and light was supposed totravel with speed c relative to the ether. If a certain frame is moving with velocity v relative to theether; the speed of light in that frame, according to Galilean transformation, is c ± v; the plus signwhen c and v are oppositely directed and minus sign when c and v have the same direction. Makingus of this result that the light has different speed in different frames; the famous Michelson-Morleyexperiment was set up to detect the motion of the earth with respect to the ether.

When Galilean transformation equations were applied to the newly discovered laws ofelectrodynamics, the Maxwell’s equations, it was found that they change their shape on transitionfrom one inertial frame to another. At first the validity of Maxwell’s equations was questioned andattempts were made to modify them in a way to make them consistent with the Galilean principle of


relativity. But such attempts predicted new phenomenon, which could not be verified experimentally.It was then realized that Maxwell’s equations need no modifications.

1.3 MICHELSON-MORLEY EXPERIMENT (1881)

The purpose of the experiment was to detect the motion of the earth relative to the hypotheticalmedium ether, which was supposed to be at rest. The instrument employed was the Michelsoninterferometer, which consists of two optically plane mirrors M1 and M2 fixed on two mutuallyperpendicular arms PM1 and PM2. At the point of intersection of the two arms, a glass plate P semi-silvered at its rear end is fixed. The glass plate P is inclined at 45° to each mirror. Monochromaticlight from an extended source is allowed to fall on the plate P, which splits the incident beam intotwo beams—beam 1 that travels along the arm PM1 and beam 2 that travels along the arm PM2.The beam 1 is reflected back from mirror M1 and comes to the rear surface of the plate P where itsuffers partial reflection and finally goes into the telescope T. The beam 2 also suffers reflection atthe mirror M2 and is received into the telescope. These interfering beams produce interference fringes,which are observed in the telescope.

Fig. 1.3.1 Michelson’s interferometer

Now suppose that at the moment of the experiment the apparatus moves together with theearth with velocity v (= 3 × 104 m/s) in its orbit along the arm PM1. Relative to the apparatus thelight traveling along the path PM1 has speed c – v and that traveling along the path M1P has speedc + v. If l is the length of the arm PM1 then the time spent by light to traverse the path PM1P isequal to

t||

12 2

2 2 2 2

2 1 2 21 1

1 /

l l l l v l v

c v c v c c cv c c c

−

= + = = − = + − + − ...(1.3.1)

Since v/c << 1, higher order terms in binomial expansion have been omitted.


For an observer stationed in ether frame the beam 2 to return to the plate P after sufferingreflection at the mirror M2 it must traverse the angular path ′ ′2PM P . Let t be the time taken by thebeam 2 to cover the distance ′ ′2PM P . During this time the plate covers a distance PP' = v t. Fromthe geometry of the Fig. (1.3.1), we have

′ ′= +2 2 22 2PM PO OM or, 2 2 2( /2) ( /2)ct vt l⊥ ⊥= +

whence

t=

1/ 22 2

2 22 2

2 2 21 1

2

l l v l v

c cc cc v

− = − = + −

...(1.3.2)

Comparing the expressions for t|| and t, we see that light beams 1 and 2 takes different timesto cover the round trips. The time difference is

t = t|| – t = 2

2

2

2

l v

c c...(1.3.3)

This time difference is equivalent to path difference

x = c t = 2

2

lv

c

Now the whole apparatus is rotated through 90°, the paths of the beams are interchanged. Therotation causes a change in path difference

(x)rot = 2

2

2lv

c...(1.3.4)

A change of path difference produces a fringe shift of unity. Therefore the expected fringeshift resulting from the rotation of the apparatus is

n = 2

2

2lv

cλ ...(1.3.5)

By using the technique of multiple reflections, Michelson and Morley made l as large as 11m.In one experiment a light source of wavelength 5900 Å was used. Substituting the values of l, , vand c we find

n =4 2

7 8

2 11m (3 10 m/s)0.37

5.9 10 m (3 10 m/s)−× × × =× × ×

The instrument was capable of measuring a fringe shift of the order of 0.01, but during therotation of the apparatus the expected fringe shift did not appear. The experiment was repeated manytimes with greater accuracy during the different parts of the day and different seasons of the year.Every time no fringe shift was detected. The result of the experiment was called null or negative.Had there been a measurable fringe shift, we could calculate the velocity of the earth relative to


ether. The negative result of the experiment contradicted the Galilean law of addition of velocity.All attempts to explain the negative result of the Michelson experiment in terms of classical mechanicsturned out to be unsatisfactory in the final analysis. The Michelson-Morley experiment showed thatall inertial frames are equivalent for the description of physical phenomena. More experiments ofthe same kind performed later perfectly confirmed the validity of the principle of relativity for allphenomena.

1.4 EINSTEIN’S SPECIAL THEORY OF RELATIVITY

After making a profound analysis of the experimental and theoretical results of physics, particularlyof electrodynamics, a virtually unknown clerk of the Swiss federal Patent Office, Albert Einstein(1879–1955) arrived at the conclusions that the very concepts of space and time over which the entireedifice of classical physics stood were no longer true. He realized that the Newtonian notions ofspace and time, that emerged from the observation of bodies moving with speeds very small comparedwith the speed of light and hence their extrapolation to bodies moving at speeds, comparable to thespeed of light c in empty space, had no claim to be right. In 1905 Einstein in his epoch makingpaper ‘on the electrodynamics of moving bodies’ created the Special Theory of Relativity, which isessentially a physical theory of space and time. The special theory of relativity is based on twopostulates, which have been confirmed by experimental tests.

1. The Principle of Relativity

This postulate is an extension of the Newtonian principle of relativity to all phenomena of nature. Itstates that the laws of physics and the equations describing them are invariant, i.e., keep their formon transitions from one inertial frame to another. In other words: all inertial frames are equivalent intheir physical properties and therefore they are equally suitable for the description of physicalphenomena.. This postulate rejects the idea of absolute space and absolute motion. No experimentwhatever can distinguish one inertial frame from the other.

2. The Universal Speed of Light

The speed of light in vacuum is the same in all inertial frames of reference, regardless of their relativemotion. Thus the speed of light holds a unique position. In contrast to all other speeds, which changeon transition from one reference frame to another, the speed of light in vacuum is an invariant quantity.

The postulates of special theory of relativity lead to a number of important conclusions, whichare in drastic conflicts with the dictates of common sense. In Newtonian mechanics space and timewere assumed to be absolute and independent of each other. According to the special theory ofrelativity space and time are not absolute, they depend on the state of motion and are inseparablefrom each other.

In order to correlate the observations carried out in different inertial frames of reference weneed transformation equations, which must be consistent with the postulates of the special theory ofrelativity. Certainly they cannot be the Galilean transformations because they contradict the secondpostulate—the constancy of speed of light. Moreover, Galilean transformation equations change the


appearance of Maxwell’s equations on transition from one inertial frame to another. We needtransformation equations, which preserve not only the form of Maxwell’s equations but also all thelaws of physics. It was Hendrik Lorentz (1853–1928) who guessed empirically the correct form oftransformation equations but Einstein gave their theoretical basis. The new transformation equationsare called relativistic or Lorentz transformation equations, which are derived in the following section.

1.5 LORENTZ TRANSFORMATIONS

Derived on the basis of the postulates of special theory of relativity, the Lorentz transformations area set of equations, which connect the space-time coordinates of an event measured in two inertialframes that are in relative motion. Consider two inertial frames S and S' with their correspondingaxes parallel and the primed frame moving relative to unprimed frame with velocity v along thecommon x–x' direction. Each frame has its own observer equipped with measuring stick andsynchronized clocks. Let the observers set their clocks at t = o = t' when their origins coincide.

Suppose that the observer in the frame S records the space-time coordinates of a particle asx, y, z, t and S'–observer records them as x', y', z', t'. Our task is to seek relations of the type

x' = f1 (x, y, z, t),

y' = f2 (x, y, z, t),

z' = f3 (x, y, z, t),

t' = f 4 (x, y, z, t) ...(1.5.1)

Since the frames have relative velocity only along x-direction, the y and z coordinates remainunchanged.

y' = y, z'= z ...(1.5.2)

In addition to the postulates of special relativity we shall assume that the space and time arehomogeneous and isotropic. This means that the length interval measured in a frame is independentof the position where it is measured and the time interval is independent of the instant when it ismeasured. A linear transformation satisfies this criterion.

In Fig. 1.5.1 a linear and a non-linear transformation are shown. A rod of length l placed alongx-axis with end coordinates x1 and x2 in the frame S is transformed by linear transformation to alength l' with end coordinates ′1x and ′2x in the frame S'. The same rod placed between points x3

and x4 in frame S is transformed to a length l' with its end coordinates ′3x and ′4 .x A linear

transformation ensures that if x2 – x1 = x4 – x3 = l then ′2x – ′1x = ′4x – ′3x = l. Whereas for non-

linear transformation this criterion is not satisfied i.e., if x2 – x1 = x4 – x3 = l then ′2x – ′1x ′4x – ′3x .This is obvious from the Fig. (1.5.1).

Thus the transformations must be linear and we can write them asx' = a11x + a12t

y' = y

z' = z

t' = a21x + a22t ...(1.5.3)


where a’s are constants. If the particle traverses a distance dx along x-axis in time dt in frame S, thenthe corresponding distance dx' and time dt' are given by

dx' = a11 dx + a12 dt ...(1.5.4)

dt' = a21 dx + a22 dt ...(1.5.5)

Fig. 1.5.1 Transformation of length by a linear and non-linear transformation

The velocity of the particle in frame S and S' are u = dx/dt, u' = dx'/dt' ...(1.5.6)

respectively. Dividing Eqn. (1.5.4) by Eqn. (1.5.5), we get

( ) +′ += =

′ + +11 1211 12

21 22 21 22

/

( / )

a dx dt aa dx a dtdx

dt a dx a dt a dx dt a

or+′ =+

11 12

21 22

a u au

a u a ...(1.5.7)

Now let us determine the constants a11, a12, a21 and a22.(i) Suppose that the particle under study is at rest in frame S then u = 0 and u' = – v. Substituting

these values in Eqn. (1.5.7), we have

− = ∴ = −1212 22

22

av a a v

a...(1.5.8)

(ii) If the particle is at rest in frame S' then u' = 0 and u = v. Substituting these values inEqn. (1.5.7), we have

+= ∴ = − = − ⇒ =

+11 12

12 11 22 11 2221 22

0a v a

a a v a v a aa v a

...(1.5.9)


Fig. 1.5.2

(iii) Instead of mechanical particle, let the observers see photon or light wave front. Accordingto the second postulate (the constancy of the speed of light in vacuum) the observers in both theframes find the speed of photon to be the same i.e., u = u' = c. Hence from Eqn. (1.5.7), we have

+ −

= =+ +

11 12 11 11

21 22 21 11

a c a a c a vc

a c a a c a

= −21 112

va a

c...(1.5.10)

Substituting the values of constants a12, a22 and a21 in Eqn. (1.5.3), we havex' = a11 (x – vt)

y' = y

z' = z

t' = a11 (t – vx/c2) ...(1.5.11)

(iv) According to the first postulate both the frames S and S' are equally suitable for thedescription of physical phenomena. Relative to frame S', the frame S is moving with velocity –v,hence the inverse transformations must look as

x = a11 (x' + v t')

y = y'

z = z'

t = a11 (t' + vx'/c2) ...(1.5.12)

Substituting the values of x' and t' from Eqn. (1.5.11) in (1.5.12), we find

=

−11 2

2

1

1

av

c

...(1.5.13)


When the value of a11 is substituted in Eqns. (1.5.11) and (1.5.12), we get the Lorentztransformations as

− −′ ′ ′ ′= = = =− −

2

2 2 2 2

/, , ,

1 / 1 /

x vt t vx cx y y z z t

v c v c...(1.5.14)

The inverse transformations are obtained by mutual interchange of primed and unprimedcoordinates and replacing v by – v. Thus

′ ′ ′ ′+ +′ ′= = = =− −

2

2 2 2 2

/, , ,

1 / 1 /

x vt t vx cx y y z z t

v c v c...(1.5.15)

It is more convenient to write Lorentz transformations in terms of = v/c and

= −β 2

1

1 as shown in the table.

Lorentz Transformation Inverse Transformation

−β′ = =γ −β−β 2

( )1

x ctx x ct

′ ′+ β ′ ′= = γ + β−β 2

( )1

x c tx x ct

y' = y y = y'

z' = z z = z'

−β′ =−β 2

/

1

t x ct = (t' – x'/c)

′ ′+ β ′ ′= =γ +β−β 2

/( / )

1

t x ct t x c

It is remarkable feature of Lorentz transformations that they reduce to Galilean transformationsin the limit of low velocity ( = v/c 0). Therefore Lorentz transformations are more general andGalilean transformations are special case of these equations.

When v > c, the Lorentz transformations for x and t become imaginary; this means that motionwith speed greater than that of speed of light is impossible.

One of the thought-provoking features of the Lorentz transformations is that the timetransformation equation contains spatial coordinate, which suggests that the space and time areinseparable. In other words, we should not speak separately of space and time but of unified space-time in which all phenomena take place.

1.6 VELOCITY TRANSFORMATION

Consider an inertial frame S' moving relative to frame S with velocity v along the commonx–x' direction. The space-time coordinates of a particle measured by S and S' observers are (x, y, z, t)


and (x', y', z', t') respectively. Let the particle move through a distance dx in time dt in frame S; thecorresponding quantities measured by S' observer are obtained by differentiating the Lorentz-transformation equations

x' = (x – vt), y' = y, z' = z

t' = (t – vx/c2)

From these equations, we havedx' = (dx – vdt), dy' = dy, dz' = dz ...(1.6.1)

dt' = (dt –vdx/c2) ...(1.6.2)

Dividing Eqn. (1.6.1) by (1.6.2), we have

′ − −= =′ − −2 2

( / )

/ 1 ( / ) ( / )

dx dx vdt dx dt v

dt dt vdx c v c dx dt

−′ =− 21 /

xx

x

u vu

vu c ...(1.6.3)

′= =

′ γ − γ −2 2

( / )

( / ) (1 ( / ) / )

dy dy dy dt


−β′ =

−

2

2

1

1 /

yy

x

uu

vu c...(1.6.4)

′= =′ γ − γ −2 2

( / )

( / ) (1 ( / ) / )

dz dz dz dt


−β′ =

−

2

2

1

1 /z

zx

uu

vu c...(1.6.5)

Inverse velocity transformation equations are

′ − β ′′ − β+= = =

′ ′ ′+ + +

2 2

2 2 2

1 1, ,

1 / 1 / 1 /

y zxx y z

x x x

u uu vu u u

vu c vu c vu c...(1.6.6)

Let us apply the transformation equation to the speed of light. If a photon moves with velocityux = c in frame S, then its velocity in frame S' will be

− −′ = = =

− −2 21 / 1 /x

xx

u v c vu c

vu c vc c

It can easily be seen that the relativistic formulae for transformation of velocity reduce to theGalilean transformation equations in the limit of low speed (v/c) 0.


1.7 SIMULTANEITY

In relativity the concept of simultaneity is not absolute. Two events occurring simultaneously in oneinertial frame may not be simultaneous, in general, in other. Assume that the event 1 occurs at pointx1, y1, z1 and at time t1 and event 2 occurs at point x2, y2, z2 and at time t2 in frame S. The space-time coordinates of these two events as measured in frame S', which is moving relative to S withvelocity v in the common x-x' direction, can be obtained from Lorentz transformations

′1x = (x1 –vt1), ′2x = (x2 – vt2)

′1t = (t1 – vx1/c2), ′2t = (t2 – vx2/c2)

The difference of space coordinates and time coordinates are in frame S' are

′2x – ′1x = (x2 – x1) – v(t2 – t1) ...(1.7.1)

′2t – ′1t = (t2 – t1) – (v/c2)(x2 – x1) ...(1.7.2)

Eqn. (1.7.2) gives the time interval between the events as measured in frame S'. It is evidentthat if the two events are simultaneous (i.e., t2 – t1 = 0) in frame S, they are not simultaneous(i.e., ′2t – ′1t 0) in frame S'. In fact

′2t – ′1t = – (v/c2)(x2 – x1) ...(1.7.3)

The events are simultaneous in S' only if they occur at the same point in S (i.e., x2 – x1 = 0).Thus simultaneity is a relative concept.

If ′2t – ′1t > 0, the events occur in frame S' in the same sequence as they occur in frame S.This always happens for events, which are related by cause and effect. That is, cause precedes theeffect, which is known as the causality principle.

If ′2t – ′1t < 0, the events occur in reverse sequence in S'. Such events cannot be related bycause and effect.

It is important to point out that the relativity of simultaneity follows from the finiteness of thespeed of light. In the limit c (classical assumption), simultaneity is an absolute concept i.e.,′2t – ′1t = t2 – t1.

1.8 LORENTZ CONTRACTION

A moving body appears to be contracted in the direction of its motion. This phenomenon is calledLorentz (or Fitzgerald) contraction. Let us consider a rod arranged along the x'-axis and at rest

relative to the frame S'. The length of the rod in frame S' is l0 = ′2x – ′1x where ′1x and ′2x are the

coordinates of the rod ends. The length l0 is called the proper length of the rod. Now consider aframe S relative to which the frame S' is moving with velocity v along x–x' direction. To determinethe length of rod in frame S, we must note the coordinates of the ends x1 and x2 at the same momentof time, say t0. The length of rod in frame S is l0 = x2 – x1. From Lorentz transformations, we have

′ = γ −1 1 0( )x x vt . ′ = γ −2 2 0( )x x vt

0 2 1 2 1( )l x x x x l′ ′= − = γ − = γ


l = (l0/) = l0 −β 21 ...(1.8.1)

Evidently l < l0. Thus the moving rod appears to be contracted.

(a) The rod is placed in frame S' (b) The rod is placed in frame S

Fig. 1.8.1 Transformation of length

If the rod is placed in frame S then its proper length is l0 = x2 – x1. Its length l in frame S' is

equal to the difference of ends coordinates ′1x and ′2x measured at the same moment of time, say

′0 .t

0 2 1 2 0 1 0 2 1( ) ( ) ( )l x x x vt x vt x x l′ ′ ′ ′ ′ ′= − = γ + − + = γ − = γ

20 0/ 1l l l= γ = −β

Thus the length contraction is reciprocal. The rod in either frame appears to be shortened tothe observer in the other frame.

1.9 TIME DILATION

According to relativity there is no such thing as universal time. The rate of flow of time actuallydepends on the state of motion of the observer. Let us see how the time interval between two eventsmeasured in one inertial frame is related to that measured in another inertial frame, which is movingrelative to the first one. Assume that an event 1 occurs at point ′0x at time ′1t in the frame S' andanother event 2 also occurs at the same point but at time ′2 .t The interval between the two events ist' = ′2t – ′1.t This time interval is measured on a single clock located at the point of occurrence ofthe events and is called the proper time interval and is usually denoted by . The same two eventsare observed from a reference frame S relative to which the frame S' is moving with velocity v. Lett1 and t2 be the time of occurrence of the same events registered on the clocks of the frame S. Ofcourse these times will be recorded on the clocks located at different points. The time interval


t = t2 – t1 measured in the frame S is called non-proper or improper time interval. From Lorentztransformations

′ ′= γ + 21 1 0( / ),t t vx c ′ ′= γ + 2

2 2 0( / )t t vx c

( )2 1 2 1t t t t′ ′− = γ −

t =

t = ∆τ

−β 21, = v/c ...(1.9.1)

Fig. 1.9.1 Transformation of time interval

Thus the time interval between two events hasdifferent values in different inertial frames, which arein relative motion. The time interval is least in thereference frame in which the events take place at the samepoint and hence registered on the same clock. Since thenon-proper time is greater than the proper time, a movingclock appears to go slow. This phenomenon is calleddilation of time. The variation of t with velocity v isshown in Fig 1.9.2.

1.10 EXPERIMENTAL VERIFICATION OF LENGTH CONTRACTION ANDT TIME DILATION

The conclusions of the special theory of relativity find direct experimental verification in many ofthe phenomena of particle physics. We shall illustrate this by an example. Muons are unstable sub-atomic particles, which decay into electron and neutrino. Their mean lifetime in a frame in whichthey are at rest is 2 µs. They are created in the upper atmosphere at a height 5 to 6 kms during the

Fig. 1.9.2 Time dilation


interaction of primary cosmic rays with the atmosphere. They are also found in considerable numberat the sea level. The speed of muons is v = 0.998 c.

Classical calculation shows that muons can travel in their lifetime a distanced = v t = (3 × 108m/s) (2 × 10–6 s) = 600 m

This distance is much smaller than the height where the muons are born. Let us explain thisparadox by relativistic calculation. The lifetime of muons is their proper life measured in their ownframe. In laboratory frame their life is t = /(1 – 2) = 31.7 × 10–6s. In this time muons can travela distance d = v t = (0.998 c) (31.7 × 10–6s) = 9.5 km. Thus muons can reach the sea level in theirlifetime.

We can arrive at the same conclusion by considering the length contraction formula. In muonsframe the distance between the birthplace of muons and the sea level appears to be contracted to

d = d0 (1 – 2) = (6 103 m) (1 – (0.998)2) = 379 m

The time required to traverse this distance t = d/v = 379 m/(0.998 × 3 × 108 m/s) = 1.26 µs.This time is less than the proper lifetime of muons.

1.11 INTERVAL

An event in a frame is characterized by space-time coordinates. Assume that an event 1 occurs atpoint x1, y1, z1 and at time t1. The corresponding coordinates for another event 2 are x2, y2, z2, t2.The quantity s12 defined by

( ) ( ) ( ) ( )= − − − − − − −2 2 2 22 212 2 1 2 1 2 1 2 1s c t t x x y y z z ...(1.11.1)

is called the interval between the events. If the events are infinitesimally close together, the intervalis defined by

ds2 = c2dt2 – dx2 – dy2 – dz2 ...(1.11.2)

In frame S' the interval is defined by

′ ′ ′ ′ ′= − − −2 2 2 2 2 2( ) ( ) ( ) ( ) ( )ds c dt dx dy dz ...(1.11.3)

A remarkable property of interval is that it is invariant with respect to Lorentz transformationsi.e.,

ds2 = ds' 2

From Lorentz transformations, we have

2, , and

1

dx cdtdx dy dy dz dz

−β′ ′ ′= = =−β

( )2

/

1

dt c dxdt

− β′ =

−β ...(1.11.4)


Substituting these values in Eqn. (1.11.3), we find

− β −β′ = − − −− β − β

2 22 2 2 2

2 2

( / ) ( )( )

1 1

dt c dx dx cdtds c dy dz

= c2 dt2 – dx2 – dy2 – dz2

= ds2

1.12 DOPPLER’S EFFECT

The apparent change in frequency of a wave due to relative motion between the source of the waveand the observer receiving it, is called the Doppler’s effect. Let a monochromatic source placed atthe origin of frame S' emit a plane wave in xy-plane in the direction making an angle ' withx'-axis. The equation of the wave in this frame is

' = ′ ′ ′ ′ ′ ′ ′ω − −cos[ ]x ya t k x k y

whereπ′ ′ ′ ′ ′ ′ ′= θ = θ =′λ

2cos , sin ,x yk k k k k

' = [ ]′ ′ ′ ′ ′ ′ ′ ′ ′ω − θ − θcos cos sina t k x k y ...(1.12.1)

The equation of the same wave in the frame S will be written as

= [ ]ω − θ − θcos cos sina t kx ky ...(1.12.2)

Fig. 1.12.1 Doppler’s effect

The phase of a wave is invariant quantity i.e., ' = . On transition from S' to S, the phase ofthe wave (1.12.1) becomes

= ′ ′ ′ ′ ′ω γ − − γ − θ − θ2[ ( / ) ( )cos sin ]t vx c k x vt k y

= ′ ω ′ ′ ′ ′ ′ ′ ′γ ω + θ − γ + θ − − θ 2

( cos ) cos sinv

k v t k x k yc

...(1.12.3)

Comparing Eqn. (1.12.3) with the phase of the wave (1.12.2), we have

= ′ ′ ′γ ω + θ( cos )k v ...(1.12.4)


k cos = ′ ′ ′γ ω + θ 2

cosv

kc

or, k cos = ′ ′ ′γ β + θ( cos )k k ...(1.12.5)

k sin = k' sin ' ...(1.12.6)

The first two equations give relativistic Doppler’s effect. Equation (1.12.4) can be transformedinto a more convenient form as follows.

= ′ω ′ ′γ ω + θ

cosvc

since k' = '/c

or = ( )′ ′γ + β θ ω1 cos ...(1.12.7)

Inverse transformation of Eqn. (1.12.7) is

' = ( )γ − β θ ω1 cos ...(1.12.8)

= ′′ ω −βω =

γ −β θ −β θ

21

(1 cos ) 1 cos

or v = 21

1 cos

−β′ν−β θ

...(1.12.9)

Eqn. (1.12.9) gives relativistic Doppler’s shift.

v' = proper frequency, v = observed frequency

Fig. 1.12.2 Relativistic Doppler’s effect

For = 0 (velocity of source coincides with that of velocity of light)

v = −β +β′ ′ν = ν−β −β

21 1

1 1...(1.12.10)

Thus > '. Thus the observed frequency is greater than the emitted frequencyFor = (velocity of source is opposite to that of light)

1

1

− β′ν = ν+ β ...(1.12.11)


In this case v < v'. Observed frequency is less than that emitted by source.For = /2, the relative velocity between the source and the observer is zero. However, even

in this case there is a shift in frequency; the apparent frequency differs from the true frequency by afactor (1 – 2). This is called transverse Doppler’s effect. In this case the observed frequency isalways lower than the proper frequency. The transverse Doppler’s shift is a second order effect anddoes not exists in classical theory.

Classical Doppler’s Effect

Retaining the terms up to first order in in relativistic expression for Doppler’s shift we get classicalDoppler’s effect. Thus

2

21(1 cos )

1 cos 1 cosneglecting β′ ′ν − β ν ′ν = → = ν +β θ

−β θ −β θ...(1.12.12)

For = 0

= ' (1 + ) or c

′λ − λ υ= −′λ (violet shift) ...(1.12.13)

and for =

= ' ( 1 – ) or c

′λ − λ υ=′λ (red shift) ...(1.12.14)

1(1 cos )

1 cos

′ ′ν = ν = ν + β θ − β θ Fig. 1.12.3 Classical Doppler’s effect

Aberration of Light

Dividing Eqn. (1.12.5) by (1.12.6), we obtain

tan = ( )sin

cos

k

k k

′ ′θ′ ′ ′γ β + θ

=2sin 1

cos

′θ −β′θ +β

...(1.12.15)


The inverse transformation is

tan ' =2sin 1

cos

θ −βθ−β

...(1.12.16)

Eqns. (1.12.15) and (1.12.16) connect the directions of light propagation and ' as seen fromtwo inertial frames S and S'. These are the relativistic equations for the aberration of light.

1.13 RELATIVISTIC MECHANICS

In Newtonian mechanics the momentum of a particle is defined as the product of its mass and velocity. p = mν (classical)

Here m is regarded as independent of velocity of particle. Newton’s laws are invariant withrespect to the Galilean transformation but not with respect to the Lorentz transformation. If momentumis defined in a classical way then the law of conservation of momentum is found to be invariantunder Galilean transformation but not under Lorentz transformation. The law of conservation ofmomentum is more fundamental than the Newton’s laws. To make this law invariant under Lorentztransformation, momentum must be redefined.

1.14 RELATIVISTIC EXPRESSION FOR MOMENTUM: VARIATION OF MASSWITH VELOCITY

Let us consider inelastic collision of two identical balls. In frame S' the two balls approach eachother with velocity u' along x-axis and after collision the stick together and the composite ball comesto rest. The same collision is observed from a frame S, which is fixed to one of the balls, say ball 2.Evidently the frame S' is moving with velocity v = u' relative to S. The second ball is at rest in theframe S. The velocity of the first ball in the frame S can be obtained from the relativistic law ofaddition of velocity

u = 2 2 2 2

2

1 / 1 /

u u u

u c u c

′ ′ ′+ =′ ′+ +

...(1.14.1)

This equation can be written as

22 22c

u u cu

′ ′− + = 0 ...(1.14.2)

whence u' = ± −

12 22 2

2c cc

u u

or u' =2 2 2

21

c c u

u u c± − ...(1.14.3)


Fig. 1.14.1 Collision of two identical particles as viewed from two inertial frames

We must choose the negative sign before the radical because it gives the classical result(u = 2u') in the limit u/c 0. Hence

u' = 2 2

2 21 /c c

u cu u− − ...(1.14.4)

and u – u' = 2 2

2 21 /c c

u u cu u

− − −

=

− + −

2 22 2

21 1 /

c uu c

u c

= − − −

22 2 2 21 / 1 1 /

cu c u c

u...(1.14.5)

Now let us apply the law of conservation of mass and momentum in frame S. Let m be themass of the ball 1 before collision. Since the ball 2 is at rest in this frame, we denote its mass by m0.After collision the composite ball comes to at rest in frame S' and hence it appears to move withvelocity u'. In frame S, we have

mu = Mu'

m0 + m = M

Eliminating M from these equations, we have

′=

′−0

m u

m u u


Making use of Eqns. (1.14.4) and (1.14.5), we get

− − = =− − − −

22 2

2 2 20 2 2 2 2

1 1 /1

1 /1 / 1 1 /

cu c

m um c u cu c u c

u

0

2 21 /

mm

u c=

− = m0 ...(1.14.6)

In general if particle with rest mass m0 moves with velocity v relative to an observer, its effectivemass (or moving mass) is given by

= = = γ− −β

0 002 2 21 / 1

m mm m

v c...(1.14.7)

The relativistic momentum is defined by

p = = = γ−β0

021

m vmv m v (1.14.8)

The variation of Newtonian momentum and relativistic momentum of a particle with velocityv is shown in the Fig. (1.14.2 ).

Fig. 1.14.2 Variation of mass and momentum with velocity

1.15 THE FUNDAMENTAL LAW OF RELATIVISTIC DYNAMICS

The fundamental equation of classical mechanics (Newton’s laws) formulated in the form

=dm

dtF

v

is not invariant under Lorentz transformation. The correct law must, therefore, be formulated in


such a way that it must be Lorentz invariant and should transform to the classical law in the limitv/c 0. If Newton’s law is formulated in the form

= −

0

2 21 /

md

dt v c

vF ...(1.15.1)

it meets both the requirements. The formula F = maa cannot be used in relativistic case because theacceleration vector a of a particle does not coincidein the general case with the direction of the force F.In the relativistic case, we have

=

+ =

( )d

mdt

dm dm

dt dt

v

vv

F

F...(1.15.2)

This equation has been graphically illustrated in the Fig. (1.15.1). The acceleration vector a isnot collinear with the force vector F in the general case. The direction of acceleration vector a coincideswith that of F only in the two cases:

(i) F is perpendicular to v. In this case |v| = constant and therefore the equation of motion becomes

−

0

2 21 /

md

dt v c

v= F

−0

2 21 /

m d

dtv c

v= F

−0

2 21 /

m

v ca = F

a = − 2 2

0

1 /v c

m

F...(1.15.3)

(ii) F is parallel to v. In this case the equation of motion may be written in the scalar form as

−

0

2 21 /

md

dt v c

v= F

− +

− −

22 2

0 2 2 2

2 2

1 /1 /

1 /

d v dm v c

dt dtc v c

v c

v v

= F

Fig. 1.15.1


+

− −

2 2

0 2 2 3/ 22 2

1 /

(1 / )1 /

v c dm

dtv cv c

v= F

a =− 2 2 3/2

0

(1 / )v c

m

F...(1.15.4)

1.16 MASS-ENERGY EQUIVALENCE

The work done by unbalanced force acting on a particle appears as increment in kinetic energy. Theincrement in kinetic energy dT due to the force F acting over the elementary path dr (= v dt) isgiven by

dT = = = =. . ( ). ( ).d

d dt m dt d mdt

F Fr v v v v v

= +. .dm mdv v v v

= +2 .v dm m dv v

= 2v dm mvdv+ ...(1.16.1)

The mass of the particle varies with velocity as

m = 0

2 21 /

m

v c−

whencem2c2 = m2v2 + m0

2c2

Taking differential of this equation we have

= +2 2 22 2 2mc dm mv dm m vdv

Canceling the common factors we havec2dm = v2 dm + mvdv ...(1.16.2)

Making use of Eqn. (1.16.2), we can write the expression for increment in kinetic energy asdT = c2 dm ...(1.16.3)

The total kinetic energy of the particle at the instant it acquires velocity v is given by

0

T2

0

Tm

m

d c dm=∫ ∫

( ) ( )2 20 0T 1m m c m c= − = γ − ...(1.16.4)


20 2 2

1T 1

1 /m c

v c

= − −

...(1.16.5)

This is the expression for the relativistic kinetic energy of a particle. For small velocities(v/c 0) Eqn. (1.16.5) reduces to the classical formula.

1/ 222

0 2T 1 1

vm c

c

− = − −

=

+ + + −

2 42

0 2 4

1 31 ....... 1

2 8

v vm c

c c

= 20

1

2m v (classical result)

The variation of relativistic and classical energy with velocity is shown in the figure. Theexpression (1.16.4) for kinetic energy can be written as

mc2 = T + m0c2 ...(1.16.6)

Einstein interpreted the term mc2 on the left hand side as the total energy E of the particle.Thus

E = mc2 = T + m0c2 ...(1.16.7)

If the particle is at rest, its kinetic energy T is zerobut it still possesses energy equal to m0c2. This energy iscalled the rest energy. In relativistic physics total energy ofa particle means the sum of kinetic energy and rest energy.The expression

E = 2

2 2002 21 /

m cmc m c

v c= = γ

− ...(1.16.8)

is one of the most fundamental laws of nature expressingthe relationship between the total energy E of a particle andits mass. Equation (1.16.8) states that a change in total energyof a particle by amount E is equivalent to change in massby amount m = E/c2 and vice-versa.

1.17 RELATIONSHIP BETWEEN ENERGY AND MOMENTUM

The total energy E and momentum p of a particle are

20E m c= γ ...(1.17.1)

Fig. 1.16.1 Variation of kinetic energywith velocity


= γ ∴ = γ0 0p m v pc m vc ...(1.17.2)

Now

2 2 2 2 2 4 2 20E ( )p c m c v c− = γ −

2 2 4 2 20 (1 / )m c v c= γ −

2 20( )m c= ...(1.17.3)

This equation gives relation between energy and momentum. The quantities on the right handside are invariant. Therefore E2 – (pc)2 is also invariant.

1.18 MOMENTUM OF PHOTON

The momentum of a particle moving with velocity v is given by

p = 2

Emv v

c= ...(1.18.1)

Photon is a quantum of light with energy E = hv and velocity c. Its momentum is

p = E h h

c c

ν= =λ ...(1.18.2)

or E = pc ...(1.18.3)

From Eqn. (1.17.3), we see that the rest mass of photon is zero. In fact all particles movingwith speed of light c have zero rest mass. Photon and neutrino are such particles. Converse is alsotrue i.e., particles having zero rest mass always move with speed of light.

1.19 TRANSFORMATION OF MOMENTUM AND ENERGY

The total energy E and momentum p of a particle are velocity dependent and hence are not invariant.In this section we shall obtain transformation equations for energy and momentum.

Let a particle move from point (x, y, z) to point (x + dx, y + dy, z + dz). The distance dl coveredin time dt is dl2 = (dx)2 + (dy)2 +(dz)2. The velocity of the particle is v = dl/dt. The elementaryinterval between the initial and the final point is

ds = 2 2 1/ 2[( ) ( ) ]cdt dl−

= −

1/ 22

21

vcdt

c

=γ

cdt


whence

γ= dsdt

c...(1.19.1)

Now the x-component of momentum of a particle is

= γ = γ =

00 0x x

m cdxp m v m dx

dt ds ...(1.19.2)

Similarly

0

0

y

z

m cp dy

ds

m cp dz

ds

= =

...(1.19.3)

The energy of the particle is given by

or

2 2 200 0

02

E

E

m cdtm c m c c dt

dt ds

m cdt

dsc

= γ = γ =

=

...(1.19.4)

Since m0, c and ds are invariants, from Eqns. (1.19.2–1.19.4) it follows that px, py, pz and E/c2

transform like dx, dy, dz and dt when we pass over from one inertial frame to another. Thus thetransformation equations for the components of momentum and energy, together with those forcoordinates and time are

2( E / ),x xp p v c′ = γ − ( )x x vt′ = γ − ...(1.19.5)

yp′ = py y' = y

zp′ = pz z' = z

2 2 2

E E,xvp

c c c

′ = γ −

2

vxt t

c

′ = γ − ...(1.19.6)

Remembering the following correlations

2, , , (E / )x y zp x p y p z c t⇔ ⇔ ⇔ ⇔

we can write out the transformation equations for energy and momentum if we know the correspondingequations for time and space coordinates.


1.20 VERIFICATION OF MASS-ENERGY EQUIVALENCE FORMULA

There are many phenomena in which Einstein’s formula E = mc2 gets its experimental verification.Of these we shall discuss only two.

Pair production: In this phenomenon a gamma ray photon while passing nearby an atomicnucleus materializes into an electron-positron pair. The presence of third particle, such as nucleus inthe present case, is necessary to conserve linear momentum. According to the law of mass-energyequivalence the minimum energy of photon for pair production must be equal to the sum of the restenergies of the created particles. The rest mass energy of electron-positron pair is 2m0c2 = 1.02 MeV.Indeed this has been verified. A gamma ray photon with energy less than 1.02 MeV cannot produceelectron-positron pair whereas that having energy greater than 1.02 MeV creates electron-positronpair and the excess energy goes into the kinetic energies of the particles. The threshold wavelengthof gamma ray photon is given by

h = 2

02ch

m c=λ

= 2

02

ch

m c

312.4 10 MeV.Å0.012 Å

1.02 MeV

−×= =

Fig. 1.20.1 Pair production and pair annihilation

When electron and positron meet, they annihilate each other creating at least two photons. Thisphenomenon is called pair annihilation.

–1e0 + +1e0 h + h

(electron) (positron) (photon) (photon)


1.21 NUCLEAR BINDING ENERGY

When nucleons (protons and neutrons) combine to form an atomic nucleus, the mass of the resultingnucleus is always less than the sum of masses of the constituent nucleons. The mass that disappearsis called the mass defect and is liberated as binding energy of the nucleus. For a nucleus with massnumber A (number of nucleons) and atomic number Z (number of protons) the mass defect is givenby

m = [Zmp + (A – Z) mp – Mnuc] ...(1.21.1)

and the binding energy of the nucleusB = (m) c2 ...(1.21.2)

Let us calculate the binding energy of 2He4, which consists of two protons and two neutrons.Mass of protons = 2 (1.00815) amu = 2 (938.7) MeVMass of neutrons = 2 (1.0080) amu = 2 (939.5) MeVMass of nucleus = 4.00260 amu = 3728.0 MeVBinding energy = (1877.4 + 1879.10 – 3728.0) MeV = 28.5 MeVBinding energy per nucleon = 28.5/4 = 7.1 MeV

SOLVED EXAMPLES

Ex. 1. Verify that the law of conservation of linear momentum is invariant under Galileantransformation.

Sol. Consider the collision two balls of masses m1 and m2 moving with velocities u1 and u2

relative to ground. After the collision the balls move with velocities v1 and v2. The statement of thelaw of conservation of linear momentum in frame S fixed to the ground reads

m1u1 + m2u2 = m1v1 + m2v2 ...(1)

The same collision is observed from a frame S', which is moving relative to S with velocity v.In frame S' the velocities of the balls before and after collision are

1u′ = u1 – v, 2u′ = u2 – v, 1v′ = v1 – v, 2v′ = v2 – v ...(2)

In frame S' the law of conservation of linear momentum is

m1 1u′ + m2 2u′ = m1 1v′ + m2 2v′ ...(3)

Making use of (2) we can transform (3) as

m1 (u1 – v) + m2 (u2 – v) = m1(v1 – v) + m2(v2 – v)

m1u1 + m2u2 = m1v1 + m2v2

which is the law of conservation in frame S.

Ex. 2. Show that the law of conservation of kinetic energy in elastic collision of two particles isinvariant under Galilean transformation.

Sol. In frame S, let us consider perfectly elastic collision of two particles of masses m1 and m2

moving with velocities u1 and u2 in x direction. After collision their velocities are v1 and v2. The


law of conservation of kinetic energy in frame reads as:

2 21 1 2 2

1 1

2 2m u m u+ =

2 21 1 2 2

1 1

2 2m v m v+ ...(1)

The same collision is observed from frame S', which moves with velocity v in x direction. Theinitial and final velocities of the particles in this frame are:

1u′ = u1 – v, 2u′ = u2 – v, 1v′ = v1 – v, 2v′ = v2 – v

Substituting the values of u1, u2, v1, v2 from above equations in (1), we have

2 2

1 1 2 21 1

( ) ( )2 2

m u v m u v′ ′+ + + = 2 2

1 1 2 21 1

( ) ( )2 2

m v v m v v′ ′+ + + ...(2)

Subtracting (1) from (2) and making use of the law of conservation of momentum in the resultingequation, we find

2 2

1 1 2 21 1

2 2m u m u′ ′+ =

2 21 1 2 2

1 1

2 2m v m v′ ′+ ...(3)

which is the law of conservation of kinetic energy in frame S'.

Ex. 3. An event occurs at x = 50 m, y = 20 m, z = 10 m, and t = 5 × 10 –8s in frame S. What arethe space-time coordinates of the event as measured by an observer stationed in frame S', which ismoving relative to S with velocity 0.6c along the common x–x' axis.

Sol. Lorentz transformations arex' = (x – vt), y' = y, z' = z, t' = (t – vx/c2)

= 1/ (1 – 2) = 1/(1 – 0. 36) = 1.25

x' = 1.25 (50 m – 0.6 × 3 × 108m/s × 5 × 10–8s) = 51.25 m

y' = 20 m, z' = 10 m

t' = 1.25 (5 × 10–8 s – 10 × 10–8 s) = – 6.25 × 10– 8 s.

Ex. 4. An event 1 occurs at x1 = 20 m, t1 = 2 × 10–8s and event 2 occurs at x2 = 60 m,t2 = 3 × 10–8s in frame S. A frame S' moves relative to S with velocity 0.6 c along the common axisx – x'. (i) What is the spatial separation of the events in frame S? (ii) What is the time interval betweenthe two events?

Sol. (i) The spatial separation of the events in frame S' is

2x′ – 1x′ = [(x2 – x1) – v(t2 – t1)]

= 1. 25 [(60 –20) m – 0. 6 × 3 × 10–8 m/s (3 – 2 ) × 10–8 s ]

= 47. 75 m.

(ii) The time interval between the events in frame S' is

2t′ – 1t′ = [(t2 – t1) – v (x2 – x1)/c2]

= 1. 25 [1 × 10–8 s – (0. 6 × 40)/ (3 × 108)]

= – 8.75 × 10–8 s.


Ex. 5. The space-time coordinates of two events 1 and 2 in a frame S are x1= 24 m,t1 = 8 × 10–8s and x2 = 48 m, t2 = 4 × 10–8s. Find the velocity of the frame S' in which both the eventsoccur simultaneously.

Sol. The time interval between the events in frame S' is

( 2t′ – 1t′ ) = [(t2 – t1) – v(x2 – x1) /c2]

For the events to occur simultaneously in frame S', 2t′ – 1t′ = 0. Hence

( ) ( )8 82 1

2 1

3 10 m/s 4 10 s0.5

48m 24m

c t tv

c x x

−× − ×−= = = −

− −

v = – 0.5c.

Ex. 6. Two events occur at the same place in a frame S and are separated by a time interval of 1s.The same events are separated by a time interval of 4 s in frame S'. What is the spatial separation of theevents in frame S'?

Sol. The spatial separation of events in frame S' is

2x′ – 1x′ = [(x2 – x1) – v(t2 – t1)]

Given that: x2 – x1 = 0, t2 – t1 = 1s, 2t′ – 1t′ = 4 s

2x′ – 1x′ = –v ...(1)

The time interval between the events in frame S' is

2t′ – 1t′ = [(t2 – t1) – v(x2 – x1)/c2]

4 = = 15

4...(2)

From (1) and (2), we have 2x′ – 1x′ = ( )22 2 151 c c− = −γ − .

Ex. 7. In a frame S, two events occur at the same time and are separated by a distance x0. Inanother frame S', which is moving along the common x-x' axis with constant velocity, the events havespatial separation 2x0. What is the time interval between the events in frame S?

Sol. Given that: t2 – t1 = 0, x2 – x1 = x0, 2x′ – 1x′ = 2x0

Now spatial separation between events in frame S'

2x′ – 1x′ = [(x2 – x1) – v(t2 – t1)]

2x0 = x0

= 2 v = c3/2

2t′ – 1t′ = [(t2 – t1) – v (x2 – x1)/c2] = – 03

xc

.


Ex. 8. How fast should a rocket ship move for its length to be contracted to 99% of its rest length?Sol.

22 2

00

1 whence 1 1 (0.99) 0.141l

l ll

= −β β = − = − =

v = 0.141 c.

Ex. 9. A rod of proper length 1.00 m makes an angle of 45° with x-axis. Determine the length of therod and its inclination with x-axis in a frame with respect to which the rod is moving with velocityv = 0 .5c along the common x–x' direction.

Sol. Let l0 be the proper length of the rod and ' its inclination with x'-axis in frame S'. Thecorresponding quantities in frame S are l and . Then

lx = lx' −β 21 or 2cos cos 1l l′ ′θ = θ −β ...(1)

y yl l′= or sin sin .l l′ ′θ = θ ...(2)

Squaring and adding (1) and (2), we have

l2 = 2 2 2 2[sin (1 )cos ]l′ ′ ′θ + −β θ

= + − = 1 1 1

1.00 1 0.8752 4 2

l = 0.94 m

From (1) and (2), we have

tan = 2

tan 11.19

1 1/ 41

′θ = =−−β

= 49°.

Ex. 10. A piece of metal in the form of a rectangular bloc of dimensions a, b, c is placed with itsedges parallel to the coordinate axes of a frame S. Its density is defined as the ratio of the rest mass to thevolume. Find the velocity of the reference frame S' moving parallel to the edge a in which the density ofbloc is 25% greater.

Sol. The density of the bloc in frame S

ρ = 00

m

abc

Let v be the velocity of frame S' relative to S. Its volume in frame S' is −β 21abc . The

density of the bloc in S' is

0 0

2 2.

1 1

m

abc

ρρ = =

−β −β


Given that: = (1 + 0.25) 0

2

11.25

1=

− β

= 0.6.

Ex. 11. A clock keeps correct time. With what speed should it move relative to an observer so thatit may seem to lose 4 min in 24 hours?

Sol. Proper time = 1436 min, non-proper time t = 1440 min. Time dilation formula is

2

2

1436 41 1

1440 14401t

∆τ∆ = ∴ −β = = −−β

− β = − = −

22 1 2

1 1 1360 360

= 1

.180

Ex. 12. A particle with mean proper lifetime of 1µ s moves through laboratory at 2.7 × 108 m/s.(i) What is the lifetime of the particle as measured in the laboratory frame? (ii) How far does it go in thelaboratory before disintegration?

Sol. (i) The lifetime of the particle in the lab frame is given by

66

2 2

1 102.29 10 s

1 1 (0.9)

st

−−∆τ ×∆ = = = ×

− β −

(ii) The distance traversed in the lab framed = v t = 2.7 × 108 m/s × 2. 29 × 10–6 s = 62 × 10 3 m.

Ex. 13. In a frame S, a muon moving with velocity v = 0.99 c travels through a distance 3.0 kmfrom its birthplace to the point where it decays. Find (i) proper lifetime of the muon (ii) the distancetravelled by muon in the frame in which it is at rest.

Sol. The proper lifetime of muon

2 2 21 1 /

dt v c

v∆τ = ∆ −β = −

32 6

8

3 10 m1 (0.99) 1.4 10 s.

0.99 3 10 m/s

−×= − = ×× ×

In muon’s frame the distance travelled by it is

2 3 2 31 (3 10 m) 1 (0.99) 0.423 10 m.d d′ = − β = × − = ×


Ex. 14. Two particles are observed to be moving with speeds 0.6 c and 0.8 c in opposite directionsin the laboratory frame. Calculate the speed of one particle relative to the other.

Sol. In the lab frame S, the speeds of the particles are

u1 = 0.6 c and u2 = –0.8 c.

Let us consider a frame S' attached to the first particle. Obviously the frame S' is movingwith velocity v = 0.6 c. In S' the first particle is at rest. Let the velocity of the second particle in

S' be 2u′ . Then according to the relativistic transformation of velocity

2u′ = 22 2

2

0.8c 0.6c0.946c.

1 / 1 ( 0.8c)(0.6c)/c

u v

u v c

− − −= = −− − −

Ex. 15. At what speed must a particle move for its mass be trebled?

Sol. Given that: m = 3 m0. From the formula =−β

0

21

mm we have

β = − = − = ∴ =0 1 2 2 2 21 1 c.

9 3 3

mv

m

Ex. 16. At what speed the total energy of a particle is n times its rest energy?

Sol. Given that m c2 = n m0 c2, m = n m0. From the formula =−β

0

21

mm we have

2

2 2 2

1 1 11 1 .

nv c

n n n

−β = = − ∴ = −

Ex. 17. At what speed the kinetic energy of a particle is n times its rest energy?Sol. Given that T = n m0 c2

(m – m0) c2 = n m0 c2

m = (n + 1) m0

( ) ( )002

21 .

11

n nmn m

n

+= + ∴β =

+−β

Ex. 18. If the total energy of a particle is n times its rest energy, what is its momentum?Sol. Given that E = n m0 c2. We know that the total energy of a particle is given by

E2 = 2 2 20( ) ( )pc m c+

pc = 2 2 2 2 2 2 20 0 0( ) ( ) ( )E m c nm c m c− = +

p = 20 1.m c n −


Ex. 19. At what momentum the kinetic energy of a particle equals n times its rest energy?Sol. Given that: T = n m0 c2

We know that

E2 = 2 2 2 2 20 0[T ] ( ) ( )m c pc m c+ = +

pc = 20 ( 2)m c n n +

p = 0 ( 2).m c n n +

Ex. 20. Show that the velocity of a particle having total energy E is given by

1/ 22201 .

m cv c

E

= −

Sol. We know that,

E2 = 2 2 2

0( ) ( )pc m c+ ...(1)

p = 2

E.v

c ...(2)

Substituting the value of p from (2) in (1), we get

E2 = 2

2 2 202

E ( )v

m cc

+

v = 2 2 1/ 201 ( /E) .c m c −

Ex. 21. Show that the momentum p of a particle can be expressed in terms of kinetic energy T as

20( 2 )pc T T m c= +

Calculate the momentum of proton having kinetic energy of 500 MeV.

Sol. Since E2 = 2 2 2 2 20 0(T ) ( ) ( )m c pc m c+ = +

2 2 2 2 20 0( ) (T ) ( )pc m c m c= + −

20T(T 2 ).pc m c= +

Ex. 22. Show that the velocity of a particle having momentum p is given by

2 2 2 1/ 20 .v pc p m c − = +


Sol. Squaring the equation =

−

0

2

21

m vp

v

c

and rearranging, we get

− =

22 2 2

021

vp m v

c

v =2 2 2

0

.pc

p m c+

Ex. 23. Calculate the momentum of electron having kinetic energy equal to 2 BeV.Sol. Total energy

E2 = 2 2 2 2 20 0(T ) ( ) ( )m c pc m c+ = + .

If the rest energy is very small in comparison to the kinetic energy, then we can write theabove formula as

E = T = pc.

The rest energy of electron is m0 c2 = 0.51 MeV. So we can neglect the rest energy of electron

in comparison to its kinetic energy (2 BeV). Thus the momentum of electron is

p = T/c = 2 BeV/c.

Ex. 24. At what velocity does the Newtonian momentum of a particle differ from its relativisticvalue by 1%?

Sol. Classical momentum = 0cp m v and relativistic momentum =−β0

21r

m vp .

r c

r

p p

p

−= 21 1− −β

n = 1 – −β 21 where n is fractional increase in momentum.

From above equation, we get

= (2 )n n−

Putting n = 0.01, we have = 0.14.

Ex. 25. Find the velocity at which the relativistic momentum of a particle exceeds its Newtonianmomentum by n fold.

Sol. 2

2

2

1 1.

11

r

c

p n c nv

p nv

c

−= = ∴ =

−


Ex. 26. An electron is accelerated through a potential difference of 1 million volts. What is thespeed of electron?

Sol. Kinetic energy of electron

T = 2

02

11

1m c

− − β

=

220

20

.1T

m c

m c

− +

Putting m0 c2= 0. 51 MeV and T = 1 MeV, we get

= 0. 9988.

Ex. 27. An electron is accelerated to energy of 2 BeV. Calculate (i) the effective mass of electron interms of its rest mass, (ii) the speed of electron in terms of the speed of light.

Sol. Total energy of electron

2 20Tmc m c= +

Dividing throughout by m0 c2, we have

20 0

T 2000 M V1 1 3915

0.51 M V

m e

m em c= + = + =

From the relation =−β

0

2,

1

mm we have

= 2 2

0 1 .1 13915

m

m

− = −

Ex. 28. What work has to be done in order to increase the velocity of a particle of rest mass m0 from0.60 c to 0.80 c? Obtain the result classically and relativistic

Sol. (i) Classically, the work done equals the increase in kinetic energy of the particle

W = 2 2 2 2 20 0 0 1 0 0

1 10.5 [(0.8 ) (0.6 ) ] 0.14

2 2m v m v m c c m c− = − =

(ii) On the basis of relativistic calculation, the required work W

= 2 22 1 2 1E E m c m c− = −


=

− = − −

2 20 02 2

2 12 2

1 10.417 .

1 1

m c m cv v

c c

Ex. 29. Calculate the effective mass of photon of wavelength (i) 5000 Å (ii) 0.1 ÅSol. (i) The relativistic energy of photon is E = mc2 and in terms of quantum picture photon’s

energy is E = hv = ch/ . Thus

mc2 =34

3610 8

6.63 10 Js4.42 10 kg

(5000 10 m) (3 10 m/s)

ch hm

c

−−

−×∴ = = = ×

λ λ × ×

(ii) The effective mass of photon of wavelength = 0.1 Å

m = 34

3110 8

6.63 10 Js2.21 10 kg.

0.1 10 m 3 10 m/s

h

c

−−

−×= = ×

λ × × ×

Ex. 30. Calculate the minimum energy of a gamma ray photon, which can produce an electronpositron pair.

Sol. In pair production a gamma ray photon materializes into two particles electron and positron.For pair production to take place the energy of gamma ray photon must at least be equal to the sumof the rest energies of the electro and positron. Thus

E = 2 m0c2 = 2 (0. 51 MeV) = 1. 02 MeV.

Ex. 31. A nucleus of mass m emits a gamma ray photon of frequency v. Show that the loss ofinternal energy of the nucleus is given by

E = 2

0

1 .2

hh

m c

νν +

Sol. The momentum of emitted photon is p = hv/c. To conserve linear momentum the nucleusrecoils in opposite direction with equal momentum. If v is the recoil velocity of the nucleus then

mv = .h

c

ν...(1)

The loss of internal energy of nucleus equals the sum of kinetic energy of recoil of the nucleusand the energy of photon. Thus

E = 21

2h mvν +

= 2

2

1 ( )

2

hh

mc

νν +


= 2

1 .2

hh

mc

ν ν +

Ex. 32. A particle of mass m0 is subjected to a force F acting along x-axis. The particle startsmoving from rest at x = 0 at t = 0. Find the position x as function of time.

Sol. The equation of motion of the particle is

dp

dt= F

0

2

21

m vd

dt v

c

−

= F ...(1)

Integrating and using the initial condition: at t = 0, v = 0, we have

0

2 21 /

m v

v c−= Ft

Solving for v, we find

v = 0

20

( / )

1 (F / )

F m tdx

dt t m c=

+...(2)

Integrating (2), we have

x = 20 0 0

F

1 (F / )

ttdt

m t m c+∫

= 22

0

0

F1 1 .

F

m c t

m c

+ −

Ex. 33. The earth receives solar energy about = 1.4 × 103 J/m2s over an area held normal to thesun-rays. Calculate the fractional loss of mass of the sun per sec. The mean radius of the earth orbit isR =1.5 × 1011m and the mass of sun is M = 1.97 × 1030kg.

Sol. The amount of energy radiated per sec by the sun

E = 4 R2


This energy is equivalent to mass m = E/c2. The fractional loss of mass is

M

m∆=

2 11 2 3 2 1

2 2 30 8 2

E 4 R 4 3.14 (1.5 10 m) (1.4 10 Jm s )

M M (1.97 10 kg)(3 10 m/s)c c

− −π σ × × × ×= =× ×

= 2 × 10–21 s–1.

Ex. 34. A particle moves relative to frame S with velocity u in xy- plane at angle to the x-axis.Find the corresponding velocity u' at angle ' in the frame S' which moves with velocity v relative to Salong the common x–x' direction

Sol. The x and y components of velocity in frame S areux = u cos, uy = u sin.

Let the velocity u' of the particle make angle ' with x'-axis in frame S'. The x and y componentsof u' can be obtained from velocity transformation formula.

xu′ = 2 2

cos

1 / 1 ( cos ) /x

x

u v u v

u v c u v c

− θ −=− − θ

...(1)

yu′ = 2 2

2 2

1 sin 1

1 / 1 ( cos ) /

y

x

u u

u v c u v c

−β θ −β=

− − θ...(2)

From (1) and (2)

tan ′θ = 2 2sin 1 sin 1

cos cos ( / )y

x

u u

u u v v u

′ θ − β θ − β= =

′ θ − θ − ...(3)

Inverse transformation is

tan = 2sin 1

cos ( / )v u

′θ − β′ ′θ +

...(4)

The velocity u' is given by

2u′ = 2 2 2 2 2

2 222

2 cos sin.

1 ( cos ) /x y

u v uv uu u

vu c

+ − θ − β θ′ ′+ = + θ

...(5)

Ex. 35. Acceleration transformation: A particle moves with acceleration a in frame S. Find theacceleration a' of the particle in the frame S' which moves with velocity v in the positive direction ofx-axis of frame S.

Sol. The x-component of acceleration in frame S' is given by

xa′ = 1

/x xdu du

dt dt dt dt

′ ′=

′ ′ ...(1)


Velocity transformation is

xu′ = 2.

1 /x

x

u v

u v c

−−

...(2)

Differentiating (2) with respect to t, we have

xdu

dt

′=

22

2 2

(1 / ) ( )

(1 / )

x xx x

x

du du vu v c u v

dt dt c

u v c

− − − − −

= 2 2

2 2

(1 / )

(1 / )x

x

v c a

u v c

−−

...(3)

The Lorentz transformation for time is

t' = 2

2 2

/

1 /

t vx c

v c

−

−

dt

dt

′=

2

2 2

1 /

1 /

xu v c

v c

−

− ...(4)

The acceleration in frame S' is

xa′ =2 2 3/ 2

2 3

(1 / ).

(1 / )x x

xx

du du dt v ca

dt dt dt u v c

′ ′ −= =′ ′ −

...(5)

Ex. 36. Derive the relativistic aberration formula from the velocity transformation equations.Sol. Consider a frame S' moving relative to S with velocity v along the common x–x' direction.

A source at origin of S' emits a light wave in the direction ' with x-axis. In frame S', we have

xu′ = cos , sinyc u c′ ′ ′θ = θ

In frame S

ux = 2

cos

1 ( / ) cos1 /x

x

u v c v

v cu v c

′ ′+ θ +=′+ θ′+

uy =

2 2

2

1 sin 1

1 ( / )cos1 /

y

x

u c

v cu v c

′ − β ′θ −β=

′+ θ′+

tan = 2sin 1

.cos

y

x

u

u

′θ − β=

′θ + β


QUESTIONS

1. What do Galilean Transformation and Galilean invariance principle mean? Show that whereas the length andacceleration are invariant this transformation, velocity is not.

2. What are Galilean transformations? Discuss the Galilean invariance of Newton’s equation of motion. What is thesignificance of Michelson–Morley experiment?

3. Discuss the theoretical background of M.M experiment. Briefly describe the experimental set up and whatconclusions of importance have been obtained from this experiment.

4. Describe the essential features of M.M. experiment and discuss the significance of this result in the developmentof special theory of relativity.

5. What are Lorentz transformations? Show that two events, which are simultaneous in one frame of reference, arenot simultaneous in other frame of reference in relative motion with the first.

6. Write down the Lorentz transformation equations. Explain the phenomenon of time dilation and length contraction.

7. State the postulates of special theory of relativity and show how Lorentz transformations have been obtainedfrom them.

8. (a) Assuming the invariance of equation of light wave front in two inertial frames of reference with uniformrelative velocity, derive the Lorentz transformations.

(b) Show that the length of a moving rod appears to be contracted.

9. Write Lorentz transformation equations. Show that two successive Lorentz transformations in the same directionare equivalent to a single transformation. Find the equivalent velocity.

10. Prove that the equation of a spherical pulse of light starting from origin at t = t' = 0,x2 + y2 + z2 – c2 t2 = 0 is invariant under Lorentz transformation.

11. Show that the differential equation dx2 + dy2 + dz2 – c2 dt2 = ds2 is invariant under a Lorentz transformations.

12. A light pulse is emitted at the origin of a frame of reference s at time t' = 0. Its distance x' from the origin after atime t' is given by (x')2 = c2 t'2. Use the Lorentz transformation to transform this equation to an equation in x andt and show that the result is x2 = c2 t2. Discuss the implications of this result.

13. State Lorentz transformation equations and use them to obtain Einstein’s addition theorem of velocities. Discussthe physical significance of the theorem.

14. Derive Einstein’s law for the addition of two velocities. What happens if one of the particles moves with thevelocity of light?

15. Derive the formula 0 2

1

1m m=

−β where = v / c.

16. Derive the Einstein’s formula expressing the equivalence of mass and energy.

17. Obtain the relativistic expression for kinetic energy of a body and show that for small speeds it reduces to theclassical expression.

18. Derive the following relations:

(i) E = mc2, (ii) E2 = (pc)2 + (m0c2)2 (iii) T = 20 2

11

1m c

− −β

(iv) T = + −2 2 20 0( ) ,c p m c m c p =

20

1T(T 2 ) ,m c

c+ (v) m0 =

2 2 21E p c

c−


PROBLEMS

[Lorentz Transformation, Space Contraction and Time Dilation]

1. An event occurs at x' = 60 m, t' = 5 × 10–8 s in frame S'. The frame s' moves with velocity 0.6 C along thecommon x-x' direction with respect to the frame s. The origins of s and s' coincide at t' = t = 0. Find the spacetime coordinates of the event in frame s. [Ans. x = 86.25 m, t = 21.25 × 10–8 s]

2. A rocket was found to be of length 100 m when measured on the earth. It then leaves the earth and moves awayat a constant speed 2 × 108 m/s. What will be its length now as measured from the earth

C = 3 × 108 m/s.

3. A rod has length 100 cm when the rod is in a satellite moving with velocity 0.8 C relative to the laboratory. Whatis the length of the rod as determined by an observer in the laboratory?

4. A rod 1 metre long is moving with a velocity 0.6 C. Calculate its length as it appears to (a) a stationary observer(b) an observer moving with the rod itself. [Ans. (a) 0.8 m (b) 1 m]

5. A rocket is 100 m long on the ground. When it is in the flight its length is 99 m to an observer on the ground.Calculate its speed. [Ans. v = 0.14 c]

6. Calculate the percentage contraction of a rid moving with velocity 0.8 c in a direction inclined at 60° to its ownlength. [Ans. 8.0%]

Hint: The component of length parallel to the direction of motion is L0 cos 60° = 1/2 L0. Only this componentchanges in length. The length of parallel component measured by an observer with respect to which the rod is

moving with velocity 0.8 c is = − =20 0

1L L 1 (0.8) 0.3L .

2x

The component of length perpendicular to the direction of motion is Ly = L0 sin 60°= 3

2L0 and this length

remains unchanged.

The moving length of rod 2 20 0 0L (0.3L ) (0.866L ) 0.92L .= + =

Percentage contraction −

× =

0 0

0

L 0.92L100 8%.

L

7. With what velocity a spaceship fly so that everyday spent on it may correspond to 3 days on earth’s surface.

Hint: t0 = 1 day, t = 3 days. 02

8gives C.

31

tt

∆∆ = β =

−β8. What is the mean life of + mesons traveling with a velocity of 0.73 C. The mean life-time for + mesons at rest

is 2.5 × 10–8 sec. [Ans. 3.7 × 10–8 sec]

9. The mean life-time of pions in the laboratory is 2.5 × 10–7 sec. Calculate its velocity if the proper mean life-timeis 2.5 × 10–8 sec.

10. The mean life-time of a muon at rest is about 2 × 10–8 sec. It is moving with velocityv = 0.99 C. What is its mean life to a laboratory observer and what distance does it traverse before decaying.

[Ans. t = 14.18 × 10–8 s, d = v.A t = 42 m]

11. Half life-time of pions is 1.8 × 10–8sec. A beam of pions is produced with speed v = 0.8 c. Calculate the distance

traversed by the beam during which its intensity reduces to half its original value. [Ans. d = 7.2 m]


[Addition of Velocities]

12. Two electrons, each of velocity 0.8 c move towards each other. Find the relative velocity of one electron withrespect to other.

13. One -particle is moving east with a velocity of 0.95 c and a second -particle is moving west with a velocity

0.85 c. Find the relative velocity of the two -particles.

′ + += = = ′ + × + 2

0.85 0.950.99c

1 0.85 0.951

xx

x

u v c cu

vu c

c

Ans.

14. In the laboratory two -particles are observed to travel in opposite directions with speed 2.8 × 10 m/sec. Deducethe velocity of one particle relative to each other?

15. An observer in a laboratory ‘sees’ two particles coming towards him from opposite directions at speeds of 0.8 cand 0.9 c respectively. What is the relative speed of the two particles as measured by an observer moving witheither one. [Ans. 0.98 c]

16. An experimenter observes a radio-active atom moving with a velocity of c/4. The atom emits a beta particlewhich as a velocity of 0.9 c relative to the atom in the direction of its motion. What is the velocity of the beta

particle as observed by the experimenter. [Ans. 0.9 c]

[Variation of Mass with Velocity : Mass-energy Equivalence]

17. A relativistic particle whose rest mass is m0, is moving with a velocity of 0.9 c. Determine its kinetic energy.[Ans. T = (m – m0)c2 = (2.3 m0 – m0) c2 = 1.3 m0 c

2]

18. Calculate the speed of a particle of rest mass 3. 33 × 10–27 g whose energy is estimated to be 2 MeV.

19. A particle moves at a speed such that its kinetic energy is equal to its rest mass energy. What is the speed of theparticle?

20. Calculate the mass and speed of (i) 1 GeV (=1000 MeV) proton (ii) 2 MeV electron. Calculate the momentum ofa photon of energy 1 × 10–12 erg.

21. An x–rays beam has a wavelength of 1Å. Calculate (i) the energy of x-ray photon(ii) the momentum of photon (iii) mass of photon (iv) rest mass of photon.

22. A -ray photon of wavelength 0.0045 Å materializes into an electron-positron pair in the neighbourhood of aheavy nucleus. What is the kinetic energy of the pair in MeV. [Ans. 1.73 MeV]

23. Calculate the binding energy in MeV for deuteron. mn = 1.008962 amu, mp = 1.008142 amu,md = 2.01470 amu.

24. The rest mass of hydrogen atom is 1.00727 amu and that of proton and electron are mp = 1.00727 amu andme = 0.005486 amu. Calculate the binding energy of the hydrogen atom.

25. A 0.50 MeV electron enters normally a region of magnetic field of 5 × 10–3 Waber/m2. Calculate the radius ofthe trajectory classically and relativistically. [Ans. 47.8 cm, 58.2 cm]

26. Deuterium is the isotope of hydrogen consisting of deuteron nucleus (containing a proton and a neutron) and anelectron revolving around it in an orbit. The ionization potential of deuterium is 13.6 eV and the energy requiredto separate the neutron and proton is 2.23 MeV. Determine the mass of the deuterium atom if the masses ofneutron, proton and electron are 1.6748 × 10–27kg, 1.6728 × 10–27 kg and 9.1 × 10–31 kg respectively.

27. Calculate the threshold wavelength for pair production.

Hint: 202 .

chm c=

λ

QUANTUM MECHANICS

UNIT


blank

CHAPTER

ORIGIN OF QUANTUM CONCEPTS

1.1 INTRODUCTION

The observation of the phenomena of interference, diffraction and polarization led to the developmentsof the wave theory of light. The negation of the concept of ether through the development of specialtheory of relativity raised serious questions regarding the nature of light waves. A large number ofexperimental evidences conclusively demonstrated that light was an electromagnetic phenomenon.The understanding of the emission and absorption of radiation by matter posed some difficulties.However this difficulty was partially resolved by making some ad-hoc assumptions regarding thestructure of matter. It was assumed that atoms and molecules, which constitute matter, consist ofelectron oscillators, begin to oscillate under the influence of some external source of excitation. Sincethe oscillating electron is an accelerated charged particle, it radiates electromagnetic radiation.Therefore, when electromagnetic wave is incident on such an atomic oscillator, electrons are set intoforced oscillations, which in turn emit electromagnetic waves of frequency equal to that of the incidentwave. To explain the phenomenon of absorption it was assumed that some kind of dissipative forceof viscous type, whose exact origin was not known at that time, act on atomic oscillator. Thetransformation of incident electromagnetic energy into other form on account of dissipative forcescauses loss of energy. At the end of 19th century, like classical mechanics, the electromagnetic theoryof radiation was regarded as the ultimate theory of radiation. At this time when the classical physicswas at the zenith of its accomplishments, some physicists were facing problems that could not beunderstood within the framework of classical physics. The most outstanding problems among themwere (i) the explanation of line spectrum emitted by elements in gaseous state (ii) the emission ofelectrons when metals are exposed to high frequency radiation—the photoelectric effect, (iii) thedistribution of energy in the spectrum of a black body. Probably, the understanding of these phenomenaindicated a different aspect of the nature of radiation. In other words, the understanding of phenomenaassociated with interaction of radiation with matter requires a fundamental change in our conceptsregarding the nature of radiation and the structure of matter. In this chapter we shall restrict ourselvesto the study of those phenomena, which have direct bearing to the nature of light. We shall see thatthe explanation of these phenomena requires that the radiation should be regarded as a stream ofparticles rather than waves. The theory, which regards radiation as stream of particles (called quanta)is called quantum theory of radiation.


Historically the quantum theory originated in an attempt to explain the spectral distribution ofthermal energy radiated by a black body. We shall, therefore, begin our study with a brief descriptionof characteristic features of the distribution of thermal energy in black body radiation.

1.2 BLACK BODY RADIATION

The thermal radiation emitted by a hot body, in general, depends on the composition and thetemperature of the body. However, there is a class of bodies, called black bodies, which emit thermalradiation whose quantity and quality depend only on their temperature. For this reason the radiationemitted by such bodies is called the thermal radiation. Such bodies are named black bodies becausethey absorb all the radiation that falls on it. Lamp black and platinum black are nearest to idealblack bodies. When a black body is maintained at constant high temperature, the emitted radiation iscalled the black body radiation. For experimental purposes a cavity having a small hole can be regardedas a perfect black body because of its identical behavior with that of the perfectly black body. Ifradiation is allowed to enter such a cavity, it is reflected back and forth at the inner walls of thecavity and at each reflection some portion of energy is absorbed. After suffering a large number ofreflections at the walls it is completely absorbed in the cavity. Therefore at lower temperature thehole appears black. When the cavity is maintained at higher temperature the radiation that comes outof the hole is similar to that emitted by a black body at the same temperature. Thus, a cavity with asmall hole acts like a black body.

Radiant Emittance

The power emitted by unit area of a black body in all directions (within the limit of a solidangle 2) is called radiant emittance and is denoted by R. The emitted radiation consists of waveshaving different frequencies (wavelengths). The power emitted by unit area of a black body withinthe frequency interval d (wavelength interval d) is called the spectral radiant emittance E or E

and is defined such that

R = ∞ ∞

ω λω = λ∫ ∫0 0

E Ed d ...(1.2.1)

The spectral emittance E(, T) or E(, T) is function of frequency (wavelength) and temperatureand describes the spectral composition of the thermal radiation.

The black body radiation is characterized by spectral energy density u(,T) or u(,T) which isdefined such that the total energy density u(T) at a temperature T for all frequencies (wavelengths)is given by

u(T) = 0 0

( ,T) ( ,T)u d u d∞ ∞

ω ω= λ λ∫ ∫ ...(1.2.2)

It can be shown that the radiant emittance of a black body is related to energy density u asR = (1/ 4) c u (,T)

This relation holds for all frequencies. Hence

R =

∞ ∞

ω ω = λ λ∫ ∫0 0

( ,T) ( ,T)4 4

c cu d u d (1.2.3)

Origin of Quantum Concepts 51

1.3 SPECTRAL DISTRIBUTION OF ENERGY IN THERMAL RADIATION

The experimental study of thermal radiation emitted from a black body was initiated by an excellentgroup of German spectroscopists Lummer, Pringsheim, Rubens and Karlbaum. The spectral energydensity u

or u

was measured at different temperatures with special spectrograph. The energy density

was plotted against the wavelength at different temperatures. The results obtained may be summarizedas follows:

1. At a particular temperature the spectral energy density increases with wavelength and attainsa maximum value and then falls to zero for longer wavelengths.

2. As the temperature is increased, the wavelength max corresponding to maximum energydensity shifts towards the shorter wavelength region. It was experimentally found by Weinthat

max

T = b = 2.898 × 10–3 m-K

This relation is known as Wein’s displacement law.

3. As the temperature is increased the total energy density u for all wavelengths increases. Itwas found that the total energy density, which is equal to the area under the curve isproportional to the fourth power of the temperature i.e.,

R =

∞

λ= λ = σ σ∫ 4

0

T , = constant4 4

c cu u d

which is the Stefan’s law.

In order to explain the dependence of spectral energy density on wavelength and temperatureit was realized that certain assumptions regarding the structure of black body on atomiclevel and its interaction with radiation were necessary.

Fig. 1.3.1 Black body Fig. 1.3.2 Spectral distribution of energy


1.4 CLASSICAL THEORIES OF BLACK BODY RADIATION

Wein’s Law : Wein, in 1893, from thermodynamic reasoning alone showed that energy density inblack body radiation is given by

( )λ λ,Tu d = − λ λ

λ2 / T1

5cc

e d ...(1.4.1)

where c1 and c2 are empirical constants. By proper choice of these constants Wein’s law can be madeto fit the experimental curve in the shorter wavelength region alone but fails in the longer wavelengthregion.

Rayleigh and Jeans Law

The British physicists Lord Rayleigh (1842–1919) and James Jeans (1877–1946) made an attemptto derive a better radiation law on the basis of the following assumptions.

1. The radiation in a cavity is electromagnetic in nature. In a metallic cavity whose wallsare perfectly reflecting, the superposition of incident and reflected waves of each frequencyresults in the formation of standing waves with nodes at the walls. The number of standingwaves (or modes) per unit volume in the frequency range and + d is given by

N()d = 2

2 3d

c

ω ωπ

...(1.4.2)

Making use of the relations 2

2 2and ,

c cd d

π πω = ω = λλ λ

Eqn.(1.4.2) can be written in terms

of wavelength as

N()d = 4

8d

π λλ

...(1.4.3)

Notice that the number of modes is proportional to the square of the frequency of theradiation.

2. The theorem of equipartition of energy is also valid for electromagnetic waves. Accordingto this theorem the average contribution of each degree of freedom to the total energy ofa system is ½ kT where k is Boltzmann constant and T is the temperature of the system.A standing wave is a system of two degrees of freedom, one corresponding to the electricfield and the other to the magnetic field. Hence the average energy of each standing wave(or mode) is kT.

The energy density of radiation in the frequency range and + d in a cavity maintainedat temperature T is

u(, T) d =ω ω

π

2

2 3Tk d

c...(1.4.4)

In terms of wavelength this relation is expressed as

u(, T) d = π λ

λ4

8Tk d ...(1.4.5)


Eqns. (1.4.4) and (1.4.5) represent the Rayleigh-Jeans formula for black body radiation. A glanceat the Rayleigh-Jeans formula, which is a rigorous consequence of classical physics, reveals that itfails to explain the experimental results in the higher frequency (lower wavelength) region. Insteadof finite energy density, it predicts infinite energy density at extremely short wavelengths ultraviolet,X-rays and gamma rays. This discrepancy between the theory and the experiment was dramaticallycalled ultraviolet catastrophe.

Fig. 1.4.1 Comparison of theoretical radiation laws with experimental curve

The failure of Rayleigh-Jeans law was taken more seriously because it was the only possiblelaw that could be derived on the basis of the then known laws of classical physics. The failure ofRayleigh-Jeans law means the failure of the basic assumptions derived from the well-established lawsof classical physics. This situation compelled the German physicist Max Planck to look beyond theframework of classical physics. He proposed a revolutionary hypothesis, according to which theemission and absorption of electromagnetic energy takes place in the form of packets (bundles), calledquanta. This concept of quantisation of energy is foreign to the classical physics.

Max Planck (1858–1947)

Max Planck was born in Germany and studied at Munich and Berlin and obtained his doctoral degreein 1879. After holding position at the university of Kiel he was appointed professor of theoreticalphysics at Berlin university in 1899 where he became acquainted with the experimental work onblack body radiation carried out by Lummer, Pringsheim, Rubens and Karlbaum and devoted himselfto the task of deriving a correct radiation formula. Before deriving the radiation law he first guessedthe correct form of the law and announced it at the meeting of German Physical Society on October14, 1900. The formula was checked by Rubens, Lummer and Pringsheim and was found to be ingood agreement with the experimental results. After about eight weeks of strenuous labor Planckwas able to derive the radiation law theoretically and presented his findings to the German PhysicalSociety on December 14, 1900. The scientific world was not ready to accept Planck’s discovery thatdestroyed one of the most fundamental principle of classical physics—the continuous emission,absorption and distribution of energy. Planck’s theory was strongly recognized only after Einsteinsuccessfully explained the phenomenon of photoelectric effect in 1905. Planck was awarded NobelPrize for his discovery in 1919.


1.5 PLANCK’S RADIATION LAW

The failure of Rayleigh-Jeans law led Planck to think that it was not possible to obtain a correctradiation law within the framework of classical physics. Planck assumed that the walls of the cavityconsist of microscopic oscillators. In thermal equilibrium the absorption and emission of radiationby these oscillators take place at equal rate. According to Planck’s hypothesis the emission andabsorption of radiation by an oscillator take place in the form of discrete packets of energy calledphotons, whose energy is proportional to the frequency of radiation. The energy of a photon offrequency is

=

where = h/2, h = 6.625 × 10–34 Js, is now known as Planck’s constant. Since an oscillator canabsorb whole number of photons, the allowed values of energy of oscillator are

0, , 2, 3, ………………,n

0, , 2, 3, ………….,n

Let N0, N

1, N

2, N

3, ..........,N

n, .........., be the number of oscillators with energy 0, , 2, 3,

........., n, .......... Obviously, the total number of oscillators N and the total energy E of the systemare given by

N = N0 + N1 + N2 + N3 + ……+ Nn + ….. =

∞

=∑

0

Nnn

E = 0 N0 + N

1 + 2N

2 + 3N

3 + .........+n N

n+ ........ =

0

Nnn

n∞

=ε∑

The number of modes of standing waves in the cavity is equal to the number of oscillators inthe walls and the average energy per mode of a standing wave is equal to the average energy of anoscillator. According to Maxwell-Boltzmann statistics the number of oscillators with energy isproportional to factor exp (– /kT). Therefore the number of oscillators with energy n is given by

Nn

= / TC n ke − ε

where C is a constant.The average energy of oscillator is given by

/ T / T

/ T/ T

0

NE

N N

n k n kn

n n nn k

n n kn n

n

n n e n e

ee

− ε − ω

− ε ∞− ω

=

ε ε ωε = = = =

∑ ∑ ∑∑ ∑ ∑

= 0

0

, whereT

nx

n

nx

n

n e

xk

e

∞−

=∞

−

=

ωω =∑

∑


0

ln nx

n

de

dx

∞−

=

= − ω

∑

( )2ln 1 ..........x xde e

dx− − = − ω + + + ∞

1ln

1 x

d

dx e− = − ω −

= ln (1 )xde

dx−ω −

= 1

x

x

e

e

−

−

ω −

= 1

1xe

ω −

=/ T 1ke ωω

−

...(1.5.1)

The number of oscillators in the frequency range and + d is

N()d =

2

2 3d

c

ω ωπ ...(1.5.2)

The energy density u(,T) in the frequency range and + d is

ωω ωω ω= ω

π −

2

2 3 / T( ,T)

1ku d d

c e...(1.5.3)

This relation can be written in terms of wavelength as

π λπλ = λλ −

2

5 2 / T

16 1( ,T)

1c k

cu d

e...(1.5.4)

Equations (1.5.3) and (1.5.4) are called Planck’s radiation laws. It agrees well with theexperimental observations for all wavelengths and at all temperatures.

In the limit 0, the average value of energy of oscillator becomes kT, which is the classicalresult.

/ T0lim

1ke ω→

ω ε = −


= 20lim

11 ....... 1

T 2 Tk k

→

ω ω ω + + + −

= 0

lim/ Tk→

ωω

= kT

Thus, the finite value of Planck’s constant saves the Planck’s formula from its failure and itssmallness forbids the discontinuity or discreteness in energy to be observed in our everyday experience.

Limiting cases of planck’s radiation law: In the lower frequency (higher wavelength) rangethe Planck’s radiation law reduces to Rayleigh-Jeans law. In the limit /kT << 1.

( )3

2 3

1

1 1T

u d dc

k

ωω ω = ωωπ + + ⋅⋅ ⋅ ⋅ ⋅ −

= 2

2 3Tk d

c

ω ωπ

which is the Rayleigh-Jeans law.In the limit of high frequencies (low wavelengths) /kT >>1.

2

3/ T3

12 3 / T

1( ) c k

ku d d c e d

c e−

ωωω ω = ω = ω ω

π

= ( )125

exp / Tc

c d − λ λ λ

which is the Wien’s law.

1.6 DEDUCTION OF STEFAN’S LAW FROM PLANCK’S LAW

The radiant emittance (energy emitted per unit area per unit time for all wavelengths) of a blackbody is given by

R4

cu=

( )0

R , T4

cu d

∞

= ω ω∫


= 3

2 2 / T0

4 1k

d

c e

∞

ωω ω

π −∫

= 4 3

2 20

T, where

T4 1x

k xdx x

kc e

∞ ω = π − ∫

= 4 4

2 2

T

154

k

c

π π

= 2 4

42 3

T60

k

c

π

= 4Tσ

where the constant has the value

2 48

2 3 2 4

watt5.57 10 .

60 m K

k

c−πσ = = ×

1.7 DEDUCTION OF WIEN’S DISPLACEMENT LAW

The wavelength max corresponding to the maximum energy density is the solution of the equation

( ),T 0d

ud

λ =λ

max

2

5 2 / T

16 10

1c k

d c

d e π λλ=λ

π = λ λ −

5 2

0, whereT1x

d x cx

d ke

π= = λ λ−

5

01x

d x dx

dx de

= λ−

− − = −

4 52

2

5 ( 1)0

( 1)

x x

x

x e x ex

e


− − =5( 1) 0x xe xe

15

x xe−− = ...(1.7.1)

The solution of this equation is the point of intersection of curves

1 and .

5x x

y e y−= − =

At the point of intersection x = 4.96 i.e.,

max

24.96

Tc

k λ

π = λ

max2

T4.96

cb

k

πλ = =...(1.7.2)

where b is a constant called Wien’s constant and has valueb = 2.898 × 10–3 m-K.

Fig. 1.7.1 Graphical solution of y = 1 – e–x and y = x/5

SOLVED EXAMPLES

Ex. 1.(a) Calculate the average energy of Planck’s oscillator for (/kT) = 0.01, 0.1, 1.0, 10.

(b) Using Planck’s radiation law, find the power radiated by a unit area of a black body within anarrow wavelength interval = 1.0 nm close to the maximum of spectral radiation density at atemperature T = 3000° K.

Sol. (a) The average energy of Planck’s oscillator is

( )

ω ω

ωωε = =− −

/ T / T

/ TT

1 1k k

kk

e e

Putting the values of (/kT) given in the problem we find

ε = kT, 0.95 kT, 0.58 kT, 0.00045 kT.


(b) Making use of Wien’s law T = b we can write the formula for the spectral energydensity as

ππ=

−

2 5

5 2 /

16 T 1(T)

1c bk

cu

b e

The power radiated per unit area in the wavelength interval is

P = 2 2 5

5 2 /

4 T 1

4 1c bk

cu c

b e ππ= ∆λ

−

Substituting the given values, we getP = 3100 W/m2.

Ex. 2. Calculate the number of modes in a cube of side 2 cm in the wavelength range 4995 Å and5005Å. What is the total radiant energy in the cavity in this wavelength interval? The cavity is maintainedat a temperature of 1500 K.

Sol. Number of modes

dN = 2 3 10

4 10 4

8 3.14 (2 10 m) (10 10 m)8V

(5000 10 m)d

− −

−× × × ×π λ =

λ ×= 3. 215 × 1011.

Energy density is given byU(,T) = dN kT d

= (3.215 × 1011)(1.38 × 10–23JK–1)(1500K)

= 6.65 × 10–9 J.

Ex. 3. A spring mass system has a mass equal to 0.10 kg and a spring constant equal to 10 N/m.The system oscillates with amplitude of 0.10 m.

(a) If the energy of the oscillator is quantized what is the quantum number n associated withthis energy?

(b) If the quantum number n changes by unity, what is the fractional change in energy?

(c) What conclusion do you draw from this example?

Sol. (a) Frequency of the oscillator is = k/m = 10 rad/s.

Energy of oscillatorE = (1/2)kA2 = 0.05 J.

The quantum number n associated with this energy is given byE = n

where n = E/() = 1031 a very large number.

(b) If n changes by unity, the fractional change in energy is given by

31E 110 a verysmall number.

E

n n

n n n−∆ ∆ ω ∆= = = =

ω


(c) This example illustrates that the energy levels of macroscopic oscillators are so close togetherthat even most delicate instruments cannot reveal the quantized nature of energy levels. All this isdue to smallness of Planck’s constant . In the limit 0, the energy levels become continuous.

1.8 PHOTOELECTRIC EFFECT

The emission of electrons by a substance under the action of light is called the photoelectric effect.This phenomenon was discovered in 1887 by the German physicist Heinrich Hertz. It is an irony offate that Hertz, who demonstrated the existence of electromagnetic waves, discovered photoelectriceffect that could not be understood in terms of the wave model of light.

The phenomenon of photoelectric effect can be studied with the help of an apparatus schematicallyshown in the Fig. (1.8.1). Within an evacuated glass jacket two electrodes A and B are enclosed andthe light radiation is allowed to enter the jacket through a quartz window. The radiation falls on theelectrode A, called cathode. The electrode B can be kept at positive or negative potential with respectto the cathode. A sensitive ammeter is put in the circuit to record the current resulting from thephotoelectrons. The potential difference between the cathode and the anode can be measured byvoltmeter. The experimental observations of photoelectric effect may be summarized as follows:

1. For a constant potential difference between the cathode and anode, the number of electronsemitted from cathode (and hence the photoelectric current) increases with increasing intensityof radiation.

2. For a constant intensity and frequency of incident radiation the photoelectric current varieswith the potential difference V between the cathode and anode and reaches a constant valuebeyond which further increase of potential difference dose not effect the photoelectric current,on the other hand, if the plate B is made more and more negative with respect of thephotocathode surface the current decreases. This negative potential (with respect to cathode) ofthe plate is called retarding potential. For a particular value of retarding potential, thephotoelectric current becomes zero. This potential is called cut-off or stopping potential V0

and is measure of maximum kinetic energy of photoelectrons and we can write

Tmax

= eV0

where Tmax is the maximum kinetic energy of the ejected electron.

3. The stopping potential V0 and hence the maximum kinetic energy T

max of photoelectrons is

independent of the intensity of incident radiation and depends only on the frequency ofradiation.

4. For each substance there exists a characteristic frequency 0 such that for radiation with frequencybelow 0 the photoelectrons are not ejected from the surface. This frequency is called thethreshold frequency and the corresponding wavelength is called threshold wavelength, 0.

5. It has been observed that as soon as the light is incident on the substance, the electrons areemitted i.e., there is no time lag between the incidence of radiation and the ejection of electron.


Failure of classical physics: The phenomenon of photoelectric effect cannot be understood interms of wave model of radiation. According to classical theory, light consists of oscillating electricand magnetic fields; the intensity of radiation is proportional to the square of the electric vector E.The force on electron exerted by incident radiation is eE and hence the kinetic energy of ejectedelectron should depend on the intensity of the radiation but the experimental results are contrary tothe prediction.

Further the existence of threshold frequency has no explanation at all in the classical theory.Also according to classical theory there should be considerable time lag between the arrival of theradiation and the ejection of electron, which is contrary to the observation. Therefore, any attemptto explain the photoelectric effect within the framework of classical physics is an impossible task.

Fig. 1.8.1 Schematic arrangement of the apparatus used for the study of photoelectric effect

Einstein’s explanation of photoelectric effect: A satisfactory explanation of photoelectric effectwas first proposed by Albert Einstein in 1905. According to Einstein, electromagnetic radiation offrequency consists of small packets, called photons, each of energy . When a photon of energy (= 2c/) is incident on the surface of a material, some of its energy is spent in making theelectron free and the rest appears as kinetic energy of the electron. The electrons at the surface ofthe material are most loosely bound and require minimum energy for their liberation. This energy iscalled the work function of the material. The maximum kinetic energy of photoelectrons, ejectedfrom the surface, is given by

Tmax

= – = (2c/) – ...(1.8.1)

The electrons, which are more tightly bound, are ejected with less kinetic energies. Thus thekinetic energy of ejected electron depends on the fact whether it is on the surface of the material orit is deeper inside the material. If 0 is the frequency of the incident radiation such that the photonenergy 0 is just sufficient to make the electron free from the material the ejected electron has zerokinetic energy. This happens when

0

= or 2c/0 =


Fig. 1.8.2 Variation of photoelectric current with intensity, frequency and accelerating potential

The frequency 0 is the threshold or cut-off frequency and the corresponding wavelength 0 iscalled threshold wavelength. Thus the threshold frequency 0 (wavelength 0) is a measure of thework function of the material. In terms of photon frequency and threshold frequency 0, Einstein’sphotoelectric equation can be written as

Tmax

= – 0

= 2c [(1/ ) – (1/0)] ...(1.8.2)

If V0 is the stopping potential corresponding to incident radiation of wavelength then

Tmax

= eV0 = 2c[ (1/) – (1/

0)] ...(1.8.3)

The frequency (wavelength) dependence of maximum kinetic energy of photoelectrons is evidentfrom above equation. By increasing the intensity of the incident radiation, we merely increase thenumber of photons and not the energy of the photons. The increase in intensity increases the probabilityof photon electron collision and hence the number of ejected photoelectrons; this explains thedependence of photoelectric current on the intensity of incident radiation. Since the energy of photonis concentrated in a small region and photon is moving at very high speed (c), the energy of photonis instantaneously transferred to electron and consequently there is no appreciable time lag betweenthe incidence of light and the emission of electron.

Millikan’s verification of Einstein equation: The linear relation between the frequency ofincident radiation and the maximum kinetic energy Tmax or eV0 of ejected electron was verified byR. A. Millikan. Using different materials as target, he illuminated them with light of differentfrequencies (wavelengths) and measured the corresponding stopping potentials. In accordance withthe expectations, a graph between and eV0 was found to be a straight line. The slope of the linegives the value of Planck’s constant and the intercept on energy axis gives the work function ofthe target material.


Fig. 1.8.3 Variation of kinetic energy of photoelectrons with frequency of incident radiation

SOLVED EXAMPLES

Ex. 4. Show that ch = 2c = 12400 eV Å = 1240 eV nm

Sol. ch = 2c =34 8

19

(6.625 10 Js)(3 10 m/s)

1.6 10 J/eV

−

−× ×

×

= 12421 × 10–10 eV m

= 12400 eV Å = 1240 eV nm.

Ex. 5. Light of wavelength 2000 Å falls on aluminium surface, which has work function of 4.2 eV.Calculate

(a) maximum kinetic energy of photoelectrons.(b) minimum kinetic energy of photoelectrons.(c) cut-off wavelength.(d) stopping potential.

Sol. (a) Maximum kinetic energy of photoelectrons

Tmax = – = (2c/) –

= − =12400 eVÅ4.2eV 2.0 eV

2000 Å (b) T

min = 0

(c) Threshold wavelength

0 = π = =ϕ2 12400 eV Å

5950 Å4.2eV

c

(d ) Stopping potential = 2 volt.


Ex. 6. In a photoelectric effect, it was observed that for light of wavelength 4000 Å, a stoppingpotential of 2.0 volt is needed and for light of wavelength 6000 Å, a stopping potential of 1.0 volt. Fromthese data, calculate the work function of the material and the Planck’s constant.

Sol. Let 1 = 4000Å, V1 = 2.0 volt2 = 6000 Å, V2 = 1.0 volt

Einstein’s photoelectric equations for the two observations are

11

2eV

cπ= − ϕλ

...(1)

22

2eV

cπ= − ϕλ

...(2)

From these two equations we have

( )1 2

1 2

1 1e V V 2 c

− = π − λ λ

( )1 2 1 2

2 1

e V V2 c

c

− λ λπ =

λ − λ

Substituting the given values in above equation, we geth = 2 = 6.4 × 10–34 J-s

The work function of the material

= 11

2 12400 eVÅeV 2.0 eV 1.1eV.

4000Å

cπ − = − =λ

Ex. 7. Which of the following materials can be used for designing photocell operable with visiblelight?

Tantalum (= 4.2 eV), Tungsten (= 4.5 eV), Aluminium (= 4.2 eV), Barium (= 2.5 eV),Lithium (2.3 eV).

Sol. For photoelectric effect to occur 0 = 2c/For tantalum 0 = 2hc/ =(12400 eV Å)/4.2 eV = 2952Å

Tungsten 0 = 2755 Å

Aluminium 0 = 2952 Å

Barium 0 = 4960 Å

Lithium 0 = 5391 Å

The wavelength interval of visible light is 4000 Å to 8000 Å. The threshold wavelength forbarium and lithium lies in the visible range and hence they can be used in photocell.


1.9 COMPTON’S EFFECT

Planck’s theory of blackbody radiation provided an indirect evidence of the quantum nature of radiation,which became more explicit in Einstein’s explanation of photoelectric effect. The most clear-cut andconclusive evidence in favour of particle nature of radiation came in 1922 from Compton’s discovery.In 1922, the American physicist Arthur Compton, investigating the scattering of X-rays by differentsubstances, observed that the scattered rays, in addition to radiation of the initial wavelength , containalso rays of a greater wavelength '. This phenomenon is known as Compton effect. The difference= '– , called Compton shift, was found to depend only on the angle made by the direction ofthe scattered radiation with that of the initial beam. The value of does not depend on the wavelength and the nature of the scattering material. Since the scattered radiation contains wavelength otherthan the incident one, the phenomenon is also called incoherent scattering.

According to classical model of radiation, it consists of oscillating electric and magnetic fields.When such a radiation is incident on matter, loosely bound electrons begin to oscillate with frequencyequal to that of the incident radiation and in doing so they radiate electromagnetic waves of thesame frequency as that of the incident one. Thus the classical picture of radiation predicts the presenceof only unmodified radiation. The presence of radiation of longer wavelength, called modifiedradiation, in the scattered radiation can never be understood on the basis of wave model of radiation.

Quantum theory of radiation, on the other hand, provides clear-cut understanding of the basicfeature of Compton scattering. According to quantum theory, a monochromatic beam of X-ray offrequency (wavelength ) consists of photons each of energy and momentum k, (k = 2/).The Compton scattering is explained by considering it as a process of elastic collision of X-ray photonswith almost free electrons. During the collision, the photon transfers some of its energy to the electronand therefore the scattered photon has less energy (that is, lower frequency or higher wavelength)and the electron recoils in some other direction. Suppose that the energy and momentum of scatteredphoton are ' and k'. The energy and momentum of electron before collision are m0c2 and zero.(The electron is assumed to be at rest.) If the electron possesses momentum pe after collision thenits energy will be [ (pec)2 + (m0c2)2]1/2.

Applying the principle of conservation of energy we have

′ω + = ω + + 2 2 2 2

0 0em c c p m c ...(1.9.1)

The law of conservation of momentum in vector form isk = pe + k' ...(1.9.2)

The first equation can be arranged as

22 2 2

0 0ep m c m cc c

′ ω ω + = − +

= [ ]20( )k k m c′− +

2 2 2 2

0( 2 ) 2 ( )ep k kk k m c k k′ ′ ′= − + + − ...(1.9.3)


Fig. 1.9.1 Compton scattering

The second equation can be written as

2 2 2 2 2 2( ) ( 2 cos )ep k kk k′ ′ ′= − = − θ +

k k ...(1.9.4)

From Eqns. (1.9.3) and (1.9.4), we obtain

m0c (k – k') = k k' (1– cos )

k k

kk

′−′

(1 cos )om c

= − θ

0

1 1(1 cos )

k k m c− = − θ

′

0 0

2(1 cos ) (1 cos )

h

m c m c

π′λ − λ = − θ = − θ

(1 cos )c∆λ = λ − θ ...(1.9.5)

where

0 0

2c

h

m c m c

πλ = = = 0.0243 Å ...(1.9.6)

The quantity c is called the Compton wavelength of electron. The Compton wavelength ofan electron is the wavelength of radiation whose photon has energy equal to the rest energy of theelectron.

Equation (1.9.5) shows that the Compton shift is independent of the wavelength of theincident radiation and the nature of the target material. It depends only on the angle of scattering. Inthe forward direction ( = 0), the Compton shift is zero. In the direction = /2, = (/m0c) = c

and in the backward direction ( = ), Compton shift is maximum equal to 2c.


When photons of the incident radiation collide with tightly bound electrons of the target atom,the energy and momentum are exchanged with the atom as a whole. Since the mass of an atom ismuch greater than that of an electron, the Compton shift in this case is negligible, and ' practicallycoincides with . This explains the existence of unmodified radiation at all angles of scattering.

Energy of Scattered Photon

The Compton shift is given by

( )0

21 cos

m c

π′λ − λ = − θ

2

0

2 2 2(2sin / 2)

E E

c c

m c

π π π− = θ′

Fig. 1.9.2 Variation of Compton’s shift with angle of scattering

E' = 2

20

E

2E1 sin

2m c

θ+

...(1.9.7)

The scattered photon will have minimum energy for = 180°. Thus

min

20

EE

2E1

m c

′ =

+ ...(1.9.8)

For = 0, E'max = E.


Kinetic energy of compton electron: The kinetic energy imparted to the recoil electron isgiven by

T = – ' ( )2 2

2c c

c hc′π π λ − λ ∆λ = − = π = ′ ′λ λ λλ λ λ + ∆λ

...(1.9.9)

In terms of initial energy E of photon, the kinetic energy of electron can be expressed as

2

20

ET E E E

2E1 sin

2m c

′= − = − θ+

22

20

22

2Esin

2T

2E1 sin

2o

m c

m c

θ

=θ+

...(1.9.10)

The Compton electron acquires maximum kinetic energy when photon is scattered at angle =180°. Hence

T max =

+

20

20

2E

,2E

1

m c

m c

= 180° ...(1.9.11)

Tmin = 0 for = 0.

Recoil direction of electron: If the Compton electron moves in the direction making angle with the direction of incident photon, the law of conservation of x and y components momentumpermits us to write

cos cosep p p′ϕ + θ =

pe sin = p' sin

Hence ′ ′θ θϕ = =

′ ′− θ − θsin E sin

tancos E E cos

p

p p

SOLVED EXAMPLES

Ex. 8. X-rays of wavelength 1.0 Å are scattered by a carbon block. The scattered radiations areobserved at 60°, 90°and 180°. Find (a) Compton shift (b) kinetic energy imparted to the recoil electron.

Sol. (a) Compton shift =c(– cos )


60 = c (1 – cos 60°) = 0.5c = 0.012 Å

90 = c (1 – cos 90°) = c = 0.024 Å

180 = c (1 – cos180°) = 2c = 0.048 Å

(b) Kinetic energy imparted to the electron

1 1T E E

chch

∆λ ′= − = − = ′λ λ λ λ + ∆λ

(i) T =

=

12400 eVÅ 0.012Å147eV

1.00Å 1.012Å

(ii) T = 290 eV

(iii) T = 568 eV

Ex. 9. For what wavelength of incident photon it shows Compton scattering in which the energy ofscattered photon is one-half that of incident photon at a scattering angle of 45°? In what region of theelectromagnetic spectrum does such a photon lie?

Sol. Given that E = E/2 , therefore ' = 2 and = Now = c (1 – cos 45°)

= c ( – 1/2) = 0.0071 Å (gamma ray)

Ex. 10. A photon with energy 1.00 MeV is scattered by a stationary free electron. Find the kineticenergy of electron if the photon’s wavelength changed by = 25% due to scattering.

Sol. Given that / = = 0.25Kinetic energy of recoil electron

( ) 0.25T E 1.00 MeV

1 1 0.25

ch ∆λ η = = = λ λ + ∆λ + η + = 0.20 MeV.

Ex. 11. A photon of energy 250 k eV is scattered at an angle =120° by a stationary free electron.Find the energy of the scattered photon.

Sol. Energy of scattered photon

′ =

+ − θ

20

EE

E1 (1 cos )

m c

Putting E = 0.250 MeV, m0c2 = 0. 510 MeV and cos120° = 0. 50 we get E' = 0.143 MeV.

Ex. 12. A photon with momentum p = 1.02 MeV/c is scattered by a stationary free electron. Itsmomentum on scattering becomes p' = 0.255 MeV/c. At what angle is the photon scattered?

Sol. Compton shift ' – = c (– cos )

22sin2c

h h

p p

θ − = λ ′


2sin2 2 c

h p p

pp

′ θ −= ′λ

= 01

2

p pm c

pp

′ − ′

=1 MeV 0.765MeV c

0.512 1.02MeV c 0.255MeV cc

×

= 0.7502

sin 0.86 sin 602

θ = =

= 120°.

Ex. 13. A photon is scattered at an angle =120° by a stationary free electron. As a result theelectron acquires a kinetic energy T = 0.45 MeV. Find the energy of the incident photon.

Sol. The energy of the scattered photon is given by

2

02 2

0

EE

2E sin 2

m c

m c′ =

+ θKinetic energy of electron

T = E – E' = E – 2

02 2

0

E

2E sin 2

m c

m c + θ

T = 2 2

2 20

2E sin 2

2Esin 2m c

θ+ θ

This is quadratic equation in E. When solved for E, we get

= + + θ

202

2TE 1 1

2 Tsin 2

m c

Substituting T = 0.45 MeV, m0c2= 0.51MeV, = 60°, we find E = 0.67 MeV

1.10 BREMSSTRAHLUNG

The quantum nature of radiation is also confirmed by the existence of a short wavelength limit ofthe bremsstrahlung X-ray spectrum. The bremsstrahlung is the radiation produced by deceleration ofthe electrons and is also called braking radiation. X-rays are produced when solid targets are bombardedwith fast electrons. An X-ray tube is an evacuated bulb with several electrodes. The electrons areproduced by thermionic emission. These electrons are accelerated under high potential differenceand then directed to fall on anode (anticathode) made of heavy metals (W, Cu, Pt etc.). Almost allenergy of electrons is liberated on the anticathode in the form of heat and only from 1 to 3% of


energy of electrons is transformed into X-rays. For this reason arrangements are made for coolingthe anticathode. The distribution of X-rays intensity at various wavelengths, with molebdenum andtungsten as targets, are shown in the Fig. (1.10.1). The important features of the curves are as follows:

Fig. 1.10.1

(1) For each accelerating potential there exists a short wavelength limit min below which noradiation is produced. The value of min depends only on the magnitude of the acceleratingvoltage and not on the nature of the target material. Duane and Hunt (1915) observedthat min is inversely proportional to the accelerating voltage. The exact relationship betweenmin and the accelerating voltage V is given by

min12400

Vλ = Å

where is in Angstrom and V in volt.(2) X-ray energy is continuously distributed among all wavelengths from min to infinity.

For this reasons the braking radiation is called continuous or white radiation. For eachtarget when the accelerating voltage is increased beyond a certain value, which is thecharacteristic of the target, the intensity-wavelength curve shows several peaks. Thewavelengths at which these peaks are observed are the characteristic properties of the targetmaterial and the radiation itself is called characteristic radiation.

The X-rays are produced from catastrophic encounters of fast electrons with atomic nuclei. Ifthis process is analyzed quantum mechanically, it turns out that there is a certain probability thatelectron-nuclear encounter will result only in deflection of the electron with no emission of radiation.This is called an elastic Rutherford scattering. Besides this, there is also a probability that electron-nuclear encounter will result in the emission of photon as well as deflection of electron. This is calledan inelastic or radiative collision. When a fast electron interacts with the target nucleus via Coulombfield, it transfers momentum to the nucleus and a photon is emitted in the process. If T' is the kineticenergy of the outgoing electron then the energy of the emitted photon is given by

= T –T'

The incoming electrons lose different amount of energy in such nuclear encounters and suffermany encounters before coming to rest. The emitted radiation, therefore, forms a continuous spectrum.


The X-ray photon of shortest wavelength is produced when the incoming electron loses its entirekinetic energy in a single encounter (T' = 0 ). Thus

max = T = eV

min

2eV

cπ =λ

where min2 /e 12400

V V

cπλ = =Å

The value of obtained making use this formula is considered to be the most accurate. In thelimit 0, min 0. This means that the existence of short wavelength limit is a quantummechanical-phenomenon. The Bremsstrahlung process is sometimes called an inverse photoelectriceffect. In a photoelectric effect, a photon is absorbed and its energy and momentum are transferredto an electron. In bremsstrahlung process, a photon is created and its energy and momentum arederived from the collision of electron with nucleus.

Fig. 1.10.2 Continuous and characteristic x-ray radiation

1.11 RAMAN EFFECT

In 1927 C.V. Raman and K.S. Krishnan were investigating the scattering of light by molecules oftransparent liquids, gases and solids. They found that in addition to the frequency of incident light,the scattered radiation had a number of other frequencies higher and lower than the incident one.This phenomenon is called Raman effect. The lower frequencies in the scattered light are called Stokes’frequencies (red satellites) and those of higher frequencies are called antistokes’ frequencies (violetsatellites). These extra frequencies are independent of the frequency of the incident light and arecharacteristic of the scattering substance. It is also found that the red satellites have greater intensitythan the violet satellites. The intensity of the violet satellites rapidly grows with rise of temperatureof the scattering substance.

The quantum physics of atoms and molecules and the quantum theory of radiation provide asimple explanation of Raman effect. This phenomenon can be considered as the inelastic collision ofphoton of the incident light with the molecule of the scatterer. Let us consider the collision of aphoton of frequency 0 with a molecule, which is initially in the energy state Ei. During the processof collision, the photon either gives up its energy to the molecule or receives energy from the moleculeand the molecule goes to the energy state Ef . If the scattered photon has frequency then the lawof conservation of energy requires that


0 + Ei = + Ef

= 0 + −

E Ei f

= 0 + ...(1.11.1)

where = (Ei –Ef)/ ...(1.11.2)

Now two cases arise:1. If Ei < Ef , is negative and the frequency of the scattered is given by

= 0 – ∆ω (Stokes’ lines) ...(1.11.3)

This happens when the molecule is initially in the ground state and the photon transfers itsenergy to the molecule. The photon energy (and hence frequency) diminishes and the molecule jumpsto the excited state. This explains the origin of Stokes’ lines. Since there are a large number of excitedstates, the transitions of molecules from the ground state to the excited states are accompanied bymany spectral lines.

At ordinary temperature, the number of molecules in the ground state exceeds those in the excitedstates, the probability of upward transitions is greater than the downward transitions. Hence theintensities of red satellites are greater than those of violet satellites.

2. Ei > Ef , is positive and the frequency of scattered photon is given by

0ω = ω + ∆ω (Antistokes’ line) ...(1.11.4)

This happens when the molecule is initially in one of its excited state. The incident photoncauses it to jump to the ground state. During this process the photon receives energy from the moleculeand hence the scattered photon has higher energy (frequency). This process causes the violet satellitesto appear.

At ordinary temperature, the number of molecules in the excited states is smaller than those inthe ground state and hence the number of downward transitions from the excited states to the groundstate is smaller. That is why the violent satellites are weaker in intensity. On heating the scatteringsubstance, the number of molecules in the excited states increases and so does the number of downwardtransitions. Thus an increase in temperature of the scattering substance causes an increase in intensityof the violet companions.

In terms wave numbers 1

2 c

ω ν = = λ π , the Raman shift is expressed as

0ν = ν ± ∆ν ...(1.11.5)

where positive sign stands for antistokes’ lines and negative sign for stokes’ lines.Experimental arrangement: The experimental arrangement for studying Raman effect is

schematically shown in the Fig. (1.11.1). The scattering substance is taken in a tube whose one endis flat and the other end extends into a horn shape. The latter is blackened and filled with mercury toprovide dark background. A spectrograph is mounted in front of the flat face of the tube. The tube


is surrounded by a jacket in which water is circulated to keep the temperature of the liquid constant.A monochromatic source with suitable filters is used to illuminate the experimental liquid. Aftersuffering scattering, the light enters spectrograph through slits. The Raman lines are recorded on aphotographic plate.

Fig. 1.11.1 Origin of Raman lines

Fig. 1.11.2 Schematic arrangement of Raman apparatus

The Raman effect is characteristic of molecules and hence it is widely used in the determinationof structure of molecules. It also provides a proof in favour of quantum nature of radiation. Ramanwas awarded Nobel Prize for this discovery in 1930.

SOLVED EXAMPLES

Ex. 14. With exciting line 2536 Å, a Raman line for a sample is found at 2612 Å. Calculate theRaman shift in m–1.

Sol. 10 10

0

1 13943000 m

2536 10 m−

−ν = = =λ ×

1

10

1 13828000 m

2612 10 m−

−ν = = =λ ×


Raman shift 1

0 115000 m .−∆ν = ν − ν =

Ex. 15. Excited by a radiation of wavelength 5000 Å, a sample gives a Raman line at 5050.5 Å.Calculate the Raman frequency in m– 1 and the position of the corresponding antistokes’ line in Å.

Sol. 6 1 6 10 2 10 m and 1 98 10 m− −ν = × ν = ⋅ ×

Raman shift 6 10.02 10 m−∆ν = ×

Frequency of antistoke’s line

6 1

0 2.02 10 m−ν = ν + ∆ν = ×

Wavelength of antistokes’ line 1

4950.05λ = =ν

Å.

1.12 THE DUAL NATURE OF RADIATION

The discovery of photoelectric effect, Compton’s effect etc. posed a serious dilemma before thecontemporary physicists. The phenomenon of interference and diffraction of light could be explainedon the basis of wave model of radiation whereas the phenomenon involving interaction of radiationwith matter viz photoelectric effect, Compton’s effect could only be understood on the basis of particlenature of radiation. In the particle picture of radiation, photon energy is expressed as . It is verydifficult to assign a meaningful significance to the frequency associated with a particle. In orderto determine photon energy ( = 2c/) we determine the speed of light and the wavelength. Thedetermination of wavelength is based on the wave nature of light. This paradoxical situation is mostforcefully encountered in Compton’s scattering experiment where X-ray wavelength is determinedwith crystal spectrometer, the interpretation and the analysis of the measurement is based on the wavemodel and the scattering of X-ray can only be understood in terms of particle model. The particleand the wave aspect of radiation are not revealed simultaneously in the same experiment. Inexperiments involving propagation of radiation including interference and diffraction, the wave aspectis shown while in experiments where radiation interacts with matter, its particle aspect is manifested.This situation has an interesting analogy, as described by Arthur Schawlow: It (the complex behaviourof light) be likened to the elephant, as blind men in fable; to one who touched the tail, the elephantwas like a rope; to another who touched a leg, it was like a tree; to others it was like a wall.

In 1928 N. Bohr suggested that the wave and the particle properties of radiation arecomplementary. It is not possible to apply both descriptions at the same time. When wave behaviouris observed easily, the particle behaviour is hardly observed and vice-versa. This idea is known asthe Bohr’s complementary principle.

The wave and the particle properties of radiation are linked by Planck’s constant , which isproduct of two variables; one is characteristic of a wave and the other is characteristic of a particle,viz.

= =

ω π

E ET

2


where E is energy of photon, is frequency of photon and T is time period of the wave (radiation).Here E represents particle characteristic and or T wave characteristic. If the magnitude of oneproperty is larger, the magnitude of the other will be smaller and vice-versa. Meaning thereby, if theparticle property is dominant, the wave property will hardly be observable. If the wave property ismore pronounced, it will be difficult to detect particle property. In case of short X-rays and gammarays, the photon energy is very high and therefore their particle property is more dominating; it isdifficult to establish their wave behaviour. On the other hand, the wavelength of radio waves verylarge and therefore their particle behaviour can hardly be demonstrated.

QUESTIONS AND PROBLEMS

1. Show that the average energy of a Planck’s oscillator is given by

.

/ T 1ke

ωε = ω −

Discuss the variation of with temperature and frequency of the oscillator.

2. Deduce Planck’s for the spectral distribution of energy in black body radiation. From Planck’s law, obtain (i)Wien’s displacement law (ii) Stefan’s law.

3. Derive Planck’s radiation law. Show that it reduces to Wien’s law and Rayleigh-Jeans law in appropriate limit.

4. Describe the experimental results obtained in the study of photoelectric effect. Give Einstein’s explanationphotoelectric effect.

5. What is Compton’s effect? Derive an expression for Compton’s shift. Discuss the dependence of Compton’sshift on the angle of scattering. Explain the existence of unmodified radiation in the scattered radiation.

6. An X-ray photon of energy E undergoes Compton scattering in the direction with the initial direction. Showthat the energy of the scattered photon is given by

′ =

+ θ22

0

EE

2E1 sin / 2

m c

and hence ′ =+

min

20

EE

2E1

m c

7. Show that the kinetic energy of the Compton electron is given by

22

20

22

20

2Esin /2

T2E

1 sin /2

m c

m c

θ =

+ θ


and hence ( )2

0max

20

2E

T2E

1e

m c

m c

=

+

8. Show that the direction of recoil of Compton electron is given by

E sin

tanE E cos

′ θϕ =′− θ

where the symbols have their usual meanings.

9. The energy required to remove an electron from sodium is 2.3 eV. Does sodium show photoelectric effect forlight of wavelength 6500 Å.

10. The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 5000Å is0.70 volt. When the incident wavelength is changed, the stopping potential is found to be 1.50 volt . What is thenew wavelength?

11. In an experiment of photoelectric effect, it is observed that for a light of wavelength 3000Å, the stopping potentialis 1.85 volt and for light of wavelength 4000Å, the stopping potential is 0. 82 volt. From these data determine (i)Planck’s constant (ii) work function of the substance (iii) threshold wavelength for the substance.

12. What is the frequency, wavelength and momentum of a photon whose energy is equal to the rest energy of anelectron?

13. An X-ray is found to have its wavelength doubled on being scattered through 90°. What is the wavelength of theX-ray?

14. If the Compton shift in an experiment is found to be 0. 0121 Å. Find the scattering angle.

15. A 300 keV photon undergoes a Compton scattering. The kinetic energy of recoil electron is 250 keV. Calculatethe wavelength of the scattered photon.

16. An X-ray quantum having a wavelength of 0.15 Å undergoes a Compton collision and is scattered through anangle of 37°. (i) What are the energies of the incident and scattered photon and of the ejected electron (ii) what isthe momentum of each photon?

17. Using the conservation laws, demonstrate that a free electron cannot absorb a photon completely.

Hint: The laws of conservation of energy and momentum permit us to write:

20E E (1) and (E / )e em c c p+ = = ...(1)

The relativistic energy of electron is

2 2 2 2

0E ( ) ( )e ep c m c= + ...(2)

In view of the conservation laws, we can write (2) as

2 2 2 2 2

0 0(E ) E ( )m c m c+ = +

2

02 E 0 which is meaningless.m c⇒ =

CHAPTER

WAVE NATURE OF MATERIAL PARTICLES

1.12.1 INTRODUCTION

The remarkable success of Bohr’s theory in providing interpretation of hydrogen spectrum stimulatedthe imagination of physicists to extend his ideas with some modifications to account for the finerdetails of complex spectra of multi-electron atoms. Sommerfeld’s theory, for instance, was an attemptof this nature. Although Sommerfeld’s theory with relativistic correction, provided an explanationfor the existence of the fine structure of H

line in the hydrogen spectrum, it could not carried further

in the interpretation of spectra of multi-electron atoms. Using instruments of higher and higherresolving power many puzzling features of the atomic spectra were recorded. The behaviour of atomsin electric and magnetic field—electro-optic and magneto-optic phenomena posed a serious challengeto theoreticians. In early days of the development of atomic physics the usual methods of understandingthese complex features of atomic spectra were to introduce empirical rules with several types ofquantum numbers without any theoretical basis. The hypothesis of spinning electron, proposed byUhlenbeck and Goudsmit (1925), was an attempt of this kind. During this period, it was realizedthat a mere patch-works on the classical principles was no longer adequate in the understanding ofatomic phenomena. The state affair indicated the need for drastic changes in our concepts ofmicroscopic systems. In 1924 a young French graduate Louis de Broglie in his doctoral thesis suggestedsuch a radical change in our concepts of micro-systems.

2.2 de BROGLIE HYPOTHESIS

The reasoning that led de Broglie to put forward his revolutionary hypothesis runs as follows. Theentire physical universe is composed of matter and radiation. In quantum theory of radiation afragment or quantum of energy is assigned a frequency, (=2) such that = . Althoughthere is no physical sense of frequency , nevertheless the theory based on this assumption workswell. From this notion de Broglie speculated that material particles, which are also fragment of energy(e.g., = mc2), might be assigned some characteristic frequency. A material particle of rest mass m0

is equivalent to energy m0c 2, therefore, according to de Broglie idea we can write

m0c2 = ...(2.2.1)

where is the frequency of some intrinsic periodic process associated with the material particle. Letus see what this periodic process appears to an observer with respect to which it is moving.

Wave Nature of Material Particles 79

Let us consider a frame S', which is moving with the particle. The frequency of the internalprocess associated with the particle is

= m0c2/

The vibratory motion associated with this periodic internal process can be represented by

20( , ) exp( ) exp

m cx t i t i t

′ ′ ′ ′ ′ψ = − ω = −

...(2.2.2)

Let S be the observer’s frame with respect to which the particle is moving along x-axis withvelocity v. Making use of Lorentz transformation for time, the equation of the periodic processassociated with the particle transforms on transition from S' to S as

( , )ψ x t =2 2

0

2

/exp , /

1

m c t vx ci v c

− − β = − β

...(2.2.3)

Fig. 2.2.1

The above equation represents a progressive wave with propagation constant k given by

k = 2 2

0

2

/

1− β

m c v c

= 0

2

1

1− β

m v

= p

...(2.2.4)

or p = k ...(2.2.5)

or 2 T

h

m= ...(2.2.6)

where T = kinetic energy of the particle.Equations (2.2.4–2.2.6) determined the wavelength of the de Broglie wave representing the

particle.


2.3 EXPERIMENTAL VERIFICATION OF de BROGLIE HYPOTHESIS

Although de Broglie speculated about the wave nature of material particles, yet none of the theoreticalphysicist of that time could think of a way to establish the physical existence of such waves. It wasElasser who in 1925 pointed out that the existence of such waves could be demonstrated by theirdiffraction by an appropriate grating. Two American physicists C.J. Davisson and L.H. Germer carriedout an experiment, which definitely showed the existence of de Broglie waves and verified the relationbetween wavelength and momentum. The schematic arrangement of their experiment is shown in theFigure 2.3.1. A beam of electrons was accelerated through a potential difference and then it wasallowed to fall on a nickel target. A detector D received the scattered electrons. The entire arrangementwas placed in an evacuated chamber. To electrons the nickel is not a smooth surface and hence theywill be scattered in all directions. From the view point of classical physics the intensity of the scatteredbeam should be independent of the direction of observation and the energy of primary electrons.These predictions were verified in the experiment.

However, in the midst of the experiment, airleaked into the chamber accidentally and the nickelsurface got oxidized. In order to reduce the oxidelayer of the target, it was taken out and was heatedstrongly in a high temperature oven. After thistreatment the target was again placed in theapparatus and the experiment was repeated. Theexperimental results were surprising. At certainangles the scattered electron intensity showedmaxima and minima. Furthermore the angularpositions of maxima and minima depended on theenergy of primary electrons.

The angular variation of scattered electronintensity for different accelerating potentials isshown in the Figure 2.3.2. The intensity at anyangle is proportional to the distance of the curveat that angle from the point of scattering.

The experimental results may be interpreted as follows. Initially, the nickel target was apolycrystalline material, which after heating became a single crystal due to arrangement of atoms ina regular lattice. According de Broglie hypothesis, the electron waves are diffracted by atomic planesin the same way as X-rays are diffracted from crystal planes.

In one particular arrangement, a beam of 54 eV electrons was allowed to be incident on a singlenickel target and the diffracted beam was observed to have maximum intensity in the direction = 50° from the incident beam. This diffracted beam arises from the crystal planes shown in theFigure 2.3.1. Evidently the angle of incidence relative to the crystal planes is = 65°. From X-raysmeasurements the spacing of these planes is found to be d = 0.91 Å. The Bragg’s equation formaximum in the diffraction pattern is

2d sin = m; m = 1, 2, ……

Fig. 2.3.1 Schematic diagram of Davisson-Germer experiment


Fig. 2.3.2 Variation of intensity of the diffracted electron beam with energy of the electron

The wavelength of electron waves calculated from this equation (for m = 1) comes out to be

= 2d sin = 2 × (0.91 Å) sin 65 = 1.65 Å

On the other hand the wavelength of electron beam from de Broglie hypothesis comes out to be

34

31 19

6.625 10 Js

2 m V 2 9.1 10 kg 1.6 10 C 54 V

h h

p e

−

− −

×λ = = =× × × × ×

101.66 10 m−= × =1.66 Å

The excellent agreement between the value obtained from diffraction experiment and that fromde Broglie relation provides experimental proof for the validity of wave nature of electrons.

G. P. Thomson’s experiment: Another verification of de Broglie hypothesis came from theG.P. Thomson’s experiment (1927). In this experiment a narrow beam of electrons was allowed topass through a thin film of polycrystalline material. The transmitted beam was received on aphotographic film.

The diffraction pattern obtained from electronbeam was similar to the powder diffraction patternof X-rays. This experiment independentlyconfirmed the de Broglie hypothesis.

Not only electrons, but all material objectsexhibit wave properties under appropriateconditions. In fact neutron diffraction technique iswidely used in the determination of crystalstructure.

Fig. 2.3.3 G.P. Thomson arrangement for gettingelectron diffraction pattern


2.4 WAVE BEHAVIOR OF MACROSCOPIC PARTICLES

It is a well-known fact that the diffraction effects of waves are more pronounced if the dimensionsof the diffracting objects are of the order of the wavelengths of the waves. For instance, if lightwaves are incident on a aperture, the diffraction effects become more and more observable when theaperture is progressively reduced. In case of electromagnetic waves, as we move from higherwavelength region (radio waves) to the lower wavelength region (gamma rays), the manifestation ofwave nature is gradually reduced and the particle nature becomes more and more pronounced. Forgamma rays particle properties become so dominant that it is very difficult to demonstrate wave-likebehavior. Moreover, the difficulty also arises because of the unavailability of diffracting element ofthe dimensions comparable to the wavelength of gamma rays.

The de Broglie wavelength of a macroscopic object moving with common velocity is so smallthat its particle property dominates the wave behavior and so the latter is not observable. (Thewavelength of a body of mass 100 g moving with velocity 1000 m/s is = h/mv = 6.6 × 10–36 m).On the other hand the wavelength of microscopic particles like electrons, protons, neutrons etc., areof the order of spacing of atoms and molecules in solids and therefore solids themselves offer asnatural diffraction gratings. Indeed the first experimental verification of wave nature of materialparticles was established by using crystal as diffracting element.

2.5 HISTORICAL PERSPECTIVE

It is instructive to recall that the particle nature of radiation was discovered in 1905, and its conversei.e., wave nature of particle was conceived in 1924 i.e., two decades later. In order to gain someinsight into the chronological order of the discoveries, we can ask why did physicists not speculatethe converse idea in the year 1905? It must be noted that Einstein proposed his revolutionary idea ofquantum nature of radiation for explaining certain experimentally observed phenomena, which couldnot otherwise explained. On the other hand, one must admire the boldness of Louis de Broglie fromthe fact that he proposed his idea in the absence of any suggestive experimental observation. Theintellectual climate at the time of de Broglie was more receptive to new ideas than the time whenEinstein published his famous paper on photoelectric effect, which was given a cold reception. However,the de Broglie notion received immediate and respectful attention at least in the minds of an influentialminority. The acceptability of de Broglie concept was facilitated as Einstein who was at the peak ofhis fame, became his strong advocate.

When de Broglie’s paper for doctoral thesis was sent to Einstein, he commented, “de Broglielifted a corner of the great veil”. The publication of de Broglie’s idea stimulated a great deal ofdiscussion among theoretical physicists. In Zurich, the well-known chemist Peter Debye suggested ayoung German physicist Erwin Schrodinger to make a careful study of de Broglie’s theory. In thecourse this study Schrodinger presented a wave equation which when applied to atomic and molecularproblems, proved to be a successful attempt in understanding the behavior of microscopic system.

The experimental verification of de Broglie concept was established after a year when Schrodingerpresented his new system of mechanics, wave mechanics. The electron diffraction experiment of Davissonand Germer had its origin in a famous suit; the parties in the suit were the General Electric Company


and Western Electric Company. The work on electron scattering continued for several years(1919–1927). Up to 1926 Davisson and Germer were unaware of the de Broglie notion. When Davissoncame to Oxford, there he heard the new theory and realized that the results of electron bombardmentin his experiment was similar to that of X-rays diffraction. Thus de Broglie hypothesis got accidentalexperimental verification.

It is more interesting to notice the fact that J.J. Thomson, who in 1897 discovered electron andshowed that it was a negatively charged particle, received Nobel Prize in 1906 and his son G. P.Thomson, who in 1927 discovered that electron is a wave, received Nobel Prize (with Davisson andGermer). In 1937, Max Jammer says, “The father was awarded the Nobel Prize for having shownthat the electron is a particle, and the son for having shown that electron is a wave.”

2.6 THE WAVE PACKET

From what has been discussed in the preceding sections we come to conclusion that as the wavelengthof electromagnetic radiation is progressively reduced, its particle nature becomes more and moredominant. On the other hand as the mass of a body is reduced, its wave nature is manifested.Therefore, it is apparent that the wave nature and the particle nature are the two aspects of the samephysical reality. Although it appears that the two aspects are so different that it is difficult to reconcilethem. The most obvious difference between particle and wave is that the former is localized whereasthe wave is extensive in time and space. Consequently the only way in which we can reconcile thetwo concepts is to localize waves. The concept of wave packet has this characteristic.

The concept of wave packet is of great importance in quantum mechanics because it provides ameans of reconciling the apparently incompatible wave and particle aspects of the behavior of matterand radiation. In fact, wave packet is a wave, which extends over a limited region. It is formed bysuperposition of a large number of waves of different wave numbers and amplitudes. On account ofits limited extension it resembles with particle and the wave behavior is implicit in its very structure.Thus the representation of a particle by a wave packet is logical.

Let us consider the superposition of two waves

1 1 1A cos ( )t k xψ = ω −

2 2 2A cos ( )t k xψ = ω −

The phase velocity of the first wave is v1 = 1/k1 and that of the second wave is v2 = 2/k2. Theresultant wave is given by

= 1 +

2

= 1 1 2 2A cos( ) A cos( )t k x t k xω − + ω −

= 1 2 1 2 1 2 1 22Acos cos2 2 2 2

k k k kt x t x

ω − ω − ω + ω + − −

= ( ) 2 cos cos2 2

∆ω ∆ − ω −

kA t x t kx ...(2.6.1)


where 1 2 1 21 2 1 2, , ,

2 2

k kk k k k

ω + ω +∆ω = ω − ω ∆ = − ω = =

The slowly varying first term of eqn. (2.6.1) represents amplitude modulated wave travelingwith velocity

vg

= /k

The dotted curve in Fig. (2.6.1) represents the modulated wave. It is evident that resultant waveis divided in groups. The velocity with which these groups travel is called group velocity v

g. If a

very large number of waves, differing in frequencies and propagation constants by infinitesimallysmall amount, are superposed, the resulting modulated wave moves with group velocity given by

0

ω = g

k

dv

dk...(2.6.2)

where the derivative is to be evaluated at the central value of k0.

If the component waves move with equal velocity, the groups also move with the same velocity.If the component waves have different speeds, the group velocity is different from those of componentwaves.

Fig. 2.6.1 Superposition of two waves of slightly different frequencies give rise to aresultant wave which consists of wave groups

In order that a group of waves may appear as a particle, it is necessary that only one group beformed. To obtain such a single group, an infinite number of waves, whose frequencies andpropagation constants differ infinitesimally from one another, are required. A more generalsuperposition of several sinusoidal waves is represented by

( )( , ) A n ni k x tn

n

x t e −ωψ =∑ ...(2.6.3)


If the component waves have continuous distribution of frequencies and propagation constantsthe above sum becomes an integral i.e.,

( )( , ) A( ) i kx tx t k e dk−ωψ = ∫ ....(2.6.4)

Representation of functions as superposition of sinusoidal functions is called the Fourier’s series.Eqn. (2.6.3) represents the Fourier’s series of function (x,t) and Eqn. (2.6.4) represents the Fourier’sintegral of function (x, t). The amplitude function A(k) represents the amplitude of the componentwave having propagation constant k. By appropriate choice of amplitude function A(k), a wave groupof any desired shape may be constructed. In Fig. (2.6.2) some wave packets along with their amplitudefunction A(k) are shown. The amplitude function A(k) is called the Fourier’s transform of the wavepacket. Notice that a wave packet having larger extension in space corresponds to a sharply peakedamplitude function and vice versa. If x is the extension of a wave packet and k is the range ofpropagation constants in which amplitudes are distributed, then it can be shown that

x . k = 1 ...(2.6.5)

It is obvious that when k is large, the corresponding x is small and vice versa. This reciprocalrelationship is very important and forms the basis of the famous Heisenberg’s uncertainty principle.

Fig. 2.6.2 Some wave groups and their Fourier’s transforms


2.7 PARTICLE VELOCITY AND GROUP VELOCITY

According to de Broglie’s hypothesis the momentum p of a particle is related to its wavelength (k = 2/) as

p = k

and according to Einstein, the energy E of a particle is related to its mass and intrinsic frequency as

E = mc2 =

The phase velocity of de Broglie waves representing the particle is

2 2 2 2/

/

mc mc mc cu

k p p mv v

ω= = = = =

...(2.7.1)

Since particle velocity v is less than the speed of light c, the phase velocity u is greater than thespeed of light c. Thus the particle and the phase waves would not accompany. Although the phasevelocity is greater than the speed of light, it does not contradict the special relativity because phasewaves do not carry energy.

Now we shall show that the velocity of the particle is equal to the group velocity of thecorresponding wave packet.

The velocity of the particle is

2 2

2 E

p pc pcv

m mc= = = ...(2.7.2)

Making use of the relations p = k and E = we have dp = dk and dE = d. Therefore(dE/dp) = (d/dk)

The group velocity of the wave packet is

E

gd d

vdk dp

ω= = ...(2.7.3)

The relativistic energy of a particle is

E2 = (pc)2 + (m0c2)2

2E

E

d pc

dp= ...(2.7.4)

From Eqns. (2.7.2), (2.7.3) and (2.7.4)

2

Egpc

v v= = ...(2.7.5)

Thus the representation of a particle by wave packet gets logical support.


2.8 HEISENBERG’S UNCERTAINTY PRINCIPLE OR THE PRINCIPLE OF INDETERMINACY

The dual nature of matter and radiation requires profound changes in our concepts built on the basisof common sense and everyday experience. The formulation of classical mechanics implies that theposition and momentum of a particle are assumed to have well defined values and can be determinedsimultaneously with perfect accuracy. But the wave particle duality compels us to abandon the ideaof simultaneous determination of position and momentum with perfect accuracy. In 1927 WernerHeisenberg, a German physicist, enunciated that it is impossible to determine both position andmomentum simultaneously with perfect accuracy. If x is the uncertainty in position and px is theuncertainty in the corresponding momentum then

px. x ≥ ...(2.8.1)

Similarly if E is the uncertainty in energy and t is uncertainty in time then

E . t ≥ ...(2.8.2)

From Eqn. (2.8.1) it is evident that if we try to measure the position of particle with utmostaccuracy i.e., x 0, the corresponding uncertainty in momentum becomes very large i.e., px

and vice versa.Let us illustrate the above assertion. Consider a particle having well-defined momentum

px (= k). Such a particle has well defined k or and is represented by a sinusoidal (monochromaticwave). A monochromatic wave has no beginning and end i.e., it is infinitely long; its amplitude isconstant for all values space coordinates x and therefore the particle may be anywhere betweenx = – to + . Thus the position of the particle is completely uncertain (x ).

Now consider a particle having well-defined position (x 0). A wave packet having verysmall extension in space describes such a particle. Fourier’s transform of this wave packet shows thatit is formed by superposition of a very large number of waves having continuous distribution of k or within a large range of k. Thus the uncertainty in k or p is very large (k ).

Fig. 2.8.1 A particle with well-defined momentum p is described by a sinusoidal waveextending from x = – to + . Here p 0 but x

Thus a particle with relatively small uncertainty in momentum has large uncertainty in position.A sinusoidal wave has well defined frequency and so is its energy (E = ). A particle described

this wave also has well-defined energy E and therefore E = 0. In order to see the constancy ofamplitude of such a sinusoidal wave, which exists from t = – to + , we have to look for a verylong time. Therefore, the uncertainty in time is infinite (t ).


Fig. 2.8.2 Some wave packets and their Fourier’s transforms, x k ≅ 1

Consider a particle, which is described by a wave packet as shown in the Fig. (2.8.2). TheFourier’s transform of the wave packet is also shown adjacent to it. Let x be the spread of the wavepacket in space and k the spread in propagation constant. It can be shown by standard mathematicaltechnique that

x k 1 ...(2.8.3)

Since p = k and p = k we have

px x ...(2.8.4)

which is the uncertainty relation.It should be carefully noted that the uncertainties in measurement of position and momentum

are not because of inadequacies in our measuring instruments. Even with ideal instrument we cannever in principle do better. This principle is the fundamental law of nature. The indeterminism isinherent in the very structure of matter. The momentum and position don’t assume well-defined valuessimultaneously.

Notice that it is the smallness of Planck’s constant that makes the uncertainty principleinsignificant in macroscopic world. In microscopic world the consequences of uncertainty principlecannot be ignored.


SOLVED EXAMPLES

Ex. 1. Show that the wavelength of electron accelerated through a potential difference V is given by

( )

h 12.3Å.

2m eV V volt= =λ

Sol. The kinetic energy of electron T = 2

2

p

m = eV

and 2 T 2 eVp m m= =

Therefore 2 T 2 eV

h h h

p m mλ = = =

Substituting m = 9.1 × 10–31 kg, e = 1.6 × 10–19 C, h = 6.6 × 10–34 Js, we have

1012.3 12.310 m

V V−λ = × = Å

For other charged particles appropriate values of m and charge q should be substituted in theabove equation.

Ex. 2. Obtain expression for the wavelength of a particle moving with relativistic speed.Sol. The relativistic momentum of a particle

0

2 21 /=

−

m vp

v c

( ) ( )( )

1/ 2 1/ 22 2 2 2

0 0

1 / 1 /

/

h v c v ch h

p m v m c v c

− −λ = = =

The momentum p of a relativistic particle can also be expressed as follows.

E 2 = p2c2 + (m0c2)2 = (T + m0c2)2

( )2

0T T 2m cp

c

+=

Hence = h/p

= ( ) 22 0 00

1

2 T 1 T/2T T 2

hc h

m m cm c=

++


=

1/ 2

20 0

T1

2 T 2

h

m m c

− +

If the particle under consideration is an electron accelerated through a potential difference of Vvolt, its de Broglie wavelength is given by

=

1/ 2

20 0

eV1

2 V 2

h

m e m c

− +

Ex. 3. Find the de Broglie wavelength of (i) electron moving with velocity 1000 m/s (ii) an objectof mass 100 gram moving with the same velocity.

Sol. (i) de Broglie wavelength of electron

34

31

6.63 10 Js

(9.1 10 kg)(1000 m/s)

h

mv

−

−×λ = = =

× 7285 × 10–10 m

= 7285 Å

(ii) de Broglie wavelength of object

34366.63 10 Js

8.63 10 m(0.1 kg)(1000 m/s)

h

mv

−−×λ = = = ×

Owing to extremely short wavelength of the object its wave behavior cannot be demonstrated.

Ex. 4. Find the de Broglie wavelength of electron, proton and -particle all having the samekinetic energy of 100 eV.

Sol. For electron

34

31 19

6.63 10 Js

2 T 2 (9.1 10 kg)(100 1.6 10 J)

h

m

−

− −

×λ = =× × × ×

= 1.23 × 10 –10 m = 1.23 Å

For proton m = 1.67 × 10–27 kg,

= 0.028 Å

For -particle m = 4 × 1.67 × 10– 27 kg

= 0.014 Å

Ex. 5. At what kinetic energy would an electron have the wavelength equal to that of yellowspectral line of sodium, = 5896 Å ?

Sol. Since 2 T

h

mλ =


2

2T

2

h

m=

λ

Substituting h = 6.63 × 10–34 Js, m = 9.1 × 10–31 kg, = 5896 × 10–10 m, we have

T = 6.93 × 10–25 J = 4.3 × 10–6 eV

Ex. 6. What is the wavelength of thermal neutron at 300 K?

Sol. Kinetic energy of thermal neutron E = 23

213 3 1.38 10 J/K 300KT 6.2 10 J

2 2k

−−× × ×= = ×

34

27 21

6.63 10 Js1.45

2 E 2 1.67 10 kg 6.2 10 J

h

m

−

− −

×λ = = =× × × ×

Å

Ex. 7. Find the de Broglie wavelength of hydrogen molecules, which corresponds to their mostprobable speed at room temperature 27°C.

Sol. The most probable speed of hydrogen molecule at temperature T is

2 Tk

vm

=

Momentum of molecule p = mv = 2 Tmk

2 T

h h

p mkλ = =

10

27 23

6.63 10 Js

2 (3.34 10 kg)(1.38 10 J/K)(300K)

−

− −

×=× × ×

= 1.26 × 10–10 m = 1.26 Å.

Ex. 8. At what value of kinetic energy is the de Broglie wavelength of an electron equal to itsCompton wavelength?

Sol. Energy of electron E = T + m0c2 = ( )2 2 20( )pc m c+

2 2

0T 2 Tm cp

c

+=

de Broglie wavelength of electron

2 20T 2 T

h hc

p m cλ = =

+

Given that 0

λ = λ =ch

m c


2 20 0T 2 T

h hc

m c m c=

+

2 20 0T 2m c m c= − ± (– sign is meaningless)

( )20T 2 1 (0.51 MeV)(0.414) 0.21 MeV.m c= − = =

Ex. 9. Find the de Broglie wavelength of relativistic electrons reaching the anticathode of an X-raytube if the short wavelength limit of continuous X-ray spectrum is equal to 0.10 Å.

Sol. Short wavelength limit of X-ray spectrum

00

eVeV

hc hcλ = ∴ =λ

Kinetic energy of electron T = eV = hc/0

Momentum of electron

20T(T 2 )m c

pc

+=


20T(T 2 )

h hc

p m cλ = =

+

0

22 0 0

00 0

(2 )2 1

hc

m chc hcm c

hc

λλ = =

λ+ + λ λ

For electron m0 c2 = 0.51 MeV, hc = 0.0124MeV Å.

0.10 A0.033A.

1.02 0.101

0.0124

λ = =×+

Ex. 10. Find the de Broglie wavelength of electron traveling along the first Bohr orbit in hydrogenatom.

Sol. The angular momentum of electron in the first Bohr orbit

2

nhmvr =

π

2

hmv

r=

π



2h

rmv

λ = = π =2 × 3.14 × 0.53 Å = 3.3 Å.

Ex. 11. Describe the Bohr’s quantum condition in terms of de Broglie wave.Sol. A stationary Bohr orbit must accommodate whole number of de Broglie wavelengths. If r

is the radius of electron orbit then2r = n ...(1)

According to de Broglie hypothesis

= h

mv...(2)

Eliminating from these equations we have

2

n hr

mvπ =

or 2

nhmvr =

πwhich is the Bohr’s quantum condition.

Ex. 12. An object has a speed of 10000 m/s accurate to 0.01%. With what fundamental accuracycan we locate its position if the object is (a) a bullet of mass of 0.05kg (b) an electron?

Sol. Momentum of bullet p = mv = (0.05 kg)(1000 m/s) = 50 kg m/sUncertainty in momentum p = 50 × 0.0001= 5 × 10–3 kg m/sMinimum uncertainty in position

3431

3

1.054 10 Js2.1 10 m

5 10 kg m/sx

p

−−

−×∆ = = = ×

∆ ×

Momentum of electronp = mv = (9.1 × 10–31 kg)(1000 m/s) = 9.1 × 10–28 kg m/s

Uncertainty in momentump = 9.1 × 10–28 × 0.0001= 9.1 × 10–32 kg m/s

Uncertainty in position

34

32

1.054 10 Js0.115 m

9.1 10 kg m/sx

p

−

−×∆ = = =

∆ ×

The uncertainty in bullet’s position is so small that it is far beyond the possibility of measurement.Thus, we see that for macroscopic objects like bullet, the uncertainty principle practically sets nolimits to the measurement of conjugate dynamic variables position and momentum. For electron, theuncertainty in its position is very large, nearly 107 times the dimensions of atom. Thus for microscopicobjects such as electrons, the uncertainty in their position is significant and cannot be overlooked.


Ex. 13. The position and momentum of 1 keV electrons are measured simultaneously. If its positionis located within 1Å, what is the percentage uncertainty in its momentum? Is this consistent with thebinding energy of electrons in atoms?

Sol. The uncertainty in position of electron

3424

10

1.054 10 Js1.054 10 kg m/s

10 mp

x

−−

−×∆ ≥ = = ×

∆

The momentum of electron inside the atom is at least equal to p = 1.054 × 10–24 kg m/s. Thecorresponding kinetic energy is

−−

−×= = = × =

× ×

2 24 217

31

(1.054 10 kg m/s)T 0.061 10 J 3.8 eV

2 2 9.1 10 kg

p

m

The ionization potential of atoms is of this order and hence the uncertainty in momentum isconsistence with the binding energy of electrons in atoms.

Ex. 14. Imagine an electron to be somewhere in the nucleus whose dimension is 10–14 m. What isthe uncertainty in momentum? Is this consistent with the binding energy of nuclear constituents?

Sol. If an electron were in the nucleus, its momentum would be uncertain by amount p given by

34

2014

1.054 10 Js1.054 10 kg m/s

10 mp

x

−−

−×∆ ≥ = = ×

∆

The momentum itself must be at least equal to p = 1.54 × 10–20 kg m/s. The correspondingkinetic energy of electron is many times greater than the rest energy m0c2 of electron and thereforethe kinetic energy of electron may be taken equal to pc.

T = pc = (1.054 × 10–20 kg m/s)(3 × 108 m/s) = 3.3 × 10–12 J

= 20 MeV

Experiments show that energy of electrons in nuclear disintegration ( decay) is very muchless than 20 MeV. Hence the uncertainty principle rules out the possibility of electrons being a nuclearconstituent.

Ex. 15. Consider a proton or neutron to be inside the nucleus. What is the uncertainty in momentumof electron? Is this consistent with the binding energy of nuclear constituents?

Sol. If a proton or neutron were inside the nucleus, the uncertainty in momentum would be

34

2014

1.054 10 Js1.054 10 kg m/s

10 mp

x

−−

−×∆ ≥ = = ×

∆

The corresponding kinetic energy T << m0c2 . Hence

2 20 2

1427

(1.054 10 kg m/s)T 3.6 10 J 0.23MeV

2 2 1.67 10 kg

p

m

−−

−×= = = × =

× ×

The binding energies of nuclei are of this order.


Ex. 16. The energy of a harmonic oscillator is given by

2

21

2 2= +p

E kxm

where p is momentum and k is force constant. Using uncertainty principle, show that the minimum

energy of the oscillator is 0, where 0 = k/mω .

Sol. According to uncertainty principle, ∆ ≥∆

px

. The momentum of oscillator is at least equal

to p where /= p x . The energy of oscillator may be written as

22

2

1E

22kx

mx= +

...(1)

For minimum energy 2

23

E0 kx x

x mkmx

∂ = = − + ⇒ =∂

Substituting the value of x2 in (1), we get

0E / .k m= = ω

Ex. 17. A nucleus exists in excited state about 10–12 sec. What is uncertainty in energy of thegamma ray photon emitted by the nucleus?

Sol. The minimum uncertainty in energy is at least equal to E, where

E.t =

Therefore

E = 34

2212

1.054 10 Js1.054 10 J.

10 st−

−×= = ×

∆

Ex. 18. The average excited atom has a life-time of about 10–8 sec. During this period it emits aphoton. What is the minimum uncertainty in the frequency of photon?

Sol. According to uncertainty principleEt

t

Minimum uncertainty in frequency of photon

81

10 rad/s.t

∆ω = =∆

Ex. 19. Making use of uncertainty principle give an estimate of radius and binding energy ofelectron in hydrogen atom in the ground state.

Sol. Energy of electron 2 2

0

E2 4

p e

m r

−= +πε

...(1)


Assuming that the uncertainties in momentum and energy are equal to the momentum and energythemselves, we can write the uncertainty principle as

= pr ...(2)

Eliminating p from these equations, we get

2 2

20

E42

e

rmr= −

πε

...(3)

In the ground state, energy is minimum. Therefore,

2 2

3 20

E 20

2 4

e

r mr r

∂ = − − =∂ πε

This gives 2

100 2

4 0.529 10 mrme

−= πε = ×...(4)

This value of r, that is ground state radius of hydrogen atom, is called Bohr radius (a0).Substituting the value of r in (3) we get ground state energy of atom

2 4

20

1 1E .

2 4

me = − πε

...(5)


1. Give reasoning that led de Broglie to speculate the wave behavior of material particles. Derive de Broglierelation.

2. Show that the wavelength of an electron beam accelerated through a potential difference of V volt is

12.3A.

V(volt)λ =

.

3. Describe Davisson-Germer experiment and interpret its results.4. State and explain Heisenberg uncertainty principle. Use this principle to show that (i) electrons cannot reside

inside the nucleus (ii) radius of the first orbit of hydrogen atom is 2

0 24 .r

me= πε

5. Show that the principle of indeterminacy can be expressed as L.∆ ∆θ ≥ where L is uncertainty in angular

momentum and is uncertainty in angular position of the particle under investigation.6. (a) An electron beam of energy 100 eV is passed through a circular hole of radius 5 10–4 cm. What is the

uncertainty introduced in the angle of emergence? (b) A lead ball of mass 200 gram is passed through a hole ofradius 25 cm, calculate the uncertainty in the angle of emergence. The velocity of ball is 20 m/s.

7. According to uncertainty principle. A particle of momentum p cannot be confined by a central force to a circleof radius r less than /p. Assume that pr = , show that the total energy of electron in hydrogen atom is

2 2

20

E .42

e

rmr= −

πε

Show that for minimum value E, r has the value 2

00 2

40.53r

me

πε= =

Å Bohr radius.

8. The accuracy in measurement of wavelength of photon is one part per million (i.e., / = 10–6). What is theminimum uncertainty x in a simultaneous measurement of the position of the photon in the case of (a) aphoton with = 6000 Å and (b) an x-ray photon with = 5 Å.

9. The atoms in a solid posses a certain minimum zero-point energy even at 0 K. Using uncertainty principle,explain this statement.

SCHRÖDINGER EQUATION

3.1 INTRODUCTION

The laws governing the behaviour of microscopic systems are radically different from those ofmacroscopic systems. Our sensory organs can have direct perception of only macroscopic objectsand the illustrative images of such objects are very useful in the description of behavior of macroscopicobjects. But these images can no longer be transferred to the description of objects of micro-world.It is therefore best approach to discard from the very beginning any tendency to construct illustrativeimages of microscopic objects and the phenomena being studied. For example, the concept of electronbeing a negatively charged sphere performing orbital and spin motion is absolutely unrealistic. Thequantum mechanics which, studies the behavior of objects of micro-world, employs abstract conceptsand the entire structure of this system of mechanics is developed on few postulates. Werner Heisenbergand P.A.M. Dirac laid the foundation of abstract formalism of quantum mechanics. The mathematicallanguage of quantum mechanics is very peculiar and the impossibility of visual representation andintricate mathematical language make quantum mechanics more difficult to understand.

The concept of state is the first basic thing on which the structure of quantum mechanics isbuilt. In order to explain it, we may well take up an example of a simple system viz hydrogen atomin magnetic field. This system consists of a proton, an electron and the applied magnetic field. Eachcomponent of the system interacts with the other according to specific laws of interaction. Therewill be various possible motions of the constituents of the system consistent with the laws ofinteraction. Each such motion is called a state of the system. This state is denoted by a complexfunction . All knowable information about the system can be derived from this function if we canobtain law describing the evolution of with space and time coordinates.

The information about the state of a system is obtained by performing measurement i.e., bymaking the system interact with an instrument (which is a macroscopic system). Consequently theresults of the measurements performed on micro-system are necessarily expressed in terms of conceptsdeveloped for characterizing macroscopic objects (coordinates, momentum, energy, angular momentumetc.). These characteristics of the micro-system are called dynamical variables or observables. Theabstract formalism of quantum mechanics is beyond the understanding of students for whom thisbook is intended. We, therefore, start with somewhat easier approach to quantum mechanics formulatedby Erwin Schrödinger in 1926.

CHAPTER


The wave like behaviour of material particles allows us to ascribe a wave associated with themoving particle, which can be called matter waves. The function describing the wave behavior of amoving particle is usually denoted by and is analogous to the displacement function of mechanicalwaves or to the components of electric and magnetic fields of electromagnetic waves. The wave functionof mechanical and electromagnetic waves of classical physics are the solution of the so-called classicalwave equation

2

2x

∂ ψ∂

2

2 2

1

v t

∂ ψ=∂

...(3.1.1)

where the function represents the displacement of the medium in case of mechanical waves andthe components of electric and magnetic field in case of electromagnetic waves. It is natural to expectthat the wave function describing the material particle be a solution of some kind of differentialequation like Eqn. (3.1.1). It is well-known fact the wave equation for mechanical waves is derivedfrom Newton’s law of motion and the wave equation for electromagnetic waves is derived fromMaxwell’s electromagnetic field equations. Unfortunately there is no fundamental equation from whichthe wave equation for matter waves can be derived. It was Erwin Schrodinger who guessed the correctequation for matter waves. Schrödinger equation is a postulate in the same sense as the postulates ofspecial theory of relativity and the laws of thermodynamics. None of these can be derived from otherprinciple. This equation is a conjecture and it must stand or fall by the test of experiments. It hasbeen found that the Schrodinger wave equation leads to the conclusions that are in complete agreementwith experiments and therefore we may regard it as a valid postulate in the development of the storyof quantum mechanics.

3.2 SCHRÖDINGER EQUATION

The wave function of a particle moving in x-direction is

( , )x tψ ( ) ( )A Ax xi t k x i k x te e− ω − −ω= = ...(3.2.1)

The wave attributes kx (= 2/) and (= 2) are related to the particle attributes px and E as

, Ex xp k= = ω ...(3.2.2)

In terms of px and E the wave function (x, t) can be expressed as

E

( , ) Aexp xp xtx t i

ψ = − −

...(3.2.3)

From Eqn. (3.2.3)

Ei

t

∂ψ = − ψ∂

Eit

∂ψ = ψ∂

...(3.2.4)

Schrödinger Equation 99

Similarly,

xip

x

∂ψ = ψ∂

xi px

∂ψ− = ψ∂

...(3.2.5)

Differentiating Eqn. (3.2.5) again with respect to x, we have

22

2x

xp

i p ixx

∂ ψ ∂ψ− = = ψ∂∂

2

2 22 xp

x

∂ ψ− = ψ∂

...(3.2.6)

For a non-relativistic free particle the total energy E of the particle moving in x-direction isequal to its kinetic energy T.

2

E T2

xp

m= =

Multiplying both sides of above equation by , we have

2

E2

xp

mψ = ψ ...(3.2.7)

Making use of Eqns. (3.2.4) and (3.2.6) we can write Eqn. (3.2.7) as

2 2

22i

t m x

∂ψ ∂ ψ= −∂ ∂

...(3.2.8)

This equation is known as time-dependent Schrödinger for a free particle.If the particle is moving in a force field described by potential energy function V, its total

energy is

E =

2

V( )2

xpx

m+

and the Schrödinger equation is

2 2

2V

2i

t m x

∂ψ ∂ ψ= − + ψ∂ ∂

...(3.2.9)

If the particle is free to in three dimensions, it is represented by wave function

( ) ( , , , ) Aexp x y zx y z t i t k x k y k zψ = − ω − − −


( )A exp x y zi

Et p x p y p z = − − − −

...(3.2.10)

From Eqn. (3.2.10), we have

Eit

∂ψ = ψ∂

...(3.2.11)

–2

2 22

andx xi p px x

∂ψ ∂ ψ= ψ − = ψ∂ ∂

...(3.2.12)

2

2 22

andy yi p py y

∂ψ ∂ ψ− = ψ − = ψ∂ ∂

...(3.2.13)

2

2 22

andz zi p pz z

∂ψ ∂ ψ− = ψ − = ψ∂ ∂

...(3.2.14)

Total energy of the particle is

2 22

E ( , , )2 2 2

y zxp pp

V x y zm m m

= + + + ...(3.2.15)

Making use of Eqns. (3.2.11) and (3.2.12,13 and 14) we can write Eqn. (3.2.15) as

2 2 2 2

2 2 2V

2i

t m x y z

∂ψ ∂ ∂ ∂= − + + ψ + ψ ∂ ∂ ∂ ∂

22 V

2i

t m

∂ψ = − ∇ ψ + ψ∂

...(3.2.16)

where 2 2 2

22 2 2x y z

∂ ∂ ∂∇ = + +∂ ∂ ∂

is Laplacian operator. Eqn. (3.2.16) is known as the time-dependent

Schrödinger equation of a particle in three dimensions.Stationary state: Time-independent Schrödinger Equation: When the potential energy V is

independent of time, the wave function (x, t) may be written as product of two wave functions, ofwhich one is function of x and the other is function of t only.

( , ) ( ) ( )x t x f tψ = ψ ...(3.2.17)

Substituting Eqn. (3.2.16) in (3.2.10) and dividing the resulting equation throughout by (x) f (t) we find


2 2

2

1 1V

( ) 2 ( )

df di

f t dt m x dx

ψ= − +ψ

...(3.2.18)

The left hand side of (3.2.18) is function of time t only and the right hand side is function of xonly. Since x and t are independent of each other, this equality can hold only if each side is equal tothe same constant. Each side has the dimensions of energy, so we write the separation constant as E.The separation constant E is number and represents the total energy of the particle. Therefore

1E

( )

dfi

f t dt= ...(3.2.19)

2 2

2

1V E

2 ( )

d

m x dx

ψ− + =ψ

...(3.2.20)

Eqn. (3.2.19) can be expressed as

0,df E

fdt i− =

which integrates to

E

( ) Cexpi t

f t = −

...(3.2.21)

where C is constant of integration. A particle whose state is described by wave function

/( , ) ( ). iE tx t x e−ψ = ψ

is said to be in stationary state because its probability density 2( , ) ( , ) or | ( , ) |x t x t x t∗Ψ Ψ Ψ is

independent of time.The time independent Schrödinger equation can be expressed as

2 2

2V E

2

d

m dx

ψ− + ψ = ψ...(3.2.22)

or H Eψ = ψ ...(3.2.23)

where

2 2 2

2

ˆH V V

2 2

p d

m m dx= + = − +

is Hamiltonian operator, an operator representing the total energy of the particle. Eqn. (3.2.22) canalso be written as

( )2

2 2

2E V 0

d m

dx

ψ + − ψ =

...(3.2.24)


[For a system whose potential energy V is a function of coordinates only, the total energy remainsconstant with time i.e. E is conserved. For a conservative system, the classical mechanical Hamiltonianfunction turns out to be the total energy expressed in terms of coordinates and conjugate momenta.]

Eqn. (3.2.23) is an eigen value equation. Thus the time-independent Schrödinger wave equationis an eigen value problem.

The time-dependent Schrödinger equation can be written as

ˆ ˆH ( , ) E ( , )x t x tψ = ψ

where

2 2

2ˆ ˆ ˆH V and E .

2i

m tx

∂ ∂= − + =∂∂

3.3 PHYSICAL SIGNIFICANCE OF WAVE FUNCTION

It is natural to ask the question regarding the physical significance of the wave function . For avibrating string, it represents the displacement of the string from equilibrium position; in case ofelectromagnetic waves it represents the electric or magnetic field at the point under consideration.But there is no physical quantity with which the wave function of matter wave may be associated.Just as the concepts of electric and magnetic field are abstraction to explain the interaction betweenelectrical charges, the concept of wave function is an abstraction to describe the dynamics ofmicroscopic particles. But such an interpretation of is of little significance. In 1926 Max Bornsuggested a useful statistical interpretation of wave function, which was inspired by Einstein’s conceptof wave like behavior of particle like photons. According to Einstein the propagation of photon inspace is described by Maxwell’s equation involving electric field E (x, y, z, t) and magnetic field B(x, y, z, t). The magnitude of field E and B provides the probability of the location of the photon. Inthe region where E and B are large, the likelihood of finding the photon is also large and vice-versa.It is therefore reasonable to associate a probability function P with wave amplitude E. The probabilityfunction P (x, y, z, t) expresses the likelihood of finding the photon and is related to the wave amplitudeE (x, y, z, t) as P (x, y, z, t) = | E (x, y, z, t) | 2

According Born, the wave function (x, y, z, t) is analogous to the electric field E and Einstein’sinterpretation can be utilized to provide a physical meaning to the wave function associated with thematerial particles. The probability of finding a particle at a point (x, y, z ) at time t is given by

ψψψ 2 *( , , , ) orx y z t where * is complex conjugate of . The probability of finding the particle in

a volume element dxdydz centered around the point (x, y, z) is given by

2 *( , , , ) orx y z t dx dy dz dx dy dzψ ψψ

Thus ||2 is probability density and itself is called probability amplitude.Since the probability of finding the particle somewhere in the universe is unity, we have

* 1dx dy dz

∞ ∞ ∞

−∞ −∞ −∞

ψψ =∫ ∫ ∫ ...(3.3.1)


The wave function, which satisfies the condition in Eqn. (3.3.1)), is said to be normalized.The probabilistic interpretation of the wave function asserts that the wave field generated by

Schrödinger wave equation is a probability field. The microscopic entities retain their status as particlesso far as the detecting devices are concerned but their distribution in space is governed by the wavefield. The probability waves show all the characteristic properties of waves viz interference anddiffraction in the same way as the waves of classical physics do.

Where do the probability waves come from? What mechanism generates them? These questionshave no answers at present. Quantum physics as it now stands asserts the existence of probabilitywaves. Richard Feynman, one of the principal contributors to the present day quantum electrodynamicshas to say about the law of probability waves: “one might still ask: how does it work? What is themachinery behind the law? No one has found any machinery behind the law. No one can explain anymore than we have explained. No one will give any deeper representation of the situation. We haveno idea about a more basic mechanism from which these results can be deduced.”

3.4 INTERPRETATION OF WAVE FUNCTION IN TERMS OF PROBABILITYCURRENT DENSITY

Consider a stream of particles moving in x-direction. Let be the wave function of the particles inthe beam. The product * represents the probability density of finding the particle at point x. Inwhat follows we shall find relation between the probability density of the particles and the probabilitycurrent density associated with the particle beam.

The time-dependent Schrödinger wave equation is

2

2 V2

it m

∂ψ = − ∇ ψ + ψ∂

...(3.4.1)

From this equation

2 V

2t mi i

∂ψ = − ∇ ψ + ψ∂

...(3.4.2)

The complex conjugate of above equation is

*

2 * *V

2t mi i

∂ψ = ∇ ψ − ψ∂

...(3.4.3)

Multiplying Eqn. (3.4.2) by * and Eqn. (3.4.3) by and adding the resulting equations, wehave

*

* 2 * * 2

2t t mi

∂ψ ∂ψ ψ + ψ = Ψ∇ ψ −ψ ∇ ψ ∂ ∂

*

2 * * 2( )

2t mi

∂ ψ ψ = ψ ∇ ψ − ψ ∇ ψ ∂

...(3.4.4)

Making use of the vector identity

. . .∇ ϕ =∇ϕ +ϕ∇A A A


we have

* * * * 2( ) ( )∇ ⋅ ψ ∇ψ − ψ∇ψ = ∇ψ ⋅∇ψ + ψ ∇ ψ −

* 2 *( )∇ψ ⋅∇ψ −ψ∇ ψ

* 2 2 *= ψ ∇ ψ − ψ∇ ψ ...(3.4.5)

In view of Eqn. (3.4.5) we can write Eqn. (3.4.4) as

* * *( ) ( )

2t mi

∂ ψ ψ = ∇ ⋅ ψ∇ψ − ψ ∇ψ ∂

* * *( ) ( ) 0

2t mi

∂ ψ ψ +∇ ⋅ ψ ∇ψ − ψ∇ψ =∂

...(3.4.6)

This equation is similar to the equation of continuity in electrodynamics viz

J 0∂ ρ +∇ ⋅ =∂t

...(3.4.7)

where is charge density and J is current density. A comparison of Eqns. (3.4.6) and (3.4.7)

allows us to interpret * as probability density and ( * *)2mi

ψ ∇ψ −ψ∇ψ as probability current

density J where J represents the rate at which probability is streaming outward across a closed surface.The Cartesian components of J are

∂ψ ∂ψ= ψ −ψ ∂ ∂

*

*J2x mi x x

∂ψ ∂ψ= ψ −ψ ∂ ∂

*

*J2y mi y y

∂ψ ∂ψ= ψ −ψ ∂ ∂

*

*J2z mi z z

For a beam of free particles moving in x-direction the wave function is

( ) Aexp E /xi t p xψ = − −

( ) * *A exp E /xi t p xψ = −

and therefore the probability current density is given by

*

*J2x mi x x

∂ψ ∂ψ= ψ −ψ ∂ ∂


( )*xp

m= ψ ψ

( )*A Axp

m=

= vx 2

A

Jx represents the flux of particles (number of particles crossing per unit area per second). Recall

that 2ψ determines the probability density of finding a particle at a point where is defined. The

interpretation of in terms of probability current density puts additional condition on that it musthave continuous and finite first derivative. It thus follows that

(i) must be finite and single valued.

(ii) and / x∂ψ ∂ must be continuous, and

(iii) must be square integrable.

These conditions, which the wave function must obey, are called the standard conditions andthe wave function is said to be well behaved.

3.5 SCHRÖDINGER EQUATION IN SPHERICAL POLAR COORDINATES

In polar coordinates the Laplacian operator is expressed as

2

2 22 2 2 2 2

1 1 1sin

sin sin

∂ ∂ ∂ ∂ ∂ ∇ = + θ + ∂ ∂ ∂θ ∂θθ θ ∂φ r

r rr r r...(3.5.1)

2 2 2 2 sin cos , sin sin , cos , , tan / x r y r z r r x y z y x= θ ϕ = θ ϕ = θ = + + ϕ =

Fig. 3.5.1 Relationship between Cartesian and polar coordinates


The relationship between Cartesian coordinates and spherical polar coordinates is defined inthe Fig. (3.5.1). The Schrodinger equation in spherical polar coordinates is

22

2 2 2 2 2 2

1 1 1 2sin (E V) 0

sin sin

mr

r rr r r

∂ ∂ψ ∂ ∂ψ ∂ ψ + θ + + − ψ = ∂ ∂ ∂θ ∂θθ θ ∂ϕ ...(3.5.2)

If the wave function depends only on radial distance r then the Laplacian operator simplifiesto

2 22

1 ∂ ∂ ∇ = ∂ ∂ r

r rr

and the Schrodinger wave equation assumes the simple form

( ) ( )22 2

1 2E V( ) ( ) 0.

mr r r r

r rr

∂ ∂ ψ + − ψ = ∂ ∂ ...(3.5.3)

3.6 OPERATORS IN QUANTUM MECHANICS

An operator is a rule or an instruction which transforms a function into another function. Linear

operators play a very important role in quantum mechanics. If Q is an operator and f (x) is an arbitrary

function then the action of Q on f (x) is represented as

Q ( ) ( )f x g x= λ ...(3.6.1)

where g (x) is another function and is constant. A linear operator is one which satisfies the followingtwo conditions:

1 2 1 2ˆ ˆ ˆQ( .........) Q Q ......f f f f+ + = + + ...(3.6.2)

ˆ ˆQ( ) Qcf c f= , where c is an arbitrary constant. ...(3.6.3)

A physically measurable property of a system is called an observable (dynamical variable).Energy, momentum, angular momentum and position coordinates are examples of observables. Inquantum mechanics every observable is represented by a linear operator.

Algebra of Operators

(i) The sum and difference of two operators ˆP and Q are defined by equations

ˆ ˆˆ ˆ(P Q) ( ) P ( ) Q ( )f x f x f x+ = +

ˆ ˆˆ ˆ(P Q) ( ) P ( ) Q ( )f x f x f x− = −

(ii) The product of two operators ˆP and Q is defined by equations

ˆ ˆˆ ˆPQ ( ) P Q ( )f x f x =


In above equation, we first operate on f (x) with the operator on the right of the operator productand then we take the resulting function and operate on it with the operator on the left of the operatorproduct.

(iii) Two operators are said to be equal if

ˆP ( ) Q ( ).f x f x=

(iv) The operator 1 (multiplication by 1) is the unit operator.

(v) The operator 0 (multiplication by 0) is the null operator.

(vi) The square of an operator is defined as the product of the operator with itself.

2ˆ ˆ ˆQ QQ=

The nth power of an operator is defined to mean applying the operator n times in succession.

(vii) Operators obey associative law of multiplication.

ˆ ˆˆ ˆ ˆ ˆP(QR) (PQ)R=

An important difference between operator algebra and ordinary algebra is that numbers obey

commutative law of multiplication but operators do not necessarily do so. That is ˆ ˆˆ ˆPQ and QP are

not necessarily equal operators.

(viii) We define the commutator ˆP, Q of operator ˆP and Q as operator ˆ ˆˆ ˆPQ QP− .

ˆ ˆ ˆˆ ˆ ˆP, Q = PQ QP −

If ˆ ˆ ˆˆ ˆ ˆPQ=QP then [P,Q]=0 and we say that ˆP and Q commute.Operators of Some Dynamical Variables: The wave function of a free particle moving in

three dimensional space is

( )( , , , ) A exp E x y zi

x y z t t p x p y p z ψ = − − − −

...(3.6.4)

Partial derivative of with respect to x is

orx xi

p i px x

∂ψ ∂ψ= ψ − = ψ∂ ∂

Similarly,

E

y

z

i py

i pz

it

∂ψ− = ψ∂∂ψ− = ψ∂∂ψ = ψ∂

...(3.6.5)


A close look at the above equations reveals that the dynamical variables px, py, pz and E are in

some sense related to the differential operators , , , andi i i ix y z t

∂ ∂ ∂ ∂− − −∂ ∂ ∂ ∂

respectively. These

symbols instruct what operation is to be carried out on the functions that follow. The operators ofposition coordinates x, y, z are the variables themselves. Similarly the operator of position dependentpotential energy V(x, y, z) is V(x, y, z) itself. For convenience we rewrite the dynamical variables andtheir corresponding operators in tabular form.

Dynamical Variables Operators

px ˆ xp ix

∂= −∂

pyˆyp i

y

∂= −∂

pzˆzp i

z

∂= −∂

p p i= − ∇

E E it

∂=∂

Kinetic energy 2

T2

p

m=

22T

2m= − ∇

The Hamiltonian (total energy) function of a mechanical system is given by

2

H V2

p

m= +

and the corresponding operator is

2

2H V2m

= − ∇ +...(3.6.6)

For a conservative system the total energy is represented by Hamiltonian function H expressedin terms of position coordinates and conjugate momenta. Therefore energy operator is Hamiltonian

operator 2

2ˆ ˆH V2m

→− ∇ + and not ∂∂

it

. Time is not an observable, but it is a parameter in quantum

mechanics. Hence there is no operator for time.In view of Eqn. (3.6.5), the time independent Schrödinger equation can be written as

H Eψ = ψor

2

2 V E2m− ∇ ψ + ψ = ψ

which is an eigen value equation.


Angular momentum operator: In classical mechanics the angular momentum of a particle isgiven by

x y z

x y z

p p p

= × =L

i j k

r p

⇒ = − = − = −L , L , Lx z y y x z z y zyp zp zp xp xp yp

The operators corresponding to these variables are

Lx i y zz y

∂ ∂= − − ∂ ∂ ...(3.6.7)

Ly i z x

x z

∂ ∂ = − − ∂ ∂ ...(3.6.8)

Lz i x yy x

∂ ∂= − − ∂ ∂ ...(3.6.9)

Notice the kind of symmetry in the expression for operator L .x By carrying a cyclic permutation

of x, y, z ( i.e., replacing x by y, y by z and z by x ) we can get operator of ˆ ˆL and Ly z .

In problems having spherical symmetry it is useful to express these operators in sphericalcoordinates, which are defined in the Figure 3.6.1. The relationships between the Cartesian and thepolar coordinates are

x = r sin cos ...(3.6.10)

y = r sin sin ...(3.6.11)

z = r cos ...(3.6.12)

r2 = x2 + y2 + z2 ...(3.6.13)

tan = y/x ...(3.6.14)

tan2 = 2 2

2

+x y

z...(3.6.15)

2 2 2

cosθ =+ +

z

x y z...(3.6.16)

From Eqn. (3.6.13)

sin cos∂ = = θ ϕ∂r x

x r...(3.6.17)

sin sin∂ = = θ ϕ∂r y

y r...(3.6.18)


cos∂ = = θ∂r z

z r...(3.6.19)

Fig. 3.6.1

Differentiating Eqn. (3.6.15) w.r.t. x, we have

2

2 2 2

2 2 sin cos2 tan sec

cos

∂θ θ ϕθ θ = =∂ θ

x r

x z r

cos cos

x r

∂θ θ ϕ=∂

...(3.6.20)

Similarly, cos sin sin,

∂θ θ ϕ ∂θ θ= = −∂ ∂y r z r

Differentiating Eqn. (3.6.14) w.r.t. x, we have

22 2 2 2

sin sinsec

sin cos

∂ϕ θ ϕϕ = − = −∂ θ ϕ

y r

x x r

sin

sinx r

∂ϕ ϕ= −∂ θ ...(3.6.21)

Similarly, 2cos

, 0∂ϕ ϕ ∂ϕ= =∂ ∂y r z

Now ∂ψ ∂ψ ∂ ∂ψ ∂θ ∂ψ ∂ϕ= + +∂ ∂ ∂ ∂θ ∂ ∂ϕ ∂

r

x r x x x

=cos cos sin

sin cossin

∂ψ ∂ψ θ ϕ ∂ψ ϕθ ϕ+ −∂ ∂θ ∂ϕ θr r r


cos cos sin

sin cossinx r r r

∂ ∂ θ ϕ ∂ ϕ ∂= θ ϕ + −∂ ∂ ∂θ θ ∂ϕ ...(3.6.22)

Similarly, r

y y r y y

∂ ∂ ∂ ∂θ ∂ ∂ϕ ∂= + +∂ ∂ ∂ ∂ ∂θ ∂ ∂ϕ

cos sin cos

sin sinsin

∂ ∂ θ ϕ ∂ ϕ ∂= θ ϕ + −∂ ∂ ∂θ θ ∂ϕy r r r

...(3.6.23)

and ∂ ∂ ∂ ∂θ ∂ ∂ϕ ∂= + +∂ ∂ ∂ ∂ ∂θ ∂ ∂ϕ

r

z z r z z

sin

cos∂ ∂ θ ∂= θ −∂ ∂ ∂θz r r

...(3.6.24)

Making use of above results we can write the operators corresponding to the Cartesian componentsof angular momentum.

Lz i x yy x

∂ ∂= − − ∂ ∂

cos sin cossin cos sin sin

sin

cos cos sinsin sin sin cos

sin

∂ θ ϕ ∂ ϕ ∂θ ϕ θ ϕ + + − ∂ ∂θ θ ∂ϕ = − ∂ θ ϕ ∂ ϕ ∂ θ ϕ θ ϕ + − ∂ ∂θ θ ∂ϕ

rr r r

i

rr r r

∂= −∂ϕ

i ...(3.6.25)

Similarly, L sin cot cosx i ∂ ∂= ϕ + θ ϕ ∂θ ∂ϕ

...(3.6.26)

L cos cot siny i ∂ ∂= − ϕ − θ ϕ ∂θ ∂ϕ

...(3.6.27)

Making use of the definitions

2 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L and L L L Lx x x x y z= = + +

we can find 2L z

2

2 22

Lz∂= −∂ϕ

...(3.6.28)


and

22 2

2 2

1 1L sin

sin sin

∂ ∂ ∂ = − θ + θ ∂θ ∂θ θ ∂ϕ ...(3.6.29)

3.7 EIGEN VALUE EQUATION

In general, each physical quantity is represented by a linear operator and for each operator one canset up an equation of the type

Q q qu qu= ...(3.7.1)

i.e., the effect of the operator is to multiply the function uq by a constant factor q. Eqn. (3.7.1) is aneigen value equation. The solutions of above equation satisfying the set of conditions can be foundnot for all values but only for selected values of the parameter q. These special values of the parameter

q are called the eigen (characteristic) values of the operator Q and the functions uq, which satisfythe above equation, are called the eigen (characteristic) functions of the operator.

When a system is in an eigenstate uq of Q , the dynamical variable Q has a definite value equalto the eigen value q. That is, the uncertainty in the value of Q is zero if the system is in one of the

eigen states of Q and the physical quantity Q is said to be quantized. The meaning of Eqn. (3.7.1)is that if the system is in the eigen state uq, the measurement of the quantity Q will yield only one

number q. The set of all eigen values of Q forms a spectrum, called eigen value spectrum. Thespectrum may be discrete or continuous or partly discrete and partly continuous. If there exists onlyone eigen function belonging to a given eigen value, the eigen value is said to be non-degenerate.It may happen that several eigen functions may belong to a single eigen value. Then this eigen valueis said to be degenerate. If uq and vq belong to the same eigen value q, then their linear combinationc1uq + c2vq, for all values of c1 and c2, is also an eigen function belonging to the same eigen value.

1 2 1 2 1 2ˆ ˆ ˆQ( ) Q Q ( )q q q q q qc u c v c u c v q c u c v+ = + = + ...(3.7.2)

Thus a degenerate eigen value corresponds to an infinite number of eigen functions. The totalityof eigen functions belonging to a degenerate eigen value forms a linear space, called eigen-space.The set of all eigen functions belonging to a given degenerate eigen value is closed under linearcombination. This implies that any linear combination of members of the set is also a member of theset. Not all the members of the set are linearly independent. From the members of the set of eigenfunctions, it is always possible to choose a subset of linearly independent eigen functions, say uq1,uq2, ………,uqr such that any eigen function belonging to the eigen value q can be expressed uniquelyas a linear combination of the type (c1uq1 + c2uq2 +………… + cruqr) with suitable coefficients c1,c2,…… ,cr. The set of independent functions uq1, uq2, ……… ,uqr is said to span the linear spaceand this set of functions is said to form the basis functions of the space. The number r is characteristicof the space. This means that out of infinite number of eigen functions belonging to a given degenerateeigen value there exists only a definite number, say r, of linearly independent functions. This numberr is called the degree of degeneracy and the eigen value is said to be r-fold degenerate.


If the system is an arbitrary state (i.e., is not an eigen function), and the measurement ofphysical quantity Q is made on a large number of identical systems (i.e., all in the same state), theresult is not a fixed value but the outcome has a range of values whose average value (also calledexpectation value) is given by

*Qq d= ψ ψ τ∫ ...(3.7.3)

where is normalized wave function of the state.The foregoing discussion may be illustrated with an example. The eigen value equation for a

free particle is

H Eψ= ψ

2 2

2E

2

d

m dx

ψ− = ψ

2

2 22 2

2 E0,

d mk k

dx

ψ + ψ = =

...(3.7.4)

The linearly independent solutions of Eqn. (3.7.4) are 1 = eikx and 2 = e–ikx. So the eigenvalue E is two-fold degenerate and the 1 and 2 are the basis functions. The following linearcombinations of 1 and 2 are also the eigen functions of the operator H.

cos , sin .2 2

ikx ikx ikx ikxe e e ekx kx

i

− −+ −= =

3.8 ORTHOGONALITY OF EIGEN FUNCTIONS

Two eigen functions m and n belonging to different eigen values m and n are said to be orthogonalif they satisfy the relation

* 0ψ ψ τ =∫ m nd ...(3.8.1)

The time-independent Schrödinger wave equation Hψ = εψ is an eigen value equation. We

shall show that the eigen functions of Hamiltonian operator are orthogonal.The Schrödinger equation is

2

2 2

2( V) 0m

m m

d m

dx

ψ+ ε − ψ =

...(3.8.2)

The complex conjugate of Eqn. (3.8.2) is

2 **

2 2

2( V) 0m

m m

d m

dx

ψ+ ε − ψ =

...(3.8.3)


Schrodinger equation for the state n is

2

2 2

2( V) 0n

n nd m

dx

ψ+ ε − ψ =

...(3.8.4)

Multiplying Eqn. (3.8.3) by n and Eqn. (3.8.4) by m* and subtracting and then integrating,

we get

2 2 ** *

2 2 2

2( ) 0

ψ ψψ −ψ + ε − ε ψ ψ = ∫ ∫

n mm n n m m n

d d mdx dx

dx dx

The quantity in the square bracket in the first term is derivative of m*dn/dx – n dm

*/dxwith respect to x. Thus

( )

** *

2

2n mm n m n m n

d dd mdx dx

dx dx dx

∞ ∞

−∞ −∞

ψ ψψ −ψ = ε − ε ψ ψ

∫ ∫

( )*

* *2

2∞ ∞

−∞−∞

ψ ψψ − ψ = ε − ε ψ ψ

∫

n mm n m n m n

d d mdx

dx dx

Physically well-behaved wave function and its derivative must vanish as x ± . So theleft hand side vanishes at both the limits. Therefore

( ) *∞

−∞

ε − ε ψ ψ∫m n m ndx = 0

Since m n, we have

*

∞

−∞

ψ ψ∫ m ndx = 0 ...(3.8.5)

Thus the eigen function belonging to different eigen values are orthogonal. The normalizationcondition for wave function (x) is

2| | 1

∞

−∞

ψ =∫ dx ...(3.8.6)

The properties of wave function expressed by Eqns. (3.8.5) and (3.8.6) can be expressed by asingle equation as

*∞

−∞

ψ ψ τ = δ∫ m n mnd ...(3.8.7)

where mn is kronecker delta having the propertiesmn = 0 for m n

= 1 for m = n


The property of wave function expressed by Eqn. (3.8.7) is known as orthonormality of wavefunction.

It should be noted that the orthonormality of eigen functions is not restricted to the eigenfunctions of Hamiltonian operator. In fact the wave functions of all hermitian operators satisfy theorthonormality condition.

3.9 COMPATIBLE AND INCOMPATIBLE OBSERVABLES

Two observables which can be measured simultaneously and precisely without influencing each other

are called compatible. The operators of such observables commute i.e., ˆˆ[P,Q] 0= . On the other hand,two observables are such that the determination of one observable introduces an uncertainty in theother, they are called incompatible. The operators of incompatible observables do not commute.

That is, ˆˆ[P, Q] 0.≠Assume that two physical quantities Q and R can simultaneously have definite values when the

system is in a common state n. The wave function n of the state in which the quantity Q has avalue qn and the quantity R, the value rn, must satisfy following two equations simultaneously.

Q n n nqψ = ψ ...(3.9.1)

R n n nrψ = ψ ...(3.9.2)

The product of operators is determined by the condition

( )ˆ ˆˆ ˆQR Q(R )n n n n n nq rΠψ = ψ = ψ = ψ ...(3.9.3)

Making use of Eqns. (3.9.1) and (3.9.2), we find that

( ) ( )ˆ ˆˆ ˆ ˆRQ R(Q ) Rn n n n n n nq r qψ = ψ = ψ = ψ ...(3.9.4)

From Eqns. (3.9.3) and (3.9.4), we see that

ˆ ˆˆ ˆQR RQΠ = = ...(3.9.5)

ˆ ˆˆ ˆQR RQ 0.− = ...(3.9.6)

Thus if two quantities can simultaneously have definite values then (i) their operatorshave common eigen functions and (ii) their operators commute.

In general, the product of operators is non-commuting i.e.,

ˆ ˆˆ ˆQR RQ≠

This can be verified taking the example of the operators

ˆ ˆQ and R x

x

∂= =∂

( )ˆ ˆQR ( )x xx x

∂ ∂ψψ = ψ = + ψ∂ ∂


( )ˆRQ xx

∂ψψ =∂

Operators ˆ ˆQ and R for which the condition

ˆ ˆˆ ˆQR RQ= ...(3.9.7)

is observed are said to be commutative operators. If the condition Eqn. (3.9.7) is not observed, theoperators are said to be non-commutative. Operators, which satisfy the condition

ˆ ˆˆ ˆQR RQ= − ...(3.9.8)

are called anticommutative operators.

3.10 COMMUTATOR

The operator ˆ ˆˆ ˆQR RQ− formed from the operators ˆ ˆQ and R is called the commutator of the given

operators and is designated by the symbol ˆ ˆQ, R i.e.,

ˆ ˆ ˆˆ ˆ ˆQ,R QR RQ = − ...(3.10.1)

The commutator of commuting operators is zero.Commutation relations: Linear operators obey following commutations rules:

ˆ ˆˆ ˆA, B B, A = −

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ[A, BC] [A, B]C B[A, C]= +

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ[AB, C] [A, C]B A[B, C]= +

Some commutation relations of quantum mechanical operators:

1. ˆ ˆ ˆ ˆˆ ˆ[ , ] [ , ] [ , ] 0= = =x y y z z x

ˆ ˆ ˆˆ ˆ ˆ[ , ] 0= − = − =x y xy yx xy yx

2. ˆ ˆ ˆ ˆ ˆˆ[ , ] [ , ] [ , ]= = = x y zx p y p z p i

( ) ( )ˆ ˆ ˆˆ ˆ ˆ[ , ]∂ψ ∂ ψ ψ = − ψ = − − ∂ ∂

x x xx

x p xp p x i xx x

∂ψ ∂ψ = − −ψ − = ψ ∂ ∂ i x x i

x x


3. ˆ ˆ ˆ ˆ ˆˆ[ , ] [ , ] [ , ] 0= = =y z xx p y p z p

( ) ( )ˆ ˆ ˆˆ ˆ ˆ[ , ] ∂ψ ∂ ψψ = − ψ = − − ∂ ∂

y y yx

x p xp p x i xy y

0 ∂ψ ∂ψ= − − = ∂ ∂

i x xy y

4. ˆ ˆ ˆ ˆ ˆ ˆ[ , ] [ , ] [ , ] 0= = =x y y z z xp p p p p p

( ) 2ˆ ˆ ˆ ˆ ˆ ˆ[ , ] ( ) 0 ∂ ∂ψ ∂ ∂ψψ = − ψ = − − = ∂ ∂ ∂ ∂

x y x y y xp p p p p p ix y y x

5. ˆ ˆ[H, ] 0p =

( )ˆ ˆ ˆˆ ˆ ˆ[H, ] H Hp p pψ = − ψ

2 2 2 2

2 22 2

d di i

m x x mdx dx

∂ ∂ = − − + − ψ ∂ ∂

3 3 3

3 30

2

∂ ψ ∂ ψ= − = ∂ ∂

i

m x x

6. ˆ ˆ ˆˆ ˆ ˆ[L , ] 0, [L , ] 0, [L , ] 0x y zx y z= = =

( )ˆ ˆ ˆˆ ˆ ˆ[L , ] L Lx x xx x x i y z x x y zz y z y

∂ ∂ ∂ ∂ ψ = − ψ = − − ψ − − ψ ∂ ∂ ∂ ∂

= 0

Inspection of above formulas shows that a component of the angular momentum and thecorresponding coordinate can have simultaneously definite values.

7. ˆ ˆ ˆˆ ˆˆ[L , ] , [L , ] , [L , ]x y zy i z z i x x i y= = =

( )ˆ ˆ ˆˆ ˆ ˆ[L , ] L Lx x xy y yψ = − ψ

( ) ∂ ∂ ∂ ∂ = − − ψ − − ψ ∂ ∂ ∂ ∂

i y z y y y zz y z y


2 2 ∂ψ ∂ψ ∂ψ ∂ψ= − − − ψ − + ∂ ∂ ∂ ∂ i y zy z y yz

z y z y

= iz

Inspection of these formulas shows that the component Lx and the coordinate y (or z)cannot be determined simultaneously. The same holds for Ly and the coordinate z (or x),and also for Lz and the coordinate x (or y).

8. =ˆ ˆ[L , ] 0,x xp similar results hold for other similar commutators.

ˆ ˆ ˆˆ ˆ ˆ[L , ] (L L )x x x x x xp p pψ = − ψ

2( ) 0 ∂ ∂ ∂ψ ∂ ∂ψ ∂ψ = − − − − = ∂ ∂ ∂ ∂ ∂ ∂

i y z y zz y x x z y

9. ˆ ˆ ˆ[L , ]x y zp i p= similar results hold for other components.

ˆ ˆ ˆˆ ˆ ˆ[L , ] (L L )x y x y y xp p pψ = − ψ

2( ) ∂ ∂ ∂ψ ∂ ∂ψ ∂ψ = − − − − ∂ ∂ ∂ ∂ ∂ ∂

i y z y zz y y y z y

2 2 2 22

2 2( )

∂ ψ ∂ ψ ∂ ψ ∂ψ ∂ ψ = − − − − + ∂ ∂ ∂ ∂ ∂∂ ∂ i y z y z

y z z y zy y

2( )∂ψ = − − ∂

iz

∂ψ = − ∂ i i

z

ˆ .zi p= ψ

10. ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ[L ,L ] L , [L ,L ] L , [L ,L ] Lx y z y z x z x yi i i= = =

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆˆ ˆ ˆˆˆ ˆ[L ,L ] L L L L L Lx y x y y x x x z x z xzp xp zp xp= − = − − −

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆˆˆ ˆL L L Lx x x z x x z xzp xp zp xp= − − +


Since Lx commutes both with x and ˆ xp we can interchange the operators Lx and x in

the second term and also the operators ˆˆ and Lx xp in the third term. The result is

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆˆˆ ˆ[L ,L ] L L L Lx y x x x z x x z xzp x p z p xp= − − +

Let us combine the first term with the third one and the second term with the fourth one.Thus

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆˆ ˆ[L ,L ] L L L Lx y x x x x z z xz z p x p p= − − −

ˆ ˆ ˆ( ) ( )= − x yi y p x i p

ˆˆ ˆˆ( )= − x yi yp xp

Lzi=

By carrying out two successive cyclic permutations on ˆ ˆ ˆ[L ,L ] Lx y zi= , we can get the

remaining two results.

11.2 2 2ˆ ˆ ˆ ˆ ˆ ˆ[L ,L ] 0, [L ,L ] 0, [L ,L ] 0x y z= = =

( ) ( )2 2 2 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ[L ,L ] L L L L L L L Lx x y z x x x y z= + + − + +

3 2 2 3 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L L L Lx y x z x x x y x z= + + − − −

Using the commutation relations ˆ ˆ ˆ ˆ ˆL L L L Lx y y x zi− = , we transform the second and fifth

term as follows:

2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L L L Ly x x y y y x x y y− = −

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL (L L L ) (L L L )Ly x y z y x z yi i= − − +

( )ˆ ˆ ˆ ˆL L L Ly z z yi= − +

Using the relation ˆ ˆ ˆ ˆ ˆL L L L L ,z x x z yi− = we can perform a similar transformation of the

third and sixth terms:

2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L L L Lz x x z z z x x z z− = −

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L Lz x z y z x y zi i= + − −

ˆ ˆ ˆ ˆ(L L L L )z y y zi= +


Making use of these transformations, we get

2ˆ ˆ[L ,L ] 0x =

We conclude that only the square of the vector L and one of its projections onto the coordinateaxes can be determined simultaneously. The other two projections are indeterminate (except when allthree components are zero). Consequently, all that we can know about the vector L is its “length”and the angle it makes with a certain axis. The direction of the vector L, however, does not lenditself to determination. The operators, which commute, can have simultaneous eigen states.

3.11 COMMUTATION RELATIONS FOR LADDER OPERATORS

The ladder operators L+ and L− are defined by

ˆ ˆ ˆL L Lx yi+ = + ...(3.11.1)

ˆ ˆ ˆL L Lx yi− = − ...(3.11.2)

L+ and L− are also called raising and lowering operators respectively. The adjoint of L+ is

L− and that of L− is L+ .

ˆ ˆ ˆ ˆL L , L L+ ++ − − += = ...(3.11.3)

The commutation relations for L+ and L− are:

2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL , L L , L , L L , L ,L 2 L , L , L 0z z z+ + − − + − ± = = − = = ...(3.11.4)

These commutation relations can be proved as follows:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL ,L L ,L L L ,L L ,Lz z x y z x z y y xi i i L L+ = + = + = +

( )ˆ ˆ ˆL L Lx yi += + = ...(3.11.5)

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL ,L L ,L L L ,L L ,L L Lz z x y z x z y y xi i i− = − = − = −

( )ˆ ˆ ˆL L Lx yi −= − − = − ...(3.11.6)

ˆ ˆ ˆ ˆ ˆ ˆL ,L L L ,L Lx y x yi i+ − = + −

( )( ) ( )( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L Lx y x y x y x yi i i i= + − − − +

( ) ( ) 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L L L L L Lx y x y y x x y x y y xi i= + − − − + + −

( )ˆ ˆ ˆ ˆ2 L L L Lx y y xi= − −


ˆ ˆ ˆ2 L ,L 2 Lx y zi = − = ...(3.11.7)

( ) ( )2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL ,L L ,L L L L L L L Lx y x y x yi i i+ = + = + − +

2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L Lx y x yi i= + − −

( )2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L Lx x y yi= − + −

2 2ˆ ˆ ˆ ˆL ,L L ,L 0x yi = + = ...(3.11.8)

Similarly, we can prove that

2ˆ ˆL ,L 0− = ...(3.11.9)

( )( ) ( )2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L L L L L L L L L Lx y x y x y x y y xi i i+ − = + − = + − −

2 2 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L ,L L L L L L Lx y x y x y z z zi = + − = + + = − +

Similarly

2 2ˆ ˆ ˆ ˆ ˆL L L L Lz z− + = − −

Hence

( )2 21ˆ ˆ ˆ ˆ ˆ ˆL L L L L L2 z+ − − += + + ...(3.11.10)

In polar coordinates the ladder operators can be expressed as

ˆ ˆ ˆL L Lx yi+ = +

sin cos cot cos sin coti ∂ ∂ ∂ ∂= ϕ + ϕ θ − − ϕ + ϕ θ ∂θ ∂ϕ ∂θ ∂ϕ

= cotϕ ∂ ∂+ θ ∂θ ∂ϕ

ie i ...(3.11.11)

ˆ ˆ ˆL L L coti

x yi e i− ϕ−

∂ ∂= − = − + θ ∂θ ∂ϕ ...(3.11.12)

3.12 EXPECTATION VALUE

When the wave function of a system is not an eigen function of operator Q representing anobservable Q of the system then the measurement of Q with identical systems will give various possible


values. The expectation value of observable Q is equal to the average value of the results of thesemeasurements.

The expectation value of a physical quantity Q represented by operator Q is defined by

ˆQ Q d∗= ψ ψ τ∫ ...(3.12.1)

where is the state of the system. If the wave function is not normalized the expectation value isgiven by

QQ

d

d

∗

∗

ψ ψ τ=ψ ψ τ∫∫ ...(3.12.2)

The expectation values of the physical quantities x, px, p, E etc., with respect to the state arecalculated from the following equations respectively:

x x d∗= ψ ψ τ∫

xp i dx

∗ ∂ = ψ − ψ τ ∂ ∫

( )p i d∗= ψ − ∇ ψ τ∫

E i dt

∗ ∂ = ψ ψ τ ∂ ∫

2 2

2

2 2

pd

m m∗ −= ψ ∇ ψ τ ∫

V V d∗= ψ ψ τ∫

Since

2

E V2

p

m= + we have

2

2E V .2

d dm

∗ ∗ = ψ − ∇ ψ τ + ψ ψ τ ∫ ∫

3.13 EHRENFEST THEOREM

The theorem states that the classical equations of motion viz

V

, Fx xx

p dpdx

dt m dt x

∂= = − =∂


are valid in quantum mechanics if we replace the physical quantities (such as x, px.) by their expectationvalues. Thus the quantum equations of motion are

xpd

xdt m

= ...(3.13.1)

V

xd

pdt x

∂= −∂

In other words, the expectation values of physical quantities obey the classical equations ofmotion.

Proof: The time derivative of expectation value of position coordinate x is

∗= ψ ψ τ∫d dx x d

dt dt

∗

∗ ∂ψ ∂ψ= ψ τ + ψ τ∂ ∂∫ ∫x d x dt t

...(3.13.2)

All changes in x with time is being determined by the change in , therefore there is no term

like ∂∂x

t in above equation. This is how Schrödinger mechanics works.

Substituting the values of andt t

∗∂ψ ∂ψ∂ ∂

obtained from Schrödinger equation in Eqn. (3.13.2)

we have

2 22 21 1

V V2 2

dx x x d

dt i m i m∗ ∗ ∗

= ψ − ∇ ψ + ψ + − − ∇ ψ + ψ ψ τ ∫ ∫

( )2

2 21

2∗ ∗−= ψ ∇ ψ −∇ ψ ψ τ∫

x x d

i m

2 2( )( )2

ix x d

m∗ ∗ = ψ ∇ ψ − ∇ ψ ψ τ ∫

2 I2

ix d

m∗= ψ ∇ ψ τ −∫ ...(3.13.3)

where I = 2( )( )2

∗∇ ψ ψ τ∫i x dm

...(3.13.4)

Making use of the identity

.( ) . .∇ ϕ = ∇ϕ + ϕ∇A A A


we can write

2.( ) .∗ ∗ ∗∇ ψ∇ψ = ∇ ψ∇ψ + ψ∇ ψx x x ...(a)

Interchanging x and *, we obtain

2.( ) . ( ) ( )∗ ∗ ∗∇ ψ ∇ ψ = ∇ψ ∇ ψ + ψ ∇ ψx x x ...(b)

In view of (a) we have

I = .( )2

∗ ∗∇ ψ∇ψ −∇ ψ ⋅∇ψ τ∫i x x dm

= ( )2

ix ds x d

m∗ ∗ψ∇ψ ⋅ − ∇ ψ ⋅∇ψ τ∫ ∫

where use of divergence theorem has been made to transform the volume integral into surface integral.The surface integral vanishes because 0 as x . So

I = ( )2

∗−∇ ψ ⋅∇ψ τ∫i x dm

Making use of (b), we have

I = 2[ ( ) ( )]2

∗ ∗ψ ∇ ψ −∇ ⋅ ψ ∇ ψ τ∫i x x dm

= 2( ) ( ) s2 2

∗ ∗ψ ∇ ψ τ − ψ ∇ ψ ⋅∫ ∫ i ix d x d

m m

= 2 ( ) 02

∗ψ ∇ ψ τ +∫i x dm

(The surface integral again vanishes)

Now Eqn. (3.13.3) becomes

2 2

2

d ix x x d

dt m∗ ∗ = ψ ∇ ψ − ψ ∇ ψ τ ∫

2 2

2 2( )

2∗ ψ= ψ − ψ τ ∫i d d

x x dm dx dx

=2 2

2 22

2∗ ψ ψ ψψ − − τ ∫i d d d

x x dm dxdx dx

= 22

∗ ψ ψ − τ ∫i dd

m dx

=1 ∗ ψ − ψ τ ∫

di d

m dx


= 1

xpm

Similarly,

xd d

p i ddt dt x

∗ ∂ψ = ψ − τ ∂ ∫

∗∗ ∂ ∂ψ ∂ψ ∂ψ= − ψ + τ ∂ ∂ ∂ ∂

∫i dx t t x

Substituting the value of andt t

∗∂ψ ∂ψ∂ ∂

from Schrödinger equation, we obtain

22 2

2xd

p ddt m x x

∗ ∗∂ ∂ψ = ψ ∇ ψ −∇ ψ τ − ∂ ∂ ∫(V ) V d

x x∗ ∗∂ ∂ψ ψ ψ − ψ τ ∂ ∂ ∫

2* 2 * 2

2d

m x x

∂ ∂ = ψ ∇ ψ −ψ ∇ ψ τ − ∂ ∂ ∫

(V ) V

x x∗ ∗∂ ∂ψ ψ ψ − ψ ∂ ∂ ∫

(V ) V dx x

∗ ∂ ∂ψ = − ψ ψ − τ ∂ ∂ ∫

Vd

x∗ ∂ = ψ − ψ τ ∂ ∫

V

x

∂= −∂

3.14 SUPERPOSITION OF STATES (EXPANSION THEOREM)

A physical quantity is represented by a linear (Hermitian) operator and for each operator one canset up an eigen value equation of the type

Q m m mqϕ = ϕ ...(3.14.1)

qm being the eigen value and m the eigen function of the operator Q representing the physicalquantity Q. The eigen values may form discrete or continuous or both kind of spectrum. In case ofdiscrete spectrum, we denote the eigen values and the eigen functions as


q1 q2 q3 q4 …………qm……..1 2 3 4 …………m……

An important property of eigen functions of the operator of a physical quantity is that theyform a complete set. This implies that any arbitrary well-behaved state wave function of the system,which obeys the same boundary conditions as the eigen functions, can be expanded as a linearcombination (superposition) of the eigen functions q. A superposition of eigen functions alsorepresents a possible state of the system. In case of discrete spectrum of eigen values, this propertyof eigen functions permits us to write

= ϕ∑ m qmm

c ...(3.14.2)

where cm are constant coefficients, in general complex. Of course, it is understood that ’s and ’sare all functions of the same set of variables.

(The particle-in-box stationary state eigen functions given by

2

( ) sin , 0 LL Ln

n xx x

πϕ = ≤ ≤

and harmonic oscillator wave functions are well-known examples of complete orthonormal functions.)Eqn. (3.14.2) may be viewed as expressing the state as superposition of eigenstates m.

This equation also states that the measurement of the quantity Q represented by Q in the state yields one of the eigen values q1, q2, q3, ….., qm. In particular if all the coefficients except one, saycm, are zero then Eqn. (3.14.2) becomes

= cmqm ...(3.14.3)

which means that the measurement of quantity Q in the state will yield only one value cm. To havethe significance of the expansion coefficients cm we assume that the eigen functions qm are normalized.If all c’s are not zero, then the measurement of quantity Q in the state does not yield a definitevalue, the outcome has a range of values whose average or expectation value is given by

ˆQ Q d= ψ ∗ ψ τ∫

* * Qm m n nm n

c c d

= ϕ ϕ τ ∑ ∑∫

* * ˆ= ϕ ϕ τ∑∑ ∫m n m nc c Q d

* *.= ϕ ϕ τ∑∑ ∫m n n m nc c q d

* . .= δ∑∑ m n n mnc c q


2| | . .=∑ n nn

c q

= 2 21 1 2 2 .....+ +c q c q ...(3.14.4)

When the system is in state , the result of any measurement of Q is one of the eigen valuesq1, q2,….., the expectation value of Q is weighted average of these eigen values. The probability ofobtaining a particular value qn equals the square of the magnitude of cn. This gives the physicalsignificance of the expansion coefficient cn in the expression Eqn. (3.14.2). Of course, the sum ofprobabilities must be unity.

2

1=∑ nn

c ...(3.14.5)

The expansion coefficient cn may be obtained by making use of the orthogonal property ofeigen functions m. The orthogonal property of eigen functions signifies that

m n mnd∗ϕ ϕ τ = δ∫or m n mnϕ ϕ = δ

where the integral is evaluated over the entire range of variables in which is defined. To determinecn, we multiply Eqn. (3.14.2) scalarly with n

* and integrate.

∗ ∗ϕ ψ τ = ϕ ϕ τ∑∫ ∫n n m m

m

d c d

∗= ϕ ϕ τ∑ ∫m n mm

c d

= δ∑ m nmm

c

= nc

∗= ϕ ψ τ = ϕ ψ∫n n nc d ...(3.14.6)

In case of continuous spectrum we shall denote the eigen value and corresponding eigen functionby q and qm respectively.

3.15 ADJOINT OF AN OPERATOR

Let ˆ ˆQ and R be two operators such that

( ) ( )* *ˆ ˆˆ ˆQ (R ) or ,Q R ,d dϕ ψ τ= ϕ ψ τ ϕ ψ = ϕ ψ∫ ∫ ...(3.15.1)


then R is said to be adjoint of Q and is denoted by ˆ ˆˆQ . ., R Qi e+ += . Eqn. (3.15.1) can be written as

( )ˆ ˆQ Qd d∗ + ∗ϕ ψ τ = ϕ ψ τ∫ ∫ ...(3.15.2)

or ˆ ˆQ Q+ϕ ψ = ϕ ψ ...(3.15.3)

In other words, as for as the value of the integral is concerned, it makes no difference whether

Q acts on or its adjoint Q+ acts on the other function .

3.16 SELF-ADJOINT OR HERMITIAN OPERATOR

An operator Q is said to be self-adjoint (or Hermitian) if ˆ ˆQ Q+ = . Dynamical variables are realquantities and therefore their operators must be Hermitian. Property of Hermitian operator is

or ˆ ˆQ (Q )d d∗ ∗ϕ ψ τ = ϕ ψ τ∫ ∫ or ( ) ( )ˆ ˆ,Q Q ,ϕ ψ = ϕ ψ ...(3.16.1)

ˆ ˆQ Qϕ ψ = ϕ ψ ...(3.16.2)

where and are two arbitrary functions.

3.17 EIGEN FUNCTIONS OF HERMITIAN OPERATOR BELONGING TO DIFFERENT EIGEN VALUES ARE MUTUALLY ORTHOGONAL

Let Q be a self-adjoint (Hermitian) operator and m and n be two eigen functions corresponding todifferent eigen values qm and qn. Then

Q

Q

m m m

n n n

q

q

ϕ = ϕ

ϕ = ϕ...(3.17.1)

The condition of self-adjointness is

ˆ ˆQ (Q )m n m nd d∗ ∗ϕ ϕ τ = ϕ ϕ τ∫ ∫ ...(3.17.2)

* * *ϕ ϕ τ = ϕ ϕ τ∫ ∫m n n m m nq d q d

*( ) 0∗− ϕ ϕ τ =∫n m m nq q d

Since qn qm, we must have

0∗ϕ ϕ τ =∫ m nd ...(3.17.3)

Thus the eigen functions belonging to different eigen values are orthogonal.


3.18 EIGEN VALUE OF A SELF-ADJOINT (HERMITIAN OPERATOR) IS REAL

Let the eigen values of operator Q in the states m and n be qm and qn respectively. That is

Q

Q

m m m

n n n

q

q

ϕ = ϕ

ϕ = ϕ...(3.18.1)

Since Q is Hermitian, we must have

ˆ ˆQ (Q )m n m nd d∗ ∗ϕ ϕ τ = ϕ ϕ τ∫ ∫ ...(3.18.2)

* * *ϕ ϕ τ = ϕ ϕ τ∫ ∫m n n m m nq d q d

*( ) 0∗− ϕ ϕ τ =∫n m m nq q d

This relation holds for all values of m and n. In particular, it must be true for m = n also. So

* *( ) 0− ϕ ϕ τ =∫n n n nq q d

Since the integral on the left hand side is not zero, we must have

*=n nq q ...(3.18.3)

The eigen values of Hermitian operator are real.

SOLVED EXAMPLES

Ex. 1. Select the acceptable wave functions from the following functions: (i) = xn (ii) = ex (iii) = e–x (iv) = sin x

(v) = exp (–x2) (vi) = tan x.

Sol. Acceptable function must be finite, continuous and single valued.(i) As x ± , ± . Hence xn is not an acceptable function.

(ii) As x , . So ex is not acceptable.

(iii) As x – , . So e– x is not acceptable.

(iv) sin x oscillates between –1 and +1, so it is acceptable.

(v) As x ± , 0, so it is acceptable.

(vi) tan x blows at x = /2, hence it is not acceptable.

Ex. 2. Obtain expressions for the following operators:

2 2 2

, , ( ) +

d d dx x x

dx dx dx


Sol.

2 + ψ = + + ψ

d d dx x x

dx dx dx

ψ = + + ψ d d

x xdx dx

22

22 ( 1)

ψ ψ= + + + ψd dx x

dxdx

2 2

22

2 1d d d

x x xdx dxdx

+ = + + +

2 ψ ψ = ψ =

d d d d dx x x x x

dx dx dx dx dx

2

2

ψ ψ= +

d dx x

dx dx

22

2

ψ ψ= +d dx x

dx dx

2 2

22

= + d d d

x x xdx dxdx

2 ψ = ψ = ψ

d d d d dx x x x x

dx dx dx dx dx

ψ = ψ + d d

x xdx dx

2 ψ = ψ + d d

x xdx dx

22

22

ψ ψ ψ= ψ + + +d d dx x x

dx dx dx

2 2

22

3 1d d d

x x xdx dxdx

= + +


Ex. 3. A particle is moving in one dimension is represented by the wave function

1/2

2

2 x +ix(x) =

1+ix

ψ π

Find the position probability density where the particle is most likely to be found.Sol. Position probability density

2

* 24

2 2P | | .

1

x

x= ψ ψ = ψ =

π +

For maximum value of P, 4 2 3

4 2

P 2 (1 ) .40 0 1

(1 )

d x x x xx

dx x

+ −= ⇒ = ⇒ = ±+

Ex. 4. Find the value of constant A which makes the function exp (– x2) an eigen function of the

operator 2

22

dAx

dx

−

. What is the corresponding eigen value?

Sol. Eigen value equation of the given operator is

22 2 2

2A exp ( ) exp ( )

dx x q x

dx

− −λ = −λ

, q is eigen value.

2 2 2 2 2( 2 4 A )exp ( ) exp ( )x x x q x− λ + λ − −λ = −λ

The function exp (–x2) will be an eigen function of the given operator if (42 x2 – Ax2 – 2)is independent of x. That is, the coefficient of x2 must vanish. Thus

42 – A = 0 or A = 42.

The expression for operator now becomes

22 2

24

dx

dx

− λ

The eigen value equation is

22 2 2 2

24 exp( ) exp ( )

dx x q x

dx

− λ −λ = −λ

where q is eigen value.

= – 2 exp(– x2)

Therefore, – 2 is the eigen value of the given operator.


Ex. 5. Normalize the wave function

( ) ( ), 0

( ), 0

x Aexp x for x

Aexp x for x

ψ = −λ >= λ <

Where is a positive constant.Sol. The normalization condition is

* 1

∞

−∞

ψ ψ =∫ dx

02

0

A exp(2 ) Aexp ( 2 ) 1x dx x dx∞

−∞

λ + − λ =∫ ∫

A = λ

Normalized wave function is

( ) ( ) for 0, and

( ) ( ) for 0.

x exp x x

x exp x x

ψ = λ −λ >

ψ = λ λ <

Ex. 6. The operator +d

xdx

has the eigen value . Obtain the corresponding eigen function.

Sol. Let be the eigen function. The eigen value equation is

+ ψ = λψ

dx

dx

ψψ + = λψd

xdx

( )ψ = − − λψd

x dx

2

exp2

ψ = − + λ

xc x

Ex. 7. Which of the following functions are eigen functions of operator 2

2

d

dx? Find the eigen value

in each state.(i) = A sin mx (ii) = B cos nx (iii) = Cx2 (iv) = D/x (v) = A e–mx

Sol.

(i)2 2

22 2

Asind d

mx mdx dx

ψ = = − ψ

So = A sin mx is eigen function and –m2 is eigen value.


(ii)2

22

(Bcos ) (Bcos )d

nx n nxdx

= −

So = B cos nx is eigen function and –n2 is eigen value.

(iii)2

22

C 2Cd

xdx

=

So = Cx2 is not an eigen function. No eigen value exists.

(iv) = D/x is not an eigen function.

(v)2

22

(A ) (A )mx mxde m e

dx− −=

Hence = A e– mx is eigen function and m2 is the eigen value.

Ex. 8. Find the eigen function and eigen value of momentum operator

ˆ .

dp =-ih

dx

Sol.We need to solve the equation

( ) ( )n n nd

i x q xdx

− ψ = ψ

ψ− =ψ

nd

i q dx

ln ln− ψ = + ni q x c

( )exp /ψ = nc iq x

The eigen value qn can be any number. However, if we impose the boundary condition thatn(x) be a periodic in some distance L then

( ) ( )exp / exp ( L) /n niq x iq x= +

( )exp L / 1niq =

Lcos 1nq

=

2

Lnn

qπ= , n is an integer.


The eigen function is given by

2

( ) expLn

i nxx c

π ψ =

So the eigen values are discrete and real. This shows that the eigen values and the eigen functionsdepend not only on the nature of the operator but also on the boundary conditions.

Ex. 9. Find the eigen function and eigen values of the operator Lzd

id

= −ϕ

.

Sol. Eigen value equation is

( )( )

ψ ϕ− = ψ ϕϕ

d

i qd

ψ = − ϕψ

d qd

i

expϕ ψ =

iqc

The function is periodic function of variable with a period of 2 i.e., ( ) = ( +2). This condition implies that the eigen value q is an integral multiple of and given by theabove formula is the eigen function.

Ex. 10. Show that ˆ = − xd

p idx

is a Hermitian operator.

Sol. Let and be orthogonal functions we have to prove that

ˆ ˆ( )∞ ∞

∗ ∗

−∞ −∞

ψ ϕ τ = ϕ ψ τ∫ ∫x xp d p d

LHS = ( )( )d d

i d i ddx dx

∞ ∞ ∗∞∗ ∗−∞

−∞ −∞

ψ ψ − ϕ τ = − ψ ϕ − ϕ τ

∫ ∫

0∞ ∗

−∞

ψ= + ϕ τ∫ di d

dx

( )∞ ∗

−∞

ψ= ϕ τ∫ d

i ddx

( )∞

∗

−∞

ψ= ϕ − τ∫ d

i ddx


ˆ( )∞

∗

−∞

= ϕ ψ τ∫ xp d

Ex. 11. 2ˆ ˆˆ ˆ3 . .d

If A x and B Show that A and B do not commutedx

= =

Sol. 2 2 2ˆ Â, B 3 , 3 (3 )

d d dx x x

dx dx dx ψ = ψ = − ψ

= 2 23 (3 )ψ − ψd d

x xdx dx

6 0= − ψ ≠x

Hence ˆ Â and B do not commute.

Ex. 12. Show that ˆ ˆˆ ˆ, , = − A B A B .

Sol. ( )ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ Â, B AB BA AB BA ψ = − ψ = ψ − ψ

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆBA BA AB BA AB AB BA ψ = − ψ = ψ − ψ = − ψ − ψ

ˆ Â, B = − ψ .

Ex. 13. Show that 1ˆ, − = n n

xx p i nx .

Sol. ( )ˆ ˆ ˆ ˆ ( ) ( ) − ψ = − − − ψ

n n n n

x xd d

x p p x x i i xdx dx

ψ = − − ψ

n nd d

i x xdx dx

1 .ni nx −= ψ

Ex. 14. Show that the function 21( ) exp

2 ψ = −

x cx x is an eigen function of the operator

22

2

−

dx

dx. Find the eigen value, normalization constant c and the expectation value of x for the state

described by the wave function, when x varies from – to + .


Sol. 2 2

2 2 3 2 22 2

1 1 1exp exp exp

2 2 2

− − = − − −

d dx cx x cx x c x x

dx dx

213 exp ( )

2cx x= −

Hence the eigen value is 3.The normalization constant c is obtained by equation

2

( ) 1∞

−∞

ψ =∫ x dx

22 2 1∞

−

−∞

=∫ xc x e dx 2

1/ 22 1

2

∞−α

−∞

π = α α ∫

xx e dx

1/ 2

2 14

π = c

1/ 44 = π

c

The expectation value of x is given by

2 2

21 1

2 32 2 0∞ ∞− − −

−∞ −∞

= = =∫ ∫x x xx cxe xcxe dx c x e dx

(integrand is odd function)

Ex. 15. Obtain the expansion of functionf (x) = x 0 x l/2

f (x) = l – x l/2 x l

in terms of one-dimensional particle in a box stationary state wave function.Sol. The function is sketched in the figure.


Let f (x) = an n (x)

where *

0 0

2( ) ( ) sin ( )

π= ψ =∫ ∫l l

n nn x

a x f x dx f x dxl l

=/ 2

0 / 2

2 2sin ( )sin

l l

l

n x n xx dx l x dx

l l l l

π π+ −∫ ∫

= 3/ 2

2 2

(2 )sin , where sin 0 for even

2 2

1 for odd

l n nn

nn

π π =π

= ±

= 2 2 2

4 1 3 1 5sin sin sin ............

3 5

π π π − + − π l x x x

l l l

The integrals can be evaluated making use of following result:

2

1sin sin cos= −∫ x

x bx dx bx bxbb

Ex. 16. The ground state and the first excited state wave functions of an atom are 0 and 1

respectively, the corresponding energies being E0 and E1. If the system has a 40% probability of beingfound in the ground state and 60% probability in the first excited state,

(i) What is the wave function of the atom?

(ii) What is the average energy of the atom?

Sol. Let the wave function of the atom be

0 0 1 1ψ = ψ + ψc c ...(1)

Here the expansion coefficients c0 and c1 have the following meanings. c02 represents the

probability of finding system in the state 0 with energy eigen value E0. Thus

= ∴ =20 00.40 0.40c c

Similarly, the square of the coefficient c1 viz. 21c represents the probability of finding the system

in the state 1 with energy eigen value E1.

= ∴ =21 10.60 0.60c c ...(2)

Of course, 2 20 1 1+ =c c

Wave function of the state 0 0 1 1 0 10.40 0.60ψ = ψ + ψ = ψ + ψc c


Average energy of the system in the state

2 20 0 1 1 0 1E E E 0.40E 0.60Ec c= + = +

Ex. 17. Show that the Hamiltonian must be Hermitian for conservation of probability.Sol. The time dependent Schrodinger equation is

Hit

∂ψ = ψ∂

...(1)

Its complex conjugate is

*

* *Hit

∂ψ− = ψ∂

...(2)

Multiplying (1) by * and (2) by and then subtracting one equation from the other we have

** * * *ˆ ˆH Hi

t t

∂ψ ∂ψψ + ψ = ψ ψ − ψ ψ ∂ ∂

( ) ( )** *1 ˆ ˆH Ht i

∂ ψ ψ = ψ ψ −ψ ψ ∂

Integrating over entire space

* * *1 ˆ ˆ(H ) (H )d d dt i

∂ ψ ψ τ = ψ ψ τ − ψ ψ τ ∂ ∫ ∫ ∫

Conservation of probability demands that * 0.dt

∂ ψ ψ τ =∂ ∫ This is possible when

* *ˆ ˆ(H ) (H )d dψ ψ τ = ψ ψ τ∫ ∫

This is the condition for hermiticity for Hamiltonian operator.

Ex. 18. Show that (i) 2 2ˆˆ ˆ ˆ ˆ, ( ) , 2x xi

x H p ii x pm x

∂ = = ∂

Sol. 2 2 2ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ, H , T V , T , V , T ,2

x y zp p px x x x x x

m

+ + = + = + = =

= 2 2 2 21 1 1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , .2 0 0

2 2 2 2

∂ + + = + + = ∂

x y z xi

x p x p x p pm m m m x m

2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , ( ) ( ). 2∂ ∂ ∂ = + = − + − = ∂ ∂ ∂

x x x x xx p x p p p x p i i i ix x x


Ex. 19. At t = 0, the Harmonic oscillator wave function in a state is given by

0 21 2

( ,0) ( ) ( )33

ψ = ψ + ψx x i x

Where 0 is ground state wave function and 2 is second excited state wave function. (i) Write down the time evolution wave function.(ii) Find the average energy of the oscillator.

Sol. Time evolution of wave function is

( )510 22 2

E 1 2( , ) ( )exp ( )exp( ) ( )exp

33

i tx t x x i t i x i t

Ψ = ψ − = ψ − ω + ψ − ω

[Energy of oscillator in ground state is E0 = ½ and in state n = 2 is E2 = (5/2).)].According to the meaning of expansion coefficients, the probability of finding the oscillator in

the ground state with energy ½ is (1/3)2 = 1/3 and that in the state n = 2 with energy(5/2) is [(2/3)]2 = 2/3.

The average energy of the oscillator is

2 20 0 2 2

1 1 2 5 11E E E

3 2 3 2 6c c

= + = ω + ω = ω

Ex. 20. Show that the average value of square of a Hermitian operator is positive.

Sol. Let Q be a Hermitian operator. The eigen value equation for Q2 is

2 2Q ψ = λ ψ

We have to show that 2 is positive. The average value of Q2 is

2 * 2 * * *ˆ ˆ ˆ ˆ ˆQ Q QQ (Q ) (Q ) ( ) ( )d d d d< > = ψ ψ τ = ψ ψ τ = ψ ψ τ = λψ λψ τ∫ ∫ ∫ ∫

= * * * 2| | .λ λ ψ ψ τ = λ λ = λ =∫ d positive number

Ex. 21. Show that complex conjugation is not a (i) linear operator (ii) Hermitian operator.

Sol. (i) Let us denote the complex conjugation by an operator A. Then

*Aψ = ψ ...(1)

Let be linear combination of 1 and 2. Therefore,

* * * * *

1 1 2 2 1 1 2 2 1 1 2 2ˆ ˆA A( ) ( )c c c c c cψ = ψ + ψ = ψ + ψ = ψ + ψ ...(2)

If A, were linear operator, one would expect the following result.

* *1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2

ˆ ˆ ˆ ˆ ˆA( ) A( ) A( ) A Ac c c c c c c cψ + ψ = ψ + ψ = ψ + ψ = ψ + ψ ...(3)


Since (1) and (2) are not the same, A is not a linear operator.(ii) A Hermitian operator satisfies the condition

* *ˆ ˆQ (Q )d dψ ϕ τ = ψ ϕ τ∫ ∫ ...(1)

For operator A,

Left hand side = * * *A d dψ ϕ τ = ψ ϕ τ∫ ∫ ...(2)

Right hand side = * *( )ψ ϕ τ = ψϕ τ∫ ∫d d ...(3)

Since * *ψ ϕ τ≠ ψϕ τ∫ ∫d d , complex conjugation is not Hermitian.

Ex. 22. If 1 2( ), ( ),...........ϕ ϕx x is a complete set of orthonormal functions, then show that the

following closure relation is satisfied

*

1

( ) ( ) ( )n nn

x x x x∞

=

′ ′ϕ ϕ = δ −∑Sol. Let (x) be an arbitrary function. We can express this function in terms given orthonormal

functions.

1

( ) ( )∞

=

ψ = ϕ∑ n nn

x c x ...(1)

where * ( ) ( )= ϕ ψ∫n nc x x dx ...(2)

Therefore,

∞

=

′ ′ ′ψ = ϕ ψ ϕ ∑ ∫ *

1

( ) ( ) ( ) ( )n nn

x x x x dx

*

1

( ) . ( ) ( )n nn

x dx x x∞

=

′ ′ ′= ψ ϕ ϕ

∑∫

( ) ( )x x x dx′ ′ ′= ψ δ −∫ ...(3)

where *

1

( ) ( ) ( )n nn

x x x x∞

=

′ ′ϕ ϕ = δ −∑ [closure relation]. ...(4)


Ex. 23. Show that the momentum p of a free particle is a constant of motion.Sol. When the operator of a physical quantity commutes with Hamiltonian H, it remains

conserved. For a free particle moving in x-direction,

2 2

2H

2m x

∂= −∂

and

∂= −∂

xp ix

Now 2 2 3 3

2 3ˆˆ H ( ) ( ) ( )

2 2xi

p x i x xx m mx x

∂ ∂ ∂ψ = − − ψ = − ψ ∂ ∂ ∂

2 2 3 3

2 3ˆ ˆH ( ) ( ) . ( )

2 2xi

p x i x xm x mx x

∂ ∂ ∂ ψ = − − ψ = − ψ ∂∂ ∂

Therefore, ˆ ˆ ˆˆ ˆ ˆ,H H H 0x x xp p p = − = .

Ex. 24. The Hamiltonian of a harmonic oscillator is = + ω2

2 2ˆ 1ˆ .2 2

pH m x

m Prove that

(i) 2ˆ ˆˆ ˆ, ( ) ,xi p

x H ii p H i m xm

= = − ω

Sol. (i) 2 2 2 2 21 1 1ˆˆ ˆ ˆ ˆ ˆ,H , , ( ) ( ) 02 2 2x x x

x x p m x x x i i xm m

∂ ∂∂ ∂

ψ = ψ + ω ψ = − − − ψ +

2 2 2

2 2( )

2

∂ ∂ = − ψ − ψ ∂ ∂

x x

m x x

2

22

ψ∂ψ = − − = ∂

xp

im x m

(ii) ψ = ˆˆ ,Hxp

+ ω ψ = + ω ψ

22 2 2 2ˆ 1 1ˆ ˆ ˆ, , 0 ,

2 2 2x

x x xp

p p m x m p xm

= 2 2 2 21( )( ) ( )

2

∂ ∂ ω − ψ − − ψ = − ω ∂ ∂ m i x x i i m x

x x

Ex. 25. Show that i [A, B] will be Hermitian if A and B are Hermitian operators.

Sol. Let ˆ ˆ ˆQ A, Bi = . Then

ˆ ˆ ˆˆ ˆQ (AB BA)i= −


Suppose that ˆ Â and B are Hermitian.

Now, ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆQ (AB) (BA) B A A B BA ABi i i+ + + + + + += − − = − − = − −

= − = = ˆ ˆ ˆ ˆˆ ˆ ÂB BA A,B Q.i i

Thus Q or ˆ Â, Bi is Hermitian.

Ex. 26. Show that Parity operator is Hermitian.

Sol. Let P be parity operator. Then

P ( ) ( )x xψ = ψ −

If parity operator is Hermitian then

* *ˆ ˆ( ) P ( ) (P )x x dx dxψ ϕ = ψ ϕ∫ ∫

or * *( ) ( ) ( ) ( )ψ ϕ − = ψ − ϕ∫ ∫x x dx x x dx ...(1)

Let us change the variable on the left hand side of (1) through the substitution x = – x.

LHS = * *( ) ( ) ( ) ( )x x dx x x dx′ ′ ′ψ − ϕ = ψ − ϕ∫ ∫ = RHS of Eqn. (1)

Ex. 27. Show that every operator can be expressed as the combination of two operators, each ofthem is Hermitian.

Sol. Let A be an operator whose adjoint is A+ . Now, let

ˆ ˆ ˆ Â A A Aˆ ˆÂ B C

2 2i i

i

+ + + −= + = +

where ˆ ˆ ˆ Â A A AˆB and C

2 2i

+ ++ −= =

Now, ˆ ˆ ˆ ˆ ˆ Â A A A A Aˆ ˆˆ ˆB B and C C

2 2 2i i

+ + ++ ++ − −= = = = =

−

Thus ˆB and C are both Hermitian. So A can be expressed as ˆ ˆÂ B Ci= + .

Ex. 28. If 1 and 2 are two degenerate eigen functions of the linear operator H, show that = c1 + c2 2 is an eigen function of H with the same eigen value as that of 1 and 2.

Sol. Given that 1 1 2 2ˆ ˆH and Hϕ = λϕ ϕ = λϕ

Now 1 1 2 2 1 1 2 2 1 1 2 2ˆ ˆH H( ) ( )c c c c c cψ = ϕ + ϕ = λϕ + λϕ = λ ϕ + ϕ = λψ


Ex. 29. Find the solution of eigen value equation 2

2

d

dx

ψ = λψ . Discuss the conditions under which

the eigen functions are well behaved and when is an eigen value.

Sol. Given equation is 2

2

ψ = λψd

dx.

For 0, the solution of the equation is

1 2exp( ) exp( )ψ = λ + − λc x c x , c1 and c2 are arbitrary constants.

(i) When < 0, let = im.

The solution is

1 2exp( ) exp( )c imx c imxψ = + −

When |x| , . The solution is not well behaved.

(ii) When > 0, 1 2exp( ) exp( )c x c xψ = λ + − λ

For x < 0, the first term of the solution is finite for all negative values of x, but the secondpart blows up for x . Hence is not well behaved.

For x > 0, the first term becomes infinite for x and the second term remains finitefor all values of x. The solution is not well behaved.

(iii) When = 0, C Dxψ = + . The solution is well behaved for C = 0, and in this case = D.

Ex. 30. Discuss the nature of solution of equation 2

2, ,

dc x c is positive constant

dx

ψ = ± ψ −∞ < < ∞

Sol. Case 1: Choose positive sign. The solution of the equation is

Aexp ( ) B exp ( )c x c xψ = ± −

For x →±∞ , is not well behaved.

However, if A = 0, B exp ( )c xψ = − is well behaved for 0 < x < .

If B = 0, A exp ( )c xψ = is well behaved for x < 0.

Case 2: Choose negative sign. The differential equation is

2

2

ψ = − ψdc

dx

Solution of equation is

Asin Bcosc x c xψ = +The solution is well behaved for all values of x. The eigen values (– c) form continuous spectrum.



1. State the physical significance of wave function. Explain the meaning of well-behaved function. Give theinterpretation of wave function in terms of probability current density j associated with particle flux.

2. Obtain time dependent and time independent Schrodinger equation. What do you mean by normalization andorthogonality of wave function?

3. What do you mean by dynamical variable and expectation value of a dynamical variable? Obtain quantummechanical operators corresponding to linear momentum, angular momentum, kinetic energy and Hamiltonian ofa system.

4. State Ehrenfest’s theorem. Prove that

V

, Fxx x

pd d dx p

dt m dt dx= = − =

5. Explain the meaning of linear operator, adjoint of an operator and Hermitian operator. Show that (i) the eigenvalues of a Hermitian operator are real (ii) the eigen functions of Hermitian operator corresponding to differenteigen values are orthogonal.

6. Explain expansion theorem. Give the meaning of expansion coefficients.

7. The wave function of a particle moving in a potential free region is given by (x) = A cos kx, where k and A are

real constants. Is it an eigen state of the operator 2ˆ ˆ ˆH, ,x xp p . If so, find the corresponding eigen values.

Ans. 2 2

,2

k

m

(x) is not an eigen state of

2 2ˆ , .xp k

8. Show that V

x

dp

dt x

∂< >= −∂

, where the symbols have their usual meanings. (All’d 1995)

9. (a) Show that the Hamiltonian must be Hermitian for conservation of probability.

(b) Two operators P and Q are non-hermitian. Make a suitable linear combination which will be Hermitian.

(c) If the operators P, Q and PQ are all Hermitian, show that [P, Q] = 0. (All’d 1995)

10. An operator A associated with a physical observable A satisfies the following eigen value equation

A ( ) ( )n n nx a xϕ = ϕ

(a) What are the possible results of observation of A?

(b) What is the average value of repeated observations when the system is in a definite eigen state n (x) ?(All’d 1996)

11. (a) Obtain the eigen functions of a one-dimensional momentum operator and normalize them by using delta-function technique.

(b) A particle of mass m is confined to move in a two dimensional square of side L. Obtain the total number ofallowed states for energy of the particle lying between E and E + dE and calculate the density of states.

(All’d 1996)

12. (a) Show that eigen vectors belonging to two different eigen values of a Hermitian operator are orthogonal.

(b) Evaluate 2, and ,z zz p xz p (1996)


13. (a) Discuss the postulates of quantum mechanics.

(b) If O1 and O2 are two mathematical operators corresponding to two physical observations, write the physical

significance of 1 2 1 2 1ˆ ˆ ˆ ˆ ˆ ˆ( ) [O ,O ] 0, ( ) [O ,O ] 0, ( ) [O ,H] 0, ( )i ii iii iv≠ = = Give explicit examples of O1 and O2

that satisfy (i), (ii) and (iii) H is Hamiltonian operator. (All’d 1997)

14. (a) Prove Ehrenfest theorem and describe its significance.

(b) What is meant by expectation value of a physical quantity? How is calculated? Calculate

<P> if 1

( , ) exp [ ( )].2

x t i kx wtφ = −π

(All’d 1997)

15. L and P are orbital angular momentum and parity operators respectively referred to the origin of a set of sphericalcoordinates. If P transforms (r,,) to (r, – , + ), show that [P, L] = 0. Hence prove that each sphericalharmonic has a well defined parity dependent on l. (All’d 1997)

16. (a) Show that V

.x

dp

dt x

∂= −∂

(b) If operators A and B are Hermitian show that I [A, B] is also Hermitian.

(c) Show that the eigen functions of a Hermitian operator having different eigen values must be orthogonal.(All’d 1998)

17. (a) Explain, in brief, the difference between Kroncker and Dirac delta functions.

(b) If 1 and 2 are eigen functions of a linear operator, show that their linear combination will also be an eigen function of the operator.

(c) For conservation of probability show that the Hamiltonian must be Hermitian.

18. (a) Evaluate 2

2,

dx

dx

(b) Prove that ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL L L (L ) and L L L (L )z z z z+ + − −= + = − . (All’d 2007)

19. (a) Test if the following are eigen functions of operator 2

2?

d

dx What are the corresponding eigen values?

(i) sin x (ii) log x (iii) exp(ax), where a is a constant?

(b) Explain why Hamiltonian of a system is always Hermitian? (All’d 2007)

20. (a) Explain with reasons which of the following wave functions are acceptable and unacceptable in quantummechanics?


(b) If the Hamiltonian can be written as following ,

H C n nnm x

n m

x p=∑ , hence prove the Ehrenfest theorem

H

x

dx

dt p

∂< >=∂

and H

x

dp

dt x

∂= −∂

(All’d 2006)

21. (a) Find ˆ ˆˆ ˆˆ ˆA, B if A and Bd d

x xdx dx

= + = − . (All’d 2006)

22. (a) Write down the Schrodinger equation for a free particle confined to move in a plane and find out the allowedenergy levels.

(b) Which of the following forms of the wave function are acceptable in quantum mechanics.

(i) sin x (ii) tan x (iii) exp(–x) (iv) exp(–x2) (All’d 2005)

23. (a) When are the energy eigen values of a quantum particle discrete in nature? (All’d 2005)

24. (a) Write Hamiltonian of a particle of mass moving with velocity v along x direction with potential energy

2

AV Bv

x= + , where A and B are constants ?

(b) Prove that (i) If two operators A and B are Hermitian and their product AB is also Hermitian. (ii) The operators A and B commute.

(c) If T ( ) ( ), prove that T Ta a ax x a +−ψ = ψ + = (All’d 2004)

25. (a) Show that momentum operator is Hermitian.

(b) Prove that 1 1, and ,n n m mx x xx p i nx x p i mp− − = = (All’d 2003)

26. (a) What is expectation value? Find expectation value of potential energy in the ground state of a linear harmonicoscillator.

(b) For operators A, B, C prove that (ABC)+ = C+B+A+ hence show that for operators A and B to commute, A, B and AB should be Hermitian. (All’d 2002)

27. (a) What is meant by free and bound states of a quantum mechanical system?

(b) Consider the momentum operator ˆxp (i) obtain its eigen functions and eigen values, (ii) normalize the eigen

functions using box normalization technique, (iii) compare the eigen values with those of a particle of massm confined inside a box of length L. In what limit do they coincide? (All’d 2001)

28. (a) Define orthogonality and normalizability of Schrodinger wave function.

(b) Write down the Hamiltonian for a particle of mass m moving along x axis with potential energyV(x) = Ax2 + Bv, where v is the velocity of the particle and A and B are constants. (All’d 2000)

29. (a) A wave function can be expanded in the form ψ =∑ n nn

a u . What do un represent and what is the physical

significance of |an|2 ? Explain in short.

(b) Show that if operators A and B are Hermitian, the operator A, B2

i is also Hermitian. (All’d 1999)

(c) If 1 and 2 are eigen functions of a linear operator, show that their linear combination will also be an eigen function of the operator. (All’d 1998)

4.1 POTENTIAL STEP OR STEP BARRIER

A potential step is described byV(x) = 0 for x < 0

= V0 for x > 0 ...(4.1.1)

This potential function is sketched in the Fig. (4.1.1).

Fig. 4.1.1 A potential step: For E < V0, region I is classically allowed and region II is forbidden

A particle with total energy E is incident on the potential step from left. The Schrodinger waveequation in the regions I and II are:

Region I (x < 0)

21

12 2

2 E0

d m

dx

ψ+ ψ =

or 2

211 12

0ψ

+ ψ =d

kdx

...(4.1.2)

where k1= 2

2 Em

...(4.1.3)

POTENTIAL BARRIER PROBLEMS

CHAPTER


Region II (x > 0)

( )202

22 2

2 E V0

md

dx

−ψ+ ψ =

or 2

222 22

0d

kdx

ψ ′+ ψ = ...(4.1.4)

where 0

2 2

2 (E V )mk

−′ =

...(4.1.5)

Solution of Eqn. (4.1.2) is

1 1 1A exp ( ) Bexp ( )ik x ik xψ = + − ...(4.1.6)

and solution of Eqn. (4.1.4) is

2 2 2Cexp ( ) Dexp ( )ik x ik x′ ′ψ = + − ...(4.1.7)

where A, B, C and D are arbitrary constants.Case I: E < V0: When the energy E of the particle is less than the height V0 of the potential

step, the kinetic energy of the particle is negative in the region II. Classical physics does not allowthe particle to enter the region II. In this case k'2 is imaginary and we may write

02 2 2

2 (E V )mk ik

−′ = =

...(4.1.8)

The solution of Eqn.(4.1.4) in the region II can be expressed as

2 2 2Cexp( ) Dexp( )k x k xψ = − + ...(4.1.9)

The second term on the right hand side of Eqn.(4.1.9) is an increasing function. This leads tothe conclusion that the probability density of finding the particle increases as x increases withoutlimit, which is physically not acceptable and therefore we must set D = 0. So the solution ofSchrodinger equation in region II becomes

2 2Cexp( )k xψ = − ...(4.1.10)

Now we use the following boundary conditions to determine the constants B and C in terms ofA.

(i) 1(0) = 2(0)

(ii) 1 2

0 0= =

∂ψ ∂ψ = ∂ ∂ x xx x

Applying above conditions to the functions 1(x) and 2(x), we obtain following equations:A + B = C ...(4.1.11)

ik1(A – B) = – k2 ...(4.1.12)

Potential Barrier Problems 149

From these equations, we get

1 2

1 2

B Ak ik

k ik

−=

+ ...(4.1.13)

1

1 2

2C A

k

k ik=

+ ...(4.1.14)

We can express the relationship between B and A and between C and A in a more convenientform by making use of following transformations:

1 2

1 2

i

i

k ik re

k ik re

δ

− δ

+ =

− =

where 2 2 21 2

1

and tank

r k kk

= + δ =

Hence

2B A ie− δ= ...(4.1.15)

The wave function 1 in the region I is

1 1 1A exp ( ) Aexp ( 2 )exp( )ik x i ik xψ = + − δ −

1 1Aexp( ) exp ( ) exp ( )i ik x i ik x i= − δ + δ + − − δ

12Aexp ( )cos( )i k x= − δ + δ ...(4.1.16)

The wave function in the region II is

2 2Cexp ( )k xψ = − ...(4.1.17)

The wave function in the region I issuperposition of two waves. The first term Aexp (ik1x) corresponds to a wave traveling tothe right and the second term B exp (– ik1x) toa wave traveling to the left. Superposition ofthese two waves give rise to a standing waverepresented by Eqn.(4.1.16). The magnitude ofthe ratio B/A is unity i.e., the incident and thereflected waves have the same amplitude. Fromthis we conclude that the wave function givenby Eqn. (4.1.16) describes the situation in which a particle incident from the left is reflected backby the potential hill. This behavior is analogous to the classical behavior of electromagnetic waves ata metal surface.

Fig. 4.1.2 Probability density in classically forbiddenand in allowed regions


The wave function in the classically forbidden region is exponentially decaying in nature andpredicts finite probability (although small) of finding the particle in this region. Notice that theprobability density shows oscillatory behavior in the region I (x < 0). This is a quantum mechanicalinterference result having no classical analogue.

The probabilities of finding the particle in the regions I and II are given by

* 2 2

I 1 1 1P 4A cos ( )k x= ψ ψ = + δ

and 2 2II 2 2P | | exp ( 2 )c k x= ψ = −

Case II (E > V0): In this case k'2 is real. The wave function in the region II is

2 2 2Cexp ( ) Dexp ( )ik x ik x′ ′ψ = + − ...(4.1.18)

The first term in above equation corresponds to a wave traveling to the right and the secondterm to a wave traveling to the left. There is nothing to reflect the wave in region II, we must,therefore, set D = 0. Appropriate boundary conditions at the junction of regions I and II are:

1 2(0) (0)ψ = ψ

1 2

0 0x x

d d

dx dx= =

ψ ψ =

Applying these boundary conditions to the wave functions 1 and 2, we haveA + B = C ...(4.1.19)

k1A – k1B = k'2 C ...(4.1.20)

From Eqns. (4.1.19) and (4.1.20), we get

1 2

1 2

B Ak k

k k

′−=

′+ ...(4.1.21)

1

1 2

2C A

k

k k=

′+ ...(4.1.22)

The current densities associated with incident wave, reflected wave and transmitted waves are

21J Ai

k

m=

...(4.1.23)

21J Br

k

m=

...(4.1.24)

22J Ct

k

m

′=

...(4.1.25)


The coefficients for energy reflection R and transmission (T) are given by

( )( )

2 21 2

2 21 2

BJR

J A

r

i

k k

k k

′−= = =

′+...(4.1.26)

( )

21 2

2 21 2

CJ 4T

J A

t

i

k k

k k

′= = =

′+...(4.1.27)

Notice thatR + T = 1 ...(4.1.28)

which is in accordance with the principle of conservation of energy.One of the peculiar results which quantum mechanics predicts is that although the particle has

enough energy to cross over the step even then there is a non-zero probability of its being reflected(R 0).

4.2 POTENTIAL BARRIER (TUNNEL EFFECT)

A potential barrier is a region in which the potential energy of a particle exceeds the total energy.Let us consider a potential barrier defined by

V(x) = 0 for x < 0

= V for 0 ≤ ≤ Lx

= 0 for x > L ...(4.2.1)

The potential barrier is sketched in the Fig. (4.2.1). Assume that a particle moving from left toright encounters the potential barrier of height V and width L on its path.

In terms of classical mechanics the behavior may be predicted as follows:(i) If the energy of the particle is greater than the height of the barrier (E > V), the particle

passes over the barrier without any hindrance. Inside the barrier, the velocity of the particlediminishes and beyond it the particle acquires its initial value.

(ii) If E < V, the particle is reflected from the barrier and is unable to penetrate through thebarrier.

Quantum mechanical treatment of this problem predicts different results. If E > V, there is afinite probability that the particle will be reflected from the barrier. If E < V, there is a finite probabilitythat the particle will penetrate through the barrier and will be found on the other side of the barrier.Thus the quantum mechanics allows the particle to leak through the barrier. This phenomenon iscalled tunnel effect. This is a purely quantum mechanical result having no classical analogue. Thusby this mechanism the alpha-particles are emitted by radioactive nuclei, although the potential barrieris such that classically they cannot be able to surmount it.


Fig. 4.2.1 One dimensional potential barrier

In the Fig. (4.2.1) the potential barrier divides the space into three regions I, II and III. TheSchrödinger wave equations in these regions are:

Region I

21

12 2

2 E0

d m

dx

ψ+ ψ =

or

2

211 12

0ψ

+ ψ =d

kdx

...(4.2.2)

where 1 2

2 Emk =

...(4.2.3)

Region II

2

222 2

2 (E V)0

d m

dx

ψ −+ ψ =

or

222

2 220

dk

dx

ψ ′+ ψ = ...(4.2.4)

where 2 2

2 (E V)mk

−′ =

...(4.2.5)

Here k'2 is imaginary and therefore we can write

2 2 2 2

2 (V E),

mk ik k

−′ = =

...(4.2.6)

Region III

2

332 2

2 E0

d m

dx

ψ+ ψ =


or

2

233 32

0ψ

+ ψ =d

kdx

...(4.2.7)

where 3 2

2 Emk =

= k1 ...(4.2.8)

Solutions of Eqns. (4.2.2), (4.2.4) and (4.2.7) are

1 1 1( ) A exp ( ) B exp ( )x ik x ik xψ = + − ...(4.2.9)

2 2 2( ) C exp ( ) D exp ( )x ik x ik x′ ′ψ = − +

2 2Cexp ( ) Dexp ( )k x k x= + − ...(4.2.10)

3 1 1( ) F exp ( ) G exp ( )x ik x ik xψ = + − ...(4.2.11a)

The term exp (ik1x) corresponds to a wave propagating in the positive direction of x-axis andexp (– ik1x) to a wave propagating in opposite direction. In region III there must be only one wavethat has penetrated through the barrier and is propagating from left to right. We must, therefore,assume G = 0.

The wave function 3 (x) then becomes

3 1( ) F exp ( )x ik xψ = ...(4.2.11b)

To find out other coefficients we use the following boundary conditions that wave functionsmust satisfy.

At x = 0,

1(0) = 2(0) ...(4.2.12)

1 2

0 0= =

ψ ψ = x x

d d

dx dx...(4.2.13)

At x = L,

2(L) = 3(L) ...(4.2.14)

32

x L x L

dd

dx dx= =

ψψ = ...(4.2.15)

These boundary conditions lead to following equations: A + B = C + D ...(4.2.16)

ik1A – ik1B = k2C – k2D ...(4.2.17)

2 2 1Cexp ( L) Dexp ( L) Fexp ( L)k k ik+ − = ...(4.2.18)

2 2 2 2 1 1Cexp ( L) Dexp ( L) Fexp ( L)k k k k ik ik− − = ...(4.2.19)


Here we are interested in transmission coefficient or transmission probability T, andreflection coefficient or reflection probability R. T is defined as the ratio of the current densityassociated with transmitted beam and that associated with incident beam. Similarly, R is defined asthe ratio of current density associated with transmitted beam to that associated with incident beam.

2 2

trans 12 2

inci 1

J ( / ) | F | | F | F FT

J ( / ) A A| A | | A |

k m

k m

∗ = = = =

...(4.2.20)

2 2ref 1

2 2inc 1

J ( / ) | B | | B | B BR

J ( / ) A A| A | | A |

k m

k m

∗ = = = =

...(4.2.21)

The conservation of energy demands that

R + T = 1 ...(4.2.22)

From Eqns. (4.2.16) to (4.2.19) the expressions for 2 2

2 2

| F | | B |and

| A | | A | come out to be

2 221 2 2

2 2 2 2 2 2 2 22 1 2 1 2 2

16 exp (2 L)| F |

| A | ( ) [1 exp (2 L)] 4 [1 exp (2 L)]

k k k

k k k k k k=

− − + + ...(4.2.23)

2 2 2 222 1 2

2 2 2 2 2 2 2 22 1 2 1 2 2

( ) [1 exp (2 L)]| B |

| A | ( ) [1 exp (2 L)] 4 [1 exp (2 L)]

k k k

k k k k k k

+ −=

− − + + ...(4.2.24)

The expression for T after simplification becomes

2 221 2

2 2 2 2 2 2 22 1 2 1 2

4| F |T

| A | ( ) sin L 4

k k

k k h k k k= =

+ + ...(4.2.25)

Substituting the expressions for k1 and k2, we have

12 22

2 22

V sin L1T 1

4E(V E)V sin L1

4E(V E)

h k

h k

− = = +

− +−

...(4.2.26)

The expression for coefficient of reflection comes out to be

2 2 222 1

2 2 22 2 2 1 22 1 2

2

( )| B |R

| A | 4( )

sin L

k k

k kk k

h k

+= =

+ + ...(4.2.27)


Substituting the expressions for k1 and k2, we obtain

12

2 22 22

2

V 4E (V E)R 1

4E (V E) V sin LVsin L

h kh k

− −= = + − + ...(4.2.28)

Classical limit is obtained by setting h 0. In this limit k2 and k1 . This impliesthat T 0 and R 1. The probability of transmission becomes zero and that for reflection isunity. This is the classical prediction.

When the barrier height and width both are large, k2L >> 1 and sin h k2L ½ exp (k2L).Under this approximation 1 can be neglected in the expression for T. Thus

( ) 222 122

4E(V E) 16E ET 1 exp ( 2 L),

V VV exp ( L)k

k

− = = − −

where 2 2

2 (V E)mk

−=

...(4.2.29a)

The exponential term in Eqn. (4.2.29a) is more dominant term than the coefficient 16E E

1V V

− and the latter is assumed to be equal to 1 in most of the application. With this approximation theexpression for transmission probability becomes

( )2T exp 2 Lk≅ − ...(4.2.29b)

The quantum mechanical analysis of potential barrier problem shows that the particle has finiteprobability of getting transmitted through the barrier even its energy is less than the height of thebarrier.

The transmission probability of the particle depends on (i) width L of the barrier and (ii) thedifference (V – E). This dependence of T on the width of the barrier and energy of the incidentparticle is displayed in the table.

E V L2 =k

2

2 (V E)m −

22 Lk T

1 eV 4 eV 0.1 nm 0.886 × 1010 m–1 1.772 0.17

1 eV 4 eV 0.2 nm 0.886 × 1010 m–1 3.544 0.03

2 eV 4 eV 0.1 nm 0.724 × 1010 m–1 1.448 0.23

Notice that when the width of the barrier is doubled, the transmission probability decreases bynearly 6 times whereas when the energy of the incident particle is doubled, the transmission probabilityincreases only by a factor of nearly 1.3 times. So the transmission probability strongly depends onthe width of the barrier.


Fig. 4.2.2 Wave functions in the three regions

Fig. 4.2.3 A potential barrier of varying width

For potential barrier of variable width, as shown in the Fig. (4.2.3), the transmission probabilityis given by

2

1

2T exp 2 (V E)

x

x

m dx ≅ − −

∫

...(4.2.30)

The emission of alpha-particle from radioactive nuclei, the passage of electron through potentialbarrier in tunnel diode and the crossing of electron through classically forbidden region between twosuperconductors are the well-known examples of tunneling phenomenon.

Case 2: E > V

In this case 2 2

2(V E)

mk = −

becomes imaginary. Let = β β = −

2 2

2, where (E V).

mk i

The expressions for R and T become

1

2 2

2 2

1 4E (E V)R 1

4E (E V) V sin L1V sin L

− −= = + − β +β

...(4.2.31)

12 2

2 2

1 V sin LT 1

4E(E V)V sin L1

4E(E V)

− β= = + −β +−

...(4.2.32)


(a) When E V, 0, sin L L and in this limit

2

2

1T

VL1

2

m=

+

...(4.2.33)

(b) Eqn. (4.2.32) shows that when E increases above V, transmission probability T becomesoscillatory due to presence of sin L. The barrier becomes transparent (T = 1) when

L , 1, 2, 3,.....n nβ = π =

or 2 2 2

2

2 (E V)L

mn

− = π

or L2

nλ= (4.2.34)

where 2 (E V)

h

mλ =

−= de Broglie wavelength of the particle.

Thus, when the width of the barrier is integral multiple of half the wavelength of the particle,the barrier becomes transparent. This phenomenon is called resonance scattering. Resonances areobtained for the values of E given by

22

2

2 (E V)

L

n mπ − β = =

or 2 2 2

2E V 1

2 VL

n

m

π= +

...(4.2.35)

Minimum value of T is obtained when

sin L 1 or L (2 1) / 2, 0,1, 2, 3,.....n nβ = β = + π =

or 2 2 2

2

(2 1)E V 1

8 VL

n

m

+ π= +

...(4.2.36)

For this value of E, T is minimum.

or

1

min1

T 14E E

1V V

− = + −

...(4.2.37)


(c) When E decreases below V, T decreases monotonically. When k2L >> 1,sin h k2L = ½ exp (k2L). In this case the expression for T becomes

12 22V sin L

T 14E(V E)

h k−

= + −

(omitting 1).

( ) 222 122

4E(V E) 16E E1 exp ( 2 L).

V VV exp Lk

k

− ≅ = − − ...(4.2.38)

The variation of T with E/V and that of T with increasing thickness of barrier L are shown inthe Fig. 4.2.4.

Fig. 4.2.4 Variation of T with E/V

Fig. 4.2.5 Variation of T with thickness L of barrier. Appearance of transmission resonances


4.3 PARTICLE IN A ONE-DIMENSIONAL POTENTIAL WELL OFFINITE DEPTH

Let us consider the motion of a particle in a one-dimensional potential well defined byV(x) = 0 x < – a

= – V0 – a < x < a

= 0 x > a ...(4.3.1)

Fig. 4.3.1 One dimensional potential well of depth V0

We shall first consider the case of the particle which is classically bound to remain within thewell, that is, – V0 < E < 0. The entire allowed region for the motion of the particle can be dividedinto three regions I, II and III. Let 1, 2 and 3 be the wave functions in the three regions. TheSchrodinger equations in the three regions are:

21

12 2

2 E, 0

d mx a

dx

ψ< − + ψ =

...(4.3.1a)

2

2112

0ψ

− ψ =d

kdx

...(4.3.1b)

where 2

2 Emk = −

, is a real positive quantity. ...(4.3.2)

2

0222 2

2 [E ( V )], 0

mda x a

dx

− −ψ− < < + ψ =

...(4.3.3a)

2222 22

0ψ

+ ψ =d

kdx

...(4.3.3b)

where 02 2

2 (E V ),is

mk

+=

a real positive quantity. ...(4.3.4)

2

332 2

2 E, 0

d mx a

dx

ψ> + ψ =

...(4.3.5)


223

2

ψ+

dk

dx3 = 0 ...(4.3.6)

The general solutions of Schrödinger equations in the three regions are:

1 1 1, A exp ( ) B exp ( )x a kx kx< − ψ = + − ...(4.3.7)

2 2 2 2 2A sin B cosa x a k x k x− < < ψ = + ...(4.3.8)

3 3 3A exp ( ) B exp ( )x a kx kx> ψ = + − ...(4.3.9)

When x – , exp (– kx) , which is unacceptable. We, therefore, set B1 = 0. Similarly,when x , exp (kx) , which is again unacceptable. We again set A3 = 0. So the above solutionsbecome

1 1, A exp( )x a kx< − ψ = ...(4.3.10)

2 2 2 2 2A sin B cosa x a k x k x− < < ψ = + ...(4.3.11)

3 3B exp ( )a x a kx− < < ψ = − ...(4.3.12)

The boundary conditions that the wave functions and their derivatives be continuous at x = ± a.That is,

1 2 2 3( ) ( ), ( ) ( )ψ − = ψ − ψ = ψa a a a

1 2 2 3( ) ( ), ( ) ( )a a a a′ ′ ′ ′ψ − = ψ − ψ = ψ

Applying these boundary conditions, we get following equations:

1 2 2 2 2 3A 0 A sin B cos B exp( ) 0k a k a ka+ + − − =

1 2 2 2 2 3A exp( ) A sin B cos B 0 0ka k a k a− − − + + =

1 2 2 2 2 2 2 3A 0 A cos B sin B exp( ) 0k k a k k a k ka− − + − =

1 2 2 2 2 2 2 3A exp( ) A cos B sin B 0 0k ka k k a k k a− + + + =This is a set of four homogeneous equations in four unknown coefficients. Solutions of physical

significance exist only if

−− −

− =−

2 2

2 2

2 2 2 2

2 2 2 2

sin cos 1 0

sin cos 0 1exp ( 2 ) 0

cos sin 0

cos sin

k a k a

k a k aka

k k a k k a k

k k a k k a k

Since exp (– 2k2a) 0, we have

2 2 2 22 2 2 2 2 2 2 2 2sin cos sin cos sin cos 0+ − − =k k a k a kk k a kk k a k k a k a

or 2 2 2 2 2 1 2 2 2sin ( sin cos ) cos ( cos sin ) 0+ − + =k k a k k a k k a k k a k k a k k a


or 2 2 2 2 2 2( sin cos )( sin cos ) 0+ − =k k a k k a k k a k k a

Dividing by cos2 k2a, we obtain

2 2 2 2( tan )( tan ) 0k k a k k k a k+ − = ...(4.3.13)

This equation is satisfied if

2 2tan = −k k a k

or 2 2cot= −k k k a ...(4.3.14)

and 2 2tan=k k k a ...(4.3.15)

We know that 2 202 2 2

2 (E V ) 2 Eand

m mk k

+= − =

2 02 2 2

2 V2 E mmk = +

2 2 22k k= − + µ ...(4.3.16)

where 2 0

2

2 Vmµ =

...(4.3.17)

Notice that is a measure of the depth V0 of the potential well. Eqns. (4.3.14) and (4.3.15)can now be written as

k = 2 22 2 2cotµ − = −k k k a ...(4.3.18)

k = 2 22 2 2tanµ − =k k k a ...(4.3.19)

Fig. 4.3.2 The permissible values of k2 and hence E corresponds to the points of intersection


The solutions (allowed values of k2) of transcendental Eqns. (4.3.18) and (4.3.19) are given by

the points of intersections of curves y = k2 tan k2a or y = – k2 cot k2a with 2 22y k= µ − which

represents a circle of radius µ). It is evident that allowed values of k2 and hence energy eigen values

(which are mutually related through the relation 2 2

20E V )

2

k

m= −

form a discrete spectrum. The number

of energy eigen values depend on the depth of potential. For shallow potential, V0 0, the radiusof the circle µ 0 and we get only one point of intersection and hence only one energy eigenvalue. With increasing strength of potential, the number of energy eigen values increases.

Without going into the details of method of calculating the wave functions corresponding todifferent energy we simply sketch their forms in Fig. (4.3.3).

Fig. 4.3.3 Wave functions of a particle in a potential of finite depth ( – V0 < E < 0)

It is important to note that the wave function as a whole is either an even function or an oddfunction of x. Here the lowest energy state is even, the next odd, the next even, and so forth, alternately.This comes about as a result of the fact that the potential V(x) is itself an even function of x.

Another important feature of the wave function is that it extends beyond the limits of the well.Therefore, there is a definite probability that the particle will be found in the classically forbiddenregion beyond the actual boundaries of the well. This phenomenon is known as barrier penetration.

For E > 0 (unbound states) the wave functions outside the well become oscillatory in naturelike that inside the well and the energy eigen values form a continuous spectrum as shown in theFig. (4.3.4).


Fig. 4.3.4 Energy states of particle (E > 0) in a potential well of finite depth

4.4 THEORY OF ALPHA DECAY

The problem of emission of -particle from radioactive nuclei is inexplicable in classical physics;quantum mechanics provides a natural explanation. In fact, the theory of -decay first given byGamow and independently by Condon and Gurney in 1928, was recognized as a spectacular triumphof newly discovered (1926) quantum mechanics.

Let us try to explain the phenomenon in terms of classical physics. Although nuclei are composedof neutrons and protons, we can think of -particle as an entity within the nucleus. When the-particle is outside the nucleus, it experiences a repulsive Coulomb force and the correspondingelectrostatic energy is

0

1 Z 2V( ) ; R

4

e er r

r= >

πε ...(4.4.1)

where R is nuclear radius and Z is atomic number of daughter nucleus. The energy V(r) given byEqn. 4.4.1 is equal to the work that is done against the Coulomb repulsion when an -particle isbrought from infinity towards the nucleus. As the -particle approaches the nuclear surface, theelectrostatic energy increases and becomes maximum at the surface r = R and is given by

2

0

1 2ZV(R)

4 R

e=πε

29 MeV for uranium nucleus ...(4.4.2)

This gives the minimum energy that an -particle must have to penetrate the nucleus. In otherwords, an -particle approaching the nucleus with kinetic energy less than V(R) cannot surmountthe repulsive Coulomb forces and will turn back. The magnitude of potential energy V(R) is calledbarrier height. The variation of potential energy V(r) of an -particle in the force field of daughternucleus with r is shown in the figure 4.4.1.


Inside the nucleus we do not know the exact shape of the potential energy curve, but we definitelyknow that the nuclear forces are strong and attractive in nature and hence the corresponding potentialenergy must be negative and the curve representing the potential energy must have a dip as shown inthe Fig. (4.4.1). The strong nuclear forces thus form a potential well. Classically an -particle insidethe well cannot escape from the nucleus unless its energy is at least equal to the height of the potentialbarrier.

Fig. 4.4.1 Potential energy curve of an alpha-particle inside and outside the nucleus

Let T be the kinetic energy of the -particle such that T < V(R). If the -particle isapproaching the nucleus, its whole kinetic energy will be converted into potential energy at pointr = R1 and the particle will come momentarily at rest and then it will turn back. The point r = R1 iscalled the classical turning point. If the -particle is inside the nucleus, it does not possess sufficientenergy to jump over the barrier height. Thus the region from r = R to r = R1 is inaccessible to the-particle and is called the thickness of the barrier. The conclusion drawn from classical view-pointthat a -particle with energy less than the barrier height cannot escape from the nucleus is not inaccord with observed facts. For instance, the energies of -particles emitted from uranium nucleusare below 10 MeV, which is much less than the barrier height 29 MeV.

Quantum mechanics, on the other hand, gives straight forward explanation of alpha activity.The central features of the quantum theory are:

(i) alpha-particle exists as a unit within the nucleus

(ii) alpha-particle is in constant motion and bounces back and forth from the barrier walls. Ineach collision with the wall there is a finite probability that the particle will leak throughthe potential barrier.

Let f be the frequency with which an -particle collides with the wall in order to escape fromthe nucleus and T be the transmission probability in each collision then the decay probability () isgiven by

= f T


In terms of the velocity v of the -particle and nuclear radius R the frequency of collision is

2R

vf =

The transmission probability T of a particle with energy T in potential barrier sketched in theFig. (4.4.2) is given by

( )2

1

2

2lnT 2 V T

x

x

mdxα= − −∫

Fig. 4.4.2

Applying this result to the problem of -decay, we have

1 2

20

2 2ZlnT 2 T

4

R

R

m edr

r α

= − − πε ∫

1R 2

20R

2 T 2Z2 1

4 T

m edr

rα

α

= − − πε

∫

Making use of the fact that at r = R1, V = T, we obtain

2

0 1

2ZT

4 R

eα=

πε

whence 2

10

2ZR

4 T

e

α=

πε ...(4.4.3)

Now

1R1

2R

2 T RlnT 2 1

mdr

rα = − −

∫

...(4.4.4)


To simplify the integration we make use of the substitutionr = R1 cos2 , dr = – 2 R1cos sin d

Doing so, we get

2

120

2 TlnT 2 R 2sin

md

βα= − ⋅ θ θ∫

where = 11cos R / R−

120

2 TlnT 2 R (1 cos2 )

md

βα= − − θ θ∫

120

2 T 12 R sin2

2

m βα = − θ − θ

212

0

2 T2 R cos (1 cos )

m βα = − θ − θ − θ

112

1 1 1

2 T R R R2 R cos 1

R R R

m −α = − − −

...(4.4.5)

Now we shall make some approximations, which are valid for thick potential barrier.

1/ 21 1

1 1 1 1

R R R Rcos sin and 1 1

R 2 R 2 R R− − π π= − ≈ − − ≈

Hence

121

2 T Rln T 2 R 2

2 R

m α π= − −

...(4.4.6)

Substituting the value of R1 from Eqn. (4.4.3) in (4.4.6), we have

2

1/ 2

0 0

4ln T ZR ZT

2

e m m e −α

= − πε ε

1/ 2ZR ZTa b −α= −

where a and b are constants defined by

2

0 0

4, / 2= =

π ε ε

me ea b m


The decay probability is given by

T T2R

vf

λ = =

ln ln ln T2R

v λ = +

1/ 2ln ZR ZT2R

va b −

α = + −

1/ 2

10 10log log 0.4343 ZR 0.4343 ZT2R

va b −

α λ = + −

...(4.4.7)

For a number of -emitters, a plot of log10 vs ZT– 1/2 is shown in the Fig. (4.4.3). A

straight line is obtained whose slope is – 0.4343b as required by the theory. The intercept on y-axisgives the value of

10log 0.4343 ZR2R

va

+ and this can be used to determine the value of nuclear radius R. The nuclear radius calculated in thisway comes out to be of the same order as obtained from scattering experiments. The correlationbetween the half-life time (or disintegration constant ) and energy of the -particles viz., mostenergetic -emitters are short lived and less energetic -emitters are long lived, is contained in thetheory.

Fig. 4.4.3

QUESTIONS

1. A particle of mass m and total energy E is incident on a one-dimensional rectangular potential barrier ofheight V > E and of finite thickness a. Show that the particle has finite probability of penetrating thebarrier and being seen on the other side. Find the transmission coefficient. (All’d 1995)


2. (a) A particle of mass m and energy E is incident on a one-dimensional step of potential of height V0from the left. Discuss the behaviour of the particle for E < V0 and explain how tunneling can beunderstood without violation of energy conservation principle.

(b) Discuss the two physical phenomena which can be understood on the basis of tunneling. (All’d 1996)

3. A beam of particles each of mass m and energy E, moving in a region of zero potential energy, approaches

a rectangular potential barrier of width a and height V0 where V0 > E. If 02

2 (V E)m −β =

, prove that the

transmission coefficient is given by 2

0 0

16 ET E 1

Vae

V− β

= − . (All’d 2007)

4. (a) A beam of particles of mass m and energy E is incident on a step potential of height V. Obtain expressionfor reflection and transmission coefficients and discuss the behaviour of the particle in theneighbourhood of E ~ V.

(b) Give three examples of quantum tunneling. (All’d 2006)

5. A beam of particles of mass m and energy E is incident on a step potential of height V0 from the left.Discuss the solution for E < V0 and explain how tunneling can be understood without violation of energyconsideration principle. Give two practical examples of quantum tunnel effect. (All’d 2005)

6. A particle of mass m free to move on a straight line is incident from x = – 8 on a potential barrier

V(x) = 0, for x < – a, and x > a,

= V0 for – a < x < a

If E < V0 then

(a) Show that there is a non-zero probability of the particle getting transmitted through the barrier andobtain an expression for the transmission coefficient.

(b) Show that for a high and wide barrier the transmission coefficient reduces to

2 2 0

2

2 (V E)T whereka m

e k− −≈ =

(All’d 2003)

EIGEN VALUES OF 2L AND zL AXIOMATIC:

CHAPTER

5.1 EIGEN VALUES AND EIGEN FUNCTIONS OF 2L AND zL

The square of angular momentum L2 and z-component of angular momentum are compatible

observables and their operators 2L and Lz commute i.e., =2ˆ ˆ[L , L ] 0,z therefore they can have

common eigen function. When we try to find solution of eigen value equation using the forms forthese operators in Cartesian coordinates, the differential equation obtained cannot be separated. Forthis reason we carry out a transformation to spherical polar coordinates. Let Y (, ) be the common

eigen function of these operators. Eigen value equations for operators 2L and Lz are

θ ϕ =λ θ ϕ2 2L Y( , ) Y( , ) ...(5.1.1)

θ ϕ = θ ϕL Y( , ) Y( , )z m ...(5.1.2)

where 2 and m are the eigen values operators 2L and Lz respectively. In polar coordinates the

operators 2L and Lz can be expressed as

2

2 22 2

1 1L sin

sin sin

∂ ∂ ∂ = − θ + θ ∂θ ∂θ θ ∂ϕ ...(5.1.3)

Lz i

∂=−∂ϕ

...(5.1.4)

In view of Eqns. (5.1.3) and (5.1.4) the eigen value Eqns. (5.1.1) and (5.1.2) become

∂ ∂ ∂ θ + +λ θ ϕ = θ ∂θ ∂θ θ ∂ϕ

2

2 2

1 1sin Y( , ) 0

sin sin...(5.1.5)

and Y( , )d

imd

θ ϕ = ϕϕ

Let us try to separate the variables and by assumingY() = () () ...(5.1.6)

FORMULATION OF QUANTUM MECHANICS


Substituting Eqn. (5.1.6) in (5.1.5), we get

22

2

sin 1sin sin

θ Θ Φ θ +λ θ=− Θ θ θ Φ ϕ d d d

d d d...(5.1.7)

The left hand side of Eqn. (5.1.7) is function of only and right hand side is function of

only, and are independent variables. This equality can hold if each side is equal to the sameconstant, say, m2.

The Eqn. (5.1.7) thus separates into two equations viz.

2

22

0Φ + Φ=ϕ

dm

d...(5.1.8)

2

2

1sin 0

sin sin

Θ θ + λ− Θ= θ θ θ θ

d d m

d d...(5.1.9)

The -Eqn. (5.1.8) integrates to

( ) C i me ϕΦ ϕ = ...(5.1.10)

where C is a constant. Since () is single valued, we must have () = ( + 2)

( 2 )ϕ ϕ + π=im ime e

2 πm ie = 1

m = 0 ± 1, ± 2, ± 3, …………. ...(5.1.11)

The constant C appearing in Eqn. (5.1.10) is determined by using the normalization condition

π

Φ ϕ ϕ=∫2

2

0

( ) 1d

22

0

C 1dπ

ϕ=∫

1C

2=

πSo the normalized solution of -equation is

1

( ) , 0, 1, 2, 3, ......2

ime mϕΦ ϕ = = ± ± ±π

...(5.1.12)

This determines the eigen function and eigen values of operator of z-component of orbital angularmomentum.

To solve the -equation, it is convenient to change the independent variable.

Eigen Values of L2 and ˆzL Axiomatic Formulation of ... 171

Let

2cos , 1 sin . 1 1= θ − = θ − < <x x x

2sin 1= = − θ =− −

θ θd dx d d d

xd d dx dx dx

In terms of new variable x, the -equation becomes

22

2(1 ) ( ) 0

1

Θ − + λ− Θ = −

d d mx x

dx dx x, m = 0, 1, 2,…. ...(5.1.13)

Equation (5.1.13) is well-known associated Legendre equation. Its solution is expressed interms of polynomials, called associated Legendre polynomials.

For m = 0, Eqn. (5.1.13) reduces to a relatively simpler equation, called Legendre equationand its solution is expressed in terms of Legendre polynomials, denoted by Pl (x). Let us writeΘ =( ) ( ).x P x The Legendre equation is

2 P

(1 ) P( ) 0d d

x xdx dx

− + λ = ...(5.1.14a)

or 2

22

P P(1 ) 2 P 0

d dx x

dxdx− − +λ = ...(5.1.14b)

Let us try power series solution of Eqn. (5.1.14) in the form of

2 3

0 1 2 30

P( ) ...... ll

l

x a a x a x a x a x∞

== + + + + =∑ ...(5.1.15)

Substituting Eqn.(5.1.15) in (5.1.14b), we have

2

0 0 0 0

( 1) ( 1) 2 0l l l ll l l l

l l l l

l l a x l l a x la x a x∞ ∞ ∞ ∞

−

= = = =− − − − +λ =∑ ∑ ∑ ∑

2

0 0

( 1) ( 1) 0l ll l

l l

l l a x l l a x∞ ∞

−

= =− − + − λ = ∑ ∑

20 0

( 1)( 2) [ ( 1) ] 0l ll l

l l

l l a x l l a x∞ ∞

+= =

+ + − + − λ =∑ ∑ (Replacing l by l + 2)

This equation is valid for all values of x (–1 < x < 1). This is possible only if coefficient ofeach power of x is zero. Hence

2( 1)

( 1)( 2)l ll l

a al l+

+ −λ=+ + (Recursion relation) ...(5.1.16)


For even values of l (0, 2, 4, …..) Eqn. (5.1.16) gives

a2 = 02!

λ− a

4 0(6 )

4!

λ − λ=−a a

6 0(6 )(20 )

6!

λ − λ − λ= −a a etc.

For odd values of l (1, 3, 5, …..), we have

3 12

3!

− λ=a a

5 1(2 )(12 )

5!

− λ − λ=a a

7 1(2 )(12 )(30 )

7!

− λ − λ − λ=a a etc.

In terms of two arbitrary constants a0 and a1, the solution of Legendre equation is written as

λ λ − λ λ − λ − λ= − − − −

2 4 60

(6 ) (6 )(20 )P( ) 1 .............

2! 4! 6!x a x x x

+ 3 5 71

(2 ) (2 )(12 ) (2 )(12 )(30 )..........

3! 5! 7!a x x x x − λ − λ − λ − λ − λ − λ+ + + +

...(5.1.17)

P(x) diverges at x = ± 1, its domain of convergence is – 1 < x < 1. The solution (5.1.17) consistsof two independent infinite series, one consisting of even coefficient a0 and the other consisting ofodd coefficient a1. It is readily seen that

→∞

+ →2lim 1r

l

l

a

a

Thus if the infinite series (5.1.17) is not terminated, it will diverge at x = 1. This is not acceptablesolution. To avoid the singularity of P(x) at x = 1, the series (5.1.17) must terminate after finitenumber of terms.

Let the series terminate for some value of integer l. The recursion relation (5.1.16) gives = l (l + 1), l = 0, 1, 2, 3, 4, 5,…. ...(5.1.18)

= 0, 2, 6, 12, 20, 30,…..

If this is so, one of the series terminates at al xl. From Eqns. (5.1.1) and (5.1.18) we see that

the condition of termination of series is the quantization condition for the eigen values of 2L . So the


eigen values of 2L are2 = l ( l + 1) 2, l = 0, 1, 2, 3, ….

or = l (l + 1) ...(5.1.19)

This means that allowed values of the observable orbital angular momentum L are

L ( 1)l l= + , l = 0, 1, 2, 3, ….. ...(5.1.20)

For even values of l, the even series becomes a polynomial but the odd series remains as aninfinite series. So we must set a1 = 0.

Similarly, for odd values of l, the odd series becomes a polynomial and even series remains asan infinite series. So we set a0 = 0.

So the acceptable solutions of Legendre equation are either even polynomials or odd polynomialsand may be represented as

λ λ − λ λ − λ − λ= − − − −

even 2 4 60

(6 ) (6 )(20 )P ( ) 1 .............

2! 4! 6!l x a x x x

or − λ − λ − λ − λ − λ − λ= + + + +

odd 3 5 71

(2 ) (2 )(12 ) (2 )(12 )(30 )P ( ) ..........

3! 5! 7!l x a x x x x

From these polynomials we have

= = − = − +even even 2 even 2 40 0 2 0 4 0

35P ( ) , P ( ) (1 3 ), P ( ) (1 10 ), etc.

3x a x a x x a x x

and = = − = − +odd odd 3 odd 3 51 1 3 1 5 1

5 14 63P ( ) , P ( ) ( ), P ( ) ( ).....etc.

3 3 15x a x x a x x x a x x x

The constants a0 and a1 are determined using the conditions =P (1) 1l .

= ⇒ = = ⇒ = − = ⇒ =0 0 2 0 4 0P (1) 1 1, P (1) 1 1/2, P (1) 1 3/8 etc.a a a

= ⇒ = = ⇒ = − = ⇒ =1 1 3 1 5 1P (1) 1 1, P (1) 1 3/2, P (1) 1 15/8 etc.a a a

Plugging the appropriate values of a0 and a1 the Legendre polynomials come out to be

P0 (x) = 1

P1 (x) = x

P2 (x) = (1/2) (3x2 –1)

P3 (x) = (1/2) (5x3 – 3x)

P4 (x) = (1/8) ( 35x4 – 30x2 + 3)

P5 (x) = (1/8) (63x5 – 70x3 + 15x) etc.


The Legendre polynomials Pl (x) are also obtained from general formula called Rodriguesformula.

( )21P ( ) . 1

2 ( !)

ll

l l l

dx x

l dx= − ...(5.1.21)

The orthogonality condition for Legendre polynomials is

1

1

2P ( )P ( )

2 1l l l lx x dxl

′ ′−

= δ+∫ ...(5.1.22)

For m = 0, the solution of Legendre equation can be written as

Θ =( ) N P ( )l l lx x ...(5.1.23)

where Nl is a constant which can be determined using the normalization condition

1

2

1

( ) 1−

Θ =∫ x dx

−

+= ⇒ =∫1

22

1

2 1N P ( ) 1 N

2l l ll

x

So the normalized solution of Legendre equation is

2 1

( ) P ( )2l l

lx

+Θ θ = (5.1.24)

For m 0, the solution of associated Legendre Eqn. (5.1.13) for positive values of m are

associated Legendre polynomials. These are denoted by P ( )ml x and defined by

2 /2P ( ) (1 ) P ( )

mm ml lm

dx x x

dx= − ⋅

2 /2 21.(1 ) ( 1)

2 ( !)

l mm l

l l m

dx x

l dx

+

+= − − ...(5.1.25)

From the very definition of P ( )ml x it is evident that m l. For m l, P ( )m

l x vanishes. Interms of normalization constant Nlm the solution of associated Legendre equation is expressed as

( ) N P ( )mlm lm lx xΘ = ...(5.1.26)

The constant Nlm is determined using the normalization condition

1

2

1

( ) 1−

Θ =∫ x dx ...(5.1.27)


This integral can be evaluated using the result

−

+=+ −∫

12

1

2 ( )!( ) .

2 1 ( )!m

ll m

P x dxl l m

...(5.1.28)


2 1 ( )!

N .2 ( )!lm

l l m

l m

+ −=+

...(5.1.29)

The normalized associated Legendre polynomials for any value of l and m (of course l = 0, 1,2, 3, ….. and |m| l or – l m l) are

+ −Θ = − = θ

+2 1 ( )!

( ) ( 1) . P ( ), cos2 ( )!

m mlm l

l l mx x x

l m...(5.1.30)

The complete normalized eigen functions of 2L are

θ ϕ =Θ θ Φ ϕ,Y ( , ) ( ) ( )ml m l

2 1 ( )! 1( 1) . P (cos ).

2 ( )! 2m m im

ll l m

el m

ϕ+ −= − θ+ π

2 1 ( )!( 1) . .P (cos ).

4 ( )!m m im

ll l m

el m

ϕ+ −= − θπ +

...(5.1.31)

For negative value of m, we have

, ,Y ( 1) Yml m l m

∗− = −

...(5.1.32)

The introduction of constant phase factor (– 1)m is a matter of convention, it makes the formof Y (, ) agree with those commonly used in literature. The functions Ylm( ) are called sphericalharmonics. The first few of these are:

00 10

1 3Y , cosY

4 4= = θ

π π

113

Y sin . ,8

ie ϕ= − θπ

1, 13

Y sin .8

ie− ϕ− = θ

π

220

5Y (3cos 1)

16= θ−

π


2115

Y (sin .cos ).8

ie ϕ= θ θπ

, − ϕ− = − θ θ

π2, 115

Y (sin .cos ).8

ie

ϕ − ϕ−= θ = θ

π π2 2 2 2

22 2, 215 15

Y sin . , Y sin .32 32

i ie e

Degeneracy with respect to m

For a general value of l, m can assume integral values from – l to + l in steps of unity i.e., m cantake on (2l + 1) values in all. Hence there are 2l + 1 different eigen functions corresponding to a

single eigen value ( 1)l l + of orbital angular momentum L. The eigen values for a given l, are

(2l+1) fold degenerate.

5.2 AXIOMATIC FORMULATION OF QUANTUM MECHANICS

This approach of quantum mechanics is based on few postulates. The postulates of thermodynamicsare stated in terms macroscopic variables such as pressure, temperature, volume, mass, energy etc.and hence readily understood. The postulates of quantum mechanics are stated in terms of microscopicand abstract concepts and hence it is difficult to form illustrative images of these concepts.

POSTULATE 1

The state of a quantum mechanical system is described by wave function (r, t). The wave function (r, t) contains all information that nature permits about the system. The collection or the totalityof wave functions of a system form an infinite-dimensional linear vector space, called Hilbert space.

If 1 and 2 are two states in which a physical quantity Q has definite values q1 and q2

respectively, then the linear combination or the superpositionc1 1 + c2 2,

where c1 and c2 are arbitrary complex numbers, also represents a possible state of the system but inthis state the physical quantity Q has not a definite value; instead the measurement of physical quantityQ yields either a value q1 or q2. Thus the superposition of states produces a new state in which Qhas indefinite value. The principle of superposition stated above can be extended to any number ofstates i.e.,

c11 + c22 + ……….. + cmm

represents a state of the system. The reverse of superposition principle is also true. That is, an arbitrarystate wave function can be expanded in terms of the states of the Hilbert space of the system. Thus

= c11 + c22 + ……….. + cmm = =

ϕ∑1

m

m mm

c ...(5.2.1)

where cm are arbitrary complex numbers.


POSTULATE 2

To every physical property there corresponds a quantum mechanical operator. The operatorcorresponding to the property Q is obtained by writing the classical mechanical expression for Q andthen making following replacements.

22ˆ ˆ ˆˆ , , H T V V( )

2qq q q p i qq m

∂→ = → − = + = − ∇ +∂

An operator Q is said to be Hermitian if its expectation value is real. For each Hermitian operator

an eigen value equation is

Q i = qi , ...(5.2.2)

where q is eigen value and i is eigen function of the operator, can be set up. The eigen values ofHermitian operator are real and two eigen functions belonging to different eigen values are orthogonali.e.,

*ϕ ϕ τ = δ∫ i j ijd ...(5.2.3)

The set of eigen functions i constitute an infinite set of linearly independent orthogonalfunctions. If these functions are normalized, they are said to form a complete orthonormal set. Thismeans that any arbitrary function of the state can be expressed as superposition of this completeset.

ψ = ϕ∑ j jj

c ...(5.2.4)

If we have a large number of identical systems, all prepared in the same state and measurementof a dynamical variable Q is made on each system, the outcome of each measurement will, in general,be different. The average or expectation value of these results will be given by

( )* ˆ ˆQ Q or Q , Qd= ψ ψ τ = ψ ψ∫ ...(5.2.5)

i.e. * *Q k j j k j

j k

c c q d= ϕ ϕ τ∑∑ ∫

*k j j kj

j k

c c q= δ∑∑2| |=∑ j j

j

c q ...(5.2.6)

Meaning of this equation is that if a measurement of the physical quantity Q represented by the

operator Q is carried out on the system represented by the state then |cj|2 gives the probability

that the result will be the eigen value qj.On the other hand if the system is in the state represented byone of the eigen function k then the measurement of Q will give only one eigen value qk.


POSTULATE 3

The operators of dynamical variables donot, in general, commute with other i.e., ˆ ˆˆ ˆQR RQ≠ . The

difference ˆ ˆˆ ˆQR RQ− is called commutator of Q and R and is denoted by ˆ ˆ[Q, R].

= −ˆ ˆ ˆˆ ˆ ˆ[Q, R] QR RQ ...(5.2.7)

POSTULATE 4

The time evolution of the state (x, y, z, t) of a system is governed by the Schrodinger equation ofmotion

∂ψ = ψ∂( , )

H ( , )r t

i r tt

...(5.2.8)

where H is the Hamiltonian operator of the system. The Schrodinger equation is a postulate and is tobe tested by agreement of its prediction with experiments.

5.3 DIRAC FORMALISM OF QUANTUM MECHANICS

In Dirac formalism the state of a dynamical system is represented by vector, called ket vector and isdenoted by symbol | >. The kets of a system form an infinite dimensional abstract linear vector space,called Hilbert space H. To every ket vector in ket space, there exists another vector, called dual vectoror bra vector in bra space and is denoted by < |. Like ket vectors, the bra vectors form a differentHilbert space H*. The state of a system is specified by direction of ket vector. Two ket vectors | >and c | >, where c is a complex number, denote the same state. A dynamical system represented bya ket vector | > can be equally well represented by corresponding bra vector < |.

PROPERTIES OF STATE VECTORS

1. If kets | > and | > represent two states then their linear combination

c1 | > + c2 | >

is also a ket representing another state where c1 and c2 are arbitrary complex numbers.

2. We can form a scalar product of a ket vector | > and a bra vector

< |, which is denoted by < | >. If the kets | > and | > represent the statesdescribed by wave functions (r) and (r) respectively then

* *| ( ) ( ) |r r d< ϕ ψ > = ϕ ψ τ =< ψ ϕ >∫

3. A bra vector < | is said to be null bra if the scalar product < | > vanishes for any| >.

< | = 0, if < | > = 0 for any | >.

4. Two bra vectors < 1 | and < 2 | are equal if

< 1 | > = < 2 | > for every | >.


5. If < | > = 0, then | > and | > are said to be orthogonal.

6. If < | > = 1, then | > is said to be normalized.

7. The scalar product of two vectors obey the rule

< | > = < | >*,

< | > = < | >* implies that < | > is real.

8. If | > = | > + | > then < | = < | + < |

If | > = c | > then < | = c* < |

9. An operator Q converts a ket | > into another ket | > i.e.,

Q | > = | > or < | Q = < |

10. The operator Q is said to be linear if

Q (c1 | 1 > + c2 | 2 > + …..) = c1 Q | 1 > + c2 Q | 2 > + ……

11. The adjoint of Q is denoted by Q+ read as Q dagger and is defined through the equation

*ˆ ˆ|Q | |Q |+< α β >=<β α > ...(5.3.1)

Expansion Postulate

If | i > denotes the complete orthonormal set of eigen vectors of a Hermitian operator, then anarbitrary state | > can be expanded in terms | i >.

| | | | |i i i i

i i

ψ >= < ϕ ψ > φ > = ϕ >< ϕ ψ >∑ ∑ ...(5.3.2)

This implies that

| | 1ϕ >< ϕ =∑ i ii

(completeness rule) ...(5.3.3)

The expansion coefficients |< ϕ ψ >i are the projections of onto i.

5.4 GENERAL DEFINITION OF ANGULAR MOMENTUM

Earlier in this chapter we defined orbital angular momentum vector through the relation L = r × p

and obtained the corresponding operator L by replacing the classical observables with their operators.

It was shown that L and its components ˆ ˆ ˆL ,L ,Lx y z satisfy the commutation relations

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL ,L L , L , L L , L , L Lx y z y z x z x yi i i = = =


2 2 2ˆ ˆ ˆ ˆ ˆ ˆL , L L , L L , L 0x y z

= = = ...(5.4.1)

The eigen values of 2L and Lz were found to be l(l + 1) and m respectively where l = 0, 1,

2, 3,….. and m = – l, – l + 1, ……….,(l – 1), l. It is worth to remember that the eigen values ofcomponents of angular momentum are integral multiple of . We know that, spin (angular momentum)has no classical analog and hence no classical expression for its representation. Therefore the waywe obtained the operator for L will not work for spin. Moreover, the eigen values of operator ofcomponent of spin observable are half-integer. Here we shall give general definition of angularmomentum and see that the magnitudes of angular momentum equal to 0, 1/2 , , 3/2 ,….come ina natural way.

We define angular momentum J with their components Jx, Jy and Jz as an observable whoseoperators satisfy the following commutation relations.

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆJ , J J , J , J J , J , J Jx y z y z x z x yi i i = = =

2 2 2ˆ ˆ ˆ ˆ ˆ ˆJ , J J , J J , J 0x y z

= = = ...(5.4.2)

We further define ladder operators ˆ ˆJ and J+ − as

ˆ ˆ ˆ ˆ ˆ ˆJ J J , J J Jx y x yi i+ −= + = − ...(5.4.3)

The ladder operators are mutually connected through the relations

ˆ ˆ ˆ ˆJ J , J J+ ++ − − += = ...(5.4.4)

The ladder operators satisfy the following commutation relations

2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆJ , J J , J , J J , J , J 2 J , J , J 0z z z+ + − − + − ± = =− = = ...(5.4.5)

where 2 2 2 2ˆ ˆ ˆ ˆJ J J Jx y z= + + ...(5.4.6)

The commutation relations can be derived in the same way as the corresponding relations fororbital angular momentum operators were derived. Further it can be shown that

2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆJ J J J J , J J J J Jz z z z+ − − += − + = − − ...(5.4.7)

( )2 21ˆ ˆ ˆ ˆ ˆ ˆJ J J J J J2 z+ − − += + + ...(5.4.8)

Since 2ˆ ˆJ , J 0,z = it is possible to find a complete set of simultaneous eigenstates of 2ˆ ˆJ and Jz .

Let one of these states belonging to the eigen value j 2 of 2J and m of Jz be | j m . Evidently

λ >=λ λ >2 2J | |j m j j m ...(5.4.9)

Jz | j m = m | j m ...(5.4.10)


Now we shall first show that there exists a lower and upper limit to the quantum number m.The value of a component of a vector cannot be greater than the value of the vector itself. So

(m)2 j 2 m j ...(5.4.11)

Thus for a fixed value of j, the value of m is bounded. We can also get the same result asfollows. The expectation value of a Hermitian operator cannot be negative. Therefore

2 2ˆ ˆ0 | J J |j x y jm m≤ < λ + λ >

2 2ˆ ˆ0 | J J |j z jm m≤ < λ − λ >

2 2 20 j m≤ λ −

2jm ≤ λ

Let the upper and lower value of m be m+ and m–..Now

ˆ ˆ ˆ ˆJ J | (J J J )|z j z jm m+ + +λ >= + λ >

J ( 1) | jm m+= + λ >

ˆ( 1) J | jm m+= + λ > ...(5.4.12)

Equation (5.4.12) shows that J | j m+ λ > is an eigenstate of Jz belonging to the eigen value

(m + 1) . The reason why J+ is called raising operator is now obvious; it raises the eigen value of

Jz by one unit (). J+ has no effect on the eigen value of 2J . Similarly we can show that

ˆ ˆ ˆJ J | ( 1) J |z j jm m m− −λ >= − λ > ...(5.4.13)

By virtue of its property displayed in Eqn. (5.4.13) J− is called lowering operator. On acting

on an eigenstate of angular momentum, it lowers the eigen value of Jzby and leaves the eigen

value of 2J intact. These results can be expressed as

+ +λ >= λ + >J | C | , 1j jm m ...(5.4.14)

J | C | , 1j jm m− −λ > = λ − > ...(5.4.15)

where C+ and C– are constants to be determined. The effect of J± is to raise (lower) the eigen value

m in steps of one unit. Since the highest possible value of m is m+ and the lowest possible value ofm is m–, the states |j, m++1> and |j, m– – 1> donot exist. Hence

J | 0jm+ +λ >= ...(5.4.16)

J | 0j m− −λ >= ...(5.4.17)


Operating both sides of Eqn. (5.4.16) by J− , we have

ˆ ˆJ J | 0j m− + +λ >=

2 2ˆ ˆ ˆ(J J J ) | 0z z j m+− − λ >=

2 2 2 2 0j m m+ +λ − − =

( 1) 0+ +λ − + =j m m ...(5.4.18)

Similarly, operating both sides of Eqn. (5.4.17) by J+, we have

ˆ ˆJ J | 0j m+ − −λ >=

2 2ˆ ˆ ˆ(J J J ) | 0z z j m−− + λ >=

2 2 2 2 0j m m− −λ − + =

( 1) 0− −λ − − =j m m ...(5.4.19)

From Eqns. (5.4.18) and (5.4.19), we have

( 1) ( 1)+ + − −+ = −m m m m

Since m+ > m– we must havem+ = – m– = j (say) ...(5.4.20)

Equation (5.4.20) defines the quantum number j. Substituting m+ = j in Eqn. (5.4.18), we getj = j ( j + 1) ...(5.4.21)

From Eqns. (5.4.9) and (5.4.21) we see that eigen value of 2J is j ( j + 1). The eigenstates

belonging to the same eigen value j ( j + 1) of 2J but different eigen values of Jz may be denoted by| j, j>, | j ,j – 1>, | j, j –2>, …………, | j, – j +1 >, | j, – j >

Successive application of lowering operator J− on the state | j , j > will ultimately lead to the

state | j , – j >. Similarly the successive application of J+ on the state | j , – j > will lead to the state

| j , j >. This means that m+ – m– = j – (– j) = 2 j

is an integer. Therefore the allowed values of j are:j = 0, 1/2 , 1, 3/2, ……………….

Thus the angular momentum can have integral and half-integral values both.

Let us find the eigenstates of 2J and zJ . The expectation value ˆ ˆJ J− + in the state | j , m > is

given by

ˆ ˆ ˆ ˆ| J J | (J ) | J |j m j m j m j m− + + +< >=< >


2| C | , 1 | , 1j m j m+= < + + >

2|C |+= ...(5.4.22)

where we have used the normalization condition < j, m + 1| j, m + 1> = 1.

Using the result 2 2ˆ ˆ ˆ ˆ ˆJ J J J Jz z− + = − − we can write Eqn. (5.4.22) as

ˆ ˆ| J J |j m j m− +< >= 2 2ˆ ˆ ˆ| J J J |z zj m j m< − − >

= j (j + 1) 2 – m2

2 – m2

= [j ( j + 1) – m (m + 1)] 2 ...(5.4.23)

From Eqns. (5.4.22) and (5.4.23)

C ( 1) ( 1) ( )( 1)j j m m j m j m+ = + − + = − + + ...(5.4.24)

Similarly we can find

C ( ( 1) ( 1) ( )( 1)j j m m j m j m− = + − − = + − + (5.4.25)

So Eqns. (5.4.14) and (5.4.15) become

J | , ( 1) ( 1) | , 1j m j j m m j m+ > = + − + + > ...(5.4.26)

J | , ( 1) ( 1) | , 1j m j j m m j m− > = + − − − > ...(5.4.27)

From Eqns. (5.4.26) and (5.4.27) we see that

ˆ ˆJ | , 0, J | , 0j j j j+ −>= − >=

All the relations obtained for angular momentum operator J are also true for orbital angular

momentum operator L and spin angular momentum operator S . The eigenstates of operator 2ˆ ˆL and Lz

are denoted by |l, m> in Dirac notation and by Ylm (,) in coordinate representation. The raising

and lowering operators ˆ ˆL and L+ − in coordinate representation are expressed as

ˆ ˆ ˆL L L coti

x yi e iϕ+

∂ ∂= + = + θ ∂θ ∂ϕ

ˆ ˆ ˆL L L coti

x yi e i− ϕ−

∂ ∂= − = − + θ ∂θ ∂ϕ

For orbital angular momentum Eqns. (5.4.26) and (5.4.27) assume the form

+ +θ ϕ = + − + θ ϕ, , 1L Y ( , ) ( 1) ( 1) Y ( , )l m l ml l m m ...(5.4.28)

− −θ ϕ = + − −, , 1L Y ( , ) ( 1) ( 1) Yl m l ml l m m ( ) ...(5.4.29)

These equations may be used to obtain the spherical harmonics for different values of l and m.


Use of Ladder operators to find the eigen functions of 2L

For a given value of l, m can only take values from – l to + l in steps of unity. From Eqns. (5.4.28)and (5.4.29) we see that

+ =,L Y 0l l ...(5.4.30)

− − =,L Y 0l l ...(5.4.31)

From Eqns. (5.4.28) and (5.4.29) we have

ϕ+

∂ ∂= + θ ∂θ ∂ϕ+ − + , 1 ,

1Y cot Y

( 1) ( 1)i

l m l me il l m m ...(5.4.32)

− ϕ−

∂ ∂= − − θ ∂θ ∂ϕ+ − − , 1 ,

1Y cot Y

( 1) ( 1)i

l m l me il l m m ...(5.4.33)

For m = 0, the solution of Legendre equation was found to be

+= θπ, 0

2 1(cos )

4l ll

Y P ...(5.4.34)

From, (5.4.34)

00 01 1

Y P (cos )4 4

= θ =π π ...(5.4.35)

10 13 3

Y .P (cos ) cos4 4

= θ = θπ π

...(5.4.36)

Making use of Eqn. (5.4.32) we can obtain Yl, 1, Yl, 2, Yl, 3,……… , Yl, l and from Eqn. (5.4.33)we can obtain Yl, – 1, Yl, – 2, Yl, – 3,…………, Yl, – l.

For l = 1, m = 1, 0, – 1, from Eqn. (5.4.32)

1,1 101

Y cot Y2

ie iϕ ∂ ∂= + θ ∂θ ∂ϕ

3cot . (cos )

42

ϕ ∂ ∂= + θ θ ∂θ ∂ϕ π

iei

3.sin .

8ϕ= − θ

πie ...(5.4.37)

From Eqn. (5.4.33)

− ϕ

− ∂ ∂= − + θ ∂θ ∂ϕ

1, 1 10Y cot Y2

iei


3.sin .

8− ϕ= θ

πie ...(5.4.38)

For l = 2, m = 2, 1, 0, –1, –2.

2

2,15

Y cot (3cos 1)166

iei

ϕ ∂ ∂= + θ θ− ∂θ ∂ϕ π

( )15sin .cos

8ϕ= − θ θ

πie ...(5.4.39)

2, 2 21Y cot Y2

iei

ϕ ∂ ∂= + θ ∂θ ∂ϕ

2 215,sin .

32ϕ= θ

πie ...(5.4.40)

2

2,0 22 1 5 1

Y .P (cos ) . .(3cos 1)4 4 2

l += θ = θ−π π

...(5.4.41)

2, 1 20Y cot Y6

iei

− ϕ

− ∂ ∂= − + θ ∂θ ∂ϕ

=6

− ϕie ( )25 1cot . . 3cos 1

4 2

∂ ∂− + θ θ − ∂θ ∂ϕ π i

( )15. sin .cos

8− ϕ= θ θ

πie ...(5.4.42)

2, 2 21

Y cot Y , 14

ie i− ϕ−

∂ ∂= − + θ − ∂θ ∂ϕ

( )15cot . sin .cos

84

− ϕ− ϕ∂ = − + θ θ θ ∂θ π

iie

i e

2 215.sin .

32− ϕ= θ

πie


5.5 PARITY

Parity is a purely quantum mechanical quantity having no classical analogue. To arrive at the conceptof parity, let us consider the behaviour of the wave function (x, y, z) upon the inversion of coordinateaxes. Inversion consists in reversing the direction of all the axes. It is not difficult to see that inversiontransforms right handed coordinate system into left handed one. Inversion results change in the signsof all the coordinates and consequently the function (x, y, z) transforms into (– x, – y, – z). Thistransition can be considered as a result of the action of the inversion operator P on (x, y, z) function.

P ( , , ) ( , , )x y z x y zψ = ψ − − − ...(5.5.1)

Applying P again, we get

ˆ ˆ ˆPP ( , , ) P ( , , ) ( , , )x y z x y z x y zψ = ψ − − − = ψ

2P ( , , ) ( , , )x y z x y zψ = ψ ...(5.5.2)

It follows that the square of the operator P equals unity.To determine the eigen values of the inversion operator, we must solve the equation

P ( , , ) ( , , )x y z x y zψ = λψ ... (5.5.3)

Now ˆ ˆ ˆPP ( , , ) P ( , , )x y z x y zψ = λ ψ

2 2P ( , , ) ( , , )x y z x y zψ = λ ψ

(x,y,z) = 2 (x, y, z) ...(5.5.4)

2 = 1

= ± 1 ...(5.5.5)

Hence the eigen values of the inversion operator are +1 and – 1. With a view to this circumstance,we can write

P ( , , ) ( , , )x y z x y zψ = ±ψ ...(5.5.6)

(– x, – y, – z) = ± (x, y, z) ...(5.5.7)

The quantity depicted by the operator P is known as the parity. Thus the wave function (x, y, z)of states with a definite parity value can be divided into two classes: (i) functions + that do notchange when the inversion operator acts on them and (ii) function – that change their sign whenthe inversion operator acts on them. The functions + and – satisfy the relations

+ + .ˆ ˆP = , P =− −ψ ψ ψ −ψ

States corresponding to the functions + are said to be even and those corresponding to thefunctions – are said to be odd. The parity of a state described by the function

= c1+ + c2–

is indeterminate.


Let us see the effect of parity operator on the spherical harmonics. The inversion transformation(x – x, y – y, z – z) in spherical coordinates is equivalent to r r , – , + .The spherical harmonics are given by

ϕθ ϕ = θY ( , ) const.P (cos )m imlm l e

The action of inversion operator on θ ϕY ( , )lm is equivalent to replacing – , and

+ . Thus

π+ϕθ ϕ = π − θ ( )PY ( , ) const.P (cos( )).m imlm l e

= | |const. ( cos ) . ( 1)m m imlP e ϕ − θ −

= − ϕ − θ −

| | | |const.( 1) (cos ) ( 1)l m m m imlP e

= const.( 1) (cos )l m imlP e ϕ − θ

= − θ ϕconst.( 1) Y ( , )llm

The parity of the state with given value of l is (– 1)l. That is all the states with even l are evenparity states and all those with odd l are odd.


1. Prove the following commutation relations for the operators L+ = Lx + iLy and L– = Lx – iLy.

L ,L L , L ,L L , L ,L 2 Lz z z+ + − − + − = =− =

Show that the operators L+ and L– are in fact angular momentum raising and lowering operators respectively.

(All’d 1995)

2. If L+ and L– are the raising and the lowering angular momentum operators show that 2L lm± ψ is an eigenstate of

Lz with eigen values m ± 2. (All’d 1996)

3. Define angular momentum raising and lowering operators. Using these operators and the property[L2, Lz] = 0 , obtain eigen values of L2 . (All’d 1998)

4. Define angular momentum raising L+ and lowering L– operators. In a representation in which L2 and Lz arediagonal, find the eigen values of the operator L– L+. (All’d 1999)

5. (a) What are angular momentum raising and lowering operators?

(b) Find L × L where L is angular momentum operator.

(c) Show that the eigen values of a Hermitian operator are real. (All’d 2000)

6. (a) Define angular momentum raising and lowering operators.

(b) Find L × L and [L2, L]

(c) Find eigen values of the operator L+ L– (All’d 2002)


7. (a) Show that L2 and [L+, L–] have simultaneous eigen functions.

(b) If an operator commutes with Lx and Ly, it will commute with L±.

(c) If b is the maximum eigen value of L± and a is eigen value of L2 then show that a = b (b + )(All’d 2003)

8. Given 2

2 22 2

1 1ˆ ˆL sin and Lsin sin

z i ∂ ∂ ∂ ∂ = − θ + = − θ ∂θ ∂θ ∂ϕθ ∂ϕ

Find the eigen values of L2 and Lz . (All’d 2003)

9. (a) Define angular momentum raising and lowering operators.

(b) Find ˆ ˆ ˆ(L L ),Lx y + +

(c) Show that momentum operator is Hermitian. (All’d 2004)

10. Find the eigen values of square of angular momentum operator

22 2

2 2

1 1L sin

sin sin

∂ ∂ ∂ = − θ + θ ∂θ ∂θ θ ∂ϕ

Use this result to find the energy of a particle of mass m moving freely on a smooth surface at a fixed distancer = a from the origin. (All’d 2006)

PARTICLE IN A BOX

6.1 PARTICLE IN AN INFINITELY DEEP POTENTIAL WELL (BOX)

Consider a particle of mass m, which is restricted to move along x-axis between the region boundedby x = 0 and x = L. Physically a bead sliding along a perfectly smooth straight wire stretched alongx-axis with rigid barriers at x = 0 and x = L or an electron confined to move along x-axis in a potentialwell defined by

V(x) = 0 for 0 < x < L

= for x < 0 and x > L ...(6.1.1)

may represent the problem under investigation. Here Vrepresents the potential energy of the electron. Obviouslyoutside the potential well, kinetic energy of the particle isnegative hence this region is inaccessible to the particle. Alsoat the boundary V is infinite, ensures that the wave function(x) representing the particle must vanish outside the well.

Let E be the energy of the particle. The time-independent Schrodinger equation for the particle is

2

2

2 Ed m

dx

ψ + ψ

= 0

or2

22

dk

dx

ψ + ψ = 0 ...(6.1.2)

where 2

2 Emk =

...(6.1.3)

Solution of Eqn. (6.1.2) is

( ) Asin Bcosx kx kxψ = + ...(6.1.4)

where A and B are arbitrary constants. The boundary conditions for this problem are(0) = 0 and (L) = 0

Fig. 6.1.1 Infinitely deep onedimensional potential well

CHAPTER


When the first boundary condition is substituted in Eqn. (6.1.4), we get B = 0. So the solution(6.1.4) becomes

(x) = A sin kx ...(6.1.5)

Substituting the second boundary condition in Eqn. (6.1.5), we findA sin kL = 0, A 0

kL = n, n = 1, 2, 3,…..

k = L

nπ...(6.1.6)

2 2

2 2

2 E

L

m n π=

2 2 2

2E

2 Ln

n

m

π= ...(6.1.7)

The value n = 0 is inadmissible as it corresponds to (x) = 0 everywhere. Since the energy ofthe particle depends on integer n, this justifies the subscript n to the energy E.

Equation (6.1.7) shows that particle can have only discrete energies i.e, energy of the particleis quantized. The discrete set of energies is called energy levels and the integer n is called the quantumnumber. The classical mechanics allows the particle to have any energy including zero. Thus thequantization of energy is a quantum mechanical result and has no counterpart in classical physics.The energy levels of the particle are shown in the Fig. ( 6.1.2 ). Evidently the energy levels are notequally spaced.

Wave function: The wave function of the particle is

( ) AsinL

n xx

πψ = ...(6.1.8)

Applying the normalization condition to the wave function we have

ψ =∫L

2

0

( ) 1x dx

L

2 2

0

A sin 1L

n xdx

π =∫

whence 2

AL

=

The normalized wave functions of the particle are

2

( ) sinL Ln

n xx

πψ = ...(6.1.9)

Particle in a Box 191

Orthogonality of wave functions: The wave functions of a particle in an infinitely deep potentialwell are orthogonal. Let us verify it. Let m(x) and n(x) be two wave functions corresponding toenergies Em and En. Then

L L

0 0

2( ) ( ) sin sin

L L Lm nm x n x

x x dx dxπ πψ ψ =∫ ∫

L

0

1 ( ) ( )cos cos

L L L

m n x m n xdx

− π + π = − ∫= mn, mn = 0, for m n and mn = 1 for m = n

Fig. 6.1.2 Energy levels, wave functions and probability density

Probability density: The probability density Pn(x) of finding the particle anywhere on thex-axis is given by

Pn(x) = (x) 2 = 22sin

L L

n xπ ...(6.1.10)

Even if we consider the time dependent wave function to calculate the probability function, itcomes out to be independent of time. The energy levels with corresponding wave functions andprobability density are shown in the Fig. (6.1.2).

Notice that the wave functions of a particle in a box are similar to the displacement functionsof a stretched string. This is to be expected because the boundary conditions are identical in both thecases. The probability density corresponding to the quantum number n = 1 is maximum at the center(x = L/2) of the potential well whereas the probability density corresponding to the quantum numbern = 2 is zero at the center of the well. This fact is at variance with the prediction of classical physicsaccording to which the probability of finding the particle is the same everywhere.

As we go higher energy levels with more nodes (the points where wave function vanishes) themaxima and minima of probability come closer together and the variations probability along thelength of the box ultimately becomes undetectable. For large quantum numbers we get the classicalresult of uniform probability density. This is in accord with the correspondence principle.


In our daily experience we are concerned with macroscopic objects only. For such objects thespacing of the energy levels is too small to be observed and therefore energy appears to be continuous.But for microscopic objects the spacing of the energy levels are considerable and hence the discretenature of energy levels becomes perceptible. These facts may be illustrated by examples.

Consider a macroscopic system, say, a bead of mass m = 10 gm confined to move along x-axisin a region of dimension L = 1 m. The energy of the bead is

2 2 266 2

2E 5.5 10 J

2 Ln

nn

m−π= = ×

The first three energy levels areE1 = 5.5 × 10–66 J, E2 = 22.0 × 10–66 J, E3 = 50.5 × 10–66 J

Evidently the spacing of the energy levels is too small to be detected. The velocity correspondingto kinetic energy 5.5 × 10–66 J is 3.3 × 10–32 m/s, the bead can hardly be distinguished from itsstationary position. Thus, because of the extremely small size of , quantization of energy isunobservable for macroscopic bodies.

Now consider a microscopic system, say, an electron which is confined to in a region ofdimension L = 1 Å. The energy of electron is

En = 2 2 2

18 2 22

6 10 J 38 eV2 L

nn n

m−π = × =

The first three energy levels areE1 =38 eV, E2 = 152 eV, E3 = 342 eV.

These energy levels are sufficiently far apart and therefore the quantization of energy of electronis perceptible. Typical atoms have this dimension and hence quantization of electron energy levels inatoms is conspicuous.

Correspondence principle: The spacing of two successive energy levels is

+π∆ = − = + −

22 2

1 2E E E [( 1) ]

2 L

n n n nm

For macroscopic bodies →∞ →∞ ∆ →, and L . E 0.m The discrete energy spectrum becomes

continuous. This is correspondence principle.

6.2 PARTICLE IN A TWO DIMENSIONAL POTENTIAL WELL

Let the potential well be defined byV = 0 for 0 < x < L1, 0 < y < L2

= for x > L1, y > L2 ...(6.2.1)

The Schrodinger equation for the particle confined to move in this two dimensional potentialwell is

2

2

2(E V) 0

m∇ ψ + − ψ =

...(6.2.2)


2 22

2 2

( , ) ( , )( , ) 0

x y x yk x y

x y

∂ ψ ∂ ψ+ + ψ =∂ ∂

where 2

2 Emk =

...(6.2.3)

To solve Eqn. (6.2.2) let(x, y) = f1 (x). f2 (y) ...(6.2.4)

Substituting Eqn(6.2.4) in (6.2.2) we have

2 221 2

2 21 2

( ) ( )1 1

( ) ( )

d f x d f yk

f x f ydx dy= − − ...(6.2.5)

The left-hand side of above equation is function of x only and the right-hand is function of yonly, x and y are independent of each other. This equation is consistent only if each side is equal tothe same constant, say – k2

1. (If we choose the separation constant to be positive, the separated equationswill have exponential solution, which will not vanish at the boundaries.) Thus

2 22 21 2

12 21 2

( ) ( )1 1

( ) ( )= − − = −

d f x d f yk k

f x f ydx dy...(6.2.6)

Equation (6.2.6) separates into two equations

2211 12

( )( ) 0+ =

d f xk f x

dx...(6.2.7)

2

22

( )d f y

dx+ 2

2 2( )k f y = 0 ...(6.2.8)

where 2 2 2 2 2 22 1 1 2ork k k k k k= − + = ...(6.2.9)

The solution of Eqn. (6.2.7) may written as

1 1 1( ) A sin Bcosf x k x k x= + ...(6.2.10)

and that of Eqn. (6.2.8) may be written as

2 2 2( ) Csin D cosf y k y k y= + ...(6.2.11)

Applying the boundary condition: f1 (x) = 0 at x = 0, we findB = 0.

Similarly applying the boundary condition f2 (y) = 0 at y = 0, we getD = 0.

So the solutions (6.2.10) and (6.2.11) become

1 1( ) A sinf x k x= ...(6.2.12)

2 2( ) Csinf y k y= ...(6.2.13)


Applying the boundary condition: f1(x) = 0 at x = L1, we have

= ⇒ = π =1 1 1 1 1 1sin L 0 L 1, 2, 3....k k n n

1

11L

nk

π= ...(6.2.14)

Similarly applying the boundary condition: f2 (y) = 0 at y = L2, we have

= ⇒ = π =2 2 2 2 2 2sin L 0 L , 1, 2, 3......k k n n

2

22L

nk

π= ...(6.2.15)

From Eqn. (6.2.9), we have

2 22 2 2 21 2

1 2 2 21 2L L

n nk k k

= + = + π

or 2

2 Em =

2 221 2

2 21 2L L

n n + π

π= +

1 2

2 2 2 21 2

, 2 21 2

E2L L

n nn n

m ...(6.2.16)

This equation gives the permitted values of energy of a particle trapped in a two dimensionalinfinitely deep potential well.

The energy levels of a square potential well of width L are given by

( )1, 2

2 22 21 2 2

E2 L

n n n nm

π= + ...(6.2.17)

whence π π π= = = =

2 2 2 2 2 2

1, 1 1, 2 2, 1 2, 22 2 2

5 8E ,E E ,E

2 L 2 L 2 L

m m m

π= =

2 2

1, 3 3, 1 2

10E E , .

2 L

etc

m

The solution of the Schrödinger equation is

1 2

1 2

( , ) Asin sinL L

n x n yx y

π πψ = ...(6.2.18)


6.3 PARTICLE IN A THREE DIMENSIONAL POTENTIAL WELL

Let the three dimensional potential well of infinite depth be defined byV (x, y, z) = 0 for 0 < x < L1, 0 < y < L2, 0 < z < L3

= outside the well ...(6.3.1)

Let a particle of mass m and energy E be confined to move inside the well. The Schrodingerwave equation for the particle is

∇ ψ + ψ = ∂ ψ ∂ ψ ∂ ψ + + + ψ = ∂ ∂ ∂

22

2 2 22

2 2 2

2 E0

0

m

kx y z

...(6.3.2)

where k = 2

2 Em

...(6.3.3)

We assume the solution of Eqn. (6.3.2) of the form(x, y, z) = f1 (x) . f2 (y) . f3 (z) ...(6.3.4)

Substituting Eqn. (6.3.4) in (6.3.2), we find

22 2231 2

2 2 21 2 3

( )( ) ( )1 1 1( )

( ) ( ) ( )+ + = −

d f zd f x d f yk say

f x f y f zdx dy dz...(6.3.5)

The first term is function of x alone, the second term is function of y alone and the third termis function of z alone. Since their sum is independent of x, y, z, this is possible only if each term isseparately constant. So we can write

2

2112

1

( )1

( )= −

d f xk

f x dx...(6.3.6)

2

2222

2

( )1

( )= −

d f yk

f y dy...(6.3.7)

2

2332

3

( )1

( )= −

d f zk

f z dz...(6.3.8)

where 2 2 2 21 2 3 2

2 Emk k k k+ + = =

...(6.3.9)

The separation constants have been assumed to be negative, otherwise the boundary conditionswould not be satisfied. Solutions of Eqns. (6.3.6), (6.3.7) and (6.3.8) may be assumed of the forms


Degeneracy223, 132, 322

E = 17 1 3123, 132, 213, 231, 321, 312

E = 14 1 6222

E = 12 1 1113, 131, 311

E = 11 1 3122, 212, 221

E3 = 9 1 3112, 121, 211

E2 = 6 1 31, 1, 1

2 2

1 111 2E E 3

2 Lm

π= = = 3 1 1

Fig. 6.3.1 Allowed energy states of a particle in a cubical box

1 1 1 1 1( ) A sin B cosf x k x k x= + ...(6.3.10)

2 2 2 2 2( ) A sin B cosf y k y k y= + ...(6.3.11)

3 3 3 3 3( ) A sin B cosf z k z k z= + ...(6.3.12)

The boundary conditionsf1(x) = f2 (y) = f3 (z) = 0 at x = y = z = 0 give B1 = B2 = B3 = 0.

The boundary conditionsf1(x) = 0 at x = L1, f2(y) = 0 at y = L2, f3(z) = 0 at z = L3 give

ππ π

= = = =31 21 2 3 1 2 3

1 2 3

, , , where , , 1,2,3, ......L L L

nn nk k k n n n

From Eqn. (6.3.9)

( )= + +1 2 3

22 2 2

, , 1 2 3E2

n n n k k km

=

22 2 2 231 2

2 2 21 2 3

2L L L

nn n

m

π+ +

...(6.3.13)

The solution of Schrödinger is

31 2

1 2 3

( , , ) Asin sin sinL L L

n zn x n yx y z

ππ πψ = ...(6.3.14)


If L1 = L2 = L3 = L (i.e., the shape of the well is cubical) then the energy levels are given by

( )1 2 3

2 22 2 2

, , 1 2 3 2E

2 Ln n n n n n

m

π= + + ...(6.3.15)

and the corresponding normalized wave functions are

( )1/ 2

31 23

8, , sin sin sin

L L LL

n zn x n yx y z

ππ π ψ = (6.3.16)

The ground state energy level is

π=

2 2

1, 1, 1 2E 3

2 L

m...(6.3.17)

Next few energy levels are

π= = =

2 2

1, 1, 2 1, 2, 1 2, 1, 1 2E E E 6

2 L

m...(6.3.18)

Notice that three sets of quantum numbers (1, 1, 2), (1, 2, 1) and (2, 1, 1) i.e., three quantumstates corresponds to the same energy state. We say that this energy level is 3-fold degenerate. Thedegeneracy of other energy levels is shown in the Fig. (6.3.1).

6.4 DEGENERACY

Consider an eigen value equation

Qu qu= ...(6.4.1)

If there are n independent eigen fuctions u1, u2, ……,un belonging to the same eigen value q,then this eigen value is said to be n-fold degenerate. The linear combination of the eigen function

= c1u1 + c2u2 + …………+ cnun

is also an eigen function of the operator Q with the same eigen value q. The eigen functions u1, u2,u3,…., un are said to be linearly independent if the equation

c1u1+ c2u2 + ………….+ cnun = 0

can only be satisfied with all c’s equal to zero. This means that no member of the set of eigen functionscan be expressed as a linear combination of the remaining members. For example, the functionsu1 = 3x,, u2 = 5x2 – x, u3 = x2 are not linearly independent since u2 = 5u3 – (1/3)u1. On the otherhand the functions u1 = 1, u2 = x, u3 = x2 are linealy independent, since none of them can be writtenas a linear combination of the other two. The degree of degeneracy of an eigen value is equal to thenumber of linearly independent eigen functions corresponding to that eigen value.

The stationary state wave functions 112, 121, 211 for the particle in a cubic box aredegenerate and their linear combination is also an eigen function of the particle with the same energyeigen value.


6.5 DENSITY OF STATES

The allowed energy levels and associated quantum states for a particle confined to move in a cubicalenclosure of side L are given by

( )1 2 3

2 2 2 22 2 21 2 3E

2 2 2n n np k

k k km m m

= = = + +

= 22 22

31 2

2 L L L

nn n

m

ππ π + +

...(6.5.1)

1 2 3

31 2( , , ) const.sin sin sinL L Ln n n

n zn x n yx y z

ππ πψ = ...(6.5.2)

where n1, n2 and n3 are non-zero positive integers. The particle described by wave function haswave vector k whose components are given by

k = 31 2, , .

L L L

nn n ππ π

...(6.5.3)

We can plot the components of the wave vector k in three dimensional space with k1, k2, k3 asCartesian axes. This space is called k-space. In k-space the allowed values k form a cubical pointlattice with spacing between points being /L. Each lattice point in k-space represents a permissiblestate of the particle. These lattice points divide the k-space into cells, each of volume (/L)3. Thecontribution to the unit cell of points lying at the corners of the unit cell is unity. Each lattice point,which corresponds to a quantum states, occupies a volume (/L)3 in k-space.

We wish to find the number of quantum states with wave vectors whose magnitude lie in theinterval k and k + dk. This number is equal to the number of lattice points in k-space lying betweentwo spherical shells, centered at the origin, of radii k and k + dk in the positive octant. The volumeof the region lying between the spherical shell of radii k and k + dk in the positive octant is

).4(8

1 2dkkπ So the number of states with wave vectors whose magnitudes lie in the range k to k + dk

is

( )

π = = ππ

2128

3 2

(4 ) V( )

2/L

k dkg k dk k dk ...(6.5.4)

where V = L3 is the volume of the enclosure. The function g(k), which represents the number ofquantum states per unit energy range at energy E, is called the density of states.

Making use of the relations

2 Ep k m= =

the expression for the density of states can be written as

2

3

V( ) 4g p dp p dp

h= π ...(6.5.5)


3/ 2

3

2 V(E) E (2 ) E Eg d m d

h

π= ...(6.5.6)

Periodic Boundary Conditions

The formula for the density of states is independent of the detailed form of boundary conditionsimposed at the surface of the enclosure. We shall show this by using an alternative boundary condition,the periodic boundary condition, which is most often used. For a cubic enclosure of side L thiscondition is expressed as

(0, , ) (L, , )

( , 0, ) ( , L, )

( , , 0) ( , , L)

y z y z

x z x z

x y x y

ψ = ψψ = ψψ = ψ

The solution of Schrodinger wave equation for a particle in a box is

1 2 3( , , ) const.exp )x y z i k x k y k zψ = + +The wave vector k is now restricted to the values

1 2 32 2 2

, ,L L L

k n n nπ π π =

Notice that the ni’s now can be positive or negative integers. Now to calculate the density ofstates, instead of taking positive octant of a sphere in k-space, we take the whole sphere. The spacingof lattice points in k-space is now 2/L. The number of states whose wave vector k has magnitude inthe range k and k + dk is given by

( )

22

3

4 V( )

22 /L

k dkg k dk k dk

π= =ππ ...(6.5.7)

Fig. 6.5.1 Three dimensional k-space. Each lattice point represents a state.


6.6 SPHERICALLY SYMMETRIC POTENTIAL WELL

A spherically symmetric potential well is defined byV(r) = 0 for r < r0

= for r = r0 ...(6.6.1)

The Schrodinger equation for the particle inside the potential well is

2

2

2(E V) 0

m∇ ψ + − ψ =

We assume that the wave function depends only on radial distance r. The Schrodinger equationsimplifies to

2

2 2

1 2 E0

d d mr

dr drr

ψ + ψ = ...(6.6.2)

or 2 22

10

ψ + ψ = d d

r kdr drr

...(6.6.3)

where k = 2

2 Em

...(6.6.4)

A great convenience results if we employ the transformation

( )

( )ψ = u rr

r...(6.6.5)

In terms of new variable Eqn. (6.6.3) transforms to

22

20+ =d u

k udr

...(6.6.6)

Its solution isu = A sin (kr + )

or A

sin( )krr

ψ = + α ...(6.6.7)

where A and are constants.At r = 0, is finite. This gives = 0. Whence

A

sin krr

ψ = ...(6.6.8)

At r = r0, = 0 whencesin kr0 = 0

0

, 1,2,3,.....n

k nr

π= =


20

2 Em n

r

π=

...(6.6.9)

En =2 2 2

202

n

mr

π

The wave function of the particle is

0

A( ) sin

n rr

r r

πψ = ...(6.6.10)

The wave function can be normalized making use of the condition

ψ =∫0

2

0

( ) 1r

r dr

0 22 2

200

Asin 4 1

rn r

r drrr

π π = ∫

02

00

22 A 1 cos 1

rr

drr

ππ − = ∫

0

1A

2 r=

π ...(6.6.11)

Hence 00

1 1( ) sin

2

πψ =π

n rr

r rr...(6.6.12)

Most probable distance: The probability of finding the particle at a distance r is

2 2 2 2

20 0

1 1P( ) ( ) 4 sin 4

2

n rr dr r r dr r dr

r rr

π= ψ π = π π

The most probable distance is obtained by using the condition

2

2

P( ) P( )0, and

d r d rive

dr dr= = −

0 0 0 0

22sin cos 0

ππ π =

n rn r n

r r r r

0

2=mp

rr ...(6.6.13)


The probability of finding the particle in the region r < rmp is given by

0/2

0

P P( )r

r dr= ∫0 2/2

2 22

0 00

1 1sin 4

2

rn r

r drr rr

π = π π ∫

= 1

.2

SOLVED EXAMPLES

Ex. 1. A particle in the ground state is located in a one dimensional potential well of width L withabsolutely impenetrable walls 0 < x < L. Find probability of finding the particle in the regionL /3 < x < 2L /3 in ground state.

Sol. Ground state normalized wave function of the particle is

2

( ) sinL L

xx

πψ =

The required probability is given by

2L / 3 2L / 3 2L / 32 2

L / 3 L / 3 L / 3

2 2 2P ( ) sin 1 cos

L L L L

x xx dx dx dx

π π = ψ = = − ∫ ∫ ∫1 3

0.61.3 2

= + =π

Ex. 2. A particle is located in a two dimensional square potential well with absolutely impenetrablewalls (0 < x < L, 0 < y < L). Find the probability of finding the particle within a region 0 < x < L /3,0 < y < L /3 with lowest energy.

Sol. Wave function in the ground state is

2

sin sinL L L

yx ππψ =

Required probability

L / 3 L / 3 L / 3 L / 32 2 2

20 0 0 0

4P sin sin

L LL

yxdxdy dxdy

ππ = ψ = ∫ ∫ ∫ ∫

L / 3 L / 3

20 0

21 21 cos 1 cos

L LL

yxdx dy

ππ = − − ∫ ∫= 0.32.


Ex 3. The wave function of a particle in one dimensional box of length L is given by

πϕ = .n

2 n xsin

L LFind the expectation value of x and x2.

Sol.

L

0

ˆx x dx= ψ ∗ ψ∫ =

L2

0

2sin

L L

n xx dx

π∫

= L

2

2 2 2

0

2sin

L L

Ln x

x x dxπ= ∫

= 2 2

2 2

L L.

3 2n−

π

Ex. 4. Show that the wave function πψ =n

n xA sin

L of a particle moving in one dimensional

potential well of width L is not eigen function of px . What can be said about the function = A exp(± ikx )?

Sol. ˆ A sinLx n

n xp i

x

∂ πψ = −∂

= A cosL L

in n xπ π−

So the function A sin nx/L is not an eigen function of px-operator.

ˆ A exp( ) [A exp( )]x np i ikx k ikx

x

∂ψ = − ± = ± ±∂

So the function A exp (±ikx) is an eigen function of px-operator with eigen value k.

Ex. 5. Show that the wave functions

πψ =n

2 n xsin

L Lare orthogonal.

Sol. L L

0 0

2sin sin

L L Lm nm x n x

dx dxπ πψ ψ =∫ ∫

=

L

0

1 ( ) ( )cos cos

L L L

n m x n m xdx

− π + π − ∫= 0

Therefore the functions are orthogonal.



1. A one dimensional potential barrier of height V extends from x = 0 to x = . A particle possessing kinetic energyE is incident from left on the potential barrier (potential step). Analyze the problem quantum mechanically for thecases (i) E < V (ii) E > V.

2. What do you mean by tunnel effect? Calculate the transmission probability of a particle incident on the potentialbarrier for case in which kinetic energy of the particle is less than the height of the barrier. Discuss the effect ofheight and width of the barrier on the transmission probability.

3. Set up Schrodinger equation for a particle is trapped in an infinitely deep potential well of width L and obtain thewave functions and energy levels of the particle.

(a) Discuss the effect of width of the well on energy eigen values of the particle.

(b) Compare the classical and quantum mechanical probability of finding the particle in the well at different

energy states.

4. A particle is confined to move in one dimensional box with perfectly rigid a walls at x = 0 and x = L. Analyze theproblem quantum mechanically. Find the quantum mechanical probability of finding the particle at x = L/4, L/3,L/2, 2L/3.

5. Give quantum mechanical treatment of motion of a particle confined to move in a two dimensional potential well.Explain the term degeneracy in this context.

6. A particle of mass m is constrained to in a box of sides L1, L2 and L3. Set up Schrodinger equation for theparticle and solve for eigen values and eigen functions. If the box is cube of side L, find the degree of degeneracyof the second, third, fourth and sixth energy levels.

7. A particle is confined to move in a cubical box of side L. Find the eigen functions and eigen values. Discuss thedegeneracy of eigen functions choosing a suitable eigen value. (All’d 1997)

8. Find the eigen values and eigen functions of a particle moving in a one dimensional square well potential withinfinite high walls. Can the particle in such a well ever have zero total energy? If not then explain clearly, whynot? (All’d 1998)

9. Find the eigen values and eigen functions of a particle moving in a one dimensional square well potential withinfinitely high walls. (All’d 2004)

10. What is meant by free and bound states of a quantum mechanical system? (All’d 2001)

11. A particle of mass m is confined to a one dimensional box of length L. Obtain expressions for the wave functionsand allowed energies as a function of the quantum number n. Generalize the results to a two dimensional box andfind energy of the ground state. What is its degeneracy? (All’d2000)

12. A particle of mass m is restricted to move in a rectangular potential box given by

V = 0 if 0 < x < a, 0 < y < b, 0 < z < c= elsewhere.

Where a, b, c are constants. Solve the Schrodinger equation for the particle and find expression for its wavefunctions and energies.

13. Show that the expression for density of states for a particle of mass m confined to move in volume V is

3/ 2 1/ 2

3

2 V( ) (2 )g m

h

πε = ε .

CHAPTER

HARMONIC OSCILLATOR

1.17.1 INTRODUCTION

Classical Treatment: A particle attached to a fixed point, say x = 0, with a force, which is proportionalto its displacement from the mean position and is directed towards the fixed point, constitutes aharmonic oscillator. Its equation of motion is

2

0

0,

mx kx

kx x

m

+ =

+ ω = ω =

...(7.1.1)

where m is mass of the particle, k is force constant and x is displacement, is called the classicalangular frequency of the oscillator. The solution of Eq. (7.1.1) is

x = A cos (t + ) ...(7.1.2)

A is amplitude and is initial phase of the particle. The potential energy of the oscillator is

V(x) = 2 21

2m xω ...(7.1.3)

and the total energy is given by

E2 21A

2m= ω ...(7.1.4)

The oscillator can be made to oscillate with any desired amplitude A and therefore the energyE may assume any value depending on the amplitude. In other words the energy is a continuousvariable. A graph showing the variation of potential energy with displacement is shown in theFig. (7.1.1). Corresponding to amplitudes A1 and A2, the energies of oscillator are 1/2 kA1

2 and1/2 kA2

2 respectively. By adjusting the amplitude between A1 and A2, the oscillator can be made tooscillate with any energy between E1 = (1/2) m2A1

2 and E2 = (1/2) m2A22. Thus there are infinite

energy levels between E1 and E2. This is what we mean by the statement that energy is a continuousvariable.

The velocity of the particle in executing simple harmonic oscillation is zero at the turning pointsx = A and x = –A and is maximum at the equilibrium point x = 0. This means that the particlespends maximum time at the turning points and minimum time at the equilibrium point. The classical


probability of finding the particle is maximum at the turning point and is minimum at the equilibriumpoint. This fact is displayed in the Fig. (7.1.2). Classical physics does not allow the particle to gobeyond the turning points.

Fig. 7.1.1 V(x) vs x graph. Fig. 7.1.2 Classical probability

Quantum Mechanical Treatment: The Schrodinger equation for the harmonic oscillator is an eigenvalue equation

H Eψ = ψ

2

2+V

2

d

m dx

ψ− ψ= Eψ where, V = 2 21

2m xω

2

2 22 2

2 1E 0

2

d mm x

dx

ψ + − ω ψ = ...(7.1.5)

We shall transform this equation into a convenient form by introducing a new independentvariable defined by

= x ...(7.1.6)

The parameter will be chosen in such a way that the new equation looks simple.

Nowd d d d

dx d dx d

ψ ψ ξ ψ= = αξ ξ

2 2

22 2

d d d d d d d

dx d d d dxdx d

ψ ψ ψ ξ ψ= α = α = α ξ ξ ξ ξ In terms of new variable Eqn. (7.1.5) becomes

2 2 2 2

2 2 2 2 4

20

ψ ω ξ+ − ψ = ξ α α

d mE m

d...(7.1.7)

Let us choose

2 ωα =

m . ...(7.1.8)

Harmonic Oscillator 207

Equation (7.1.7) now becomes

2

22

2E0

d

d

ψ + − ξ ψ = ω ξ ...(7.1.9)

Introducing the dimensionless parameter defined by

2Eβ =ω ...(7.1.10)

Eqn. (7.1.9) becomes

( )2

22

0ψ + β − ξ ψ =ξ

d

d...(7.1.11)

Asymptotic solution () : The wave function () must satisfy the condition(±) = 0

In the limit , may be neglected. Eqn. (7.1.11) assumes the form

2

22

0ψ − ξ ψ =ξ

d

d...(7.1.12)

The solution to this equation is

21exp

2 ψ = ± ξ

We omit the positive sign because it does not satisfy the condition (± ) 0. Hence theasymptotic solution of Eqn. (7.1.11) has the form

21exp

2 ψ = − ξ

...(7.1.13)

Let us verify that Eqn. (7.1.13) satisfies (7.1.12). From Eqn. (7.1.13)

( )2

2 22

, 1 for .d d d

d dd

ψ ψ ψ= −ξψ = − ψ + ξ = ξ − ψ = ξ ψ ξ→∞ ξ ξξ

This ensures that Eqn. (7.1.13) is an asymptotic solution of Eqn. (7.1.11). We may now assumethat the solution of Eqn. (7.1.11) is of the form

21

( ) H( ) exp2

ψ ξ = ξ − ξ ...(7.1.14)

where, H() is unknown function to be determined. Substituting Eqn. (7.1.14) in (7.1.11) we obtain

2

2

H( ) H( )2 ( 1)H( ) 0

d d

dd

ξ ξ− ξ + β − ξ =ξξ

...(7.1.15)


The unknown function H() obeying the differential Eqn. (7.1.15) is known as Hermite function.Solution of Eqn. (7.1.15) is obtained in the form of power series. We assume the solution of the form

2

0 1 20

H( ) .... ...n nn n

n

a a a a a∞

=ξ = ξ = + ξ + ξ + + ξ +∑ ...(7.1.16)

Now,

1 1

1 20

H0 2 ... ....n n

n nn

dna a a na

d

∞− −

== ξ = + + ξ + + ξ +

ξ ∑

0

H2 2 n

nn

dna

d

∞

=ξ = ξ

ξ ∑ ...(7.1.17)

2

2 222

0

H( 1) 0 0 2 ... ( 1) ...n n

n nn

dn n a a n n a

d

∞− −

== − ξ = + + + + − ξ +

ξ∑ ...(7.1.18)

20

( 2)( 1)∞

+=

= + + ξ∑ nn

n

n n a

Substituting Eqns. (7.1.16), (7.1.17), (7.1.18) in (7.1.15), we have

20

( 1)( 2) ( 1 2 ) 0∞

+=

+ + + β − − ξ = ∑ nn n

n

n n a n a ...(7.1.19)

Equation (7.1.19) holds for all values of . Hence the coefficient of each power of mustvanish separately. Hence

22 1

( 1)( 2)++ −β=

+ +n nn

a an n

...(7.1.20)

This equation is called the recurrence formula for the coefficients an. Since the recurrenceformula determines the coefficients an+2 in terms of an, the power series (7.1.16) contains only witheven or only odd powers of

For n = 0, 2, 4, 6, ….

2 0 4 0 6 01 (1 )(5 ) (1 )(5 )(9 )

, ,2! 4! 6!

a a a a a a− β − β − β − β − β − β= = =

In this way all the even coefficients are expressed in terms of a0.For n = 1, 3, 5,…

3 1 5 1 7 13 (3 )(7 ) (3 )(7 )(11 )

, ,3! 5! 7!

a a a a a a− β − β − β − β − β − β= = =


In this way all the odd coefficients are expressed in terms of a1. Thus the power series of Eqn.(7.1.16) contains only two arbitrary constants a0 and a1. Since even coefficients (a2, a4, a6,…) arerelated to a0 and odd coefficients (a3, a5, a7,…) to a1, we can split the solution (7.1.16) in even andodd series as follows:

2 4

06

1 (1 )(5 )1

2! 4!H( )

(1 )(5 )(9 )...

6!

a

−β −β −β + ξ + ξ + ξ = − β −β −β ξ +

+

3 5

17

3 (3 )(7 )

3! 5!

(3 )(7 )(11 )....

7!

−β −β −β ξ + ξ + ξ + − β −β −β ξ +

a ...(7.1.21)

or H() = a0 (even series) + a1 (odd series) ...(7.1.22)

Let us see whether the solution () expressed in the form 21( ) H( ) exp( )

2ψ ξ = ξ − ξ qualifies to

be a physically acceptable solution or not. Any acceptable solution must vanish at infinity i.e., 0 as . Let us examine the asymptotic behavior of H() and exp(2). A suitable way tocompare H() and exp(2) is to express them in power series. H() has already been found in powerseries. So

( )2 4 6 2

2exp( ) 1 .... ..........1! 2! 3! ( 2) / 2!/ 2 !

n n

nn

+ξ ξ ξ ξ ξξ = + + + + + + + ∞+

0, 2, 4 0, 2, 4

1 2;

( / 2)!!

2

n nn n

n n

b bn n= =

= ξ = ξ =

∑ ∑...(7.1.23)

The ratio of successive coefficients in this series is

2

!2 22

21 !

2

nn

n

nb

nb n n+

→∞

= = →

+ + ...(7.1.24)

The ratio of successive coefficients in H() is

2 2 1

( 1)( 2)+ + −β=

+ +n

n

a n

a n n

2→∞

→n n

...(7.1.25)

Thus for large values of n, H() behaves like exp(2) and the solution () becomes

() = H () exp 2 2 2 21 1 1exp( ) exp exp

2 2 2 − ξ ≈ ξ − ξ = ξ

Obviously for , does not remain finite and hence it cannot be an acceptable solutionso long as H() is of the form expressed by Eqn. (7.1.22). There is a simple way out of this dilemma.

If all the coefficients ,sna beyond the certain value of n vanish in the series representing H() then

0 as because of the term exp 21

2 − ξ

. In other words if H() terminates as polynomial


with finite number of terms instead of infinite series then it is acceptable. From the recursion formula

22 1

( 1)( 2)++ −β=

+ +n nn

a an n

...(7.1.26)

it is evident that our requirement is met if = 2n +1 for some value of n then an + 2 = an + 4 = an + 6

=….= 0. This restriction on implies that

2E2 1

1E

2n

n

n

β = = +ω

= + ω

...(7.1.27)

Thus the harmonic oscillator can have only a discrete set of energies given by Eqn. (7.1.27). Itis remarkable to observe that the lowest energy state corresponds to n = 0 and has energy 1/2 h,called zero-point energy. The restriction = 2n + 1 takes care of only one sequence of coefficients,either the sequence of even n starting with a0 or the sequence of odd n starting with a1. If n is even,only even powers of appear in the polynomial while if n is odd, only odd powers of appear. Ifthe restriction = 2n + 1 is satisfied, only one of the series (either even series or odd series) terminatesas a polynomial and the other remains as infinite series and H() can be written as

H() = a0 (polynomial) + a1 (infinite series) ...(7.1.28)

or H() = a0 (infinite series) + a1 (polynomial) ...(7.1.29)

If H() represented by Eqn. (7.1.28) is to be an acceptable function a1 must be chosen equalto zero. Similarly H() expressed by Eqn. (7.1.29) will be acceptable solution if a0 = 0. So theacceptable forms of H() are

2 4

01 ( 1)( 5)

H( ) 1 ........2! 4!

a β − β − β −ξ = − ξ + ξ +

...(7.1.30)

or3

13

H( ) ....................3!

a β −ξ = ξ − ξ +

...(7.1.31)

For example if we put n = 4, = 9 the even series becomes polynomial

H() = 2 40

41 4

3 − ξ + ξ

a

and for n = 3, = 7 the odd series becomes a polynomial

H() = 31

2

3 ξ − ξ

a

It is customary to choose the arbitrary constants a0 or a1 such that the coefficient of the highest

power of in the polynomial is 2n or ( 1) / 22 β− . The resulting polynomials are called Hermitepolynomials. (For example, if = 9 or n = 4 the highest power of is 4, therefore we choose

40

( 1)( 5)2 or

4!a

β − β − = 0 04

16 12.3

= =a or a Then the Hermite polynomial becomes

H4() = 164 – 482 + 12


Similarly for = 7 or n = 3 we choose 31 1

2( ) 2 or 12

3a a− = = − then

H3 () = 8 3 – 12

Some Hermite polynomials are tabulated below.H0() = 1

H1() = 2

H2() = 42 – 2

H3() = 83 – 12

H4() = 164 – 48 2 + 12

H5() = 325 – 1603 + 120

Higher order Hermite polynomials can be determined from the recurrence relationHn + 1() = 2 Hn() – 2 nHn – 1 (), n 1 ...(7.1.32)

The Hermite polynomials are defined by Rodrigues formula

ξ −ξξ = −

ξ

2 2H ( ) ( 1) ( )

nn

n n

de e

d...(7.1.33)

Putting n = 0, 1, 2, …. We can find H0(), H1(), H2()....etc.The wave functions (x) of the harmonic oscillator are

2 2 21 1

2 2( ) N H ( ) N H ( )x

n n n n nx e e x− ξ − αψ = ξ = α ...(7.1.34)

ωξ = α =

mx x

The multiplicative constant Nn can be determined using the normalization condition

2

( ) 1n x dx∞

−∞

ψ =∫

2 2 21

N exp( )H ( ) 1n n d∞

−∞

−ξ ξ ξ =α∫

2N2 ( !) 1nn n⋅ π =

α

2exp( )H ( )H ( ) 2 ( !)nm n mnd n

∞

−∞

−ξ ξ ξ ξ = π δ ∫

This gives

2 !

α=πn n

Nn

=

1/ 4

/ 2 1/ 2

1 1

2 ( !)n

m

n

ω ⋅ π ...(7.1.35)

Some of the normalized wave functions of harmonic oscillator are given below.

( )2 21

20

− ααψ =π

xx e


=1/ 4

21exp

2

ω ω − π

m mx (ground state)

( ) ( )2 21

21 2

2

− ααψ = απ

xx x e

( ) ( ) 2 2122 2

2 4 28

− ααψ = α −π

xx x e

( ) ( ) 2 2123 3

3 8 1248

− ααψ = α − απ

xx x x e

( ) ( )

2 212H

2 !

xn nn

x en

− ααψ = ξπ

where 2,ω ωξ = α = α =

m mx x

Fig. 7.1.3 Wave functions of harmonic oscillator


Fig. 7.1.4 Ground state classical and quantum mechanical probability

Fig. 7.1.5 Quantum mechanical probability of oscillator in state 10. As n becomes very large (n ) thequantum mechanical probability becomes identical with the classical probability

Probability of finding the harmonic oscillator within classical limits: The ground state wavefunction of harmonic oscillator is

21

21/ 4

0 ( )

mxm

x e

ω − ω ψ = π

In the ground state the amplitude A of the oscillator is given by

( )2 21 1 1say

2 2m A A

mω = ω ⇒ = ± =

ω λ

The probability of finding the oscillator within the classical limits is

P0 = 2 2

0 0

0

( ) 2 ( )A A

A

x dx x dx−

ψ = ψ∫ ∫


= 2 2

1/

0

2λ

−λ λ

π ∫ xe dx

= 2

1

0

2 −

π ∫ze dz where x = z

=

1 2 4 6

0

21 ......

1! 2! 3!

z z zdz

− + − + π

∫

=

13 5 7

0

2......

3 10 42

z z zz

− + − + π

= 2 1 1 1

1 .........3 10 42

− + − + π

= 0.83.

The ground state probability of finding the oscillator outside the classical limits is 17%.Correspondence Principle: The position and velocity of oscillator at time t are given by

x = A cos t

= – A sin t =

1/ 22

2A 1

A

x ω −

Let t be the time spent by oscillator in traversing a distance x. The classical probability Pc offinding the oscillator within the region x is defined as the fraction of time that the oscillator spendswithin this region. Thus

2

2

/ 1P ( ) .

T 2 / 2 A1

A

ct x x

x dxx

∆ ∆ ν ∆= = =π ω π

−

At x = ± A, Pc . The classical probability Pc is minimum at x = 0. Between the classical turningpoints Pc has non-zero value. For n = 0 (ground state) the quantum mechanical probability P offinding the oscillator between classical turning points differs considerably from the classical probability.For n > 0, quantum mechanical probability Pquantum shows peaks between the points x = ± A. Withincreasing n, the number of peaks of quantum mechanical probability increases and hence they becomecrowded. In the limit of large quantum numbers (n ) the peaks of quantum mechanical probabilitymerge together and Pquantum approaches the classical probability. This is the Bohr correspondenceprinciple.



1. Give quantum mechanical treatment of one dimensional harmonic oscillator. Obtain the energy levels and wavefunctions of the oscillator.

2. Set up Schrodinger equation for harmonic oscillator. Write down the expressions for the energy eigen values andwave functions. Sketch first four eigen functions and the corresponding probability of finding the particle. Comparethe classical and quantum mechanical probability. What happens when the quantum numbers become very large?

3. Write down Schrodinger equation for harmonic oscillator and the ground state wave function. Find the wavemechanical probability of finding the particle in the ground state within the classical limits.

4. The one dimensional motion of a particle of mass m is described by the following equation

ψ− + ω ψ = ψ2 2

2 2 22

1E .

2 2

dm x

m dx

All symbols have their usual meanings. Find the energy eigen values of the particle. (All’d 1995)

5. (a) The eigen function and the energy of the nth quantum state of a one dimensional harmonic oscillator are given

by

1/ 2

2 21 1( ) H ( )exp , E

2 22 !n n nn

x x x nn

α ψ = α − α = + ω π

where 2 2 2and H ( ) ( 1) exp( ) exp( ).

nn

n n

m d

d

ωα = ξ = − ξ −ξξ

Sketch the wave function and the probability density for n = 2 state and discuss how the quantum behaviouris different from the classical one.

(b) Generalize the above results to obtain the wave function and energy eigen values of a two dimensionalharmonic oscillator. (All’d 1996)

6. The one dimensional motion of a particle of mass m is described by the Hamiltonian

2 2

21 22

H2

dc x c x

m dx= − + +

where c1 and c2 are constants. Find the eigen values of the particle. Do not use perturbation theory to solve theproblem. (All’d 1998)

7. A linear harmonic oscillator in its nth quantum state is characterized by a wave function

2C exp( /2) ( ), where H ( ) is Hermite polynomial andn n n nH

mx

ψ = −ξ ξ ξ

ωξ =

Find 2x and the expectation value of the potential energy if the oscillator is in the first energy level.

Given: 2 1.3.5.........(2 1)exp( )

2n

n

nd

∞

−∞

−ξ −ξ ξ = π∫ . (All’d 1999, 2004)


8. Show that the wave functions for a linear harmonic oscillator have definite parity. Explain the origin of zero pointenergy in a quantum oscillator. (All’d 2000)

9. The wave function for a one dimensional harmonic oscillator is expressed as

1 / 4

2 212 2

( ) exp( )H ( ), where2 ( !)

n nn

mkx x x

n

α ψ = − α α α = π

(a) Sketch the wave function and the probability density function for n = 1 state as a function of x .

(b) How is the behaviour of a quantum oscillator different than that of a classical oscillator.

(c) Write down the energy and wave function for a two dimensional harmonic oscillator by generalizing theresults of a one dimensional harmonic oscillator. What is the degeneracy of the first excited state of a twodimensional oscillator. (All’d 2001)

10. What is expectation value? Find expectation value of the potential energy in the ground state of a linearharmonic oscillator. (All’d 2002)

11. For a linear harmonic oscillator, solve the Schrodinger equation showing clearly necessary steps to obtain firstthree eigen functions and eigen values of the oscillator. Obtain normalization constant for the eigen functions.

(All’d 2002)

12. (a) Obtain the wave equation for a linear harmonic oscillator. Solve the equation to find eigen functions and the eigen values.

(b) The generating function for Hermite polynomial is

2 2 2exp(2 ) or exp[ ( ) ]zx z x z x− − − …………

Use this to evaluate the normalization constant. (All’d 2003)

13. Sketch the wave function and the probability density as a function of x for the ground state and first excited stateof a linear harmonic oscillator and answer the following:

(i) What is the parity of the wave function in each case?

(ii) In what ways is the quantum behaviour different from the classical behaviour of the oscillator?

(iii) What is the origin of zero point energy in a quantum oscillator?

(All’d 2005)

14. (a) The wave function for a one dimensional harmonic oscillator is given by

2

2 2

( ) N exp( / 2)H ( ), , ,

H ( ) ( 1) exp( ) exp( )

n n n

nn

n n

mx x x x

d

d

ωψ = −λ λ λ = ξ = λ

ξ = − ξ −ξξ

Sketch the wave functions and the probability densities for the first three states. What is the parity of eachstate?

(b) Generalize the result of one dimensional harmonic oscillator to express the energy eigen values and eigenfunctions of a two dimensional isotropic harmonic oscillator. What is the degeneracy of the first excitedstate? (All’d 2006)


15. (a) The generating function for a Hermite polynomial is

∞

=

ξξ = ξ − − ξ =∑2 2

0

H ( )S( , ) exp[ ( ) ]

!nn

n

s s sn

Show that ′′ ′ξ − ξ + ξ =H ( ) 2 H 2 H ( ) 0.n n nn

(b) Discuss the origin of zero point energy in a harmonic oscillator. (All’d 2007)

16. The ground state wave function of a linear harmonic oscillator of mass m is

2 210 2( ) Aexp( )x xψ = − α

Where A is normalization constant and mωα =

. Calculate the expectation value of 2 21

2V m x= ω for this

state.

CHAPTER

RIGID ROTATOR

8.1 INTRODUCTION

A rigid rotator is a system of two particles always remaining at fixed separation and capable of rotatingabout an axis passing through their center of mass and perpendicular to the line joining them. If theplane, containing the particles, is fixed in space i.e., the orientation of axis of rotation remains fixedthen the system is said to be a rigid rotator with fixed axis. If the axis of rotation is free to take anyposition in space, it is called rigid rotator with free axis. A rigid diatomic molecule can be treated asa rigid rotator with free axis. In this section we shall set up Schrodinger equation for rigid rotatorfind the eigen values and eigen functions of the equation.

Let m1 and m2 be the masses of the particles, r1 and r2 be their distances from the center ofmass and r their separation. From the definition of center of mass

m1 r1 = m2 r2 ...(8.1.1)

r1 + r2 = r ...(8.1.2)

From these equations we can find r1 and r2 in terms of r. Thus

2 11 2

1 2 1 2

,= =+ +m m

r r r rm m m m

...(8.1.3)

The moment of inertia of the rigid rotator about an axis passing through the center of mass andperpendicular to the line joining the particles is

2 2 2 21 21 1 2 2

1 2

m mI m r m r r r

m m= + = = µ

+ ....(8.1.4)

where = 1 2

1 2+m m

m m is called reduced mass of the system.

The kinetic energy of particle of mass m moving in space in Cartesian coordinates is given by

T = ( )2 2 2 21 1

2 2mv m x y z= + + ....(8.1.5)

Rigid Rotator 219

In spherical polar coordinates (r, ) the expression for kinetic energy is

( )2 2 2 2 2 21T sin

2m r r r= + θ + ϕ θ ...(8.1.6)

If r is fixed (i.e., the particle is moving on the surface of a sphere) then 0=r and the expressionfor kinetic energy simplifies to

2 2 2 21

T ( sin )2

mr= θ + ϕ θ ...(8.1.7)

The kinetic energy of rotation of a rigid rotator is equal to the sum of the kinetic energies ofthe constituent particles.

2 2 2 2 2 2 2 21 2 1 1 2 2

1 1T T T ( sin ) ( sin )

2 2m r m r= + = θ + ϕ θ + θ + ϕ θ ...(8.1.8)

Fig. 8.1.1 Polar coordinates of particles constituting rigid rotator

For a rigid rotator moving in free space, potential energy is zero, hence the total energy of therotator is

2 2 2 2 2

1 1 2 21

E T ( )( sin )2

m r m r= = + θ + ϕ θ

2 2 2 2 2

1 1 2 21

E I( sin ),2

I m r m r= θ + ϕ θ = + ...(8.1.9)

Comparison of Eqns. (8.1.8) and (8.1.9) shows that a rigid rotator behaves like a single particleof mass I moving on the surface of a sphere of fixed radius, equal to unity.

The Schrodinger equation for a particle, in polar coordinates, is expressed as

2

22 2 2 2 2 2

1 1 1 2sin (E V) 0

sin sin

mr

r rr r r

∂ ∂ψ ∂ ∂ψ ∂ ψ + θ + + − = ∂ ∂ ∂θ ∂θθ θ ∂ϕ

To write the equation for rigid rotator, we must replace m by I and put r = 1 and V = 0 inabove equation. Doing so, we obtain


2

2 2 2

1 1 2IEsin

sin sin

∂ ∂ψ ∂ ψ θ + + ψ θ ∂θ ∂θ θ ∂ϕ

= 0 ...(8.1.10)

Equation (8.1.10) can also be obtained as follows. The Hamiltonian of a rigid rotator, in absenceof potential field, is

2L

H T2I

= = , L = angular momentum ...(8.1.11)

The corresponding operator is

2 2 2

2 2

L 1 1H sin

2I 2I sin sin

∂ ∂ ∂ = = − θ + θ ∂θ ∂θ θ ∂ϕ

...(8.1.12)

The energy eigen value equation for rigid rotator is

H Eψ = ψ

or2 2

2 2

1 1sin E

2 sin sinI

∂ ∂ ∂ − θ + ψ = ψ θ ∂θ ∂θ θ ∂ϕ

...(8.1.13)

which is the same as Eqn. (8.1.10).We assume the solution of Eqn. (8.1.13) to be of the form

() = ( ) ( )Θ θ Φ ϕ ...(8.1.14)

Substituting Eqn. (8.1.14) in (8.1.13), we get

2

22

sin 1sin sin 0

θ Θ Φ θ + β θ + = Θ θ θ Φ ϕ d d d

d d d...(8.1.15)

where 2

2IEβ =

(8.1.16)

Transposing the dependent terms on the right hand side in Eqn. (8.1.15), we get

2

22

sin 1sin sin

θ Θ Φ θ + β θ = − Θ θ θ Φ ϕ d d d

d d d...(8.1.17)

The left hand side of Eqn. (8.1.17) depends on alone whereas right hand side on alone andboth the sides remain equal for all values of independent variables and ; this can happen onlywhen each side is equal to the same constant, say m2. So the Eqn. (8.1.17) separates into two equations:

2

2

1sin 0

sin sin

Θ θ + β − Θ = θ θ θ θ

d d m

d d...(8.1.18)

Rigid Rotator 221

2

22

0d

md

Φ + Φ =ϕ ...(8.1.19)

Solution of Eqn. (8.1.19) is of the form

C ime ϕΦ = ...(8.1.20)

Since 2( ) ( 2 ), we have 1 0, 1, 2, 3,......ime m± πΦ ϕ = Φ ϕ + π = ⇒ = ± ± ±The constant C in Eqn. (8.1.20) can be obtained making use of normalization condition

2

0

1π

∗ΦΦ ϕ =∫ d ...(8.1.21)

Thus 2

2 2

0

C .C 0im ime e dπ

π ϕ − π ϕ ϕ =∫

1

C2

=π

So the solution of Φ equation can be written as

1

, 0, 1, 2,.....2

ime mϕΦ = = ± ±π ...(8.1.22)

Now let us return to Eqn. (8.1.18). This equation can be transformed into a convenient formby change of independent variable to x as follows:

2cos , sin 1= θ θ = −x x

sin sinΘ Θ Θ= = − θ ⇒ = − θθ θ θ

d dx d d d d

d d dx dx d dx

2 2sin sin (1 )Θ Θ Θθ = − θ = − −θ

d d dx

d dx dx

Making use of these results, Eqn. (8.1.18) becomes

2

22

(1 ) 0, 1 11

Θ − + β − Θ = − < < −

d d mx x

dx dx x ...(8.1.23)

Equation (8.1.23) is similar to the famous associated Legendre’s equation:

2

22

F(1 ) ( 1) F 0

1

d d mx l l

dx dx x

− + + − = −

...(8.1.24)


Writing the constant appearing in Eqn. (8.1.23) as = l (l + 1) where l is another constant,we have

2

22

(1 ) ( 1) 01

d d mx l l

dx dx x

Θ − + + − Θ = −

...(8.1.25)

= l (l + 1) = (2IE)/ 2 ...(8.1.26)

Equation (8.1.25) has single-valued and finite solutions only for certain values of parameter given by

= l (l + 1) = 0, 2, 6, 12, 20,….. ...(8.1.27)

or l = 0, 1, 2, 3, 4,…… ...(8.1.28)

Substituting the value in Eqn. (8.1.27), we find

2( 1)

E , 0, 1, 2, 3,.......2I

l ll

+= = (8.1.29)

This gives the possible values of energy that a rigid rotator can have. Thus the energy of rotatorforms a discrete spectrum. It is customary to write the energy of a rigid rotator in the form

El = Bch l (l +1) ...(8.1.30)

where B = h/82Ic. The separation of adjacent energy levels increases linearly with l.El – El–1 = Bch [l (l+1) – l (l –1)] = 2Bchl

Wave functions of rigid rotator: For m = 0, the associated Legendre equation assumes the form

2(1 ) ( 1)d d

x l ldx dx

Θ − + + Θ

= 0 ...(8.1.31)

Acceptable solutions of Eqn. (8.1.30) are expressed in the form of polynomials, known asLegendre polynomials, which are represented by Pl (x) and defined by

21

P ( ) ( 1)2 !

ll

l l l

dx x

l dx= − ...(8.1.32)

It is a simple matter to obtain Legendre polynomials from Eqn. (8.1.32). Some of them aregiven below:

l = 0, P0 (x) = 1

l = 1, P1(x) = x

2

21

2, P ( ) (3 1)2

l x x= = −

( )33

13, P ( ) 5 3

2l x x x= = −

( )4 24

14, P ( ) 35 30 3

8l x x x= = − +

Rigid Rotator 223

Higher order polynomials can be obtained from the following recurrence formula:

1 12 1

P ( ) P ( ) P ( )1 1l l l

l lx x x x

l l+ −+= −+ + ...(8.1.33)

Legendre polynomials form an orthogonal system in the interval 1 1− ≤ ≤x i.e.,

1

1

2P ( )P ( )

2 1l l llx x dxl

∗′ ′

−

= δ+∫ ...(8.1.34)

The square of the norm of Legendre polynomials has the following value:

1

1

2P ( )P ( )

2 1l lx x dxl

∗

−

=+∫ ...(8.1.35)

The finite solutions of Eqn. (8.1.24) in the interval [–1, +1], for the positive value of m, are

the associated Legendre polynomials, which are denoted as P ( )ml x and defined by

( ) /22P ( ) 1 P ( )mmm

l lm

dx x x

dx= − , m l≤

( )2 / 2 21(1 ) . . 1

2 ( !)

m l lmm l l

d dx x

dx l dx

= − −

...(8.1.36)

The associated Legendre polynomials satisfy the condition

( )( )

1

1

!2P ( )P ( )

2 1 !m ml k lk

l mx x dx

l l m−

+= δ

+ −∫ ...(8.1.37)

The recurrence relation for the associated Legendre polynomials is:

( )( )

1 1

1P ( ) P ( ) P ( )

2 1 2 1m m mk l l

l m l m l mx x x x

l l+ −− − + += +

+ +...(8.1.38)

and ( )( ) ( )( )2 1

1 1

1 11 P ( ) P ( ) P ( )

2 1 2 1m m ml l l

l m l m l m l mx x x x

l l+

+ −− − + + + +

− = −+ +

…(8.1.39)

The solutions of Eqn. (8.1.25) can be written as

( ) N P ( )m ml lm lx xΘ = ...(8.1.40)

where Nlm is normalization constant and can found making use of condition in Eqn. (8.1.37). Thenormalized solution of Eqn. (8.1.25) is

( )lm xΘ =2 1 ( )!

. P ( )2 ( )!

ml

l l mx

l m

+ −+


=2 1 ( )!

.2 ( )!

l l m

l m

+ −+

2 / 2(1 ) P ( )m

mlm

dx x

dx−

= (–1)m ( )2 / 2 22 1 ( )! 1(1 ) 1)

2 ( )! 2 ( !)

m lm l

m l l

l l m d dx x

l m dx l dx

+ −⋅ − ⋅ ⋅ − +

...(8.1.41)

The inclusion of phase factor (–1)m is a matter of convention. Finally, the complete solution of-Eqn.(8.1.13) can be written as

1

( , ) ( ) ( )2

ψ θ ϕ = Θ θ Φ ϕ =π

2 1 ( )!

2 ( )!

l l m

l m

+ −⋅+

ϕime P (cos )ml θ

=2 1 ( )!

4 ( )!

l l m

l m

+ −⋅π +

ϕime sin θm P (cos )(cos )

m

lm

d

dθ

θ ...(8.1.42)

For negative value of m, the solution of Schrodinger equation for rigid rotator is obtained fromthe formula

*

, ,( , ) ( 1) ( , )ml m l m− ψ θ ϕ = − ψ θ ϕ ...(8.1.43)

The functions () defined by Eqn.(8.1.42) are called spherical harmonics and are usuallydenoted by Ylm (). These functions are the solutions of Eqn. (8.1.13). In fact the eigen functionsof Hamiltonian operator for a particle moving on a sphere or of rigid rotator always come out to bethe spherical harmonics.

The physical interpretation of () is that |()|2 d, where d = sin d d, representsthe probability of finding the axis of the rotator pointing in the solid angle element d = sin d d

about the direction . A rigid rotator is a simple model of rigid diatomic molecule.


1. What do you mean by rigid rotator? Set up Schrodinger wave equation for rigid rotator. Obtain its wavefunction and energy levels.

CHAPTER

PARTICLE IN A CENTRAL FORCE FIELD

9.1 REDUCTION OF TWO-BODY PROBLEM IN TWO EQUIVALENT ONE-BODY PROBLEM IN A CENTRAL FORCE

The potential energy of a particle in a centrally symmetric field depends only on the distance fromthe particle to the center of the force i.e., V = V(r). The Hamiltonian operator of the particle has theform

( )2

2H V2

rm

= − ∇ +

...(9.1.1)

Consider a system consisting of two particles moving under their mutual interaction, which isdescribed by a spherically symmetric potential. Owing to the central symmetry of the force field, itis expedient to solve the problem in spherical polar coordinates. The well-known two-body problemin central force is hydrogen atom in which electron and nucleus move under their mutual interaction.In classical mechanics, the energy of a system consisting of two interacting particles is

H2 21 2

1 2

V( )2 2

p pr

m m= + + ...(9.1.2)

and the Hamiltonian operator is

2 2

2 21 2

1 2

H V( )2 2

rm m

= − ∇ − ∇ +

where m1 and m2 are the masses, p1 and p2 are the momenta of the particles. The Schrodinger equationof the system is

Hψ = Eψ

2 2

2 21 2

1 2

V( ) E2 2

rm m

− ∇ − ∇ + ψ = ψ

...(9.1.3)


Let r1 (x1, y1, z1) and r2 (x2, y2, z2) be the radius vectors of the particles. The radius vector r0

(x0, y0, z0) of the center of mass is given by

x0 = 1 1 2 2 1 1 2 2 1 1 2 20 0

1 2 1 2 1 2

, ,m x m x m y m y m z m z

y zm m m m m m

+ + += =

+ + +

Let us introduce the relative radius vector r (x, y, z) defined byr = r2 – r1

or x = x2 – x1, y = y2 – y1, z = z2 – z1 ...(9.1.4)

Now 0 1

1 1 0 1 1 2 0

∂∂ ∂ ∂ ∂ ∂ ∂= + = − +∂ ∂ ∂ ∂ ∂ ∂ + ∂

x mx

x x x x x x m m x

222

12

1 1 2 01

∂ ∂ ∂ ∂= = − + ∂ ∂ + ∂∂

m

x x m m xx

= 22 2 2

1 12 2

1 2 0 1 2 0

2m m

m m x x m mx x

∂ ∂ ∂− + + ∂ ∂ +∂ ∂ ...(9.1.5)

Similarly,

22 2 2 22 2

2 2 21 2 0 1 22 0

2m m

m m x x m mx x x

∂ ∂ ∂ ∂= + + + ∂ ∂ +∂ ∂ ∂ ...(9.1.6)


2 2 2 2

2 2 2 21 2 1 2 1 21 2 0

1 1 1 1 1 ∂ ∂ ∂ ∂+ = + + +∂ ∂ ∂ ∂ m m m m m mx x x x...(9.1.7)

Similarly, 2 2 2 2

2 2 2 21 2 1 2 1 21 2 0

1 1 1 1 1 ∂ ∂ ∂ ∂+ = + + +∂ ∂ ∂ ∂ m m m m m my y y y ...(9.1.8)

2 2 2 2

2 2 2 21 2 1 2 1 21 2 0

1 1 1 1 1 ∂ ∂ ∂ ∂+ = + + +∂ ∂ ∂ ∂ m m m m m mz z z z...(9.1.9)

Adding Eqns. (9.1.7), (9.1.8) and (9.1.9), we have

2 2 2 21 2 0

1 2 1 2

1 1 1 1∇ + ∇ = ∇ + ∇µ +rm m m m

...(9.1.10)

where 2∇r is the Laplacian operator with respect to the components of vector r (x, y, z) and 20∇ is

the Laplacian operator with respect to the components of the vector r0 (x0, y0, z0). is the reducedmass of the system. The Hamiltonian operator of the system is

Particle in a Central Force Field 227

2 2

2 20

1 2

H V( )2( ) 2 r r

m m= − ∇ − ∇ +

+ µ

...(9.1.11)

Fig. 9.1.1 Two body problem in central force

The Hamiltonian thus breaks up into the sum of two independent Hamiltonians, one of whichcontains the total mass of the system and the position vector of the center of mass, and the secondcontains the reduced mass and the vector of mutual arrangement of the particles. The Schrodingerequation of the system is

2 2

2 20

1 2

V( ) E2( ) 2 r r

m m− ∇ ψ − ∇ ψ + ψ = ψ

+ µ

...(9.1.12)

We shall seek the solution of Eqn. (9.1.12) in the form of the product of two functions = 0 (x0, y0, z0) r (x, y, z) ...(9.1.13)

Substituting Eqn. (9.1.13) in (9.1.12), we obtain

2 22 20 0

1 2 0

V( ) E2( ) 2 r r

r

rm m

− ∇ ψ + − ∇ ψ + = + ψ µψ

...(9.1.14)

The sum of these expressions at any values of r0 and r must equal to the constant quantity E.This is possible only if each of the expressions equals its own constant and the sum of these constantsis E. Consequently we arrive at two differential equations.

(i)2

20 0 0 0

1 2

E2( )m m

− ∇ ψ = ψ+

...(9.1.15)

(ii) 2

2 V( ) E2 r r r r rr− ∇ ψ + ψ = ψµ

...(9.1.16)

with E0 + Er = E ...(9.1.17)

Equation (9.1.15) is the Schrodinger equation for a free particle having the mass (m1 + m2)


and describes the translational motion of the entire system. It is evident that E0 is the kinetic energyof motion of the system as a whole.

Equation (9.1.16) describes the motion of a fictitious particle of mass moving in a centralforce characterized by potential energy V(r). It differs from the Schrodinger equation for a particlein a central force field only in containing the reduced mass of the system of particles instead of themass of the one particle. The energy Er is the internal energy of the system.

9.2 HYDROGEN ATOM

Hydrogen atom is a system consisting of electron and proton moving under their mutual interaction.The Schrodinger equation for the system is

( )22

2E V 0

µ∇ ψ + − ψ =

...(9.2.1)

where E is the internal energy of the system and V2

0

Z

4

e

r= −

πεis the electrostatic potential energy of

the system. Since the potential energy is the function of r only, the task of finding the solution ofSchrodinger equation becomes easier in spherical polar coordinates. The Schrodinger equation inpolar coordinates is

( )

22

2 2 2 2 2

2

1 1 1sin

sin sin

2E V 0

rr rr r r

∂ ∂ψ ∂ ∂ψ ∂ ψ + θ + + ∂ ∂ ∂θ ∂θθ θ ∂ϕ µ − ψ =

...(9.2.2)

where the wave function is function of polar coordinates r . Multiplying Eqn. (9.2.2) byr2 sin2 , we obtain

( )

22 2

2

2 2

2

sin sin sin

2 sinE V 0

rr r

r

∂ ∂ψ ∂ ∂ψ ∂ ψ θ + θ θ + + ∂ ∂ ∂θ ∂θ ∂ϕ

µ θ − ψ =

(9.2.3)

Let us assume that the wave function (r, ) can be written as the product of functions

( ) ( ) ( )R , and .r Θ θ Φ ϕ

( ) ( ) ( ) ( ), , Rr rψ θ ϕ = Θ θ Φ ϕ ...(9.2.4)

From above relation, we have

2 2

2 2

R, R , R

d d d

r dr d d

∂ψ ∂ψ Θ ∂ ψ Φ= ΘΦ = Φ = Θ∂ ∂θ θ ∂ϕ ϕ ...(9.2.5)


Substituting these values in Eqn. (9.2.3) and dividing the resulting equation by R ,Θ Φ we get

( )2 2 2 2

22 2

sin R sin 1 2 sinsin E V 0

R

d d d rr

r dr d d

θ ∂ θ ∂ Θ Φ µ θ + θ + + − = ∂ Θ ∂θ θ Φ ϕ

Fig. 9.2.1 Spherical polar coordinates of a point

or ( )2 2 2 2

22 2

sin R sin 2 sin 1sin E V

R

d d r dr

r dr d d

θ ∂ θ ∂ Θ µ θ Φ + θ + − = − ∂ Θ ∂θ θ Φ ϕ ...(9.2.6)

Left hand side of this equation is function of r and whereas the right hand side is function of only. This equality can hold only if each side is equal to the same constant. Usually the separation

constant is denoted by 2lm . Thus, we have

2

22 l

dm

d

Φ = − Φϕ ...(9.2.7)

and ( )2 2 2

2 22

sin R sin 2 sinsin E V

R ld d r

r mdr d

θ ∂ θ ∂ Θ µ θ + θ + − = ∂θ Θ ∂θ θ ...(9.2.8)

Dividing Eqn. (9.2.8) by sin2 and transferring the r-dependent terms on left hand side and-dependent terms on the right hand side of equality sign, we have

( )22

22 2

1 R 2 1E V sin

R sinsinlmd r d

rr dr d

∂ µ ∂ Θ + − = − θ ∂ Θ θ ∂θ θθ

Again the equality of two functions of independent variables demands that each side be equalto the same constant. The equations obtained by equating both sides to a constant were already solved


in classical physics where the separation constant was chosen of the form l (l +1) and the two equationsobtained were

( ) ( )2

22

1 R 2E V 1

R

d rr l l

r dr

∂ µ + − = + ∂ ...(9.2.9)

( )2

2

1sin 1

sinsinlm d d

l ld d

Θ − θ = + Θ θ θ θθ ...(9.2.10)

For the sake of convenience we write the three equations together

22

20l

dm

d

Φ + Φ =ϕ ...(9.2.11)

( )2

2

1sin 1 0

sin sinlmd d

l ld d

Θ θ + + − Θ = θ θ θ θ ...(9.2.12)

22

2 2 2

1 R 2 ( 1)E V R 0

2

d d l lr

dr drr r

µ + + − − = µ

...(9.2.13)

Notice that an extra term 2

2

( 1)

2

l l

r

+µ

appears as addend in potential energy V(r). It is often called

centrifugal potential energy since its negative gradient is equal to the centrifugal force experiencedby the particle while moving in a circular orbit of radius r. The kinetic energy associated with the

rotational motion is 2 2

2

L ( 1)

2I 2

l l

r

+=µ

. So this term may be interpreted as the centrifugal energy of the

particle.

The solution of Eqn. (9.2.11) is

( ) A lime ϕΦ ϕ =

where A is constant. In order that the function ( )Φ ϕ be single valued it must satisfy the condition

( ) ( 2 ) exp( 2 ) 1li mΦ ϕ = Φ ϕ + π ⇒ π =

cos2 sin 2 1l lm i mπ + π =

cos2 1lmπ =

0, 1, 2, 3,.........lm = ± ± ± ...(9.2.14)


The separation constant ml is now called the magnetic quantum number. The constant A can bedetermined making use of the normalizing condition.

2

0

1( ) ( ) 1 A

2d

π∗Φ ϕ Φ ϕ ϕ = ⇒ =

π∫

The normalized ( )Φ ϕ function then becomes

1

( ) exp( )2

limΦ ϕ = ϕπ ...(9.2.15)

The method of finding the solution of Eqn. (9.2.12) is quite complicated. The finite and well-

behaved solutions are found only if the constant l is an integer and equal to or greater than . .,lm i e

orl ll m m l≥ ≤

This condition can be expressed in the form

0, 1, 2, 3,..............,lm l= ± ± ± ± ...(9.2.16)

The constant l is now called orbital quantum number or azimuthal quantum number.Before solving the Eqn. (9.2.12), it is convenient to change the independent variable to x

through the transformation x = cos . The resulting equation is known as the associated Legendreequation and its acceptable solutions are expressed in the form of polynomials, called associated

Legendre functions. Since these polynomials depend on l and ml, and are written as , ( )ll mΘ θ . The

normalized associated Legendre polynomials, for positive value of ml are given by

( ) / 22 2,

( )!2 1 1( ) ( 1) . . 1 . ( 1)

2 ( )! 2 ( !)

lll

l l

m lmm lll m m l l

l

l ml d dx x x

l m l dxdx

−+Θ = − − − +

where x = cos

or ( )!2 1

( ) ( 1) . . .P (cos )2 ( )!

lll

mm llm l

l

l ml

l m

−+Θ θ = − θ+

...(9.2.17)

For negative value ml, we have

*

, ,( ) ( 1) ll l

ml m l m

Θ θ = − Θ ...(9.2.18)

Some of these polynomials are tabulated below. The product ()() is called the sphericalharmonics Yl, ml (, ).


, ( )l ll mΘ θNormalized associated Legendre Functions

l ml ( )Θ θ Y (,)

0 0 100 2

Θ = 001

Y4

=π

1 0 101

6 cos2

Θ = θ 103

Y .cos4

= θπ

1 ±1 1 11

3 sin2±Θ = θ 1, 1

3Y .sin .

8ie± ϕ

± = θπ

2 0 220

110(3cos 1)

4Θ = θ− 2

205

Y (3cos 1)16

= θ −π

2 ±1 2 11

15 sin cos2±Θ = θ θ 2 1

15Y .(cos .sin ).

8ie± ϕ

± = θ θπ

2 ±2 22 2

115 sin

4±Θ = θ 2 22 2

15Y .sin .

32ie± ϕ

± = θπ

Solution of Radial Equation: For bound state, the energy E is negative so the radial equationbecomes

22

2 2 2 20

1 R 2 ( E) 2 Z ( 1)R 0

4

d d e l lr

dr drr r r

µ − µ + + + − = πε ...(9.2.19)

Equation (9.2.19) can be written in a convenient form making use of transformation

= r ...(9.2.20)

where is a constant and is so chosen that resulting equation look simpler. When Eqn. (9.2.20) issubstituted in Eqn. (9.2.19), we get

22

2 2 2 2 20

1 R 2 E 2 Z 1 ( 1)R 0

4

d d e l l

d d

− µ µ +ρ + + − = ρ ρ ρρ α πε α ρ ...(9.2.21)

Now is chosen to make the first term in square bracket equal to 1/4. So

2

8 E− µα=

...(9.2.22)


In the second term in square bracket the coefficient of 1/ is put equal to .

1/ 22

0

Z

4 2E

e µ λ= πε − ...(9.2.23)

Equation (9.2.21) now becomes

2

2 2

1 R 1 ( 1)R 0

4

d d l l

d d

λ +ρ + − − = ρ ρ ρρ ρ ...(9.2.24)

For large , the first term of Eqn. (9.2.24) reduces to d2R/d2 and Eqn. (9.2.24) becomes

2

2

R 1R 0

4

d

d− =

ρ...(9.2.25)

The solution of Eqn. (9.2.25) isR() = e± /2

Positive sign leads to an unacceptable solution. So we choose the negative sign.

/ 2R( ) e −ρρ = ...(9.2.26)

To determine the nature of solution near origin (for small ) we put

1R( ) F( )ρ = ρ

ρ...(9.2.27)

where F() is unknown function. Substituting Eqn. (9.2.27) in (9.2.24), we have

2

2 2

F( ) 1 ( 1)F( ) 0

4

d l l

d

ρ λ ++ − − ρ = ρρ ρ ...(9.2.28)

For l = (l 0) the last term l (l +1)/2 is large near origin (0), Eqn. (9.2.28) reduces to

2

2 2

F( ) ( 1)F( ) 0

d l l

d

ρ +− ρ =ρ ρ ...(9.2.29)

Solution of Eqn. (9.2.29) can be obtained in formF() = constant s ...(9.2.30)

Substituting Eqn. (9.2.30) in Eqn. (9.2.29), we haves ( s – 1 ) – l ( l + 1) = 0

This gives s = – l or s = l + 1

1 1F( )

R( ) ors

s l l− − −ρ ρρ = = = ρ =ρ ρρ ρ

Near origin 1

10,

l+ρ→ →∞ρ

this is not acceptable.


However near origin 0, l 0, this is acceptable. So we can assume the solution ofEqn. (9.2.24) of the form

/ 2R( ) . .L( )le−ρρ = ρ ρ ...(9.2.31)

where L() is unknown function to be determined. Substituting Eqn. (9.2.31) in (9.2.24), we get

2

2

L( ) L( )2( 1) ( 1) L( ) 0

d dl l

dd

ρ ρρ + + −ρ + λ − + ρ = ρρ...(9.2.32)

Let us assume the power series solution of Eqn. (9.2.32) of the form

20 1 2

0

L( ) ......... rr

r

a a a a∞

=

ρ = + ρ+ ρ + = ρ∑ ...(9.2.33)

Substituting Eqn. (9.2.33) in (9.2.32) and equating the coefficient of r equal to zero, we get

11

( 1)( 2 2)r rr l

a ar r l+

+ + − λ=+ + + ...(9.2.34)

For large value of r

1 1+→∞→r

rr

a

a r

The ratio of successive coefficients of series

2 1

1 ............ .......2! ! ( 1)!

r r

er r

+ρ ρ ρ ρ= +ρ+ + + + +

+

is 1 1 1

1r

rr

b

b r r+

→∞= →

+

So for large value of the function L() behaves like e. Hence

/2 /2R( ) . . .l le e e−ρ ρ ρ

ρ→∞ρ →ρ →ρ →∞

This form of R() is not acceptable. If infinite series L() terminates after finite number ofterms i.e., it becomes a polynomial, then it will be valid solution of Eqn.(9.2.32). From the recursionrelation (9.2.34) we see that this requirement is met if r reaches some integer, say n' (= r), given by

n' + (l + 1) – = 0

or = n' + l + 1

then 1 1( )r na a ′+ += and all higher coefficients become zero and L() becomes a polynomial of degree

n'. Since n' us a non-negative integer, so is . The integer is denoted by n and is called principalquantum number

n = = n' + l + (9.2.35)


Since n' > 0

n – l – 1 > 0

l < n – 1 ...(9.2.36)

This puts restriction on the values that l can take on for a given value of n. Putting the value of in (9.2.23), we have

2 4 2

2 20

1 1 ZE

2 4

e

n

µ=− πε , n = 1, 2, 3,….. ...(9.2.37)

2

2

ZE (13.6eV)

n=−

Thus, the energy of electron in hydrogen atom is quantized. The integer n is called principalquantum number.Radial wave functions: Laguerre polynomials are defined by

( )L ( ) . .qq

q qq q

d de e e e

ddρ −ρ ρ −ρ

ρ = ρ = ρ ρρ ...(9.2.38)

Laguerre Polynomiala

0L ( ) 1ρ =

1L ( ) 1ρ = − ρ

22L ( ) 2 4ρ = − ρ + ρ

2 33L ( ) 6 18 9ρ = − ρ + ρ − ρ

2 3 44L ( ) 24 96 72 16ρ = − ρ + ρ − ρ + ρ

2 3 4 55L ( ) 120 600 600 200 25ρ = − ρ + ρ − ρ + ρ − ρ

2 3 4 5 66L ( ) 720 4320 5400 2400 450 36ρ = − ρ + ρ − ρ + ρ − ρ + ρ

Associated Laguerre polynomials are defined by

L ( ) L ( )p

pq qp

d

dρ = ρ

ρ= ( ). .ρ −ρ

ρ ρ ρ

p qq

p q

d de e

d d...(9.2.39)

Polynomials L ( )pq ρ satisfy the differential equation

( ) ( )2

2L ( ) 1 L ( ) L ( ) 0p p p

q q qd d

p q pdd

ρ ρ + + −ρ ρ + − ρ =ρρ

...(9.2.40)


Comparison of Eqn.(9.2.32) with = n and Eqn. (9.2.40), we havep = 2l + 1 and q = n + l

In view of this correspondence we can write the solution of Eqn. (9.2.32) as 2 1L ln l++ .

The solution of radial equation can be written as

2 1/ 2( ) N . . .L ( )llnl nl n lR e +−ρ

+ρ = ρ ρ ...(9.2.41)

where

12 1 1

0

( )!( 1)

( 1 )! . (2 1 )! !

n ll k k

n lk

n lL

n l k l k k

− −+ ++

=

+= − ρ− − − + +∑

The normalization constant Nnl can be determined making use of following property of Laguerrepolynomial.

21 3

0

2 1. L ( ) .( !)

( )!p p

qq p

e d qq p

∞−ρ + − + ρ ρ ρ= −∫ ...(9.2.42)

Normalization condition for Rnl (r) is

2 2

0

( ) 1,nlR r r dr∞

=∫ ...(9.2.43)

Remembering that

2 o

0 0 20

2Z, where Bohr radius 4 . 053A.r r a

na e

ρ = α = = πε =

µ

the normalization condition Eqn. (9.2.43) assumes the form

22 2 2 1 2

30

1N . . ( ) . 1l l

nl n le L d∞

−ρ ++ρ ρ ρ ρ=

α∫

3

3

N 2( ) (2 1) 1. . ( )! 1

( 1)!nl n l l

n ln l

+ − + + + = − −α

3/ 2

30

2Z ( 1)!N .

2 ( ) !nl

n l

na n n l

− −= ± +

In above expression negative sign is chosen to make R10 (r) positive. The radial wave functionis given by


3/ 22 1/ 2

30

2Z ( 1)!R ( ) . . . . ( )

2 ( )!

llnl n l

n le L

na n n l

+−ρ+

− −ρ =− ρ ρ +

...(9.2.44)

Some of the radial wave functions are:

0

3/ 2Z /

100

ZR ( ) 2 . r ar e

a−

=

0

3 / 2Z / 2

200 0

Z ZR 2 1 .

2 2r ar

ea a

− = ⋅ −

0

3/ 2Z / 2

210 0

Z 1 ZR ( )

2 3r ar

r ea a

− = ⋅ ⋅ ⋅

0

3 / 2 2Z / 3

300 0 0

Z 2Z 2 ZR ( ) 2 1

3 3 27r ar r

r ea a a

− = ⋅ − − ⋅ ⋅

0

3 / 2Z / 3

310 0 0

Z 4 2 Z ZR ( ) 1

3 3 6r ar r

r ea a a

− = ⋅ ⋅ ⋅ − ⋅

0

3 / 2 2Z / 3

320 0

Z 2 2 ZR ( )

3 27 5r ar

r ea a

− = ⋅ ⋅ ⋅

The complete wave function is given by

( , , ) R ( ).Y ( , )l lnlm nl lmr rψ θ ϕ = θ ϕ

0

1/ 23/ 2 1

30 0 0

2Z ( 1)! 2Z 2Z( , , ) L Y ( , )

(2 ) ( )!l l

lZr na l

nlm n l lmn l r r

r ena na nan n l

− ++

− − ψ θ ϕ = ⋅ ⋅ ⋅ ⋅ θ ϕ + ...(9.2.45)

The wave function , , ( , , )ln l m rψ θ ϕ represents a quantum state of electron and is characterized

by a set of quantum numbers n, l, ml. For n =1 there is only one state 100, for n = 2 there are fourstates 200, 210, 211, 21±1. In spectroscopy the states corresponding to a given value of l aredenoted according to the following scheme.

l = 0, s-state; l = 1, p-state; l = 2, d-state, l = 2, f-state etc.


Complete wave functions of hydrogen-like atom

n l ml state ( ) ( ) ( )R rψ = Θ θ Φ ϕ

1 0 0 1s 0

3/ 2Z /

1000

Z 1 r aea

− ψ =

π

2 0 0 2s 0

3 / 2Z / 2

2000 0

Z 12

4 2r ar

ea a

− ψ = −

π

2 1 0 2p 0

3 / 2Z / 2

2100 0

Z 1.cos .

4 2r ar

ea a

− ψ = θ

π

2 1 ±1 2p ( )0

3/ 2Z / 23/ 2

21 1 00 0

Z 1sin .

8r a ir

a e ea a

−− ± ϕ±

ψ = θ

π

Electron Probability Density: The complete wave function describing the behaviour of electron inhydrogen atom is

( , , ) ( ) ( ) ( )l l lnl m nl lm mr R rψ θ ϕ = Θ θ Φ ϕ

The probability density of electron around the point (r,,) is

2 2 2 2Rψ = Θ Φ ...(9.2.46)

Now 2 ∗Φ = Φ Φ = 1

2π2

measuresΦ the probability of finding the electron at a particular azimuth angle . Here we see

that likelihood of finding the electron is independent of angle . 2measuresΘ the electron probability

density in a direction . For s-electron l = 0, ml = 0,2 1

2Θ = which is independent of . According

to quantum mechanics the electron charge density may be thought of as being spread over the space

and 2Θ gives the angular dependence of charge density. Evidently, for s-electron the distribution of

charge or charge cloud is spherical. For p-electron l = 1, ml = 1, 0, –1, the corresponding charge

distribution that is given by 2Θ is of dumb-bell shape. For other states, the charge distribution is

complicated.

9.3 MOST PROBABLE DISTANCE OF ELECTRON FROM NUCLEUS

The radial function R(r) is plotted against r for 1s, 2s and 2p electron. The function 2

R is a measure

of probability density of finding the electron at a distance r. The probability of finding the electronin a volume d at point (r, ) is given by


2 2 2P( , , ) | ( , , ) | | ( , , ) | sinl lnlm nlmr r d r r dr d dθ ϕ = ψ θ ϕ τ = ψ θ ϕ θ θ ϕ ...(9.3.1)

The ground state wave function of electron is

0/

1003

0

1( ) r ar e

a

−ψ = ⋅π ...(9.3.2)

The probability of finding the electron between r and r + dr, irrespective of coordinates and is

0

22 / 2

300 0

1P( ) sin r ar dr d d e r dr

a

π π−

= ϕ θ θ ⋅ ⋅ π ∫ ∫

02 / 23

0

4 r ae r dra

−= ⋅ ⋅ ...(9.3.3)

The most probable is the value of radial distance r given by condition

02 /23

0

0

4P( ) 0r ad d

r r edr dr a

r a

− = =

=

...(9.3.4)

Thus, the maximum probability of finding 1s electron is at a distance r = a0 = 2 o

02

40.53A.

me

πε=

Fig. 9.3.1 Angular probability function in s-state [|00|2]

Fig. 9.3.2 Angular probability function in p-state


Fig. 9.3.3 Radial wave function and radial probability


9.4 DEGENERACY OF HYDROGEN ENERGY LEVELS

We have seen that each quantum state is characterized by a set of three quantum numbers n, l, ml.For n =1, l = 0, ml = 0 , the corresponding state is denoted by 100. This state is the ground state.For n = 2, l = 0, 1. For l = 0, ml = 0 and for l = 1, ml = 1, 0, –1. Thus there are four states namely200, 211, 210, 21±1. Since total energy of electron depends only on principal quantum numbern, all the four states corresponding to n = 2, have the same energy. This energy level is said to befour-fold degenerate. Here the degeneracy is due to the symmetry of Coulomb potential. In atomsother than hydrogen, the energy E depends on n and l both because the Coulomb potential is modifieddue to screening effect. Even in hydrogen atom this degeneracy is removed by applying externalmagnetic field.

In addition to the above degeneracy there is also another degeneracy, which arises because thestates having the same n and l but different ml have also the same energy. Since there are 2l +1different values of ml for each value of n and l, each level is 2l +1 fold-degenerate. This degeneracyis common to all central fields i.e., to all potentials that are function of radial distance r only. Thisdegeneracy is removed by applying a non-central field such as magnetic field. Magnetic field causesenergy levels of different ml to have different energies. This splitting of energy levels by an externalmagnetic field is responsible for the phenomenon of Zeeman Effect.

9.5 PROPERTIES OF HYDROGEN ATOM WAVE FUNCTIONS

The operator of the square of angular momentum is

2

2 22 2

1 1L sin

sin sin

∂ ∂ ∂ = − θ + θ ∂θ ∂θ θ ∂ϕ

and that of the z-component of angular momentum is

Lz i∂= −∂ϕ

Now 2 2L ψ = −2

2 2

1 1sin

sin sin

∂ ∂ ∂ θ + θ ∂θ ∂θ θ ∂ϕ RΘΦ

22

2 2R sin

sin sin

Φ ∂ ∂Θ Θ ∂ Φ = − θ + θ ∂θ ∂θ θ ∂ϕ

22

2R sin

sin sinlm Φ ∂ ∂Θ = − θ − Φ Θ θ ∂θ ∂θ θ

2R= − Φ

2

2

1sin

sin sinlm ∂ ∂Θ θ − Θ θ ∂θ ∂θ θ


2R ( 1)l l= − Φ − + Θ

( ) 21l l= + ψ

Thus the eigen value of operator L is l(l+1) 2 . This means that the measurement of the square

of angular momentum will yield a value given by

( ) ( )2 2 2L 1 or L 1l l l l= + = +

Since the magnitude of angular momentum is determined by quantum number l therefore l iscalled orbital angular momentum quantum number. The quantum number l can assume only definitevalues, i.e., the magnitude of angular momentum is quantized.

Now, L Rz i

∂ψ = − ΘΦ ∂ϕ

Ri∂Φ= − Θ∂ϕ

Rli m= − ΘΦ

lm= ψ

(i) For l = 1, L = 2 , Lz = , 0, –

(ii) For l = 2, L = 6 , Lz = 2 , 1 , 0, 1− , – 2

Fig. 9.5.1 Allowed orientations of angular momentum vector

Thus the z-component of angular momentum can have only discrete values. In other words theangular momentum vector L can have only certain orientations in space. In vector model of atom,the angular momentum vector performs precessional motion around the z-direction in such a way


that its projection onto the z-axis has fixed value and the average value of x and y components becomezero. The quantization of direction angular momentum vector (and any vector associated with it,such as magnetic moment) is known as space quantization. For p-electron l = 1,

|L| = 1(1 1) 2 and 1, 0, 1.lm+ = = −

The vector L for this electron has only three orientations. Similarly for d-electron l = 2,

| | 2(2 1) 6 and 2,1, 0, 1, 2.lm= + = = − − L

The vector L of this electron can have only five orientations. The possible directions of angularmomentum vector L are, in general, given by

( )cos

1lm

l lθ =

+

SOLVED EXAMPLES

Ex. 1. Verify that the spherical harmonics Y1,1 and Y2,1 are orthogonal.

Sol.

2

1, 1 2, 1

0 0

3 15Y Y sin . sin cos . .sin

8 8i id e e d d

π π∗ − ϕ ϕτ = θ θ θ θ θ ϕ

π π∫ ∫ ∫

= 2

2

0 0

3 5sin .cos .sin .

8

π π

θ θ θ θ ϕπ ∫ ∫d d

= 2

0

3 5(1 cos )cos .sin . 2

8

π − θ θ θ θ π π ∫ d

= 0. (To evaluate the integral put cos = x)

Ex. 2. The ground state function of H-atom is ( / )0Aexp r aψ = − , where A is constant. Using the

normalization condition find the value of A.

Show that 03ar

2< >=

Sol. Normalization condition

2| | 1ψ τ =∫ d

02 /2 2

0

A .4 1r ae r dr∞

− π =∫


02 /2 2

0

4 A 1r ar e dr∞

−π =∫

3

2 20

0

4 A 12

xax e dx

∞−

π = ∫ (Put 2r/a0 = x)

30

1A

a=

π(The value of integral is 3 )

* 2 30 0

0 0

3( ) ( ).4 (4)

4 2

∞ ∞−< >= ψ ψ π = = Γ =∫ ∫ xa a

r r r r r dr x e dx

Ex. 3. For hydrogen atom

1/ 2

100 030

1= exp( r/a )

a

ψ − π

, find the probability of finding the electron

in a sphere of radius r = a0.

Sol. 0 0

2 2 2100 03 2

00 0

4 5P .4 exp( 2 / ) 1 0.32

a a

r dr r r a dra e

−= ψ π = − = + =∫ ∫ (Evaluate the integral by method ‘integration by parts’.)

Ex. 4. For hydrogen atom ψ = − θ 0( / ) . .cos210 A exp r 2a r , find A.

Sol. Normalization condition

∞ π π

θ= ϕ=

= − θ θ ϕ ∫ ∫ ∫2

2 4 20

0 0 0

1 A exp( / ) .cos .sin .r a r dr d

2

2 4 20

0 0 0

1 2 A exp( / ) cos sinr r a dr d d∞ π π

= π − θ θ θ ϕ∫ ∫ ∫

2 4

0

0

21 2 A exp( / ) 2

3r r a dr

∞ = π − π ∫

2 5 40

0

21 2 .2 . . exp( )

3

∞

= π π −∫A a x x dx [The value of integral is (5)]


1/ 2

50

1A

32 a

= π

1/ 2

210 050

1cos .exp( / 2 )

32r r a

a

ψ = θ − π

Now,2 2

* 2 2 5 20

0 0 0 0 0 0

.sin . A exp( / ). cos sinr r r dr d d r r a dr d d∞ π π ∞ π π

< >= ψ ψ θ θ ϕ = − θ θ θ ϕ∫ ∫ ∫ ∫ ∫ ∫

= 2 6 5 2 60 0 0

0

2 4A .2 exp( ) A (6) 5

3 3a x x dx a a

∞ ππ − = Γ =∫


1. Establish Schrodinger equation for hydrogen atom and obtain its energy levels and wave functions.

Write the ground state wave function for hydrogen atom and calculate the most probable distance of the electronfrom the nucleus. Sketch radial wave functions and radial probability in 1s, 2s, 2p, 3s, 3p states

2. Spherical harmonics are defined by

+ − −θ ϕ = − ⋅ ⋅ − π +

2 | |/ 2 | |2

| |

2 1 ( | |)! (1 )Y ( , ) ( 1) ( 1)

4 ( | |)! ! . 2

m m lm

lm l m l

l l m x d dx

l m l dx dx

where x = cos .

Find Y00, Y10, Y11, Y1 – 1, Y20, Y21, Y2 – 1, Y22, Y2 – 2.

3. Laguerre polynomials are defined by

( ).L ( ) .p q

p qq p q

d de e

d dρ −ρ

ρ = ρ ρ ρ

Find 1 1 31 2 3L ,L ,L .

4. The wave functions of hydrogen-like atoms are given by

− +

+

− − ψ θ ϕ = θ ϕ +

0

1/ 23

Z / 2 13

0 0 0

2Z ( 1)! 2Z 2Z( , , ) . . .L . ( , )

(2 ) ( )!l l

r na lnlm n l lm

n l r rr e Y

na na nan n l

Find 100, 210, 21 1.±ψ ψ ψ


5. Write the radial equation of the hydrogen atom in dimensionless form and explain all the symbols used . Solvethe equation to obtain an expression for energy eigen values. (All’d 1995)

6. Write down the Schrodinger equation for a positronium atom which consists of a positron and an electron.Reduce the equation to two equivalent one body problem and discuss the significance of each one of them. If theground state wave function for hydrogen atom is given by

ψ θ ϕ = −

π03

0

1( , , ) exp( / ).r r a

a

What would be the corresponding wave function for positronium atom? (All’d 1996)

7. The radial part of the wave function for n = 2, l = 1 state of hydrogen atom is given by

ψ = −

3/ 2

21 00 0

2( ) exp( / 2 ),

3

rr r a

a a a0 is Bohr radius.

(a) What is the parity of the radial wave function and of total wave function for the above state?

(b) Plot the probability distribution function as a function of r and obtain the most probable distance betweenthe proton and the electron.

(c) Calculate the size of the hydrogen atom [< r2 >]1/2 for this state. (All’d 1996)

8. (a) The radial equation for the hydrogen atom is

2

2 2 2

1 R 2 ( 1)E V( ) R 0

m l lr r

r rr r

∂ ∂ + + − − = ∂ ∂

Write it in dimensionless form. Find its solution in the limit r 0 and r .

(b) Substitute R (r) = (r)/r in the above equation to get the following equation:

2 2 2

2 2

( 1)V( ) E

2 2

l lr

m r mr

∂ +− + + χ = χ ∂

Explain physical significance of the term + 2

2

( 1).

2

l l

mr(All’d 1998)

9. (a) Show that the probability of finding the electron in the ground state of hydrogen atom is maximum at a distance equal to the Bohr radius.

(b) Explain briefly the Lamb shift with reference to the first excited state of the hydrogen atom. (All’d 1998)

10. Solve the radial equation for the hydrogen atom

L ( ) (2 2 )L ( ) [ ( 1)]L( ) 0l l l′′ ′ρ ρ + + − ρ ρ + λ − + ρ =

where

1/ 22

20

Z 8 Eand

4 2E

e µ − µ λ = α = πε −

to find the energy levels. What functions are L(). (All’d 1999)


11. The normalized ground state wave function for the electron in the hydrogen atom is

3/ 2

00

1 1( , , ) exp( / )r r a

a

ψ θ φ = − π

(a) Sketch the wave functions and the probability density versus r.

(b) Find the radius at which electron is most likely to be found.

(c) Find the probability of locating the electron between r = a0/2 and r = 3a0/2.

Where a0 is Bohr radius. (All’d 2000)

12. The ground state of the hydrogen atom is described by the function

030

1( , , ) exp( / )r r a

aψ θ φ = −

π

(a) Calculate the probability of finding the electron in the range a0/2 < r < 2a0.

(b) Sketch the radial probability density as a function of r. At what value of r, it is maximum? What would bethe corresponding probability for a classical orbit?

(c) Calculate the average radius of the hydrogen atom? (All’d 2001)

13. (a) Solve the radial equation of the hydrogen atom in the limit r 0 and r .

(b) The normalized ground state wave function of the hydrogen atom is

030

1( , , ) exp( / )r r a

aψ θ φ = −

π

Find the expectation value of r and most probable radius of the orbit in the ground state. (All’d 2002)

14. Solve the radial equation

2

2 2 2 2

1 R 2 E 2 V( ) ( 1)R 0

m m r l lr

r rr r

∂ ∂ + + − − = ∂ ∂

of the hydrogen atom, where symbols have their usual meanings and show that the energy values are exactly thesame as those obtained by Bohr. (All’d 2003)

15. Solve the radial equation for the hydrogen atom and compare your result with those obtained by Bohr.(All’d 2004)

16. In a hydrogen atom the wave function describing the electron in 1s state is given as

100 030

1( , , ) exp( / )r r a

aψ θ φ = −

π

(a) Calculate average distance of the electron from the nucleus.

(b) Probability P(r)dr as a function of r and compare it with the prediction of Bohr model. (All’d 2005)

17. Find solution of the radial equation of hydrogen atom and show that the result obtained agrees with that of Bohr.(All’d 2006)


18. (a) The complete wave function of the hydrogen atom for 2p state is

210 0300

1 1cos exp( / 2 )

4 2

rr a

aa

ψ = θ − π

Prove that the wave function is normalized.

(b) Find the expectation value of the distance of the electron from the nucleus in the hydrogen atom in 2p state.(All’d 2007)

19. Write down the Hamiltonian for hydrogen atom and reduce it to relative and center of mass coordinates.

(All’d 2007)

20. A particle is moving in free space at a fixed distance r = a i.e., on the surface of a smooth sphere. Write theHamiltonian and Schrodinger equation for such a system and solve it to find the energy eigen values.

(All’d 2004)

21. A positronium atom consists of an electron and a positron interacting via coulomb force. Reduce the Schrodingerequation to two equivalent one body problem. (All’d 2000)


UNIT


blank

CHAPTER

PRELIMINARY CONCEPTS

1.1 INTRODUCTION

The main objective of statistical mechanics is to predict the properties of a macroscopic system fromthe knowledge of the behaviour of particles constituting the system. In a physical system containinga very large number of particles (atoms and molecules or other constituents) it is usually impossible,for practical reasons, to apply the basic physical laws (classical or quantum) directly to each particle.Instead, it is often advantageous to take a statistical approach, in which one describes the distributionof particles in various states in a statistical manner. The existence of a very large number of particlesof the system can be used to advantage in the statistical description. The theory of random processesand quantities form the mathematical tools for this approach.

In statistical mechanics the description of a state of a many particle system is given by statinghow the particles are distributed in various allowed microstates. Depending on the nature of theparticles, three kinds of statistics or distribution laws are used to describe the properties of the system.The three statistics are:

1. Maxwell-Boltzmann or classical statistics.

2. Bose-Einstein statistics.

3. Fermi-Dirac statistics.

Fremi-Dirac statistic and Bose-Einstein statistics are quantum statistics.

1.2 MAXWELL-BOLTZMANN (M-B) STATISTICS

M-B statistics is applicable to the system of identical, distinguishable particles. The particles are sofar apart that they are distinguishable by their position. In the language of quantum mechanics, theapplication of classical statistics is valid if the average separation between particles is much greaterthan the average de Broglie wavelength of the particle. In this situation the wave functions of theparticles don’t overlap. The particle may have any spin. The classical statistics put no restriction onthe number of particles that occupy a state of the system. M-B statistics can be safely applied todilute gases at room and higher temperature.


1.3 BOSE-EINSTEIN (B-E) STATISTICS

B-E statistics is applicable to the system of identical, indistinguishable particles, which have integralspin (0, 1, 2,….). Particles with integral spin are called bosons. Bosons don’t obey Pauli’s exclusionprinciple. So any number of bosons can occupy a single quantum state. The particles are close enoughso that their wave functions overlap. Examples of bosons are photons (spin 1), phonons (quantum ofacoustical vibration), pions, alpha particle, helium atom etc.

1.4 FERMI-DIRAC (F-D) STATISTICS

F-D statistics is applicable to the system of identical, indistinguishable particles, which haveodd-half-integral spin (1/2, 3/2, 5/2,…). Particles with odd-half-integral spin are called fermions andthey obey Pauli’s exclusion principle. Hence, not more than one fermion can occupy a quantumstate. The F-D statistics is valid if the average separation between fermions is comparable to theaverage de Broglie wavelength of fermions so that their wave functions overlap. Examples of fermionsare electrons, positrons, µ-mesons, protons, neutrons etc.

In the limit of high temperature and low particle density the two quantum statistics(B-E and F-D) yield results identical to those obtained using the classical statistics.

1.5 SPECIFICATION OF THE STATE OF A SYSTEM

A system consisting of micro-particles (such as atoms and molecules) is described by the laws ofquantum mechanics. In quantum mechanical description the most precise possible measurement on asystem always shows this system to be in someone of a set of discrete quantum states characteristicof the system. The microscopic state of a system is described completely by specifying the particularquantum state in which the system is found. Each quantum state of an isolated system is associatedwith a definite value of energy and is called an energy level. There may be several quantum statescorresponding to the same energy of the system. These quantum states are then said to be degenerate.Every system has a lowest possible energy. There is usually only one possible quantum state of thesystem corresponding to this lowest energy; this state is said to be the ground state of the system.(Exceptions may be there.)

For illustration we take an example. Consider a particle of mass m restricted to move inside a

box of sides Lx, Ly, Lz located at the origin of cartesian axes such that 0 L ,0 L ,x yx y≤ ≤ ≤ ≤0 Lzz≤ ≤ . Schrodinger equation for the particle is

2

2

2 E0

m∇ ψ + ψ =

2 2 22 2

2 2 2 2

2 E0,

mk k

x y z

∂ ψ ∂ ψ ∂ ψ+ + + ψ = =∂ ∂ ∂

...(1.5.1)

Preliminary Concepts 253

The solution of the Eqn. (1.5.1) subject to the boundary conditions: ψ = = = =0 at 0, L , 0,xx x y

= = =L , 0, Ly zy z z is found to be

π ππψ =( , , ) Asin sin sin

L L Ly zx

x y z

n y n zn xx y z ...(1.5.2)

where nx, ny, nz are positive integers and each can take on values 1, 2, 3,….. The allowed energiesof the particle comes out to be

π = + +

2 222 2

2 2 2E

2 L L L

y zx

x y z

n nn

m ...(1.5.3)

If Lx = Ly = Lz = L, then the energy of the particle is given by

( )2 2

2 2 22

E2 Lx y zn n n x y zn n n

m

π= + +

...(1.5.4)

and the state of the particle is given by the wave function

( , , .) sin sin sinL L L

y zxn y n zn x

x y z Aπ ππ

ψ = ...(1.5.5)

The triad nx, ny, nz defines a quantum state of the particle. For ground state, nx = ny = nz = 1.

This state is represented as 111( , , ).x y zψ The energy in this state is π=

2 2

111 2

3E

2 L

m. A single quantum

state corresponds to the lowest energy level. When only one quantum state belongs to an energylevel, that energy level is said to be non-degenerate. The ground state is thus non-degenerate. Ifmany different quantum states belong to a single energy level, that energy level is said to bedegenerate. The degeneracy of an energy level is given by the number of ways that the integer

( + +2 2 2x y zn n n ) can be written as the sum of squares of the three positive integers. Some of the lower

energy levels of a particle in box are given below.

π π= = = =

2 2 2 2

111 112 121 2112 2

3 6E , E E E ,

2 L 2 L

m m

π= = =

2 2

122 212 221 2

9E E E

2 L

m,

π= = =

2 2

113 131 311 2

11E E E ,

2 L

m

π=

2 2

222 2

12E ,

2 L

m


π= = = = = =

2 2

123 132 213 231 321 312 2

14E E E E E E ,

2 L

m

π= = =

2 2

223 232 322 2

17E E E

2 L

m

Notice that the second, third and fourth energy levels are 3-fold degenerate, the fifth energylevel is non-degenerate, sixth energy level is 6-fold degenerate and the seventh energy level is 3-folddegenerate and so on.

DegeneracyE223, E132, E322 ______________________ 3E123, E132, E213, E231, E321, E312 ______________________ 6E222 ______________________ 1E113, E131, E311 ______________________ 3E122, E212, E221 ______________________ 3E112, E121, E211 ______________________ 3E111 ______________________ 1

Fig. 1.5.1 Energy levels of a particle in a box

1.6 DENSITY OF STATES

The allowed energy levels and associated quantum states for a particle confined to move in a cubicalenclosure of side L are given by

( )π= = + +2 2 2

2 2 22

E2 2 L

x y zp

n n nm m

...(1.6.1)

This equation can be written as

+ + = =π

22 2 2 2

2 2

2 LE R

x y z

mn n n ...(1.6.2)

where =π

2

2 2

2 L ER

m...(1.6.3)

A quantum state (microstate) of the particle is given by

π ππ

ψ =, ,( , , ) const. sin sin sinL L Lx y z

y zxn n n

n y n zn xx y z ...(1.6.4)


When we plot the positive integers nx, ny, nz along the x, y, z axes of the Cartesian coordinatesystem in three dimensional space, the resulting space is called number space. In this space eachtriad nx, ny, nz is represented by a point. When all the triplets (nx, ny, nz), formed by the allowedvalues of integers nx, ny, nz are plotted in number space, we get a lattice of points. In number space,Eqn. (1.6.2) represents a sphere of radius R given by Eqn. (1.6.3). Each quantum state (microstate)which is described by Eqn. (1.6.4) is represented by a point n this number space. Now draw a sphere

of radius =π

2

2 2

2 L ER

m in number space. The number of lattice points which lie on the surface of

this sphere lying in the positive octant is equal to the number of quantum states with energy E. Weare interested in the number of quantum states with energy less than E. This number is denoted by (E) and is equal to the number of lattice points lying within the positive octant of sphere of radiusR. Obviously,

31 4(E) R ,

8 3

πΦ = L3 = V

or (E) = π 3/ 2

3

4 V(2 E)

3m

h...(1.6.5)

The number of quantum states with energy lying in the range dE about E is

∂Φ πΩ = =∂

3/ 2 1/ 23

2 V(E) E E (2 ) E E

Ed d m d

h...(1.6.6)

The density of states g(E) is defined as the number of quantum states in the unit energy rangeabout E and is given by

π= 3/2

3

2 V(E) (2 ) Eg m

h...(1.6.7)

The number of quantum states in the energy range dE about E is

π= 3/23

2 V(E) E (2 ) E Eg d m d

h...(1.6.8)

Assuming that the particle has only translational energy, we have

= =2

E , E2

p pd dp

m m

With the help of these results we can transform the expression for density of states in terms ofmomentum. Thus the number of quantum states such that the magnitude of momentum of the particlelies in the range p and p + dp is

2

3

V( ) 4g p dp p dp

h= π ...(1.6.9)


If we take E = (3/2) kT, T = 300 K, m = 10–22 g, L = 10 cm, dE = 0.01 E, we find that

= 28(E) E 10g d

So even for a system as simple as a particle in a box, the density of states can be very large atroom temperature.

1.7 N-PARTICLE SYSTEM

For an N-particle system, the density of states is tremendously large. To see this, consider a systemof N non-interacting particles in a cube of side L. The energy of the system is

= =

π = = + + ∑ ∑N N2 2

2 2 22

1 1

E E2 L

i xi yi zii i

n n nm

...(1.7.1)

A quantum state of the system is defined by the set of 3N integers

1 1 1 2 2 2, , , , , , ....................................................., , ,x y z x y z N x N y N zn n n n n n n n n

Using the concept of 3N dimensional number space we can calculate the number (E) of quantumstates with energy less than E. This calculation is some what difficult and we state the result.

πΦ =N 3N/2

3N

V (2 E)(E)

(3N/2)!

m

h...(1.7.2)

The number of states within the energy interval dE at E is

( )−∂Φ πΩ = = = ⋅ ⋅

∂

3N2

N 3N/2 1

3N

V 3N (2 )(E) E (E) E E E E

E 2 (3N/2)!m

d g d d dh

...(1.7.3)

If E = (3/2)kT, T = 300 K, m = 10–22 g, L = 10 cm, N = 6.02 × 1023, E = 0.01 E, we have

N(E) E 10g d =

This shows that as the number of particles in the system increases, the density of quantum statesbecomes so high that they form continuum. For a system consisting of 1023 particles the allowedstates are so crowded that it is impossible to enumerate and work with individual states. The best wecan do is to work with density of states g(E), which is the number of states per unit energy range.For a large system the density of states may be taken to be a smooth rapidly increasing function ofenergy.

1.8 MACROSCOPIC (MACRO) STATE

Consider a system containing a very large number N of particles in a vessel of fixed volume V atpressure, temperature T. The total energy of the system is E. The state of the system specified byparameters, which can be measure in laboratory such as pressure P, temperature T, volume V calledthe macroscopic description of state of the system. These parameters refer to the system as a whole.If the system is in equilibrium, the macroscopic characteristics P,V, T, E don’t vary with time.


1.9 MICROSCOPIC (MICRO) STATE

The most complete description of many particle system is given by specifying the positions andmomenta of its constituent particles. The state of the system characterized by positions and momentaof all its particles is called the microscopic or microstate. The state of a particle moving in space isspecified by 3 spatial coordinates (x, y, z ) and 3 momentum coordinates (px, py, pz). These 6 numbersx, y, z, px, py, pz completely determine the state of the particle. If there are N-particles in the system,the state of the entire system is specified by 6N numbers: of which 3N are spatial coordinates and3N are momentum coordinates. In equilibrium the macroscopic variables P, V, T characterizing thesystem are independent of time. However, the particles of the system are in random motion and themicroscopic states of the system undergo continuous change in course of time. So there are anenormously large number of microscopic states corresponding to each macrostate. In other words, amacrostate is realized through an enormously large number of microstates. The aim of the statisticalmechanics is to establish a relation between macrostate and microstates.

The statistical treatment of a thermodynamic system may be developed using either classical orquantum mechanics. In what follows we shall use quantum mechanics at most points. Objects of realworld obey quantum mechanics. Objects described by quantum mechanics don’t usually have arbitraryinternal energy. Bounded systems exist only in certain well-defined energy well-defined quantumstates. In other words, the energy of a system restricted to certain region of space is quantized. Theallowed energies of the system are called energy levels. In such systems one also finds that somedistinct states have the same energy, such states are said to be degenerate. The number of distinctstates corresponding to the same energy level is called the degree of degeneracy of the level. For asmall system we may identify the various quantum states and their energies without too much difficulty.For a large system the situation is quite different. The energy levels of a large system are very muchclose together and the mean separation between the energy levels is extremely small and so they maybe assumed to form continuum.

The most detailed description of a state of a N-particle system is given by a specification of thestate of each of the N-particles. We can make a chart showing which particles are in each of thevarious quantum states having energy 1, which ones in the states with energy 2 and so on. Thisdescription specifies a state of the system, which we call a microscopic state.

Usually, the microstates themselves are not very useful. If the particles are identical, it is of noconcern precisely which particles are in which energy states. Instead, we want to know how manyparticles are in energy state, without regard to which particle they are. Thus a different and moredirectly useful kind of the state description consists in specifying only the number of particles ineach of the possible energy level.

The specification that there aren1 particles in energy level 1 with degeneracy g1

n2 particles in energy level 2 with degeneracy g2

………………………………………………….ni particles in energy level i with degeneracy gi

is a description of a macrostate of the system. A macrostate is a less detailed specification of thesystem than the microstate. The number of ways in which this macrostate may be achieved is


called microstates of that macrostate. The quantity is also called statistical weight or thermodynamicprobability of that macrostate. The larger is, the greater the probability of finding the system inthat macrostate. If the volume V, the number of particles N and the total energy E of the system iskept constant, the equilibrium state of the system will correspond to that macrostate in which ismaximum. The principal objective of statistical mechanics is to determine the possible distributionof particles among the various energy levels and quantum states. If the distribution of particles of asystem among its quantum states is known, the macroscopic properties of the system can be determined.

If a system is composed of N identical and distinguishable particles, the total number ofmicrostates corresponding to a macrostate specified by the set of occupation numbersn1, n2, …. is given by

( ) ( )Ω = 1 21 2

1 2

!.......

! !.....n nN

g gn n

...(1.9.1)

In a gas containing N molecules, the molecules are distinguishable if the mean separation betweenthe molecules is much larger than their de Broglie wavelength. In deriving the above formula forthe number of microstates, it is assumed that there is no restriction on the number of particles thatcan occupy a quantum state.

If the system is composed of N indistinguishable bosons, the total number of microstatescorresponding to a macrostate specified by the set of occupation numbers n1, n2, …. is given by

+ −+ − + −

Ω = ⋅ = Π− − −

1 1 2 2

1 1 2 2

( 1)!( 1)! ( 1)!..........

!( 1)! !( 1)! !( 1)!i i

i i i

n gn g n g

n g n g n g...(1.9.2)

If the system is composed of N indistinguishable fermions, the total number of microstatescorresponding to a macrostate specified by the set of occupation numbers n1, n2, …. is given by

Ω = ⋅ = Π− − −

1 2

1 1 1 2 2 2

!! !........

!( )! !( )! ! ( )!i

i i i i

gg g

n g n n g n n g n...(1.9.3)

SOLVED EXAMPLES

Ex. 1. Two particles are to be distributed in an energy level, which is 3 fold-degenerate. Find thepossible microstates if the particles are (i) distinguishable (ii) indistinguishable bosons (iii)indistinguishable fermions.

Sol.(i) If the particles are distinguishable, they can be labeled as A and B. Here N = 2, n1 = 2,

g1 = 3. The number of possible microstates is:

1 21 2

1 2

N!( ) ( ) ....

! !.....n ng g

n nΩ =

22!

(3) 92!

Ω = =


A B A B A B

B A B A B A

AB AB AB

Fig. 1.9.1 Distribution of two distinguishable particles A and B in a triply degenerate energy level

It is worth to note that if there are more than one particle in a given quantum state, aninterchange of order in which the labelled objects appear does not produce a new microstate. So theAB and BA are the same.

(ii) If the particles are indistinguishable bosons, the number of possible microstates is

+ −

Ω = = =−

1 1

1 1

( 1)! 4!6

!( 1)! 2!2!

n g

n g

• •• •

• •• •

• •• •

Fig. 1.9.2 Distribution of two indistinguishable particles (bosons) in a triply degenerate energy level

The number of microstates is 6.(iii) The particles are indistinguishable Fermions.

The number of microstates is Ω = = =−

1

1 1 1

! 3!3.

!( ) 2!1!

g

n g n

• • • • • •

Fig. 1.9.3 Distribution of two indistinguishable particles (fermions) in a triply degenerate level

Ex. 2. Four particles are to be distributed among four energy levels 1 = 1, 2 = 2, 3 = 3, 4 = 4units having degeneracies g1 = 1, g2 = 2, g3 = 2, g4 = 1 respectively. The total energy of the system is 10units. Find the possible distribution (macrostates) and the microstates corresponding to most probablemacrostate. Assume that the particles are: (i) distinguishable, (ii) indistinguishable bosons and (iii)indistinguishable fermions.

Sol.(i) Particles are distinguishableThe possible macrostates are:

2, 0, 0, 2 = 2, 0, 0, 2, 1, 1, 1, 1 = 1, 1, 1, 1, 0, 3, 0, 1 = 0, 3, 0, 1, 1, 0, 3, 0 =1, 0, 3, 0, 0,2,2,0 =0,2,2,0. Distribution of particles (circles) is shown in the table.


4 = 4 o o o o ---------- --------3 = 3 ------- o ------- o o o o o2 = 2 -------- o o o o ---------- o o1 = 1 o o o -------- o --------

The number of microstates in above macrostates is given by

Ω = 1 21 2, ...... 1 2

1 2

N!........

! !.......n n

n n g gn n

Ω = =2 0 0 22,0,0,2

4!1 .2 .2 .1 6

2!0!0!2!

Ω = =1 1 1 11,1,1,1

4!1 .2 .2 .1 96

1!1!1!1!

Ω = =0 3 0 00,3,0,1

4!1 .2 .2 .1 32

0!3!0!1!

Ω = =1 0 3 01,0,3,0

4!1 .2 .2 .1 32

1!0!3!0!

Ω = =0 2 2 00,2,2,0

4!1 .2 .2 .1 96

0!2!2!0!

The most probable macrostates are 1, 1, 1, 1 and 0, 2, 2, 0.The possible microstates corresponding to the macrostate 1,1,1,1 are shown in the table. Since

the particles are distinguishable, they have been labelled A, B, C and D.Table 1.9.1: Microstates associated with macrostate 1, 1, 1, 1 for non-degenerate level

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

4 A A A A A A B B B B B B C C C C C C D D D D D D3 B B C C D D A A C C D D A A B B D D A A B B C C2 C D B D B C C D A D A C B D A D A B B C A C A B1 D C D B C B D C D A C A D B D A B A C B C A B A

The macrostate 1, 1, 1, 1 has 24 microstates as shown in the table. If the degeneracies of thesecond and third levels are considered, it is found that each of the 24 microstates has 4 microstates.The microstates corresponding to the first microstate are shown below:


Similarly each of the 24 microstates gives 4 microstates. Thus the total number of microstatescorresponding to the macrostate 1,1,1,1 is 96.

(ii) Particles are bosonsThe possible microstates are:

2, 0, 0, 2, 1, 1, 1, 1, 0, 3, 0, 1, 1, 0, 3, 0, 0, 2, 2, 0

i gi macrostatesni ni ni ni ni

4 1 2 1 1 0 03 2 0 1 0 3 22 2 0 1 3 0 21 1 2 1 0 1 0

The number of microstates associated with a macrostate n1, n2, …. is

+ −Ω = Π

−( 1)!

!( 1)!i i

i i i

n g

n g

+ − + − + − + −Ω = =

− − − −2, 0, 0,2(2 1 1)! (0 2 1)! (0 2 1)! (2 1 1)!

12!(1 1)! 0!(2 1)! 0!(2 1)! 2!(1 1)!

Similarly

Ω = Ω = Ω = Ω =1,1,1,1 0,3, 0,1 1, 0,3, 0 0, 2,2,04, 4, 4, 9

The most probable macrostate is 0, 2, 2, 0. The various microstates associated with thismacrostate are shown in the figure.

Fig. 1.9.4 Microstates associated with macrostate (0,2,2,0)


(iii) Particles are fermionsThere are two possible macrostates. They are 1, 1, 1, 1 and 0, 2, 2, 0. The number of

possible microstates associated with the first macrostate is

Ω = Π = =

− − − − −1, 1, 1, 1! 1! 2! 2! 1!

4!( )! 1!(1 1)! 1!(2 1)! 1!(2 1)! 1!(1 1)!

i

i i i

g

n g n

Similarly, the number of microstates associated with the second macrostate is

Ω =0,2,2,0 1

(four microstates corresponding to macrostate 1, 1, 1, 1) (one microstate 0, 2, 2, 0)

Fig. 1.9.5 Microstates associated with macrostate (1, 1, 1, 1) and (0, 2, 2, 0)

Ex. 3. A system consists of 4 distinguishable particles, labelled 1, 2, 3, and 4. These particles areto be distributed in two non-degenerate energy levels 1 and 2. Find the possible macrostates andcorresponding microstates. There is no restriction on the number of particle that can be put in a quantumstate.

Sol. For non-degenerate levels g1 = g2 =…… = 1, so the number of microstates correspondingto a macrostate is given by

Ω =1 2

N!

! !...n n

The possible macrostates are 4, 0, 3, 1, 2, 2, 1, 3 and 0, 4. The number ofmicrostates corresponding to these macrostates is:

Ω = = Ω = = Ω = = Ω = = Ω = =4, 0 3, 1 2, 2 1, 3 0, 44! 4! 4! 4! 4!

1, 4, 6, 4, 14!0! 3!1! 2!2! 1!3! 0!4!

So there are five macrostates and 16 microstates in total. The macrostate 2, 2 has maximumnumber of microstates (= 6) and hence it is the most probable macrostate of the system. Thesemicrostates are shown in the table.


Energy Levels Macrostate n1, n2 Microstates 1 2

1 2 3 4 × 4, 0 1

1 2 3 41 2 4 3 3, 1 41 3 4 22 3 4 1

1 2 3 41 3 2 41 4 2 3

2, 2 62 3 1 42 4 1 33 4 1 2

1 2 3 42 1 3 4

1, 3 43 1 2 44 1 2 3

× 1 2 3 4 0, 4 1

Ex. 4. A system consisting of 4 identical distinguishable particles has total energy 12 units. Theparticles are to be distributed in 4 non-degenerate energy levels with energies 1 = 1, 2 = 2, 3 = 3 and4 = 4 units . Find the possible macrostates and their microstates. Assume that any number of particlescan be put in the allowed energy levels.

Sol. The possible macrostates are:

1, 0, 1, 2, 0, 2, 0, 2, 0, 1, 2, 1, 0, 0, 4, 0o o o o o 4 = 4o o o o o o o 3 = 3

o o o 2 = 2o 1 = 1

Fig. 1.9.6 Distribution of four particles in four energy levels

The number of microstates belonging to the above macrostates is:

1, 0, 1, 24!

121!0!1!2!

Ω = =

0, 2, 0, 24!

60!2!0!2!

Ω = =


0, 1, 2, 14!

120!1!2!1!

Ω = =

0, 0, 4, 04!

10!0!4!0!

Ω = =

Ex. 5. A system composed of 6 bosons has total energy 6 units. These particles are to be distributedin energy levels 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6. Each level is triply degenerate.Calculate the thermodynamic probability (statistical weight) of all the macrostates and of the system.

Sol. The possible macrostates are 5,0,0,0,0,0,1, 4,1,0,0,0,1,0, 4,0,1,0,1,0,0,3,2,0,0,1,0,0,, 4,0,0,2,0,0,0, 3,1,1,1,0,0,0, 2,3,0,1,0,0,0, 3,0,3,0,0,0,0, 2,2,2,0,0,0,0,1,4,1,0,0,0,0, 0,6,0,0,0,0,0. There are 11 macrostates.

The thermodynamic probability (statistical weight) of a macrostate is given by

+ −

Ω = Π−

( 1)!

!( 1)!i i

i i i

n g

n g

I II III IV V VI63, 135, 135, 180, 90, 270,Ω = Ω = Ω = Ω = Ω = Ω =

VII VIII IX X XI180, 100, 216, 135, 28Ω = Ω = Ω = Ω = Ω =The thermodynamic probability of the system is

= 63 135 135 180 90 270 180 100 216 135 28 1532.ii

Ω = + + + + + + + + + + =∑Note: For fermions, the macrostates 1, 2, 3, 5, 10 and 11, in which there can be more than one

particle in a quantum state, are not allowed.

1 2 3 4 5 6 7 8 9 10 11

6 0

5 0

4 0 0

3 00 0 0

2 0 0 000 00 0

1 0 00 0 000 00 0000 000000

0 00000 0000 0000 000 0000 000 00 000 00 0


Ex. 6. A system consisting of 6 fermions has total energy 6 units. These particles are to be distributedin 5 energy levels 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4 units. Each energy level is triply degenerate. Findthe possible microstates.

Sol. The possible macrostates are 5. They are:

3,2,0,0,1, 3,1,1,1,0, 2,3,0,1,0, 3,0,3,0,0, 2,2,2,0,0

The thermodynamic probability (the number of microstates) of a macrostate is given by

Ω = Π−!

!( )!i

i i i i

g

n g n

The number of microstates associated with first macrostate is

Ω = ⋅ ⋅ ⋅ ⋅ =− − − − −I

3! 3! 3! 3! 3!9

3!(3 3)! 2!(3 2)! 0!(3 0)! 0!(3 0)! 1!(3 1)!

Similarly, the number of microstates associated with other macrostates can be calculated. Theycome out to be

II III IV V27, 9, 1, 27.Ω = Ω = Ω = Ω =The thermodynamic probability of the system is = I = 9 + 27 + 9 + 1 + 27 = 73.

CHAPTER

PHASE SPACE

2.1 INTRODUCTION

The specification of the state of a particle in classical mechanics involves the concept of phase space.To understand the meaning of phase space, consider a particle moving in one dimension, along thex-axis, say. In classical mechanics the state of motion at any instant is specified by specifying itsposition coordinate x and momentum coordinate px. Now imagine a two dimensional conceptual spacewith x and px as orthogonal axes. We call this space phase space. At any instant t, the state of theparticle is represented by a point (x, px) in the phase space and this point is called phase point orrepresentative point. As the particle moves on the straight line, x and px take on different values andthe corresponding representative point traces a trajectory in phase space. Thus the evolution ofsuccessive states of the particle will be represented by a trajectory in phase space. A point on thetrajectory in phase space represents a definite state of motion of the particle.

Consider a one dimensional harmonic oscillator of mass m with total energy E. Let q and pdenote the instantaneous position and momentum of the particle. The total energy of the oscillator isgiven by

2

2 21E, frequency of oscillator

2 2

pm q

m+ ω = ω =

This equation can be put in the form

( )2 2

21

2 E2E/

q p

mm+ =

ω

The form of this equation shows that if we plot the instantaneous position q and momentum p

on q-p plane for one cycle of motion, we get an ellipse with semi-major axis, 2

2Ea

m=

ω and

semi-minor axis, 2 Eb m= . Each point on the ellipse represents same state of the oscillator. The

area of the ellipse is

= π = π ⋅ = π ωω 2

2E EA 2 E 2ab m

m

Phase Space 267

According to quantum mechanics energy of oscillator is given by E = 1

,2

n + ω

0, 1, 2,......n = . The area of the ellipse then becomes

1 1

A 22 2

n n h = + π = +

The area between two successive ellipses (which represent two successive energy levels) is

A h∆ = = Planck’s constant.

Usually the spatial coordinate of a particle is denoted by q and the corresponding momentumby p. Then the state of the particle is specified by stating its position and momentum coordinates(q, p). To specify the state of the particle more precisely, it is convenient to subdivide the ranges ofthe variables q and p into arbitrarily small discrete interval. One can choose fixed intervals of sizeq for the subdivision of q and fixed interval of size p for the subdivision of p. The phase space isthen subdivided into small cells of equal size and of two dimensional volume (i.e., area)qp = h0

where h0 is a constant having the dimensions of angular momentum. The state of the system canthen be specified by stating that its coordinates lies in some interval between q and q + q and itsmomentum lies in some interval between p and p + p i.e., by stating that the representing point (q,p) lies in a particular cell of phase space. The specification of the state of the system clearly becomesmore precise as one decreases the size chosen for the cells into which phase space has been dividedi.e., as one decreases the magnitude chosen for h0. Of course, h0 can be chosen arbitrarily small inthis classical description.

Fig. 2.1.1 Subdivision of phase space Fig. 2.1.2 Trajectory of a phase pointin cells of size q p = h0 in phase space

A particle moving in three dimensions requires 3 position coordinates (x, y, z) and 3 momentumcoordinates (px, py, pz) to specify its state of motion. The corresponding phase space has 6 dimensionswith spatial coordinates x, y, z, and momentum coordinates px, py, pz as orthogonal axes and the stateof the particle is specified by a point (x, y, z, px, py, pz). If we denote the position and momentumcoordinates by q1, q2, q3, p1, p2, p3 respectively then the state of the particle is denoted by point(q1, q2, q3, p1, p2, p3). Each point in this 6-dimensional phase space represents a possible state of


motion of the particle. If the particle under consideration is a molecule of a gas, the corresponding6-dimensional phase space is called -space. ( stands for molecule). The state of a system consistingof N-particles at any instant will be represented by N phase points in -space.

Like-wise the state of a N-particle system is specified by 3N position coordinates x1, y1, z1; x2,y2, z2; ……….xN, yN, zN and 3N momentum coordinates px1, py1, pz1;………..pxN

, pyN, pzN

. Insteadof denoting position coordinates by x, y, z and momentum coordinates by px, py, pz let us denotethem by generalized position coordinates q1, q2, q3 and by generalized momentum coordinates p1,p2, p3 respectively. So the state of the N-particle system is specified by 3N generalized positioncoordinates q1, q2, q3, …… q3N – 2, q3N – 1, q3N and corresponding generalized momentum coordinatesp1, p2, p3, ……, p3N – 2, p3N – 1, p3N. These 6N coordinates along with the equations of motion viz

Hamilton’s equations ∂ ∂= = − =∂ ∂

H H, , 1,2,3, ....... k k

k k

q p kp q

, 3N, where H is Hamiltonian of the

system, completely determine the behaviour of the system.We now imagine a 6N dimensional space with 6Nrectangular axes, one for each of the spatial coordinatesq1, q2,…….., q3N and for each momentum coordinatep1, p2,………,p3N. This 6N dimensional phase space of theN-particle system is called -space. The state of the entiresystem (gas) at any time t is completely specified by a phasepoint in -space. In course of time the spatial andmomentum coordinates undergo continuous change, thecorresponding phase point traces a trajectory in -space; themotion of the phase point is governed by Hamilton’sequations. Each phase point on the trajectory in -spacerepresents a possible microstate of the entire system.

Once again the -space can be subdivided into littlecells of volume q1q2….. q3Np1p2 ….. p3N = (h0)3N.The state of the system can then again be specified by statingin which particular range or cell in phase space, thecoordinates q1, q2,…..,q3N, p1, p2,……, p3N of the systemcan be found.

2.2 DENSITY OF STATES IN PHASE SPACE

Consider a particle moving in x-direction. Classical physics puts no restriction on the accuracy withwhich the simultaneous specification of position and momentum can be made. That is the uncertaintiesq and p in specification of position and momentum can be made as small as we like. In otherwords, the size of the cell q p = h0 into which the phase space is subdivided can be chosen arbitrarilysmall. But quantum mechanics imposes a limitation on the accuracy with which a simultaneousspecification of coordinate q and its corresponding momentum p can be made. According to Heisenberguncertainty principle, the uncertainties in q and p are such that q p h, where h is Planck’sconstant. The subdivision of the phase space into cells of volume less than h is physically meaningless.

[q] stands for 3N spatial coordinatesand [p] for 3N momentum coordinates.

Fig. 2.1.3 Trajectory of phase point in-space

Phase Space 269

For a particle free to move in three dimensions, if x and px denote the uncertainty in positionand momentum then

x .px h

Similar relations hold for other components also.

y . py h

z . pz h

Hence x . y . z . px . py . pz h3 ...(2.2.1)

The product x . y . z . px . py . pz = h3 represents an element of volume in phasespace. Two particles whose representative points lie in such an elementary cell cannot be distinguishedand hence the representative points in this cell represent a single quantum state. It follows from abovethat different quantum states shall corresponds to different elements of volume in the phase spaceonly if the size of these elements is no less than h3. Therefore a volume equal to h3 in the phasespace may be allotted to each microscopic state of the particle. In other words, each elementary cellof phase space represents a microstate of the particle. The state of a particle is specified by stating inwhich particular cell the coordinates x, y, z, px, py, pz of the particle lie. The process of dividing thephase space into cells finite size is termed quantization of phase space. The number of cells d inan element of phase volume d is

3 3 3

V.x y z x y zdx dy dz dp dp dp d dp dp dpd

dh h h

ΓΩ = = =

The total number of cells (microstates or quantum states) in the entire phase space of the particleis given by

3 3

Vx y z

x y z

dx dy dz dp dp dpdp dp dp

h hΩ = =∫ ∫ ∫ ∫ ∫ ∫ ∫∫∫ ...(2.2.2)

where V = dx dy dz∫∫∫ is the volume in coordinate available to the particle.

In many applications we are interested in a quantity g(p)dp which represents the number ofstates (cells) of the particle moving in volume V such that the magnitude of momentum lies in theinterval p and p + dp irrespective of its direction. To calculate this number we change the volumeelement in momentum space dpx dpy dpz in spherical polar coordinates. Thus

= θ ϕ2 sinx y zdp dp dp p dp d

Hence

22 2

3 30 0

V V( ) sin (4 )

p dp

p

g p dp p dp d p dph h

+ π π

= θ ϕ = π∫ ∫ ∫In terms of energy E = p2/2m the number of states in the energy range dE about E is given by

3/ 2 1/ 23

2 V(E) E (2 ) E Eg d m d

h

π= ...(2.2.3)


If the particle has internal degree of freedom such as spin, there will be (2s + 1) spin statescorresponding to each momentum or energy states. Therefore, the number of states then becomes

3/ 2 1/ 2

3

2 V(E) E (2 1) (2 ) E Eg d s m d

h

π= + ...(2.2.4)

For spin 1/2 particles (such as electrons) s = 1/2 , the number of states in the energy range dEat E is

3/ 2 1/ 23

4 V(E) E (2 ) E Eg d m d

h

π= (for spin 1/2 particles) ...(2.2.5)

The function g(E) is called the density of states and is defined as the number of quantumstates per unit energy range at energy E and is given by

3/ 2 1/ 2

3

2 V(E) (2 1) (2 ) Eg s m

h

π= + ...(2.2.6)

2.3 NUMBER OF QUANTUM STATES OF AN N-PARTICLE SYSTEM

In 6N-dimensional -space of an N-particle system, a quantum state of the system is assigned anelement of hypervolume h3N. The number of cells (quantum states or microstates) in a volume element

dq1…dq3N, dp1…….. dp3N at point (q, p) in -space is d = 1 3N 1 3N3N

..... . .....dq dq dp dp

h. The total

number of accessible states in -space is

1 3N 1 3N3N

1.... . ......dq dq dp dp

hΩ = ∫

= 1 3N 1 3N3N

1..... . .......dq dq dp dp

h ∫ ∫

= N1 3N3N

1.V . .....dp dp

h ∫

= N

3 3 31 2 N3N

V. .........d p d p d p

h ∫

= N

3N3N

Vd p

h ∫

= π⋅N 3N/2

3N

V (2 E)

(3N/2)

m

h,

2

E2

p

m= , (for large N) ...(2.3.1)

CHAPTER

ENSEMBLE FORMULATION OF

3.1 ENSEMBLE

The method of ensemble in statistical physics was introduced in 1902 by the American physicist J.W.Gibbs. Consider a system consisting of N molecules with total energy E enclosed in a vessel of volumeV. The macroscopic state of the system is described by pressure P, volume V and energy E. Withpassage of time the coordinates and momenta of molecules and hence the microscopic states of thesystem undergo continuous change. Meaning thereby, a macrostate is realized through an enormouslylarge number of microstates. Now imagine a large number (possibly infinite) of systems, which areexactly identical in structure to the system of interest, but suitably randomized in microscopic statessuch that they represent at one time the possible states of the actual system attained in the course oftime. This mental collection of similar non-interacting, independent systems is called an ensemble.All the members of an ensemble, which are identical in feature like N, V and E are called the elementsor systems. These elements, although identical in structure are randomized in the sense that theydiffer from one another in the coordinates and momenta of the individual molecules i.e., the elementsdiffer in their unobservable microscopic states.

The state of an N-particle system (element, or component) of an ensemble can be specified bythe 3N canonical coordinates q1, q2, ….., q3N, and 3N canonical momenta p1, p2, ….., p3N of the Nmolecules. The 6N dimensional space spanned by the 3N spatial and 3N momentum coordinates iscalled -space of the system. An element of the ensemble is represented by a point and the ensembleis represented by a distribution of points in -space usually a continuous distribution.

Instead of denoting the spatial and momentum coordinates of an N-particle system by q1,….,q3N,and p1,….., p3N, it is convenient to denote them by q1, ……, qf, p1,……, pf respectively. Of course,f = 3N. The system is said to have f degrees of freedom. In 2f (= 6N) dimensional -space having 2frectangular axes, one for each of spatial coordinates q1, ……, qf and one for each of correspondingmomentum coordinates p1,….., pf, the state of the system is represented by a point. In this space theensemble of system looks like cloud of points. The ensemble may be conveniently described by adensity function (p, q, t) where (p, q) is an abbreviation for p1, p2, ….., pf, q1, q2, ….., qf, sodefined that (p, q, t) dfp dfq is the number of representative points which at time t are contained inthe infinitesimal volume element dfp dfq of -space centered about the point (p, q). An ensemble is



completely specified by (p, q, t). It is to be emphasized that the members of an ensemble are themental copies of a system and do not interact with one another.

Each element of the ensemble is a quantum mechanical system of N interacting molecules in acontainer of volume V. The value of N and V along with the force law between the molecules, aresufficient to determine the eigen values and the quantum states of the Schrodinger equation alongwith their associated degeneracy. These energies are the only energies available to the system.

3.2 DENSITY OF DISTRIBUTION (PHASE POINTS) IN -SPACE

The state of an ensemble can be described in terms of the density (q, p, t) with which therepresentative points are distributed in -space. For an ensemble consisting of very large number ofsystems, the distribution of representative points is continuous and the density of phase points in-space can be treated as a continuous function. With passage of time, the microstates of the systemsundergo change and their representative points move in -space from one region to another. So thedensity of phase points is function of q1,…., qƒ, p1,……., pƒ and time t and may be written as

= (q1,….., qƒ, p1,……, pƒ, t ) = (q, p, t) ...(3.2.1)

The meaning of is such that dq1,…., dqf dp1,……, dpf or d ...(3.2.2)

represents the number of systems in infinitesimal hypervolume d = dq dp = dq1,…, dqf .dp1,…..,dpf

located at point (q, p). The number of systems dM in hypervolumre d isdM = d ...(3.2.3)

and the total number of systems in the ensemble is

M = ρ Γ∫ d ...(3.2.4)

Where the integration is over the accessible phase space of the ensemble. The average value ofa physical quantity Q (q, p) is given by

Q( , ) ( , , )

Q( , , )

q p q p t d

q p t d

ρ Γ=

ρ Γ

∫∫

...(3.2.5)

The concept of ensemble is introduced to facilitate the calculation of average value, a physicalsystem.

3.3 PRINCIPLE OF EQUAL A PRIORI PROBABILITY

To specify the microscopic state of a system, the phase space of the system is subdivided into smallcells of equal size. Each cell represents a microscopic state of the system. According to the postulateof equal a priori probability, an isolated system in equilibrium is equally likely to be in any of itsaccessible microscopic states satisfying the macroscopic conditions of the system.

Ensemble Formulation of Statistical Mechanics 273

A many particle isolated system, according to quantum mechanics, possesses discrete energylevels and discrete quantum states. It is found that many distinct quantum states correspond to thesame energy levels. The number of different quantum states having the same energy is called thedegeneracy of that energy level. The particles constituting the system are distributed among the variousenergy states. The specification of the macroscopic parameters, such as total energy E, volume V,the total number N of particles of the system, defines a particular macroscopic state of the system.Let the allowed energy levels be denoted by 1, 2, …….., i and the occupation number of theseenergy levels by n1, n2, ….., ni. Let the system obey the constraints

N and Ei i ii i

n n= ε =∑ ∑ ...(3.3.1)

There can be a large number of different ways in which the total energy E of the system can bedistributed among N-particles constituting the system. Each of these different ways specifies a particularmicroscopic state of the given system. To a given macroscopic state there may be a large number ofmicroscopic states. According to the principle of equal a priori probability, when a system is instatistical equilibrium, all the microstates are equally probable.

There is no direct proof of this postulate. It does not contradict any known laws of mechanics.All calculations based on this postulate have yielded results that are in very good agreement withobservations. The validity of this postulate can therefore be accepted with great confidence as thebasis of our theory.

3.4 ERGODIC HYPOTHESIS

In statistical mechanics we often deal with average or the mean of a quantity. The average of a physicalquantity can be determined in two ways:

(i) One could consider an ensemble of a large number of identical systems and average thephysical quantity over all these systems at one instant of time to determine its ensembleaverage.

(ii) A system could be followed over a very long period of time, during which the physicalquantity of the system takes different values. The average of the physical quantity overthe long period gives the time averaged value of the quantity.

According to ergodic hypothesis the mean over the ensemble is equal to the mean over time. Sofar, there is no proof of the validity of this statement in the general case and is taken as one of thebasic assumptions of statistical physics. The ergodic hypothesis and the principle of equal a prioriprobability are the main postulates that are employed for studying the properties of an ensemble.

3.5 LIOUVILLE’S THEOREM

Consider an isolated system specified by spatial and momentum coordinates q1, ……., qf, p1,……..,pf.In 2f dimensional -space the ensemble of systems appears as a cloud of points. In course of time thephase points move in -space because of change in position and momentum coordinates. This will


result in change in the distribution density (q, p, t) of phase points. Let us define (q, p, t) suchthat (q, p, t) = dq1,…..,dqf.dp1,…….,dpf represents the number of phase points (systems) inhypervolume d = dq dp located at point (q, p). The Liouville’s theorem gives the rate of change ofdensity (q, p, t) at a fixed point in -space.

Consider an element of hypervolume of phase space located between q1 and q1 + dq1, q2 andq2 + dq2, ……, qf and qf + dqf and p1 and p1 + dp1, p2 and p2 + dp2,…………,pf and pf + dpf. Thevolume of this element is d = dq1,………,dqf . dp1,……..,dpf. The coordinates and momenta of thephase points vary according to the Hamilton’s equations of motion

H H,i i

i i

q pp q

∂ ∂= = −∂ ∂

...(3.5.1)

where H = H (q1, …., qf, p1,…….,pf) is Hamiltonian of the system. The change in q’s and p’s resultsa change in the number of phase points in the element of hypervolume. In time dt, the change in thenumber of phase points within this hypervolume of phase space is

∂ρ Γ ∂ dt d

t

This change is equal to the difference in the number of phase points entering and leaving thisvolume in time dt. The number of phase points entering this volume in time dt through the facelocated at q1 = constant is

ρ 1 2 1( , , )( ) ........... ............f fq p t q dt dq dq dp dp

The number of phase points leaving through the opposite face located at q1 + dq1 = constant is

∂ ρ + ρ ∂ 1 1 1 2 1

1

( ) ....... .........f fq q dq dt dq dq dp dpq

The net number of phase points entering the hypervolume element in time dt

= ∂ ρ ρ − ρ + ∂

11 2 1 1 1 2 1

1

( )... ..... ...... ....f f f f

qq dtdq dq dp dp q dq dtdq dq dp dp

q

= – ( )∂ ρ∂

1 1 11

..... .......f fq dtdq dq dp dpq

= –∂ ρ

Γ∂1

1

( )qdt d

q

The total net increase in time dt of the number of phase points in the hypervolume of phasespace is obtained by summing the net number of phase points entering the hypervolume through allthe faces labeled by q1,…..,qf . p1,……..,pf. Thus one obtains

( ) ( )1 1

f f

i ii ii i

dt d q p dt dt q p= =

∂ρ ∂ ∂Γ = − ρ + ρ Γ ∂ ∂ ∂

∑ ∑


or ( ) ( )=

∂ρ ∂ ∂= − ρ + ρ ∂ ∂ ∂ ∑

1

f

i ii ii

q pt q p ...(3.5.2)

Fig. 3.5.1 A volume element in phase space

Equation (3.5.2) can be written as

1

fi i

i ii i i ii

q pq p

t q p q p=

∂ ∂∂ρ ∂ρ ∂ρ= − + + ρ + ∂ ∂ ∂ ∂ ∂ ∑

...(3.5.3)

Making use of Hamilton’s equations, we have

2 2H H0i i

i i i i i i

q p

q p q p p q

∂ ∂ ∂ ∂+ = − =∂ ∂ ∂ ∂ ∂ ∂

...(3.5.4)

In view of Eqn. (3.5.4), Eqn.(3.5.3) reduces to

=

∂ρ ∂ρ ∂ρ = − + ∂ ∂ ∂ ∑

, 1

f

i ii iq p i

q pt q p

...(3.5.5)

Making use of Hamilton’s equations of motion we can write Eqn. (3.5.5) as

=

∂ρ ∂ρ ∂ ∂ρ ∂ = − − ∂ ∂ ∂ ∂ ∂ ∑

, 1

H Hf

i i i iq p it q p p q

...(3.5.6)


Equation (3.5.5) or (3.5.6) is known as the Liouville theorem. It gives the rate of change ofdensity at a fixed point in -space. In view of the following results:

,

f fi

ii i p ti i

qq

q q t t

∂∂ρ ∂ρ ∂ρ = = ∂ ∂ ∂ ∂ ∑ ∑

and ,

f fi

ii i q ti i

pp

p p t t

∂∂ρ ∂ρ ∂ρ = = ∂ ∂ ∂ ∂ ∑ ∑

we can write Eqn. (3.5.5) as follows:

=

∂ρ ∂ρ ∂ρ+ + = ∂ ∂ ∂ ∑

1,

0

f

i ii iiq p

q pt q p

, , ,

0q p p t q tt t t

∂ρ ∂ρ ∂ρ + + = ∂ ∂ ∂

0d

dt

ρ = ...(3.5.7)

Thus, the total derivative of density (q, p, t), which is a measure of the rate of change of inthe immediate vicinity of a moving phase point (q and p changing) in -space, is zero. In other words,the density of a group of phase points remains constant along their trajectories in the -space. Thedistribution of phase points moves in -space like an incompressible fluid. Gibbs called this conclusionthe principle of the conservation of density in phase.

From Eqn. (3.5.7) we can obtain another fundamental principle of statistical mechanics. Considera region in -space which, although finite, is small enough for the density to be treated as uniformthroughout; if the hypervolume of the region is , the number M of the phase points in this regionwill be given by

M = ...(3.5.8)

On differentiating this expression with respect to time t, it is seen that

( )

( M)d d d

dt dt dt

ρ δΓδ = δΓ + ρ ...(3.5.9)

If it is supposed that the boundaries of the region under consideration are permanentlydetermined by the phase points that were originally on the surface, then no phase points can enter orleave this region. In other words, the points on the outer surface act like a continuous thin skin bywhich all the points in the region are enclosed. The hypersurface enclosing the region changes itsshape and moves about in gamma space due to the flow of phase points. Further, since each phase

point represents a definite system, these points can neither be created nor destroyed. So ( M) 0.d

dtδ =

Eqn. (3.5.9) then becomes


( )d d

dt dt

ρ δΓδΓ + ρ = 0 ...(3.5.10)

Since 0d

dt

ρ = , we have

( )d

dt

δΓ= 0 ...(3.5.11)

This means that the volume or extension-in-phase in -space of the particular region, occupiedby a definite number of phase points, does not change with time. Since every finite arbitraryextension-in-phase may be regarded as composed of infinitesimal parts, the result may be generalized.This theorem, mathematically expressed by Eqn. (3.5.11) is called the principle of conservation ofextension in phase.

3.6 STATISTICAL EQUILIBRIUM

An ensemble is said to be in statistical equilibrium if the density of phase points is independent oftime at all points in -space i.e.

,

0q pt

∂ρ = ∂ for all q’s and p’s.

Consider an ensemble of conservative systems for which energy E is constant in time and isfunction of q’s and p’s. Thus

= (E) , E = E (q, p) and E

0d

dt= ...(3.6.1)

Therefore E E

. , .E Ei i i i

d d

q d q p d p

∂ρ ρ ∂ ∂ρ ρ ∂= =∂ ∂ ∂ ∂ ...(3.6.2)

According to Liouville’s theorem

∂ρ ∂ρ ∂ρ = − + ∂ ∂ ∂ ∑

,

f

i ii iq p i

q pt q p ...(3.6.3)

Making use of Eqn. (3.6.2) in (3.6.3), we have

∂ρ ρ ∂ ∂= − + ∂ ∂ ∂

∑,

E E

E

f

i ii iiq p

dq p

t d q p ...(3.6.4)

Since E = E (q, p) and dE/dt = 0, we have

E E E

0f

i ii ii

dq p

dt q p

∂ ∂= + = ∂ ∂ ∑ ...(3.6.5)


From Eqns. (3.6.4) and (3.6.5)

∂ρ = ∂ ,

0q pt

for all q’s and p’s. ...(3.6.6)

Thus, an ensemble is in statistical equilibrium if density of phase points (or the probability offinding the phase points) in the various regions of -space is independent of time. This means thatevery portion of the phase space continues to contain the same number of phase points at all times.Under these conditions, the average values of the properties of the systems in the ensemble alsodo not change with time.

THERMODYNAMIC FUNCTIONS

3.7 ENTROPY

Entropy is a very important thermodynamic function, which connects thermodynamics to statisticalmechanics. It is known from thermodynamics that when a system, with constant volume and energy,is in equilibrium the entropy is maximum. On the other hand, according to statistical mechanics,such a system is in equilibrium when the total thermodynamic probability is a maximum. It appears,therefore, as suggested by Boltzmann, that there should be a relationship between entropy andthermodynamic probability. The thermodynamic probability is defined as the number of microstatescorresponding to the given macrostate. Let the entropy S and be related through the expression

S = f () ...(3.7.1)

Consider two systems having entropies S1 and S2 and thermodynamic probabilities 1 and 2

respectively. In view of Eqn. (3.7.1), we haveS1 = f(1) and S2 = f(2).

Since the entropy is an additive quantity, the entropy of the combined system is equal toS12 = S1 + S2. The thermodynamic probability is a multiplicative quantity, therefore, the jointthermodynamic probability of the combined system is 12 = 1 2 and S12 = f (1 2).

S1 + S2 = S12

f (1) + f (2) = f (12) ...(3.7.2)

Differentiating Eqn. (3.7.2) with respect to 1, we have

ƒ' (1) = [ƒ' (1 2)] 2 ...(3.7.3)

Similarly, differentiating Eqn. (3.7.2) with respect to 2, we have

ƒ' (2) = [ƒ' (1 2)] 1 ...(3.7.4)

From Eqns. (3.7.3) and (3.7.4)

1 2

2 1

( )

( )

f

f

′ Ω Ω=

′ Ω Ω


1 ƒ' (1) = 2 ƒ' (2) = k, (= a constant, say)

df (1) = ΩΩ

1

1

dk

f (1) = k ln 1 + C1

Similarly, we can havef (2) = k ln 2 + C2

C1 and C2 are constants. General form of these relations isf () = k ln + C

or S = k ln + C ...(3.7.5)

At absolute zero, any system is in most ordered state and this state has only one microstate i.e., = 1 and this state is assigned zero entropy S = 0. The constant C in Eqn. (3.7.5) comes out to bezero. So, we have

S = k ln ...(3.7.6)

where k is Boltzmann constant. Equation (3.7.6) is taken as the statistical definition of entropy.

3.8 FREE ENERGY

In a mechanical system, such as a spring, the work done the system is stored in the system as potentialenergy and this energy may be recovered as work. In the similar way one can store energy inthermodynamic system, which can be recovered in the form of work. The energy, which can bestored and recovered, is called free energy. The four kinds of free energy that can be stored inthermodynamics system are: (i) Internal Energy E, (ii) Enthalpy H = E + PV, (iii) HelmholtzFree Energy F = E – TS and (iv) Gibb’s Free Energy G = E – TS + PV.

The energy of a system also depends on the number of its constituent particles. When a particleleaves a system, it takes away a definite amount of energy with it. When it enters a system, it addsenergy to it. To take into account the change in energy contributed by a particle we introduce aquantity, chemical potential µ, which is defined as the change in energy of the system associatedwith unit change in number of particles.

E

N

∂ µ = ∂ ...(3.8.1)

If a system is to be in equilibrium state, the temperature T, pressure P and chemical potential µmust be the same throughout the system.

The law of conservation of energy for a system with variable number of particles can be writtenas

= − +µE T S P V Nd d d d ...(3.8.2)

where dN is the change in number of particles. The first term on the right hand side represents thechange in energy due to transfer of heat energy to the system, second term represents change inenergy due to work done by the system and the third term represents the change in energy due to


change in number of particles in the systemFor an isolated system at constant volume, Q = TdS = 0 and dV = 0. For such a system

E Nd d= µ

S,V

E

N

∂ µ= ∂ ...(3.8.3)

Hence the chemical potential represents the variation of the energy of an isolated system of constantvolume brought about by a unit change in number of particles.

If U = constant and N = constant then Eqn.(3.8.2) becomesT dS = P dV

∂ ∴ = ∂ N, E

S P

V T ...(3.8.4)

If V = constant, and E = constant, then Eqn. (3.8.2) becomesT dS = – dN

∂ µ ∴ = − ∂ V, E

S

N T...(3.8.5)

3. Helmholtz Free Energy

Helmholtz free energy is defined byF = E – TS ...(3.8.6)

Therefore,dF = dE – T dS – S dT ...(3.8.7)

Substituting dE from Eqn. (3.8.2) in (3.8.7), we have

d F = –Pd V – Sd T + d N

Whence ∂ ∂ ∂ µ = − = − = ∂ ∂ ∂ T, V V, N V, N

F F F, P , S

N V T ...(3.8.8)

Thus Helmholtz function F plays a very important role in calculation of thermodynamic quantitiesof a system.

3.9 ENSEMBLE FORMULATION OF STATISTICAL MECHANICS

There are three kinds of formulations of statistical physics. They are:(i) Microcanical Ensemble.

(ii) Canonical Ensemble.

(iii) Grand Canonical Ensemble.


3.10 MICROCANONICAL ENSEMBLE

Consider a system containing N-particles in a volume V with total energy of particles E. The wallsof the container are perfectly insulating. Thus, the system is completely isolated from outside world.There can be no exchange of energy and matter from the surroundings. Therefore E, V and N havefixed values. The macroscopic state of the system is specified by E, V, and N. A collection of a verylarge number of such identical isolated systems is called a microcannical ensemble. The systemsof the ensemble are also called elements or components. Truly isolated systems can never be realizedin the laboratory.

In 6N dimensional -space the microstate of the N-particle system is represented by a point.The locus of all points in -space satisfying the condition E = constant, defines a surface called theergodic surface of energy E. As the state of the system evolves in time according to Hamilton’sequations of motion, the representative point traces out a path in -space. This path always stays onthe same energy surface because by definition energy is conserved for an isolated system.

We cannot specify exactly the energy of a system. However, we can certainly specify the energywithin a narrow range, say E and E + dE. We can then select two neighboring ergodic surfaces, oneat E and the other at E + E. In 6N dimensional -space the microcanonical ensemble, whose eachmember has energy between E and E + E, is represented by points that lie between two ergodicsurfaces of energies E and E + E. A microcanonical ensemble may be represented by distributionof points in -space characterized by a density function (p, q, t) defined in such a way that(p, q, t) d3Np d3Nq gives the number of representative points contained in the volume elementd3Np d3Nq located at point (p, q) in -space at the instant t. For microcanonical ensemble the densityfunction satisfies the condition

(p, q) = 1 if E < H(p, q) < E + E ...(3.10.1)

= 0 otherwise

Fig. 3.10.1 Microcanonical ensemble (E V N fixed)

The average of a physical quantity Q(p, q) is defined by

ρ=

ρ∫∫

3N 3N

3N 3N

Q( , ) ( , )R

( , )

d pd q p q p q

d pd q p q...(3.10.2)


3.11 CLASSICAL IDEAL GAS IN MICROCANONICAL ENSEMBLEFORMULATION

The statistical description of a system in this approach is given in terms of statistical weight (EVN)which leads to the thermodynamic description in terms of entropy S through Boltzmann equation

S = k ln (EVN). ...(3.11.1)

All other thermodynamic properties of the system can be derived from these relations. We shall considerideal gas as an example in this formulation.

The number of accessible microstates of an isolated N-particle system occupying a volume Vwith total energy lying in the range E and E + dE is

− Ω = ⋅ π ⋅

N3N / 2 [(3N / 2) 1]

3

V (3N / 2)(E,V,N) (2 ) E

(3N / 2)!m

h...(3.11.2)

For large N this results simplifies to

π

Ω =3N / 2N

3N

(2 E)V(E, V,N)

(3N / 2)!

m

h...(3.11.3)

The entropy of the gas is

= ΩS lnk

π =

N 3N / 2

3

(2 E)Vln . .

(3N / 2)!

mk

h

Using Stirling’s approximation ln n! = n ln n – n, we can simplify the expression for entropyas follows.

( ) = π − −

N N3/ 2

3

V 3N 3N 3NS ln 2 E ln

2 2 2k m k

h

= π +

NN 3/ 2N3/ 23

V 2 3S ln (2 E) N

3N 2k m k

h

π = +

3/ 2

3

4 EV 3S N ln . N

3N 2

mk k

h ...(3.11.4)

This expression for the entropy S does not satisfy the additive property. For example, if weincrease E, V, N by a factor , the entropy of the new system does not become S. The aboveexpression for entropy gives the entropy of new system S', where

η π η ′ = η + η = η η+ η ≠η η

3/ 2

3

V 4 E 3S N ln . N N ln S S

3 N 2

mk k k

h


This result is known as Gibbs paradox. The origin of Gibbs paradox lies in the classical assumptionthat identical particles are distinguishable. The expression for the number of quantum states wasderived making use of this assumption. In fact, when two identical particles are permuted, the resulting2! states produce no observable effects. Similarly, the permutation of N-particles gives N! states whichare not distinct states. So the expression for is larger by a factor N!. In quantum mechanics, identicalparticles are inherently indistinguishable. The correct expreesion for the number of quantum states, taking the indistinguishability of identical particles into consideration, should be

π Ω =

N 3N/2

3

1 V (2 E). (indistinguishable particles)

N! (3N/2)!

m

h...(3.11.5)

The expression for entropy now modifies to

π = Ω= +

3/ 23/2

3

41 V E 5S ln N ln N

N N 3 2

mk k k

h ...(3.11.6)

For an ideal gas = 3E N T

2k ...(3.11.7)

Therefore

π= + +2

V 3 2 T 5S N ln N ln N

N 2 2

mkk k k

h...(3.11.8)

Equation (3.11.8) is known as Sackur-Tetrode equation.The thermal de Broglie wavelength of particle is

λ = π2 T

h

mk ...(3.11.9)

The expression for entropy in terms of is

= +

λ 3

V 5S N ln N T

2Nk k ...(3.11.10)

The other thermodynamic quantities can be calculated from the expression of entropy.Helmholts free energy

F = E – TS

= π −

3/2

2

2 TNN T ln

V

mkek

h...(3.11.11)

The pressure of the gas

∂ ∂ = = − ∂ ∂ E, N T, N

S FP T

V V

= N T

V

k...(3.11.12)


Chemical potential

∂ µ = − = ∂ π

3/22

E, V

S NT T ln

N V 2 T

hk k

mk...(3.11.13)

3.12 CANONICAL ENSEMBLE AND CANONICAL DISTRIBUTION

Consider a system, with fixed number of particles N and volume V, immersed in a large heat reservoirat temperature T. The wall of the container in which the system is enclosed are heat conducting.When the equilibrium is reached, the temperature of the system attains the value T. The macroscopicstate of the system is specified by temperature T, volume V and number of particles N, all of whichhave fixed values. The energy of the system assumes different values due to exchange of heat withthe heat reservoir. Now imagine a large number of such systems, say M , which are in thermalcontact with each other and immersed in a large heat reservoir kept at temperature T. The aggregateof all these systems is called canonical ensemble. Thus, the entire ensemble is at the same temperatureT. Each system of the ensemble has the same values of N, V, and T. Now, the entire canonical ensembleis isolated from the surrounding. All the systems (called elements or components) of the canonicalensemble are at the same temperature but the different systems have different energies.

Fig. 3.12.1 Canonical ensemble

In a system consisting of a gas of identical molecules, the individual molecules may be treatedas a system and the rest of the molecules as a heat bath. Energy of different molecules (systems) havedifferent values.

Gibbs Canonical Probability Distribution

The Gibbs canonical distribution gives the probabilities of occurrence of different energy states ofsystems (components) constituting the canonical ensemble. To obtain an expression for this probabilityconsider a system A placed in a heat reservoir, which we denote by A'. Our system of interest A and


the heat reservoir A' constitute a composite system A*. The composite system is totally isolated andhence it is a microcanonical system. The system A and A' are free to exchange energy but thetemperature of the heat reservoir remains constant because of its large size. In thermal equilibriumthe temperature of A is the same as that of the heat reservoir. The energy of A is not fixed. Themacroscopic state of the system is specified by T, V, N.

We assume that the system A of the ensemble possesses a discrete set of microstates labeled by

1, 2, 3, …, r ,….and in these states the system has energy ε ε ε ε1 2 3, , , ........, , ........r It is possible that

many distinct microstates (quantum states) have the same energy. First we shall consider the non-degenerate system. We also assume that all energy levels are discrete. Let , ' and * denote theenergies of the systems A, A' and A*. The system A has energy means that its energy lies anywherebetween and + . Similar statements apply to energies of A' and A*. Since A*is enclosed withina heat insulating walls, we have

+ ' = * = constant

or ' = * – ...(3.12.1)

A(), R(' ) and A*(*) represent the number of microstates of system A, reservoir R and

composite system A*. The total number of microstates of the composite system A* is

* *A A R A R( ) ( ) ( ) ( ) ( )∗ ′Ω ε = Ω ε Ω ε = Ω ε Ω ε − ε

When the system A is in one of the accessible state with energy r, A (r) = 1, from above relationit follows that the number of microstates of A* is equal to the number of microstates of reservoirwith energy * – r. Thus Ω ε =Ω ε − ε* * *

R( ) ( )r

According to the fundamental postulate of statistical mechanics, all the accessible states of an isolatedsystem are equally probable. Therefore, the probability that the system A is in the state r with energyr is proportional to the corresponding number of micro-states accessible to the system A*.

∗∝ Ω ε − εR ( )r rp ...(3.12.2)

The numberΩ ε − ε*R ( )r is a rapidly varying function of energy. It is more convenient to work

with more slowly varying function ∗Ω ε − εRln ( )r . Expanding ∗Ω ε − εRln ( )r about the value ∗ε , we

have

∗ ∗

′ε =ε

∂ Ω Ω ε − ε = Ω ε − ε − ′∂ε *

RR R

lnln ( ) ln ( )r r higher order terms

Now we assume that our system is a very small in comparison to the reservoir, so that r << '. Thehigher order terms in the above expansion may be omitted. Therefore,

∗Ω ε − ε = Ω ε −βε*R Rln ( ) ln ( )r r ...(3.12.3)

where ∂ Ω

β =′∂ε

Rln...(3.12.4)


The derivative is evaluated at fixed energy ∗′ε = ε . Equation (3.12.3) can be written as

R Rln ( ) ln ( )r r∗ ∗Ω ε − ε − Ω ε =−βε

or R R( ) ( )exp( )r r∗ ∗Ω ε − ε = Ω ε −βε ...(3.12.5)

R( )∗Ω ε is constant (= C say) and is independent of r. Making use of Eqn. (3.12.5) in (3.12.2),

we get

Cexp( )r rp = −βε ...(3.12.6)

The constant C can be determined by the normalization condition that the system must haveprobability unity of being in some one of its states i.e.,

=∑ 1rr

p

where the summation is over all states of A irrespective of energy. That is

C exp( ) 1rr

−βε =∑

This gives

1C

exp( )rr

=−βε∑

Equation (3.12.6) then becomes

= −βε = −βε−βε∑ 1

1 1exp( ) exp( )

exp( )r r r

rr

pZ

...(3.12.7)

where 1Z exp( )rr

= −βε∑ ...(3.12.8)

The summation is performed over the accessible quantum states of the single system of theensemble. Eqn. (3.10.7) is a very general result of fundamental importance in statistical mechanics.It says that if a system is in equilibrium at temperature T, the probability that a system selected atrandom will be found in the state with energy r is Cexp( )r rp = −βε . The exponential factor

exp(-βε )r is called Boltzmann factor and the corresponding probability distribution is known as thecanonical distribution.

The quantity Z1 defined by 1Z exp( )rr

= −βε∑ is called the partition function of a single system

(component).The quantity can be shown equal to 1/kT, k is Boltzmann constant and T is absolute temperature

of the system.


It is possible that many different states may have the same energy. Let g(r), called degeneracy,be the number of distinct quantum states having the same energy r. Then the expression for thepartition function becomes

−βε= ε∑1

.

Z ( ) rr

diff energy levels

g e...(3.12.9)

The probability p(r) that a system of canonical ensemble be in a state with energy r is thengiven by

1

1( ) ( )exp( )

Zr r rp gε = ε −βε ...(3.12.10)

−βε

−βεε

ε =ε∑

( )( )

( )

r

r

rr

rr

g ep

g e...(3.12.11)

So far we have not disclosed the identity of the system of canonical ensemble. The onlyassumption about A we made was that the system was distinguishable and has size much smaller thanthe size of the reservoir. In a solid the atoms (microscopic system) are distinguishable by virtue offixed position. In a dilute gas the molecules are distinguishable because of large inter-molecularseparation.

The expression in Eqn.(3.12.10) may be extended to a system composed of molecules, whichare distinguishable. Suppose that the gas is in equilibrium at temperature T. If the gas is dilute enoughso that the molecules are distinguishable, we can focus attention on a particular molecule of the gasand regard it as a small system (component) in thermal contact with a heat reservoir consisting of allthe remaining molecules of the gas. The probability of finding the molecule in any one of its quantumstate r where its energy is r is then given by the canonical distribution. This result may also beapplied to atoms in a solid which are distinguishable by virtue of their fixed (localized) positions atlattice sites.

Another interpretation of probability p(r) is that if < nr > is the mean number of moleculesoccupying the state with energy level r then

( )N

rr

np ε = , N = total number of molecules in the gas

Therefore 1

( ) ( ).N Z

rrr r

np g e−βεε = = ε ...(3.12.12)

whence N

( )Z

rr rn g e−βε= ε ...(3.12.13)

A system in canonical ensemble in equilibrium with a reservoir has access to all its possiblestates, and the probability varies exponentially with the energy of the state. The Boltzmann probability


is a probability per quantum state. To get probability per unit range of energy, f() we must multiplypr by the density of states.

ε ε = ε ε ε( ) ( ). ( )f d p g d ...(3.12.14)

It is true that p() falls rapidly as energy rises but initially, g() rises even more rapidly, so thatthe energy distribution has a peak far above the ground state energy , even though the individualstates near the peak have extremely low probabilities.

If the energy spectrum of the system (component) is continuous the classical partition functionof a distinguishable single system is given by

= −ε∫single systemZ exp( / T)k dq dp ...(3.12.15)

Later we shall see that the partition function of indistinguishable N-particle system is given by

( )

=

N

singleN

ZZ

N!...(3.12.16)

Average Energy of particleIf a molecule is found with probability p(r) in a state r of energy r, then its mean energy is

given by

−βε

−βε

εε = ε =

∑∑ ∑

r

r

rr

r rr

r

e

pe

(single particle) ...(3.12.17)

Now, 1Z

r r rr

r r r

e e e−βε −βε −βε ∂∂ ∂ε = − = − = − ∂β ∂β ∂β ∑ ∑ ∑

Therefore 1 1

1

Z ln Z1

Z

∂ ∂ε = − = −

∂β ∂β ...(3.12.18)

Hence total energy of the N-particle system is

21 1lnZ lnZ

E N N N TT

k∂ ∂

= ε = − =∂β ∂

...(3.12.19)

3.13 THE EQUIPARTITION THEOREM

Consider a system whose state is described classically in terms of f coordinates q1, …..,qf and fcorresponding momenta p1, ……,pf. Let the energy of the system be function of these coordinates.For most of the systems, which we shall be dealing with, the energy can be written as

E(q, p) = i (pi) + ' (q1,…., qf, p1,….., pf) ...(3.13.1)


Where the first term is function of the particular momentum pi only and the second term maydepend on all coordinates and momenta except pi. For example, the energy of a harmonic oscillatorcan be written as

= ε + ε = + α2

2kinetic potential

1E

2 2

pq

m

The first term depends on momentum p only and the second term on coordinate q, is forceconstant. Suppose that the system under consideration is in thermal equilibrium with a heat reservoirat temperature T. The probability of finding the system with its coordinates lying in the range q andq + dq and momenta in the range p and p + dp is given by canonical distribution

E( , )

E( , )P( , )

q p

q p

eq p

e dq dp

−β

−β=∫ ...(3.13.2)

The mean value of energy i (pi) is given by

E( , )

E( , )

( )q pi i

i q p

e p dq dp

e dq dp

−β

−β

εε = ∫

∫...(3.13.3)

where the integrals extend over all possible values of all coordinates [q] and momenta [p]. Substitutingthe expression for E from Eqn. (3.13.1) in (3.13.3), we have

( )

( )

i

i

ii

e dq dp

e dq dp

′−β ε +ε

′−β ε +ε

εε = ∫

∫

i

i

i i

i

e dp e dq dp

e dp e dq dp

′−βε −βε

′−βε −βε

ε= ∫ ∫∫ ∫

The primes on the last integrals indicate that these integrals extend over all the coordinates [q]and momenta [p] except pi. The primed integrals in the numerator and denominator are equal andcancel out. Therefore

−βε

−βε

εε = ∫

∫i

i

i i

i

i

e dp

e dp

=

−βε

−βε

∂ − ∂β ∫∫

i

i

i

i

e dp

e dp


= ∞

−βε

−∞

∂ −∂β

∫ln iie dp ...(3.13.4)

If i is a quadratic function of pi theni = a pi

2 where a is a constant.

[For example the kinetic energy of a particle is quadratic function of momentum = p2/2m.]Then the integral in Eqn. (3.13.4) becomes

∞ ∞ ∞−βε − β −

−∞ −∞ −∞

= = = ββ∫ ∫ ∫

2 21, wherei ia p ax

i i ie dp e dp e dx x p

Hence ∞ ∞

−βε −

−∞ −∞

= − β+

∫ ∫21

ln ln ln2

i axie dp e dx

The last integral does not involve at all and its derivative with respect to is zero. So we areleft with

1 1 1

ln T2 2 2i k

∂ ε = − − β = = ∂β β ...(3.13.5)

Thus, we arrive at the conclusion that if a system is in thermal equilibrium at temperature T,then each independent quadratic term in its energy contributes a mean value of energy equal to1/2 kT to the total energy of the system. This statement is known as the theorem of equipartition ofenergy.

3.14 ENTROPY IN TERMS OF PROBABILITY

The entropy is connected with the fact that in most thermodynamic states the system is not in adefinite quantum state, but is spread over a large number of states according to some probabilitydistribution.

To define entropy in terms of probability distribution consider an ensemble of very large numberM( M ) of systems. Let m1 systems be in energy state 1, m2 systems in energy state 2, and soon. The statistical weight M of the ensemble or the number of ways in which the systems aredistributed among the various energy states is given by

M1 2

M! M!

! !..... !ir

r

mm m m

Ω = =Π ...(3.14.1)

The probability pr that a system chosen at random will be in the state r with energy r is

M

rr

mp = ...(3.14.2)


whence Mr rm p= . The entropy of the ensemble is

M MS ln ln M! ln ! M ln M M lnr r r rr r

k k m k m m m

= Ω = − = − − −

∑ ∑

= M ln M ln M ln M M ln Mr r r rr r

k m m k p p

− = −

∑ ∑

= ( )( )M ln M M ln M M lnr r rk p p p − + ∑= ( )M lnr rk p p− ∑ ...(3.14.3)

The entropy is an extensive quantity and therefore entropy of a single system is

S = MS

M

lnr rr

k p p= − ∑ ...(3.14.4)

3.15 ENTROPY IN TERMS OF SINGLE PARTICLE PARTITION FUNCTION Z1

By definition, the entropy of a single system is given by

S lnr rr

k p p= − ∑

= ( )1ln Zr rr

k p− −βε −∑1Z

r

re

p−βε

=

=

β ε + ∑ ∑1lnZr r r

r r

k p k p =∑ 1rr

p

= 1lnZT

kε + ...(3.15.1)

or –kT ln Z1 = E – TS ...(3.15.2)

CHAPTER

DISTRIBUTION FUNCTIONS

4.1 MAXWELL-BOLTZMANN DISTRIBUTION

Consider a system of N identical distinguishable non-interacting particles confined to move freely ina vessel of volume V. The system is in equilibrium at temperature T. The total energy of the systemis E. A monatomic ideal gas containing N molecules in a container of volume V is an example ofour system. A macrostate of the system is specified by specifying how the total energy E of thesystem is distributed among N particles.

Let 1, 2, ….. i, … be the allowed single-particle energy levels and g1, g2, ....., gi,..... thedegeneracies of these energy levels respectively. The description of a macrostate is given by specifyingthe number of particles in each energy level i. Let

n1 be the number of particles in energy level 1 with degeneracy g1

n2 be the number of particles in energy level 2 with degeneracy g2

…… ………… ….. …..ni be the number of particles in energy level i with degeneracy gi and so on.The set of numbers n1, n2, …., ni defines a macrostate. Our objective is to determine

n1, n2, ….,ni. In classical (Maxwell-Boltzmann) statistics all the particles are assumed to be identicaland distinguishable from one another. We wish to find the number of different arrangements ormicrostates corresponding to a macrostate. The required number of accessible microstates is equal tothe number of ways a given macrostate is realized by arranging the particles in different states. Nowconsider the i th energy level, which is gi fold degenerate. Each of the ni particles can be placed inthe gi states in gi different ways. The number of different ways of arranging ni particles in gi statesis equal to the product

gi . gi . gi . gi. ……….. (ni factors) = .inig

This number inig gives number of microstates available to ni particles, each having energy i.

The number of microstates available to the system such that ni particle are in energy level 1, n2 arein energy level 2 …and so on, is given by the product

Distribution Functions 293

1 21 2. ........... i in nn n

i ii

g g g g=Π

Now we must take into account the possible permutations of the particles among the differentenergy levels. The number of permutations possible for N particles is N!. In other words, N particlescan be arranged in N! different sequences. Of these permutations some are irrelevant. When morethan one particle is in an energy level, permuting them among themselves has no significance in thissituation. Thus the ni particles in the i th level contribute ni! irrelevant permutations. If there are n1

particles in level 1, n2 particles in level 2 and so on, there are n1! n2! n3 !....irrelevant permutations.The number of ways N particles can be divided into groups of n1, n2, n3, …. or the thermodynamicprobability of a macrostate n1, n2, …., ni is

1 2 3

N! N!or

! ! !........... !ii

n n n nΠ

The total number of distinct ways (accessible microstates) in which N particles can be distributedamong possible energy levels such that n1 particles are in level 1 with degeneracy g1, n2 are in level2 with degeneracy g2..... and so on, is

1 2

MB 1 21 2 3

N!( ) ( ) ...... N!

! ! !....... !

inn n i

i i

gg g

n n n n

Ω = = Π

...(4.1.1)

According to the principle of equal a priori probabilities, all the microstates are equally probable.The most probable macro state specified by n1, n2, …..is one which corresponds to the maximumnumber of micro states. So to obtain most probable distribution we must maximize MB taking careof the conditions that the total number of particles and the total energy of the system is constant.These two restrictions are expressed as

n1 + n2 + n3 +………. = Nii

n =∑ ...(4.1.2)

n1 1 + n2 2 + ……….= Ei ii

n ε =∑ ...(4.1.3)

Mathematically it is more convenient to maximize ln MB than . So we first simplify ln MB.

Now, ln MB = ln N! + ( )−∑ ln ln !i i ii

n g n

Making use of Stirling formula ln n! = n ln n – n, we have

MBln N ln N N ln lni i i i ii i i

n g n n nΩ = − + − +∑ ∑ ∑


= NlnN ln lni i i ii i

n g n n+ −∑ ∑ ...(4.1.4)

Since ni = N.

The most probable distribution is one for which MB or ln MB is maximum. In other words,a small change ni in any of the ni's has no effect on the value of MB. We assume that ni arecontinuous, so the condition of maximum ln MB′Ω becomes

MBln

0in

∂ Ω=

∂ for each ni

If the change in ln MB corresponding to change ni in ni is ln MB then

MBln ln ln lni i i i i ii i i

g n n n n nδ Ω = δ − δ − δ∑ ∑ ∑ = 0

since N is constant. Now δ = δ1ln i i

i

n nn

. So

MBln ln ln 0i i i i ii i i

g n n n nδ Ω = δ − δ − δ =∑ ∑ ∑ ...(4.1.5)

Since the total number of particles is constant, δ =∑ 0ii

n .

Therefore, MBlnδ Ω = δ − δ =∑ ∑ln ln 0i i i ii i

g n n n ...(4.1.6)

In order to incorporate the conditions of conservation of number of particles and of energy inabove equation we use Lagrange’s method of undetermined multipliers. From Eqns. (4.1.2) and (4.1.3)we have

δ = δ +δ + =∑ 1 2 ............ 0ii

n n n ...(4.1.7)

ε δ =ε δ +ε δ =∑ 1 1 2 2........... 0i ii

n n n ...(4.1.8)

Multiplying Eqn. (4.1.7) by – and Eqn. (4.1.8) by –, where and are quantitiesindependent of ni, and adding them to Eqn. ( 4.1.6 ) we have

( ln ln ) 0i i i ii

n g n− + − α − β ε δ =∑ ...(4.1.9)

Equation (4.1.9) will hold if the quantity in parentheses vanishes for each value of i. Hence

– ln ni + ln gi – – i = 0

−β ε−α= ii in g e e ...(4.1.10)


The ratio ni /gi represents the average number of particles per state of the system and is calledenergy distribution function fMB(). The distribution function represents the average number of particlesin each of state of energy or the probability of occupancy of each of state of energy .

MB( ) i

i

nf e e

g−βε−αε = = ...(4.1.11)

In other words fMB () gives the probability that a particle, selected randomly from the system,will have its energy .

This is the classical (MB) distribution function. It is applicable to system whose constituentparticles are distinguishable and don’t obey Pauli’s exclusion principle. It is the most probabledistribution of particles among the accessible energy levels at equilibrium for a system of constanttotal energy. For a system comprising of large number of particles the most probable distributiondescribes the actual behaviour of the system. In deriving MB distribution function no assumptionregarding the nature of energy was made so it is valid for translational, rotational, vibrational andelectronic.

Later we shall prove that the quantity appearing in the distribution function is related to thetemperature T of the system through the relation

1

Tkβ = ...(4.1.12)

where k is Boltamann constant. So the MB distribution can be written as

/ TMB( ) kf e e−α − εε = ...(4.1.13)

The constant may be expressed in terms of total number of particles N.

Since / TN i k

i ii

n e g e− ε− α= =∑ ∑

Therefore / T

Ni k

ii

eg e

− α− ε

=∑ ...(4.1.14)

The quantity / Ti ki

i

g e−ε∑ plays a fundamental role in statistical mechanics. It was first introduced

by Boltzmann who called it “Zustandsumme” or sum over states. It is called single particle partitionfunction Z1.

/ T

1Z i ki

i

g e− ε= ∑ ...(4.1.15)

In terms of single particle partition function Z1 the MB distribution function is given by

/ T

1

N N,

Z Zi k

i in g e e− ε − α= = ...(4.1.16)


If the energy levels of the system are very close together, such as those of molecules in a gas,the g () is replaced by g () d, which represents the number of states with energies between and + d. The number of particles occupying the states in the energy range d about is

MB( ) ( ) ( )n d f g dε ε = ε ε ε ...(4.1.17)

For a system of free particles each of mass m, enclosed in a vessel of volume V, the number ofstates in the energy range d about is

3/2 1/2

3

2 V( ) . (2 )sg d g m d

h

πε ε = ε ε ...(4.1.18)

where gs = (2s + 1) is spin degeneracy, s = spin of the particle. For electron s = 1/2, gs = 2. Forspin 0 particle gs = 1. The function g() is called density of states.

The expression for partition for a gas, with continuous energy levels, is obtained by replacingthe summation sign by integral sign.

1Z ( )g e d−βε= ε ε∫ ...(4.1.19)

The value of parameter e – can be determined as follows:

1 3 / 2 1/ 2

30

N N N

Z ( ) 2 V(2 )s

eg e d

g m e dh

− α∞−β ε

−β ε

= = =ε ε π ε ε

∫ ∫

3 / 2

3/2 1/23

0

N( ), ,

2 V(2 ) x

s

x dx d

g m x e dxh

∞−

β= = βε = β ε π

∫

3 / 21/ 2

03

N 1where 3 / 2

2 2Vx

s

x e dxm

gh

∞− β= = Γ = π π

∫

=

3/ 22N

V 2s

h

g m

β π

...(4.1.20)

So e – is temperature dependent ( = 1/kT).Substituting the value of e – in the expression for MB distribution we have

3/22

MBN 1

( ) ,V 2 Ts

hf e

g m k−β ε βε = β = π

...(4.1.21)


The number of particles occupying the energy states in the interval d at in a N-particlesystem in equilibrium at temperature T is

( )1/ 2

MB 3/ 2

2 N 1( ) ( ) ( ) ,

T/n d f g d e d

k−β επε ε = ε ε ε = ε ε β =

π β ...(4.1.22a)

= 3/2

1/212 N exp ( )

Td

k

π ε −βε ε π

...(4.1.22b)

Evaluation of

The total energy of the N-particle system is

3 / 2

3 / 20 0

2 NE ( )

( / )n d e d

∞ ∞−βεπ= ε ε ε = ε ε

π β∫ ∫

3/ 23/ 2 5 / 2

0

2 N 1, ,

( / )xx e dx x dx d

∞−π= ⋅ = βε = β ε

π β β ∫

2N 1(5 / 2)

= Γ βπ

2N 1 3

4

= π βπ

3N

2=

β ...(4.1.23)

From kinetic theory we know that, the total translational energy of an ideal gas containing Nmolecules is

3

E N T2

k= ...(4.1.24)

Comparison of these two expressions gives

1

Tkβ = ...(4.1.25)

4.2 HEAT CAPACITY OF AN IDEAL GAS

The total energy of one mole of an ideal gas is

A A3 3

E N T RT, R N2 2

k k= = = = universal gas constant ...(4.2.1)


The molar heat capacity at constant volume is

E 3

C R.T 2v

v

∂ = = ∂ ...(4.2.2)

4.3 MAXWELL’S SPEED DISTRIBUTION FUNCTION

Now we shall derive Maxwell speed distribution for a perfect classical gas. To obtain the requireddistribution function we must convert the energy distribution into appropriate speed distribution. Ina perfect gas of free molecules possessing no internal degree of freedom (such as monatomic molecule)all the energy resides in the form of translational kinetic energy of molecules. So

21,

2mv d mv dvε = ε =

In terms of velocity, the Maxwell distribution function becomes

( ) 21/ 22 / 2 T123/ 2

2 N( )

( T)mv kn v dv mv e mv dv

k−π=

π

23 / 2

2 / 2 T4 N2 T

mv kmv e dv

k−

= π π ...(4.3.1)

or 2

3 / 22 / 2 T( )

( ) 4N 2 T

mv kn v mf v dv v e

k−

= = π π ...(4.3.2)

The function f(v) gives the fraction of allmolecules having speeds in the interval dv about v.In other words, f(v) represents the probability thata molecule selected at random from the gas willhave its speed in the interval dv about v.

Now suppose that we wish to know howmany molecules of the gas have velocities such thatthe x-component of velocity is in a range dvx aboutvx, the y-component is in a range dvy about vy, andz-component is in a range dvz about vz. This numberof molecules is in a rectangular volume elementdvx dvy dvz in velocity space centered on the value(vx, vy, vz). We shall call this number n(vx, vy, vz)dvx dvy dvz .

Fig. 4.3.1 Velocity space for molecules


The Maxwell-Boltzmann energy distribution function is

3 / 22/ TN

( )V 2 T

k

s

hf e

g mk−ε

ε = π ...(4.3.3)

For monatomic spin less molecule gs = 1.Therefore

2 2 23 / 22

( ) / 2 TN( , , )

V 2 Tx y zm v v v k

x y zs

hf v v v e

g mk− + +

= π ...(4.3.4)

Let us find the expression for the density of states g(vx, vy, vz). We know that

) 3

V( , , )x y z x y z x y z

s

g p p p dp dp dp dp dp dpg h

=

, ,( )x y z x y zg v v v dv dv dv3

3

Vx y z

s

mdv dv dv

g h= ...(4.3.5)

Now ( , , ) ( , , ) ( , , )x y z x y z x y z x y z x y zn v v v dv dv dv f v v v g v v v dv dv dv=

2 2 2

3/2( ) / 2 T

N2 T

x y zm v v v kx y z

me dv dv dv

k− + +

= π ...(4.3.6)

Expression for n (vx)

The number of molecules whose x-component of velocity lies in a range dvx about vx regardless ofwhat values the y-component and z-component of velocity may have, is given by

( ) ( , , )

y z

x x x y z x y z

v v

n v dv n v v v dv dv dv∞ ∞

= −∞ = −∞

= ∫ ∫

2 2 23/2

( )/2 TN

2 Tx y zm v v v k

x y zm

e dv dv dvk

∞ ∞− + +

−∞ −∞

= π ∫ ∫

2 2 21 1 12 2 2

3/2/ T / T / T

N2 T

y z xmv k mv k m v ky z x

me dv e dv e dv

k

∞ ∞− − −

−∞ −∞

= π ∫ ∫

− π π π

212

3/2 1/2 1/2/ T2 T 2 T

= N2 T

xm v kx

m k ke dv

k m m


21/2

/ 2 TN2 T

xmv kx

me dv

k−

= π ...(4.3.7)

Table of some useful definite integrals.

∞

= −α∫ 2

0

( ) exp( )nf n x x dx

odd n f(n) even n f(n)

1α1

20

πα

1

2

3α2

1

22

πα3

1

4

5α3

14

πα5

3

8

7α4

36

πα7

15

16

A plot of the distribution functions n(v) and n(vx) as functions of velocity coordinates is shownin the Fig. (4.3.2).

Fig. 4.3.2 Velocity distribution functions at different temperatures

Average velocity < v >

The average velocity < v > of a particle in M-B distribution is given by

23 / 2

3 /2 T

0 0

0

( ) 4 N2 T

N( )

m v kmv n v dv v e dv

kv

n v dv

∞ ∞−

∞

π π

= =∫ ∫

∫


8 Tk

m=

π...(4.3.8)

Where we have used the standard result: 23

20

1, where .

2 T2x m

x e dxk

∞− α = α =

α∫

Root mean square speed vrms

The root mean square speed is defined by

2

3 / 22 2 4 / 2 T

0 0

1 1( ) 4 N

N N 2 Tmv k

rmsm

v v n v dv v e dvk

∞ ∞−

= = π π ∫ ∫

3 Tk

m= ...(4.3.9)

Where we have used the standard result: 24

20

3,

T8x m

x e dxk

∞−α π= α =

αα∫ .

Most probable speed vmp

The most probable speed corresponds to the maximum value of n(v). Now

2

3/22 / 2 T( ) 4 N

2 Tmv km

n v v ek

− = π π

The maximum value of n(v) corresponds to the speed that satisfies the equation

=( )0

dn v

dv

( )22 / 2 T 0mv kdv e

dv− =

2 T

mpk

v vm

= = ...(4.3.10)

Therefore vmp : < v > : vrms = 2 T 8 T 3 T

: :k k k

m m mπ

=π8

2 : : 3

= 1 : 1.13 : 1.22


Fig. 4.3.3 Three kinds of velocities associated with Maxwell distribution

4.4 FERMI-DIRAC STATISTICS

Many properties of solid materials such as thermal and electrical conductivity, magnetic properties,specific heats etc. are related to the electron energy states. The understanding of these propertiesrequires the distribution of electrons, which are fermions, in various states. The distribution functionfor electrons must satisfy two requirements: (1) electrons are indistinguishable particles i.e., it is notpossible to label them as electron number 1, number 2 and so on (2) electrons obey Pauli’s exclusionprinciple, which states that no quantum state may be occupied by more than one electron. Thedistribution function that results from these two considerations was first developed by Fermi andDirac in 1927. In fact all the particles, which have half odd integral spin (1/2, 3/2, 5/2, …), obeyexclusion principle, are described by F-D statistics. These particles are called fermions.

Consider a system composed of N fermions with total energy E. An assembly of non-interactingfermions is called fermi gas or electron gas if the fermions are electrons. The assembly of free electronsin a metal is a well-known example of fermi gas.

Let the allowed energy levels and the associated degeneracies of the system be 1, 2, ….. i

and g1, g2, ….., gi respectively. The occupation numbers of these energy levels are n1, n2, ……, ni

respectively. The most detailed description that is possible in principle is to specify the number ofelectrons (always 0 or 1) in each of the gi states of energy i. (Because of indistinguishability ofelectrons, it is meaningless to specify which electrons are in these states.) A macro state is specifiedby set of numbers n1, n2, …,ni. A microstate is specified by specifying which of the gi statescorresponding to each i are occupied by electrons.

First suppose that the electrons are distinguishable. Consider a macro state n1. n2, ….,ni.Here ni represents the number of electrons in the energy level i. We want to calculate the number ofways in which ni electrons can be placed in gi states associated with the energy level i. The firstelectron can be placed in any one of the gi states. This can be done in gi ways. For each of thesechoices, the second electron can be placed in (gi – 1) ways, for each of these, there are ( gi – 2) waysfor the third electron and so on. For the last electron there would be gi – (ni – 1) ways. The totalnumber of possible ways of placing ni electrons in the gi states would be the product of all these


factors. This is equal to

− − − +( 1)( 2)..............( 1)i i i i ig g g g n

or =−

!

( )!i

i i

g

g n

Since the electrons are indistinguishable, permuting the ni electrons among the various statesdoes not produce a physically different state of the system. There are ni! ways of permuting theparticles among themselves in any given arrangement of electrons. These ni! ways don’t count asseparate arrangements. So, we over counted the number of possible arrangement by a factor ni!. Thenumber of physically different ways of putting ni electrons into gi states of energy i is

!

!( )!i

ii i i

g

n g nΩ =

− ...(4.4.1)

The number of microstates corresponding to a given macrostate is obtained by multiplying thefactors given in above expression. Thus, the number of microstates corresponding to themacrostate specified by the set of occupation numbers n1, n2, ......, ni is

1 2

1 21 1 1 2 2 2

! !........ ....

!( )! !( )! ii

g g

n g n n g n= ⋅ = Ω Ω = Π Ω

− − ...(4.4.2)

Ω = Π−!

!( )!i

i i i i

g

n g n...(4.4.3)

gi

| | | | | | | | | | ni, gi

1 2 ………………………… ni

| | | | | | | n2 = 3, g2 = 6

| | | | | | n1 = 2, g1 = 5

Fig. 4.4.1

According to the fundamental principle of statistical mechanics, for a system of given totalenergy all the microstates are equally probable. The most probable state of the system is the mostprobable macrostate, which corresponds to the maximum number of microstates . Maximizing isequivalent to maximizing ln . In calculation it is more convenient to use ln than . So we firstcalculate ln .

Ω = − − − ∑ln ln ! ln ( )! !i i i ii

g g n n


Using Stirling’s approximation ln n! = n ln n – n we have

Ω = − − − − + − − + ∑ln ln ( ) ln( ) ( ) lni i i i i i i i i i i ii

g g g g n g n g n n n n

Ω = − − − − ∑ln ln ( ) ln( ) lni i i i i i i ii

g g g n g n n n ...(4.4.4)

The most probable distribution is one for which a small change ni in any of ni has no effecton the value of . We assume that ni are continuous, so the condition of maximum ln becomes

∂ Ω =

∂ln

0in

for each ni.

If the change in ln corresponding to change ni in ni is ln then

ln ln ( ) ln 0i i i i

i

g n n nδ Ω = − − δ = ∑ ...(4.4.5)

Note that ni are not independent, but they are related through the conditions that the total numberof particle is constant.

ni = N = constant ...(4.4.6)

And the total energy of the system is constant.

ni i = E = constant ...(4.4.7)

From these two equations, we have ni = 0 ...(4.4.8)

and i ni = 0 ...(4.4.9)

In order to incorporate the conditions of conservation of number of particles and of energy inEqn. (4.4.5) we use Lagrange method of undetermined multipliers. To do so, we multiplyEqn. (4.4.8) by – and Eqn. (4.4.9) by – and add them to Eqn. (4.4.5). Here and areindependent of ni.

ln ( ) ln 0i i i i ii

g n n n− − − α−βε δ = ∑

ln ( ) ln 0i i i ig n n− − − α − βε =

1i

ii

gn

eα + β ε=+

...(4.4.10)

The ration ni/gi is called F-D distribution function and is denoted by fFD.

FD1

( )1

i

i

nf

g e eα β εε = =

+ ...(4.4.11)


The function fFD () is the average number of electrons per quantum state of energy . It alsorepresents the probability that a state of energy is occupied. The quantity is 1/kT, where T istemperature of the system. The value of is determined by the normalization condition

N1i

ii

i i

gn

e eβ εα= =

+∑ ∑

It is customary to express in terms of another constant µ, called chemical potential. (In solidstate physics µ is called Fermi level F).

FF or

T Tk k

εµα = −βµ = −βε α = − = − ...(4.4.12)

In terms of chemical potential µ and Fermi energy F the FD distribution becomes

FFD ( ) ( ) / T

1 1( )

1 1kf

e eβ ε − µ ε − εε = =+ + ...(4.4.13)

For systems with continuous energy levels, the number of fermions occupying the states in theenergy range d at is given by

FD( ) ( ) ( )n d g f dε ε = ε ε ε ...(4.4.14)

The total number of fermion in the system is given by

FDN ( ) ( ) ( )n d g f d= ε ε = ε ε ε∫ ∫ ...(4.4.15)

For electron gas, the density of states is given by

3/ 2 1/ 2

3

2 V( ) (2 1) (2 ) ,g d s m d

h

πε ε = + ε ε

1 / 2 for electron.s = ...(4.4.16)

4.5 BOSE-EINSTEIN STATISTICS

There are many systems, which are composed of weakly interacting identical and indistinguishableparticles with integral spin. These particles don’t obey Pauli’s exclusion principle. Thus any numberof particles can occupy the same quantum state. The statistical behaviour of such particles is governedby a different kind of statistics, called Bose-Einstein statistics, named in honor of S.N. Bose andA. Einstein who independently derived it. Particles obeying Bose-Einstein statistics are called bosons.Examples of bosons are photons (spin 1), H2 molecule, helium 4He, meson etc.

Consider a system composed of N bosons. The macrostate state of the system is specified byspecifying the occupation numbers n1, n2, …….in energy levels 1, 2, ….. having degeneraciesg1, g2,……….The ith level with energy i and degeneracy gi contains ni particles. There is norestriction on the number of bosons that a quantum state can accommodate. This level can be picturedas ni particles in a row divided arbitrarily into gi states by gi – 1 partitions.


o o | o | | o o o o | o o | | o | o o o | og = 1 2 3 4 5 6 7 8 9ni =14

Fig. 4.5.1 Distribution of 14 particles in 9 states separated by 8 partitions

The number of different ways i, the ni bosons can be placed in the gi quantum states withoutany limit to the number of particles in a state is equal to the number of independent permutations ofparticles and partitions. There are a total of (ni + gi –1) particles plus partitions, which can be arrangedin (ni + gi –1)! ways. Since the permutation of particles among themselves and permutations ofpartitions among themselves don’t produce a different arrangement, we must, therefore, divide(ni + gi – 1)! by ni! . (gi – 1)!. Thus

+ −

Ω =−

( 1)!

! ( 1)!i i

ii i

n g

n g...(4.5.1)

The total number of different ways to arrange n1, n2, …. bosons in the energy levels1, 2, ……, if there are g1, g2, ….. states in each level, is

1 1 2 2

1 1 2 2

( 1)! ( 1)!...........

!( 1)! !( 1)!

n g n g

n g n g

+ − + −Ω = ⋅

− −

+ −=Π

−( 1)!

!( 1)!i i

i i i

n g

n g...(4.5.2)

Since ni >> 1, gi >> 1, 1 may be omitted in the above expression. Thus

+

Ω = Π( )!

! !i i

i i i

n g

n g...(4.5.3)

If the number N of the particles and the total energy E of the system is constant then we have

1 2 ........ Nii

n n n+ + = =∑ ...(4.5.4)

1 1 2 2 ............ Ei ii

n n nε + ε = ε =∑ ...(4.5.5)

Since N and E are constants, the sum of changes in occupation numbers and the sum of changesin energies of the energy levels must be zero.

δ =∑ 1 0i

n ...(4.5.6)

0i ii

n δε =∑ ...(4.5.7)

The most probable distribution corresponds to the maximum value of subject to the restrictionsexpressed by Eqns. (4.5.6) and (4.5.7). It is more convenient to maximize ln than . So we firstsimplify ln .


ln ln( )! ln ! ln !i i i in g n gΩ = + − −∑Using Stirling approximation we have

Ω = + + − + − + − + ∑ln ( ) ln( ) ( ) ln lni i i i i i i i i i i in g n g n g n n n g g g

= + + − − ∑ ( ) ln( ) ln lni i i i i i i in g n g n n g g ...(4.5.8)

The condition of maximum ln is

δ Ω =ln 0

1 1

ln( ) ( ) ln 0( )i i i i i i i

i i i

n g n g n n nn g n

+ + + ⋅ − − ⋅ δ = +

∑

or ln( ) ln 0i i i in g n n+ − δ = ∑ ...(4.5.9)

To incorporate the conditions expressed by Eqns. (4.5.6) and (4.5.7), we use Lagrange methodof undetermined multipliers. Multiplying Eqn. (4.5.6) by – and Eqn. (4.5.7) by – and addingthem to Eqn. (4.5.9) we have

+ − − α − βε δ = ∑ ln( ) ln 0i i i i in g n n

Since the change ni’s are arbitrary, we must have

+ − − α − βε =ln( ) ln 0i i i in g n

This simplifies to

α+βε=−1i

ii

gn

e...(4.5.10)

BE1

( )1i

i

i

nf

g eα + β εε = =

−= β ε −µ −( )

1

1ie, = 1/kT ...(4.5.11)

Equation (4.5.10) represents the BE distribution function. Eqn. (4.5.11) represents number ofbosons per quantum state at energy i or the occupation probability of state with energy of a systemin thermal equilibrium at temperature T.

In a system in which the number of bosons is not conserved, the condition ni = N = constantdoes not apply. For example, the number of photons in a cavity increases with increasing temperature.This in contrast to an ideal gas contained in a vessel. Removal of this condition is equivalent tosetting = 0 (µ = 0). For such system the B-E distribution becomes

β ε=−1i

ii

gn

e

BE1

( )1

feβ ε

ε =−

(for photons = ) ...(4.5.12)


The parameter may be determined from the condition ni = N. It increases monotonicallywith temperature for FD and BE statistics both. This can be seen as follows. The number of bosonsoccupying the states with energy between and + d is given by

BE( ) ( ) ( )n d f g dε ε = ε ε ε ...(4.5.13)

The total number of bosons in the system is

BE

0

N ( ) ( )f g d∞

= ε ε ε∫3/ 2 1/ 2

3 / T0

2 V(2 )

. 1k

m d

h e e

∞

α επ ε ε=

−∫ ...(4.5.14)

Putting /kT = q, in the above integral, we have

3 / 2 1/ 2

30

2 V(2 T)N

1q

mk q dq

h e e

∞

απ=

−∫ ...(4.5.15)

With increasing T, the factor multiplying the integral in the above expression (4.5.15) increases.Since N is constant, the integral must decrease. This implies that increases with rise in temperature.Thus is an increasing function of T. Since N is finite, the integral must always converge and so must always be non-negative.

CHAPTER

APPLICATIONS OF QUANTUM STATISTICS

FERMI-DIRAC STATISTICS

5.1 SOMMERFELD’S FREE ELECTRON THEORY OF METALS

According to free electron model, the valence electrons of atoms constituting the metal are free tomove within the limits of the metal. The positive ion cores produce a constant average potential inwhich free electrons move. The potential energy of interaction of electrons with the ion cores isconstant throughout the solid and may be assumed to be zero for convenience. The free electronsdon’t leave the boundaries of the metal because of the electrostatic force. The potential energy ofelectrons may be assumed to be infinitely great at the boundaries. Thus the electrons in a metal maybe treated as a gas which is composed of non-interacting spin 1/2 fermions confined in box. Becauseof this analogy the assembly of free electrons in a metal is called Fermi gas. Quantum mechanicaltreatment of motion of electron in a box shows that energy of electron is quantized. The energy ofelectron, which is free to move in a cubical box of side L is given by

( )2 2

2 2 222 L

x y zn n nm

πε= + +

...(5.1.1)

where nx, ny, nz are integers, each can take on values 1, 2, … .The set of integers nx, ny, nz and spinquantum number 1/2 define a state of electron. It is found that more than one quantum state correspondto a single energy level. The different quantum states belonging to a energy level are called degeneratestates and the number of such states is called the degeneracy of that energy level.

The density of states (the number of quantum states per unit energy interval) at energy isgiven by

3/ 2 1/ 2.

3

2 V( ) (2 1) (2 )g s m

h

πε = + ε ...(5.1.2)

For electron spin s = 1/2 , and (2s + 1) = 2. The free electrons in a metal are distributed amongvarious available quantum states according to Pauli’s exclusion principle. The Fermi-Dirac distribution

function gives the probability that a state with energy is occupied at temperature T.


Fermi-Dirac Distribution for Electron Gas and Fermi Energy

F-D distribution is

FD ( ) / T

1( )

1kf

e ε −µε =

+ ...(5.1.3)

At T = 0 K, for the energy states < where µ is chemical potential, the exponential termexp ( – )/kT exp (– ) 0 and therefore,

fFD = 1

At T = 0 K, for energy states > µ , the exponential term exp ( – µ)/kT exp () . So

fFD = 0.

The value of chemical potential at T = 0 K is dented by µ0. Thus at T = 0 K, all the energystates below the = 0 are occupied and those above it are empty. The energy = 0 is calledFermi energy F(0). Thus, the Fermi energy is the maximum value of energy that a fermion can acquireat 0 K. In other words, the energy of the top most filled level in a Fermi gas at absolute zero is theFermi energy.

At T ≠ 0 K, the probability the energy level = 0 = F is occupied is

FD F 0

1 1( )

21f

eε = µ = ε = =

+

That is, at Fermi energy one half the energy states will be occupied. The variation of F-Ddistribution with energy at different temperatures is shown in the Fig. (5.1.1).

Fig. 5.1.1 Dependence of Fermi distribution function fFD and number of electrons with temperature

From the figure, it is evident that at temperature T above 0 K, some of the states just below F

that were occupied at T = 0 K are now empty and some of those states just above the F are now

Applications of Quantum Statistics 311

occupied. The explanation for this is that as the temperature of the Fermi gas is raised, a small fractionof the electrons occupying the states just below the Fermi level gain thermal energy and get excitedto the states just above the Fermi level.

For a free electron gas the number of states in the energy interval d at is

3/ 2 1/ 2

3

4 V( ) (2 )g d m d

h

πε ε = ε ε ...(5.1.4)

The number of electrons occupying the energy states between the energy interval d at is

FD( ) ( ) ( )n d f g dε ε = ε ε ε

( )

1 exp ( )/ T

gd

k

ε= ε+ ε − µ ...(5.1.5)

The Fermi energy is in general a function of temperature. Its value is determined by the condition

FD( )

N ( ) ( ) ( )1 exp ( ) / T

gn d f g d d

k

ε= ε ε = ε ε ε = ε+ ε − µ∫ ∫ ∫

where the integral is taken over all the energy states available to the electrons of the system.Substituting the value of g() in above expression we have

1/ 23/2

30

4 VN (2 )

1 exp( ) / Tm d

kh

∞π ε= ε+ ε − µ∫ ...(5.1.6)

At T = 0 K,

fFD = 1 for < µ0 = F0

= 0 for > µ0 = F0

So the limits of integration can be taken from 0 to F0. Then

F03 / 21/ 2

30

4 V(2 )N

md

h

επ= ε ε∫ (since fFD = 1)

0

3/ 23 / 2F3

4 V(2 ) 2

3m

h

π = ε

0

2/32

F3N

8 Vh

m

ε = π

...(5.1.7)

( )2 / 3

19 2 N3.646 10 eV m

V− = × ⋅

...(5.1.8)


For copper N/V = 8.5 × 1028 m–3, the Fermi energy is

0

19 2 28 3 2/3F (3.646 10 eV.m )(8.5 10 m ) 7.0 eV− −ε = × × =

The energy of an electron confined to move in a cubical box of side L is

2 2

28 L

n h

mε =

The quantum number of electron occupying the highest energy state = F0 is

0F

max

8L

mn

h

ε=

For a box of size L = 1 cm, nmax = 43 × 106. Thus, we see that the quantum numbers of theoccupied states may from 1 to 43 million. The existence of such a huge number of states allows usto treat the energy levels as continuous.

In order to express density of states g() in terms of Fermi energy F0 we multiply the expressions

for g() and F0. Doing so we get

0

1/23/2F

3N( )

2g ε = ε

ε ...(5.1.9)

For 1 mol of copper, N = NA = 6.02 × 1023 mol–1, F0 = 7 eV, we have

0

231/ 2 23

F 3/ 2

3 6.02 10( ) (7eV) 1.3 10 states/eV

2(7eV)g

× ×ε = = ×

With this huge number of states per unit energy range, it is clear that we may consider theenergy to be virtually continuous.

Fermi temperature TF

The Fermi temperature TF of a Fermi gas is defined as

0FFT

k

ε= ...(5.1.10)

where k is Boltzmann constant. Fermi temperatures for typical metals are of the order of 105 K,which is quite high. No metal remains in solid state at this temperature.

For T << TF or kT << F0 the F-D distribution is called degenerate and for T >> TF the

distribution is non-degenerate. The parameter = – µ/kT is negative for degenerate systems andpositive for non-degenerate systems. This means that > 0 at low temperature and < 0 at hightemperature.

At T = 0 K the system is completely degenerate, at T << TF the system is degenerate and atintermediate temperature it is slightly (weakly) degenerate.


Table 5.1.1 Fermi energy, Fermi temperature and Fermi velocity of some metals

Element N/V cm – 3 F eV TF K vF cm/s

(10 22) (104) (10 8)

Li 4.6 4.7 5.5 1.3

K 1.34 2.1 2.4 0.85

Cu 8.5 7.0 8.2 1.56

Au 5.9 5.5 6.4 1.39

Degeneration of Fermi Gas

A many particle system is said to degenerate if its behaviour shows deviations form the expectedclassical behaviour. At T = 0 K, a Fermi gas is completely (strongly) degenerate. In the temperaturerange 0 < T << TF the gas is degenerate. At T < TF it is weakly (slightly) degenerate. At T > TF itis non-degenerate.

(i) Completely Degenerate Fermi Gas (T = 0 K)

Let us calculate the energy, entropy and pressure of completely degenerate Fermi gas. The averageenergy of electrons at 0 K is

FD1 1

( ) ( ) ( )N N

n d f g dε = ε ε ε = ε ε ε ε∫ ∫3/2

3/23

F0

4 V(2 )

1 exp ( )/ TN

dm

kh

∞π ε ε=+ ε−ε∫

F03/2 3/2

30

4 V(2 )

Nm d

h

επ= ε ε∫ (At T = 0 K, fFD = 1 for < F0

)

0

3/2 5/2F3

4 V 2(2 )

5Nm

h

π = ε ...(5.1.11)

Making use of the result

0

3/2 3/2F3

4 V 2N (2 )

3m

h

π = ε ...(5.1.12)

We can express < > as

00 F3

5ε = ε ...(5.1.13)


The total energy of the system is

0F3

E (0) N5

= ε ...(5.1.14)

Entropy of Fermi gas at absolute zero is zero.

S0 = 0

Pressure of the gas is given by the relation

2

PV E3

=

At T = 0

0F

E2 2 NP

3 V 5 V= = ε ...(5.1.15)

Thus a Fermi gas exerts very high pressure ( 610 Atm.≈ ) even at T = 0.

(ii) Degenerate Fermi Gas (T << TF)

At T << TF the chemical potential > 0 ( < 0) and at T > TF, µ < 0 ( > 0). The variation ofchemical potential with temperature is shown in the Fig. (5.1.2)

Fig. 5.1.2 Variation of chemical potential with temperature

The total number of electrons in a Fermi gas is

1/ 23/ 2

FD 3F0 0

4 VN ( ) ( ) (2 )

1 exp ( ) / Tf g d m d

kh

∞ ∞π ε= ε ε ε= ε

+ ε − ε∫ ∫ ...(5.1.16)

The result of integral of this type can be obtained making use of the following standard result:

( )2

1 12F F2 1 2

F F10

1 12( T) 1 (2 )

1 exp[( ) / T] 1 2

p np pn

n nn

d dk n

k p d

∞ ∞+ +

−=

ε ε = ε + − ζ ε + ε − ε + ε ∑∫ ...(5.1.17)


In present case, p = 1/2, n = 1, 2, 3,.. (2) =π =

2

1.645.6

The function (n) is called Riemann zeta function, see appendix.Making use of this result in Eqn. (5.1.16), we have

N 3/ 2 2 1/ 2F F

2 1 3C. 2( T) 1 (2)

3 2 4k − = ε + − ζ ε

23/ 2 2F 2

F

2C 1 ( T)

3 8k

π= ε + ε

...(5.1.18)

where 3/2

3

4 VC (2 )m

h

π= ...(5.1.19)

In the limit T 0

0

3 / 2F

2N C

3= ε or

3/2F(0)

3C N

2−= ε ...(5.1.20)

Therefore 0

2/32/3 2

F3N 3N

2C 8 Vh

m

ε = = π ...(5.1.21)

Since kT/F is small, we see that F does not change rapidly with temperature. Therefore, we

can set F = F0 in the second term on the right hand side of Eqn. (5.1.18). After putting N =

0

3/2F

2C ,

3ε

in Eqn. (5.1.18) we get

0

0

22

3/2 3/2F F

F

2 2 TC C 1

3 3 8

k π ε = ε + ε

From this we get

00

2/322

F FF

T1

8

k− π ε = ε + ε

...(5.1.22)

Using the result (1 + x) – 2/3 = 1 – 2x/3 we can write

0

0

22

F FF

T1

12

k π ε = ε − ε

...(5.1.23)

Thus, the Fermi energy (chemical potential) of a Fermi gas decreases with increasing temperature.


Total energy of electron gas at low temperature

The total energy of electron gas is given by

3/2

FDF0 0

E ( ) ( ) C1 exp( )/ T

f g d dk

∞ ∞ ε= ε ε ε ε = ε+ ε − ε∫ ∫ where g() = C 1/2

5/ 2 2 2 1/ 2F F

2C 152( T) (1/ 2)( /6)

5 4k

= ε + π ε

= 22

5/ 2F

F

2C 5 T1

5 8

k π ε + ε ...(5.1.24)

Substituting 0

3/2F

3C N

2−= ε and F and recalling that F does not change rapidly with temperature

we can express the total energy of electron gas as

0 00 0

2 22 2

3/2 5 / 2F F

F F

2 3 T 5 TE N 1 1

5 2 12 8k k−

π π = ε ε − + ε ε

E 0

0

2 42 2

FF F

3 5 T TN 1 .....

5 12 16k k

π π ≈ ε + − + ε ε ...(5.1.25)

The pressure of the electron gas is given by

22

F(0)F

2 E 2 N 5 TP 1 ............

3 V 5 V 12 T

π = = ε + − ...(5.1.26)

The total energy E and pressure P of a degenerate Fermi gas increase with temperature in thesame manner. The pressure of degenerate Fermi gas is greater than that of an ideal classical gas.This is because fermions obey Pauli’s exclusion principle. A quantum state can accommodate at mostonly one fermion and they are prevented to occupy already occupied low lying energy levels. Femionstend to remain as far apart as possible and hence exert larger pressure. This behaviour of fermions iscontrary to that of bosons, which do not obey Pauli’s exclusion principle. Many bosons can occupy asingle energy level i.e., they can assemble in low energy states. This is why a Bose gas exerts lesspressure than a classical gas.

E E EFermi gas Classical gas Bose gas> >

P P PFermi gas Classical gas Bose gas> >


Fig. 5.1.3 Variation of E of FD and BE gas with temperature

5.2 ELECTRONIC HEAT CAPACITY

According to classical theory, when a system of particles is heated, all the particles absorb heat andcontribute to the heat capacity. Thus the classical theory applied to electron gas in a monovalentmetal predicts electronic contribution to heat capacity equal to 3R/2. But experimental results arefound to be less than 1% of the classical value. This anomaly is removed if one uses Fermi-Diracdistribution function to the electron gas. When a metal is heated only a small fraction of electrons,which are within an energy kT below the Fermi level are excited thermally to vacant states abovethe Fermi level. Electrons, which are deeply situated below the Fermi level don’t participate in thermalexcitation because the energy kT is not enough to take them to the vacant levels above the Fermilevel. This explains why electronic contribution to heat capacity is very small.

The total energy of electron gas is

E 0

0

22

FF

3 5 TN 1

5 12

k π ≈ ε + ε

...(5.2.1)

The heat capacity of electron gas is

0

2 2

F

E NC T

T 2vek ∂ π = =

∂ ε ...(5.2.2)

Let us calculate the value of Cve for copper. For copper F0 = 7 eV. Substituting

NA = 6.02 × 1026 (kmol)–1, k = 1.38 × 10 –23 J/K, Room temperature T = 300 K, kT = 0.026 eV.

0

2 223 26 1

AF

T (3.14) 0.026 eVC N (1.38 10 J/K)(6.02 10 kmol )

2 2 7.0 eVvek

k − −π= ⋅ ⋅ = ⋅ ⋅ × ×ε

= 1460 J (kmol) –1 K –1.

We know that at very low temperature, the lattice heat capacity varies as T3 (Debye T3 law)while the electronic heat capacity varies linearly with temperature T. At very low temperature thelattice heat capacity decreases very rapidly and the electronic heat capacity dominates. At high


temperature the lattice heat capacity dominates over the electronic contribution. The total heat capacityis given by

Cv = A T + B T3 where A and B are constants.

Fig. 5.2.1 Variation of heat capacity with temperature

5.3 THERMIONIC EMISSION (RICHARDSON-DUSHMANN EQUATION)

The emission of electrons from a substance when it is heated to a high temperature is called thermionicemission. The thermionic current density depends on the temperature T, work function and the natureof the emitting surface. The expression, which represents the dependence of thermionic current ontemperature and work function of emitting material, was first derived by Richardson making usethermodynamic principle and later by Dushmann using quantum statistics developed by Fermi andDirac.

According free electron model, the electron are distributed among various available quantumstates according to Pauli’s exclusion principle. At T = 0 K, all the states up to fermi level F arefilled and those above it are empty. The work function denotes the energy required to liberate theelectron at Fermi level from the metal. In order to liberate electron from a metal, the energy impartedto it must exceed (F + ). For an electron to escape, it must arrive at the emitting surface withmomentum p suitably directed. We take the emitting material in the shape of a rectangular box withemitting surface perpendicular to x-axis. For electron emission to take place, the x-component ofmomentum px must be greater than the critical value pxc given by

F2 ( )x xcp p m≥ = ε + ϕ ...(5.3.1)

Let n(px)dpx represent the number of electrons per unit volume having x-component ofmomentum in the range dpx at px. The thermionic current density J is given by

J ( ) . . . ( )

xc xc

x x x x x x

p p

en p dp e v p n p dp

m

∞ ∞

= =∫ ∫ ...(5.3.2)

where vx is x-component of velocity of electron.The number of quantum states in volume element dx dy dz dpx dpy dpz of phase space is

3

2.x y zdx dy dz dp dp dp

h The presence of factor 2 accounts for the fact that for a given momentum


state px, py, pz, there can be two spin states: spin up and spin down. The number of quantum states inunit volume of coordinate space and in volume element dpx dpy dpz of momentum space is denotedby g (px, py, pz)dpx dpy dpz and will be given by

g(px, py, pz) dpx dpy dpz = 3

2 .x y zdp dp dp

h

The number of electrons per unit volume with momentum between px and px + dpx, py andpy + dpy, pz and pz + dpz is

FD3( , , ) 2 ( )x y z

x y z x y z

dp dp dpn p p p dp dp dp f

h= ⋅ ε

= 3F

12

1 exp[( ) / T]x y zdp dp dp

k h⋅

+ ε − ε ...(5.3.3)

where = 2 2 2( )/2x y zp p p m+ + is the energy of electron.

Fig. 5.3.1 Energy of electron in terms of Fermi energy F and work function

The number of electrons with x-component of momentum between px and px + dpx, irrespectiveof the values that py and pz can assume, is given by

3F

2 1( )

1 exp ( ) / Tx x x y zn p dp dp dp dpkh

∞ ∞

−∞ −∞

=+ ε − ε∫ ∫ ...(5.3.4)

At any temperature F( ) / T 1kε − ε >> so 1 may be neglected in the denominator of the FDdistribution. Under this approximation we have

2 2 2

F3

2( ) exp / T

2x y z

x x x y z

p p pn p dp dp k dp dp

mh

∞ ∞

−∞ −∞

+ + = − + ε

∫ ∫

2 22F

3

2exp .exp . exp . exp

T 2 T 2 T 2 Ty zx

x y z

p ppdp dp dp

k mk mk mkh

∞ ∞

−∞ −∞

ε = − − − ∫ ∫


( )( )2

F3

2exp exp 2 T 2 T

T 2 Tx

xp

mk mk dpk mkh

ε = ⋅ − ⋅ π π ...(5.3.5)

where we have used the standard result: 2

0

1exp( )

2x dx

∞π−α =α∫ . In view of this result we have

2F

3

4 T( ) exp exp

T 2 Tx

x x xpmk

n p dp dpk mkh

επ = ⋅ ⋅ − ...(5.3.6)

The thermionic current now becomes

2F

3

4 TJ exp exp

T 2 Txc

xx x

p

pe mkp dp

m k mkh

∞ επ = ⋅ ⋅ ⋅ ⋅ − ∫

2

2 F2

3

4 2T expT

xcpmek m

kh

ε − π =

22 / T

3

4T kmek

eh

−ϕ π=

2AT exp( / T)k= −ϕ ...(5.3.7)

where 2

6 2 23

4A 1.20 10 A K

mekm

h− −π= = × ...(5.3.8)

From above relation

2

Jln ln A

TT k

ϕ= −

10 102

Jlog log A 0.434

TT k

ϕ= −

A plot of log10 J/T2 against 1/T is a straight line. The intercept on y-axis gives log10 A and theslope of the line gives the work function of the emitting material.

The value of A determined from experiment does not agree with the theoretical value obtainedby putting the values of constants in the expression for A. A correction needs in the expression forthe thermionic current density. When an electron leaves the emitting surface, the latter becomes positiveand pulls the outgoing electron back to the material. If r denotes the fraction of electrons reflected


back to the material, the thermionic current density will be given by

2J (1 )A T exp

Tr

k

ϕ = − − ...(5.3.9)

Fig. 5.3.2 Variation of log10 J/T2 with 1/T

5.4 AN IDEAL BOSE GAS

A Bose gas is a many particle system consisting of non-interacting bosons which are indistinguishableparticles having integral spin (0, 1, 2 .....). The appropriate distribution function which describes thebehaviour of bosons is Bose-Einstein distribution function.

For an ideal BE gas of N bosons in an enclosure of volume V, the mean occupation number nr

in r th single particle state (the most probable number of particles with energy r ) is

,( )

exp 1 exp ( ) 1r r

r rr r

g gn ε = =

α +βε − β ε −µ − = 1/kT ...(5.4.1)

where = – µ = – Tk

µ...(5.4.2)

The parameter µ is called chemical potential and can be determined as a function of N andtemperature T by the condition

1 2

1 20

N .......exp[ ( )] 1 exp[ ( )] 1r

r

g gn

∞

== = + +

β ε − µ − β ε − µ −∑= n1 + n2 + …………. ...(5.4.3)

where the sum is over the discrete energy levels.The total number of particle in the system is

1/ 23/2

3 ( )0

2 VN (2 )

1s

dg m

h e

∞

β ε−µπ ε ε=

−∫ ...(5.4.4)

Where gs = 2s + 1 is spin degeneracy, s = spin of the boson. For spinless boson s = 0 and gs = 1


The mean occupation number is always positive or zero i.e., nr 0 for all values of . We takethe energy scale such that the ground state energy 1 is zero. The occupation number of the groundstate is

1

1 1 1, 1, 01

gn g

e−βµ= = ε =

− (ground state is non-degenerate) ...(5.4.5)

The condition n1 0 implies that µ 0. Thus µ of an idealB-E gas is always negative. From Eqn. (5.4.3) we see that left handside i.e., N is constant, so must be the right hand side. This impliesthat as T is lowered (or is increased), µ which is negative, mustincrease. Thus, the maximum value of µ is zero, µmax = 0.

Let us define a function by equation

η = µβexp( )

is called fugacity (absolute activity) of system. The valueschemical potential µ and fugacity satisfy the inequality

−∞ ≤ µ ≤ ≤ η = µβ ≤0, 0 exp 1 ...(5.4.6)

We define a temperature TC , called critical temperature, which corresponds to the maximumvalue of chemical potential i.e., µ = 0. This temperature is given by

C

3 / 2 1/2

3 / T0

2 V(2 )N

1s k

m dg

h e

∞

επ ε ε=

−∫ ...(5.4.7)

To find the value of integral, we change the variable from to x by substitution x = /kTC.Doing so, we get

3 / 2 1/ 2C

20

2 T 2N V

1s x

mk x dxg

h e

∞ π = π − ∫

The value of the quantity in square bracket is represented by function F3/2(1) whose value isequal to 2.61.(See appendix). Therefore,

3/2C

3/22

2 TN V F (1)s

mkg

h

π =

= 3/ 23

0

VF (1)sg

λ...(5.4.8)

= = ∴ =λ 3/ 23

0

VN 2.61 F (1) 2.61sg ...(5.4.9)

where 0C2 T

h

mkλ =

π

Fig. 5.4.1 Variation chemicalpotential µ with temperature T


is the thermal de Broglie wavelength of the particle at T = TC. The critical temperature TC is givenby

2 / 32

CN

T , .2 2.61 V

h

mk

ρ = ρ = π (for spinless particle s = 0, gs = 1) ...(5.4.10)

For 1 kmol of Helium gas N = NA = 6.023 × 1026 kmol–1. m = 6.65 × 10–27 kg,N/V = 2.2 × 1028 m–3,

TC = 3.13 K.

The Eqn. (5.4.4) which gives the total number of particles in the system is not valid whenT < TC.

Let us see why? The expression for density of states viz. 3 / 2 1/ 23

2 V( ) (2 )g m

h

πε = ε contains a

factor 1/2. For ground state 1 = 0, g(0) = 0 and hence n1 = 0. The occupation number correspondingto ground state energy ( = 0) does not contribute to the total number of particles. In fact g() for

ground state is not zero but 1. At high temperature this replacement of by ∫ does not introduce

any significant error because the ground state is thinly populated and the contribution of this termmay be omitted. At very low temperature T < TC we cannot overlook the first term in the sum givenby Eqn. (5.4.3). Instead we must explicitly retain the first term as such and write the remainingterms as integral as given below.

1/ 23/ 2

3 ( )2 0

2 V1 1N (2 )

1 1 1i s

i

n g m de e h e

∞∞

−βµ −βµ β ε−µ=

π ε= + = + ε− − −

∑ ∫ ...(5.4.11)

N = N0 + Nexc ...(5.4.12)

The first term in Eqns. (5.4.11 and 5.4.12) represents the number of particles in the groundstate (1 = 0). These particles do not contribute to the energy and momentum of the system. Thesecond term represents aggregate of particles occupying the higher energy state > 0. Only theseparticles contribute to the energy and momentum of the system.

At higher temperature (T > TC) the number of particles in the ground state ( = 0) is negligiblysmall and hence may be omitted.

Below TC, the chemical potential is very close to zero (µ 0) and the number of particles inthe excited states ( > 0) is given by

1/ 2

3/ 23 1

0

2N V (2 ) , 0

1exc s

dg m

h e

∞

− βεπ ε ε= µ→

η −∫ , µβη = →1e ...(5.4.13)

3 / 2 1/ 2

2 10

2 T 2V ,

1s x

mk x dxg x

h e

∞

−

π = = βε π η − ∫ ...(5.4.14)


Nexc

(3/ 2) 1

3 10

V 1

(3/ 2) 1s x

xg dx

e

∞ −

−

= Γλ η −

∫ = 3/2 3/ 23 3

V VF ( ) F (1)s sg gη =

λ λ

...(5.4.15)


3/2

C

N T

N Texc

=

...(5.4.16)

Equation (5.4.16) gives the fraction of particles occupying the states with energy > 0. Thefraction of particles occupying the ground state is given by

3/20N T

1N Tc

= −

...(5.4.17)

A plot of N = 0/N as a function T/TC is shown in the Fig. 5.4.2. From the figure, it is evidentthat for T > TC, the number of particles in the ground state is negligible. As T falls below TC, thenumber of particles in the ground state suddenly increases rapidly. The process of dropping particlesrapidly into the ground state with zero energy is known as Bose-Einstein condensation. Theparticles in this state possess zero energy and zero momentum. They contribute neither pressure nordo they possess viscosity. (Viscosity is related to transport of momentum.) B-E condensation is secondorder phase transition. A B-E gas at temperature below TC is called degenerate and TC is known asdegeneracy temperature or condensation temperature. The ordinary condensation of vapour into liquidtakes place in conventional coordinate space whereas Bose-Einstein condensation occurs in momentumspace.

Fig. 5.4.2 Fraction of particles in the ground state and excited state as function of temperature

5.5 DEGENERATION OF IDEAL BOSE GAS

Degeneration of a system refers to the state in which it shows significant deviation from the propertiesideal Boltzmann gas. This can be illustrated by calculating energy, pressure, entropy, specific heatetc., of the Bose gas. We shall calculate these quantities for T < TC and T > TC. Below TC, Bose gasis strongly degenerate.


Strongly Degenerate Bose Gas ( T < TC )

Total Energy E: In the degenerate state the thermodynamic properties of the system, such as totalenergy, pressure, entropy are less than those of classical Boltzmann gas.

The total number of particles in the system is

1/ 2

3/23 1

0

2 VN (2 ) ,

1sg m d

h e

∞

− βεπ ε= ε

η −∫ βµη = e . ...(5.5.1)

Total energy of the system is

0

E ( )n d∞

= ε ε ε∫

3/ 23/ 2

3 10

2 VE (2 )

1sg m

h e

∞

− βεπ ε=

η −∫ ...(5.5.2)

Putting = x in Eqns. (5.5.1) and (5.5.2), we have

3 / 2 1/ 2

2 10

2 TN 2 V

1s x

mk x dxg

h e

∞

− = π

η − ∫ ...(5.5.3)

and

3/2 3/2

2 10

2 TE (2 V) T

1s x

mk xg k dx

h e

∞

− = π

η − ∫ ...(5.5.4)

In terms of thermal de Broglie wavelength

2 T

h

mkλ =

π

we can express N and E as follows.

1/ 2

3/ 23 10

V 2 VN F ( )

1s sx

xg dx g

e

∞

−= ⋅ = ηλπλ η −∫ ...(5.5.5)

where

(3 / 2) 1

3 / 2 10

1F ( )

(3 / 2) 1x

xdx

e

∞ −

−η =

Γ η −∫ ...(5.5.6)

Now,

3 / 2 3 / 2

2 10

2 TE (2 V)( T)

1s x

mk xg k dx

h e

∞

− = π η − ∫

= 3/ 2

3 10

V 3 1. T.2 (5/ 2) 1

s x

xg k dx

e

∞

−⋅Γλ η −∫


= 5/ 23

V 3T F ( )

2sg k⋅ ⋅ ⋅ ηλ

...(5.5.7)

where

(5 / 2) 1

5 / 2 10

1F ( )

(5 / 2) 1x

xdx

e

∞ −

−η =

Γ η −∫ ...(5.5.8)

Below condensation temperature TC, 1, the total energy of the system is

3/ 230 5/ 2

3/ 2 C

F (1)3 3 T 1.342E N T N T

2 F (1) 2 T 2.612k k

λ = ⋅ ⋅ = ⋅ ⋅ λ

3 / 2

C

TE 0.77N T

Tk

=

...(5.5.9)

At T = TC, 0 classical3

E 0.514 N T 0.514E2

k = =

...(5.5.10)

Thus, at T = TC, the total energy of a Bose gas is one-half of the ideal Boltzmann gas.Pressure P : The pressure of Bose gas is given by

3 / 2 3 / 2

C C

2 E 2 3 T N T TP . N T (0.5134) 0.5234

3 V 3V 2 T V T

kk

= = = ...(5.5.11)

At T = TC, 0

0N T

P 0.5134V

k= ...(5.5.12)

Thus, at T = TC, the pressure of Bose gas is nearly one-half of the ideal gas. This is becausethe particles in the ground state have zero momentum and contribute nothing to the pressure.

Specific Heat : The specific heat is given by

( )

3/2 3/2

C CV

E T TC 0.77 N T. 1.926 N

T T T Tv k k ∂ ∂ = = = ∂ ∂

...(5.5.13)

Entropy : It is given by

3/2 3/2T T

C C0 0

C 1.926N T TS T T 1.28 N

T T T Tv k

d d k

= = =

∫ ∫ ...(5.5.14)

At T = 0, S = 0.Non-degenerate Bose Gas T > TCEnergyFor T > TC, << 1. In this condition

3/23

VN N .F ( )exc sg= = η

λ...(5.5.15)


Total energy from Eqn. 5.5.7

= ⋅ ⋅ ηλ 5/23

3 VE T F ( )

2sg k

From Eqns. (5.5.15) and (5.5.7)

3/25 / 2 C

33/ 2

F ( ) T3 V 3E N T N T 1 0.462 .....

2 F ( ) 2 Tk k

η = ⋅ ⋅ = − − η λ ...(5.5.16)

Pressure P

3/2CT2 E N T

P 1 0.462 ......3 V V T

k = ⋅ = − − ...(5.5.17)

Specific Heat Cv

3/2CTE 3

C N T 1 0.462T T 2 Tv

V

k ∂ ∂ = = − − ∂ ∂

3/ 2CT3

N 1 0.2312 T

k = + +

...(5.5.18)

Entropy S

C

T 3/2C

C CCT

C T T3 TS S(T ) S(T ) N ln 0.154 1

T 2 T Tvd

k = + = + + − +

∫ ...(5.5.19)

Fig. 5.5.1 Variation of CV with temperature


5.6 BLACK BODY RADIATION: PLANCK’S RADIATION LAW

The thermal radiation in a cavity maintained at temperature T is a well-known example of B-E system.In quantum picture the radiation is regarded as an assembly of particles, called photons, each ofwhich has spin 1. The energy and momentum p of photon are given by = and p = (/c) = /c.

To obtain the energy of photon gas we need to know the number of quantum states g(p) dpavailable to photons with momentum in the range dp at p. Since photon has zero rest mass, theexpression for density of states

3/ 2 1/ 2

3

2 V( ) (2 )g d m d

h

πε ε = ε ε ...(5.6.1)

as such cannot be applied. To apply it to photons we must convert it in terms of momentum p throughsubstitutions = p2 /2m and d = (2p dp/2m). Making use of this substitution we get

2

3

4 V( )g p dp p dp

h

π= ...(5.6.2)

This expression for the number of quantum states accessible to photons needs correction.Electromagnetic waves are purely transverse and there can be two sets of waves polarized in mutuallyperpendicular planes or right and left circular polarizations. Thus, a photon of definite momentumcan be in two possible states. The net effect is to multiply the above expression for the density ofstates by two. Thus, the number of photon states in which the photon has momentum in the range pand p + dp is

2

3

8 V( )g p dp p dp

h

π= ...(5.6.3)

In terms of frequency , the number of states in the frequency range d at is

2

3

8 V( )g d d

c ch

π ω ω ω ω =

22 3

Vd

c= ω ωπ

...(5.6.4)

According to B-E distribution the mean number of photons per quantum state at energy is

1 1 1

( ) ,T1 1

fke eβε β ωε = = β =

− − ...(5.6.5)

The number of photons in the frequency range and + d is

2

2 3

V( ) ( ) ( )

1

dn d f g d

c eβ ωω ωω ω= ω ω ω= ⋅

π − ...(5.6.6)


The energy of photon gas in the frequency range and + d is

3

2 3

V( ) . ( )

1

dd n d

c eβ ω

ω ωε ω ω = ω ω ω= π −

...(5.6.7)

The energy density in the frequency range and + d is

3

2 3( ,T) .

1u d d

c eβ ωωω ω = ω

π −

...(5.6.8)

or 2

5

16 1( , T)

[exp(2 / T)] 1

cu d d

c k

πλ λ = λπ λ −λ

...(5.6.9)

This is the Planck’s radiation law. Stefan’s law and Wien’s law both have been derived fromthis law.

5.7 VALIDITY CRITERION FOR CLASSICAL REGIME

The mean occupation number n () of energy state , according to classical (M-B) statistics, is

/ T

/ T( ) ki

ik

gn g e e

e e−α −ε

α εε = = ...(5.7.1)

and that according to quantum statistics is

/ T( )

1ik

gn

e eα εε =

± ...(5.7.2)

+ sign for F-D and – sign for B-E statisticsThe quantum statistics become identical with the M-B statistics in the limit

/ T 1ke eα ε >> ...(5.7.3)

for all values of . For = 0 the Eqn. (5.7.3) reduces to

α >> 1e ...(5.7.4)

Equation (5.7.4) will certainly hold for > 0. So the equation represents the criterion for thevalidity of classical statistics. Now the parameter is determined from the condition

N = ( ) ii i i

i i

n g e e−βε−α= ε∑ ∑

3/2 1/2

30

2 VN ( ) (2 )i

i ii

e g e m e dh

∞−βεα −βεπ= ε → ε ε∑ ∫

= 3/23

2 V 1(2 )

2m

h

π π


3 / 2

2

V 2 T

N

mke

hα π =

...(5.7.5)

So, in view of Eqn. (5.7.5) the criterion (5.7.4) becomes

3/2

2

2 TV1

N

mk

h

π >> ...(5.7.6)

If a is the average separation between the particles of the system then each particle may beallotted a cubical volume a3. This must be equal to V/N. So (V/N)1/3 gives the mean distance betweenthe molecules. Now the thermal de Broglie wavelength of particles is given by

3

, T22 3 T

h h hk

p m mkλ = = = ε =

ε

So

3/2 3/2

3 2 2

2 T1 3 T mkmk

h h

π = ≈ λ

...(5.7.7)

In view of Eqn.(5.7.7), the condition (5.7.6) becomes

>> λ

3

1a

>> λa ...(5.7.8)

Thus, the classical statistics is valid if the average separation between particles is much greaterthan the mean de Broglie wavelength of the particles. This condition will be satisfied when (i) thetemperature is large, (ii) number density is very small (i.e., the gas is dilute), and (iii) mass of particleis not too small. When these conditions are not met, the particles are close together and their wavefunction overlap and they are no longer distinguishable.

Let us illustrate this by example. Consider helium gas at N.T.P., the average separation betweenmolecules is

1/ 31/ 3 38

23

V 22.4 1032 10 cm

N 6.6 10a − × = = = × ×

= 32 Å

The de Broglie wavelength of molecule is

34

10

27 23

6.6 10 Js0.8 10

3 T 3 6.8 10 kg 1.38 10 J/K 300K

h

mk

−−

− −

×λ = = = ×× × × × ×

m = 0.8 Å

Since a >> , classical (M-B) statistics can be applied to helium gas at room temperature.


Now consider liquid helium gas at 10 K. The average separation between molecules is

1/ 3

23 3 1/3 8V(5 10 cm ) 4 10 cm 4

Na − − = = × ≈ × =

Å

de Broglie wavelength of molecule is

3410

27 23

6.6 10 Js4 10 m 4

3 T 3 6.8 10 kg 1.38 10 J/K 10 K

h

mk

−−

− −

×λ = = = × =× × × × ×

Å

Since a , quantum (B-E) statistics must be applied to liquid helium.For conduction electrons in metals the average separation between electrons is

( )1/ 3

1/323 3 8V10 cm 2 10 cm 2

Na − − = = ≈ × =

Å

The de Broglie wavelength of electron is

34

10

31 23

6.6 10 Js62 10 m 62

3 T 3 9.1 10 kg 1.38 10 J/K 300 K

h

mk

−−

− −

×λ = = = × =× × × × ×

Å

Since the condition >> a is satisfied, quantum (F-D) statistics is most appropriate for thetreatment of conduction electrons in metals.

5.8 COMPARISON OF M-B, B-E AND F-D STATISTICS

1. The distribution functions for the three statistics giving the mean number in of particlesoccupying a state with energy i are given by

−βε−α= ii in g e e (M-B)

βεα=

−1i

ii

gn

e e (B-E)

βεα=

+1i

ii

gn

e e(F-D)

In the classical limit gi >> ni, B-E and F-D both distribution functions approach the M-Bdistribution function. M-B statistics is a classical statistics, B-E and F-D statistics are quantumstatistics.

2. M-B statistics applies to systems comprising of distinguishable particles, whereas B-E andF-D statistics apply to indistinguishable particles.

3. Spin of particles constituting the system is not a criterion for the applicability of M-Bstatistics. B-E statistics is applicable to particles having integral spins 0, 1, 2, ..... Such


particles are called bosons. Examples of bosons are: photons, phonons, hydrogen molecule,liquid helium, mesons, etc. F-D statistics applies to particles having half-integral spins

1 3 5, ,

2 2 2….. Such particles are called fermions. Examples of fermions are: electrons, protons,

neutrons, …etc.

4. M-B and B-E statistics put no restriction on the number of particles that can occupy aquantum state. F-D statistics permits at the most one particle that a quantum state canaccommodate.

5. To specify a microstate of a system, the classical phase space is divided into cells whosevolume may be taken as small as we please. A cell in this phase space represents a microstateof the system. In quantum mechanical description the phase space is divided into cellswhose volume is not less than h3, h being Planck’s constant. A cell of volume h3 representsa quantum state (microstate) of the system.

6. At high temperature both quantum statistics (B-E and F-D) approach the M-B statistics.

7. The variation of three distribution functions f() = ni /gI, with energy at differenttemperatures is shown in the figures. According to M-B and B-E distribution functions,at a given temperature, particles like to occupy the lower enrgy states. At lower energythe occupation number is larger for B-E than for M-B distribution function. As the energyof the system increases, the occupation number decreases. According to F-D distributionfunction, at T = 0 K, all the quantum states with energy less than Fermi energy are occupiedby electrons in an electron gas and those above the Fermi level are empty. As temperatureincreases, the electrons in the energy states a little below the Fermi level are excited toempty energy states a little above the Fermi level. At very high temperature the F-Ddistribution becomes more and more like M-B distribution.

Fig. 5.8.1 M-B statistics Fig. 5.8.2 F-D statistics


Fig. 5.8.3 B-E statistics

8. The values of gi/ni for the three distribution functions are:

M-B ii

i

ge

nα+βε=

B-E 1 ii

i

ge

nα+βε+ =

F-D 1 ii

i

ge

nα+βε− =

When the number of quantum states in an energy level is much larger than the number ofindistinguishable particles (bosons or fermions) i.e., gi >> ni , the term 1 may be omitted. In thissituation, both the quantum statistics reduce to M-B statistics and hence M-B statistics can safely beused to the system. Thus, the classical limit is reached when gi >> ni. This condition may be put inother forms as

e >> 1

or

3/ 2

2

V 2 T1

N

mk

h

π >>

or a >>

where V = volume of the system, N = total number of particles in the system, T = temperature ofthe system, a = average separation of particles and = de Broglie wavelength of the particles.

This condition is satisfied in a gaseous system when the pressure is not too large and thetemperature is not too low under normal conditions.

CHAPTER

PARTITION FUNCTION

6.1 CANONICAL PARTITION FUNCTION

The canonical partition function of a single molecule of an N-particles system occupying a volumeV at temperature T is defined by

/ T(V,T) r k

r

z e−ε=∑ ...(6.1.1)

Henceforth, we shall denote single particle partition function by lower case letter z and that ofN-particle system by upper case letter Z. The summation in Eqn. (6.1.1) is performed over all discretequantum states of the particle. If the degeneracy of the energy level r is g(r), Eqn. (6.1.1) takes theform

/ T(V,T) ( ) r kr

energy levels

z g e−ε= ε∑ ...(6.1.2)

where summation is only over all different energies r. The quantity z is very useful quantity forcalculating the macroscopic properties of any thermodynamic system in equilibrium. The evaluationof the sum in Eqn. (6.1.1) requires the knowledge of single particle quantum states of the constituentparticles of the system.

There are many problems in which the Hamiltonian can be written as a sum of simplerHamiltonians. The most obvious example is the case of a dilute monatomic gas where the moleculesare on the average far apart and hence their intermolecular interaction can be neglected. Themolecules of the gas have kinetic energies only. The total Hamiltonian is in this case is expressed as

N 2

1

H2

i

i

p

m==∑ ...(6.1.3)

N is the number of particles (molecules) in the gas.Another example is the decomposition of Hamiltonian of a polyatomic molecule into its various

degrees of freedom viz translational, rotational, vibrational, electronic etc.

H H H H Htrans rot vib ele= + + + ...(6.1.4)

Partition Function 335

There are many other problems in physics in which the Hamiltonian by a proper and clever choiceof variables can be written as a sum of individual terms as shown above. In all such cases the partitionfunction of a molecule comes out to be a product of partition functions corresponding to each degreeof freedom. For example, consider a system comprising of distinguishable particles a, b, c, ….. Let

the energy states of these particles be , , ...a b cj k lε ε ε The superscripts denote the particles and the

subscript the energy states. In this case the partition function of the system becomes

..... / T

, , ...

Z (N,V,T)a b cj k l k

j k l

e − ε + ε + ε + = ∑

/ T / T / T

a b cj k lk k k

j k l

e e e−ε −ε −ε=∑ ∑ ∑

= . . ......a b cz z z ...(6.1.5)

This is a very important result. It shows that if we write the N-particle Hamiltonian as a sum ofindependent terms, then the calculation of Z of the system reduces to a calculation of partition functionz of a single particle. Since z requires knowledge only of energy levels of a single particle, its evaluationis quite simple. If the constituent particles are identical (i.e., of the same kind), za = zb = zc..... = z,the partition function of the system is

NZ z= (distinguishable particles) ...(6.1.6)

6.2 CLASSICAL PARTITION FUNCTION OF A SYSTEM CONTAINING NDISTINGUISHABLE PARTICLES

In classical approximation, the energy of a N-particle system depends on 3N generalized coordinatesq1, ….q3N, and 3N momentum coordinates p1, …..p3N. The phase space is divided into cells ofvolume h3N. To evaluate the partition function from Eqn. (6.1.1) we first take the sum over the

number 1 3N 1 3N3N

.... . ......dq dq dp dp

h of cells of phase space at point (q1,….q3N, p1……p3N) and then

take the sum (or integral) over all such volume elements. Thus in classical approximation the partitionfunction of a N-particle system is given by

( , ) / T 1 3N 1 3N

3N

..... . ......Z ..... .q p k dq dq dp dp

eh

−ε= ∫ ∫ ...(6.2.1)

Notice that the transition from quantum partition function to classical partition function can beaccomplished by following replacement :

1 3N 1 3N

3N

..... . ......

r

dq dq dp dp

h→∑ ∫ ...(6.2.2)


For a system of particles possessing only translational kinetic energy, the evaluation of integralin Eqn. (6.2.1) is simple.

2 2 23 3 3 31 2 N

1 N 1 N3N

( .....1...... exp ......

2

p p pZ d q d q d p d p

mh

β + + += −

∫ ∫

where d3q1 = dx dy dz and d3p1 = dpx dpy dpz etc.

22N3 3N1

N 1 N3N

VZ exp ................ exp p

2 2

ppd p d

m mh

∞ ∞

−∞ −∞

ββ= − −

∫ ∫

= 3/ 2N

3N

V 2..........N factors

m

h

π β

= 3N / 2N

3N

V 2 m

h

π β

ZN =

3N / 2N

2

2V

m

h

π β

...(6.2.3)

We have already mentioned that the evaluation of partition function for many-particle systemreduces to the evaluation of partition function for a single particle. Consider a N-particle systemwhose constituent particles have translational kinetic energy r = p2/2m only. Let us illustrate this. Inthe phase space of a single particle, in the volume element dx dy dz dpx dpy dpz (= d3q d3p) there are

3

x y zdx dy dz dp dp dp

h or

3 3

3

d q d p

h possible quantum states. Therefore, the partition function for a

single particle is

23 3

3

1...... exp .

2

pz d q d p

mh

β = − ∫ ∫

The integration over the ordinary space coordinates gives the volume V occupied by the system ofparticles. So

2 2 2

3

(Vexp

2x y z

x y z

p p pz dp dp dp

mh

∞ ∞ ∞

−∞ −∞ −∞

β + + = − ∫ ∫ ∫

2 22

3

Vexp exp exp

2 2 2y zx

x y z

p ppdp dp dp

m m mh

∞ ∞ ∞

−∞ −∞ −∞

β ββ = − − − ∫ ∫ ∫


1/ 2 1/ 2 1/ 2

3

V 2 2 2m m m

h

π π π= β β β

z 3 / 2

2

2 TV

mk

h

π = =

3

V

λ,

2 T

h

mkλ = =

πde Broglie wavelength of the particle.

The partition function for the indistinguishable N-particle system is

3N / 2

N N2

2 TZ V

mkz

h

π = = ...(6.2.4)

2

3 2 3lnZ N lnV ln ln , 1/ T

2 2

mk

h

π = + − β β = ...(6.2.5)

6.3 THERMODYNAMIC FUNCTIONS OF MONOATOMIC GAS

Mean energy

lnZ 3N 3E N T

2 2k

∂= − = =∂β β ...(6.3.1)

Helmholtz free energy

3 / 2

2

2 TF TlnZ N Tln V

mkk k

h

π = − = − ...(6.3.2)

The entropy of the system

S lnZ Ek = + β

= 2

3 3 2 3N ln V ln T ln

2 2 2

mkk

h

π + + + ...(6.3.3)

= 3

N lnV ln T2

k + + σ

...(6.3.4)

where π σ = + 2

3 2ln 1

2

mk

h...(6.3.5)

The application of Eqn. (6.3.4) in calculating the change in entropy when two samples of thesame gases of equal volume and at the same pressure and temperature are mixed leads to the famous


paradox known as Gibbs paradox. This equation is valid for a gas of distinguishable molecules andneeds correction when it is to be applied to a system of indistinguishable molecules.

6.4 GIBBS PARADOX

Consider a vessel divided into two compartments by a removable partition. The two compartmentsare then filled with two different gases. The number of molecules and volumes of the two gases areN1, V1 and N2, V2 as shown in the Fig. (6.4.1). The gases are at the same temperature and pressure.Now the partition is removed. On removing the partition, the gases are mixed owing to diffusion of

Fig. 6.4.1 Mixing of two different gases initially separated by a partition wall

molecules. The diffusion is an irreversible process. (By putting the partition back in its original position,the gases don’t separate and the original state of gases is not achieved.) In irreversible process ofmixing of the two different gases the entropy increases. Let us calculate the increase in entropy. Theentropy of a gas composed of N identical distinguishable molecules is given by

2

3 3 2 3S N ln V ln T ln

2 2 2

mkk

h

π = + + + ...(6.4.1)

Before mixing the entropy of the combined system is

12 1 2S S S= + = 11 1 2

23 3 3N lnV lnT ln

2 2 2

m kk

h

π + + +

+

2

2 2 2

23 3 3N ln V ln T ln

2 2 2

m kk

h

π + + +

...(6.4.2)

After mixing the entropy of the combined system is

12 1 2S S S′ ′ ′= + = 11 1 2 2

23 3 3N ln (V V ) ln T ln

2 2 2

m kk

h

π + + + +

+

2

2 1 2 2

23 3 3N ln (V V ) ln T ln

2 2 2

m kk

h

π + + + +

...(6.4.3)


The change in entropy is

1 2 1 21 2

1 2

V V V VS N ln N ln

V Vk + +

∆ = +

...(6.4.4)

If we take N1 = N2 = N and V1 = V2 = V then S = 2 k N ln 2 = positive number. ...(6.4.5)

The entropy in the process increases and this result is in agreement with the experiments.Now suppose that the two compartments of the vessel are filled with the same gases such that

N1 = N2 = N, V1 + V2 = V. The two samples of the gases are at the same temperature and pressure.If the partition is removed, the increase in entropy of the combined system calculated as before comesout to be equal to

S = 2 N k ln 2

That is, the entropy increases in the process of mixing of two identical samples of the samegases. This conclusion is not correct. Further if we put a large number of partitions in the vessel andremove them one by one, then by increasing the entropy in each act of removing the partition, we

Fig. 6.4.2 Mixing of two identical samples of the same gases

can increase the entropy of the combined system by any amount. The mixing of two identical samplesof the same gases is a reversible process because by inserting the partition to its initial position, weget the state of the gases, which are in no way different from that we had before mixing. The totalentropy of the system should not change on removal of the partition. This is known as Gibbs paradox.The origin of this paradox lies in the use of expression for entropy derived from the formula ofpartition function

3N / 2

N N2

2 TZ V

mkz

h

π = = (distinguishable particles) ...(6.4.6)

In the derivation of this expression it was assumed that the particles of the system aredistinguishable and the interchange of positions of two molecules would lead to a physically distinctstate of the system. But this is no so. In quantum mechanical treatment of a gas, the molecules arecompletely indistinguishable. A calculation of partition function and entropy, assuming theindistinguishability of molecules, would not give rise to Gibbs paradox. A way out to the Gibbsparadox is to apply a correction to the expression of partition function (6.4.6). We know thatN-molecules can be arranged in N! ways by permuting among themselves. If the molecules areconsidered indistinguishable, then these N! possible permutations of such molecules would not lead


to physically distinct states. So the number of distinct states over which the summation is made incalculation of classical partition function is large by a factor N!. The correct partition function willbe that which takes into account the indistinguishability of molecules. This is obtained by dividingthe expression (6.4.6) of partition function by N!.

N3/ 2N

2

1 2Z V

N! N!

z m

h

π = = β (indistinguishable particles) ...(6.4.7)

2

3 2ln Z N ln V ln ln N!

2

m

h

π= + − β

Using Stirling’s approximation ln N! = N ln N – N, we obtain

2

3 2ln Z N ln V ln N ln N N

2

m

h

π= + − + β

= 2

V 3 2N ln ln 1

N 2

m

h

π+ + β ...(6.4.8)

Entropy

S ln Z Ek= +β , E = (3/2) N k T

Therefore

3S lnZ N

2k k= +

= 2

V 3 3 2 5N ln lnT ln

N 2 2 2

mkk

h

π + + + ...(6.4.9)

Using this formula for calculating the increase in entropy in mixing of two samples of thesame gases, we obtain

1 2 2

V 3 3 2 5S S S 2N ln ln T ln

N 2 2 2initialmk

kh

π = + = + + + ...(6.4.10)

1 2 2

2V 3 3 2 5S S S 2N ln lnT ln

2N 2 2 2finalmk

kh

π ′ ′= + = + + + ...(6.4.11)

Change in entropy on mixing the gasesS = 0


6.5 INDISTINGUISHABILITY OF PARTICLES AND SYMMETRY OF WAVEFUNCTIONS

Let us consider a system consisting of two identical particles labelled 1 and 2. If the particle is restrictedto move in a certain region, the quantum mechanical treatment of the particle allows it to have discretequantum states and discrete energy levels. Let r (1) and Er (1) denote the r th quantum state (wavefunction) and energy of particle 1. Similarly s (2) and Es (2) denote the sth state and energy ofparticle 2. Now suppose that both the particles 1 and 2 are present simultaneously in the same region.

If the average separation between the particles is much greater than their de Broglie wavelengthi.e., the wave functions of the particles donot overlap, the particles are said to be distinguishable andthe wave function of the two particles is simple product of individual particles. Thus

(1, 2) = r (1) s (2) ...(6.5.1)

r (1) means that the particle 1 is in the state r. This may be generalized to a system of N-particles,where N is very large. The wave function of an N-particle system is

(1, 2, 3…..) = r (1) s (2) t (3) ……..z (N) ...(6.5.2)

By distinguishable particles we mean that any interchange of particles among the occupied statesviz., r (2) s (1) leads to a new state for the system without any change in the total energy of thesystem. A distribution function, which gives the distribution of particles among the various energylevels, derived on the assumption that the particles are distinguishable, is called classical or Maxwell-Boltzmann distribution.

If the average separation between the particles is less than the de Broglie wavelengths of theparticles, then their wave functions overlap. The particles are said to be indistinguishable and quantumstatistics is appropriate for their description. The wave function of the whole system must satisfycertain symmetry requirements.

If the system is composed of particles having integral spin (0, 1, 2, …), it must be describedby a wave function that must be symmetric with respect to interchange of two particles. That is, thewave function should not change its sign on interchanging two particles.

s (1, 2) = s (2, 1)

The superscript s stands for symmetry. For a system of two particles, the wave function of thesystem is obtained from the linear combination of single particle wave functions r (1) and s (2).Thus

1

(1, 2) (1) (2) (2) (1)2

sr s r sψ = ϕ ϕ + ϕ ϕ ...(6.5.3)

where 2 is normalization factor. We can verify that the wave function Eqn. (6.5.3) of the systemremains unchanged on interchanging the particles.

1

(2,1) (2) (1) (1) (2) (1, 2)2

s sr s r sψ = ϕ ϕ + ϕ ϕ = ψ


Thus, the wave function s (1, 2) given by Eqn. (6.5.3) satisfies the symmetry requirement. Asystem consisting of particles of integral spins are described by symmetric wave function and thestatistical behaviour of the system is described by quantum statistics called Bose-Einstein statistics.The particles with integral spin are called bosons. If nr represents number of bosons in any quantumstate then nr = 0, 1, 2, 3, …. There is no restriction on the number of bosons that can be in a quantumstate.

A system consisting of particles having half-integral spins (1 3

, ,......)2 2

must be described by

wave function that must be anti-symmetric with respect to interchange of two particles. That is, thewave function must change sign without a change in its magnitude. For a two-particle wave functionthis requirement can be expressed as

A (1, 2) = – A (2, 1) ...(6.5.4)

For a two-particle system, the wave function A(1, 2), which is antisymmetric is obtained fromthe linear combination of single particle wave functions as follows.

A 1

(1, 2) (1) (2) (2) (1)2

r s r sψ = ϕ ϕ − ϕ ϕ ...(6.5.5)

= (1) (2)1(1) (2)2

r r

s s

ϕ ϕϕ ϕ

= A (2,1)−ψ

For an N-particle system, the wave function of the system is given by Slater determinant

A

(1) (2) .... (N)

(1) (2) ..... (N)1(1, 2, 3,....N)

N(1) (2) ..... (N)

r r r

s s s

z z z

ϕ ϕ ϕϕ ϕ ϕ

ψ =

ϕ ϕ ϕ

...(6.5.6)

where N is normalization factor. The particles with half-integral spin are called fermions.Putting r = s in A (1, 2) given by Eqn.(6.5.5) we have A (1, 2) = 0. That is if we put the

two fermions in the same state then the wave function vanishes. In other words, no two fermions canbe in the same quantum state. This statement is called Pauli exclusion principle. If nr is the numberof fermions in any quantum state then nr = 0, 1 for all r.

6.6 PARTITION FUNCTION FOR INDISTINGUISHABLE PARTICLES

Let the wave functions and energies of two non-interacting distinguishable particles 1 and 2 ber (1), r (1) and s (2), s (2) respectively. If both the particles are present simultaneously in aregion, the combined wave function and energy of the system are given by


(1, 2) = r (1) s (2)

12 = r (1) + s (2)

The canonical partition function of the system is

( )12 (1) (2)

, ,

Z r s

r s r s

e e−β ε +ε−βε= =∑ ∑

= −βε−βε =∑ ∑ (2)(1)1 2.sr

r s

e e z z

where summation extends over all quantum states of the individual particles 1 and 2. If the particlesare identical r (1) = s (2) and z1 = z2 = z (say). The partition function of the system becomes

Z = Z2 (two particle system)

Generalization of this result to N-particle system givesZ = zN (identical distinguishable N-particle system)

If the particles 1 and 2 are indistinguishable the wave function (1, 2) of the system must beeither symmetric or anti-symmetric. The total energy 12 of the system can be expressed in 2! = 2ways as

12 = r (1) + s (2) or 12 = r (2) + s (1)

These two ways of writing the total energy corresponds to a single wave function s (1, 2) or

A (1, 2). The expression for partition function 12

,

Zr s

e−βε=∑ for the system, contains two terms

12 = r (1) + s (2) and 12 = r (2) + s (1)

corresponding to the same energy. Actually Z should contain only one term. To obtain the correctexpression for partition function for a system containing indistinguishable particles Z should be dividedby 2!.

2

Z2!

z= (two particle system)

This result may be generalized to a system of N indistinguishable particles. Thus

( )N

ZN!

singlez= (N identical indistinguishable particle system) (6.6.1)

In the classical limit, when the number of quantum states is much greater than the number ofparticle available i.e., gi >> ni, the difference between bosons and fermions may be neglected andMaxwell-Boltzmann statistics along with the expression for partition function

=N

ZN!

z

may be used without any appreciable error.


If the energy levels of all the particles are the same, then the partition function of a system of Nidentical, indistinguishable particles, satisfying the condition that the number of available quantumstates is much greater than the number of particles, is

N N/ T1

Z(N,V,T)N! N!

r k

r

ze−ε

= =

∑ ...(6.6.2)

The presence of factor 1/N! is in accordance with the rule of correct Boltzmann counting. Eqn.(6.6.1) is an extremely important result since it reduces a many body problem to a one-body problem.

The partition function for the indistinguishable N-particle system whose constituent particleshave only translational kinetic energy is

N 3N/ 2N

2

V 2 TZ(T,V,N)

N! N!singlez mk

h

π = = ...(6.6.3)

6.7 MOLECULAR PARTITION FUNCTION

To a first approximation, the internal degrees of freedom (vibrational, rotational, electronic, nuclear)may be assumed to be independent of each other and the total energy E may be expressed as the sumof translational, rotational, vibrational, electronic and nuclear energies.

E = Et + Er + Ev + Ee + En

The partition function of a molecule is

E (E E E E E )i t r v e n

i i

z e e−β −β + + + += =∑ ∑

= E E E EEt v e nr

i i i i i

e e e e e−β −β −β −β−β ∑ ∑ ∑ ∑ ∑

= zt. zr .zv .ze .zn ...(6.7.1)

Thus, the partition function of molecule is the product of the translational, rotational, vibrational,electronic and nuclear partition functions.

The partition function of a gas of N molecules is

N N N N N1

Z . . . .N! t r v e nz z z z z = ...(6.7.2)

6.8 PARTITION FUNCTION AND THERMODYNAMIC PROPERTIES OFMONOATOMIC IDEAL GAS

Consider a monoatomic gas dilute enough so that intermolecular interactions can be neglected. Thiscondition is achieved at pressure below 1 atmosphere and at temperature greater than room temperature.The number of available quantum states far exceeds the number of particles of the gas. Under thiscondition the molecules of the gas have only translational kinetic energy viz., = p2/2m.The partition function for a single molecule is


3/ 2 1/ 2

30 0

2 V( ) (2 )z g e d m e d

h

∞ ∞−βε −βεπ= ε ε = ε ε∫ ∫

Making the substitution x = we have

3 / 21/ 2

30

2 V 2 xmz x e dx

h

∞− π= β ∫

3 / 2

3

2 V 2(3/ 2)

m

h

π= Γ β , 3/2 = ½ .

3 / 2 3 / 2

3 2

V 2 2 TV

m mk

h h

π π = = β

3 / 2

2

2 TV

mkz

h

π =

The partition function for the entire gas is

3N / 2N N

3N 2

V 2 TZ

N!N!

z mk

h h

π = = ...(6.8.1)

The same result can also be obtained as follows.The translational energy states of a molecule confined to move in a cube of length L are given

by

2

2 2 2, , 2

E ( )8 Lx y zn n n x y z

hn n n

m= + + ...(6.8.2)

The translational partition function of a single molecule is

( ), ,, , 1

expx y z

x y z

n n nn n n

z∞

=

= −βε∑

= 2 2 2

2 2 22 2 2

1 1 1

exp exp exp8 L 8 L 8 L

x y z

x y zn n n

h h hn n n

m m m

∞ ∞ ∞

= = =

β β β− − − ∑ ∑ ∑

=

32

22

1

exp8 Ln

hn

m

∞

=

β− ∑ ...(6.8.3)


The expression on the right hand side of Eqn. (6.8.3) can not be expressed in terms of anysimple analytic function. So we shall evaluate it in classical approximation. For a microscopic particlethe successive terms in the expression for energy differ very little and therefore the summation inEqn.(6.8.3) can be replaced by integral. So

32

20

exp8 L

hz dn

m

∞ β = − ∫ ...(6.8.4)

Making use of the standard results

1/ 22

/ 20

1.3.......( 1) 1.exp( ) . , 2, 4, 6...

2(2 )n

n

nx x dx n

∞ − π −α = = αα ∫

( 1) / 2

2.4......( 1), 3, 5, 7....

(2 ) n

nn

+−= =

α

0 1 2 3 2

1 1 1 1I ( ) , I ( ) , I ( ) , I ( )

2 2 4 2

π πα = α = α = α =α α α α α

we get

3 / 2

2

2 T.V

mkz

h

π = ...(6.8.5)

6.9 THERMODYNAMIC FUNCTIONS IN TERMS OF PARTITION FUNCTION

Helmholtz Free Energy F

Translational contribution to Helmholtz free energy of an N-particle system is given by

F = – kT ln Z ...(6.9.1)

Making use of Stirling approximation N! = (N/e)N in Eqn.(6.8.1) we have

3 / 2

2

eV 2 TlnZ Nln

N

mk

h

π = ...(6.9.2)

Therefore

3 / 2

2

eV 2 TF N T ln

N

mkk

h

π = − ...(6.9.3)


Translational energy of the system is

3 / 2

2

eV 2 3N 3E ln Z Nln N T

N 2 2

mk

h

∂ ∂ π = − = = = ∂β ∂β ββ ...(6.9.4)

The pressure of the gas is

3 / 2

2T, N

T, N

F V 2 T N TP N T ln

V V N V

e mk kk

h

∂ ∂ π = − = = ∂ ∂ ...(6.9.5)

This gives the equation of state PV = NkT

Entropy of the gas is given by

3 / 2

2V, N

V, N

F eV 2 TS N T ln

T T N

mkk

h

∂ ∂ π = − = ∂ ∂

2

V 3 3 2 5N ln ln T ln

N 2 2 2

mkk

h

π = + + + ...(6.9.6)

Equation (6.9.6) is known as Sackur-Tetrode equation. This equation shows that the entropy Stends to infinity as T tends to zero. This clearly violates the third law of thermodynamics. Had weused the original definition of partition function viz Z = exp (– r) this difficulty would not havecrop up. The problem stems from the replacement of the sum by integral. In fact this replacement isnot justified near T = 0 where the contribution to energy from the ground state is significant. Whereasin the evaluation of integral for partition function we take = 0 or p = 0 in the ground state. Athigher temperature the contribution from ground state is insignificant and so the replacement of sumby integral causes no appreciable error.

6.10 ROTATIONAL PARTITION FUNCTION

A rigid diatomic molecule in which two atoms of masses m1 and m2 are separated by a fixed distancer0 is an example of rigid rotator. The moment of inertia of molecule is

2 21 20 0

1 2

Im m

r rm m

= = µ+

The rotational energy levels of the molecule are

2

E J(J 1)2Ir = +

...(6.10.1)

where J = 0, 1, 2, …. are rotational quantum numbers. The Jth energy level is (2J + 1) fold degenerate.Hence the partition function of the molecule is


2J(J 1)

2I

J 0

(2J 1)z e

β− +∞

=

= +∑

...(6.10.2)

= J(J 1)

T

0

(2J 1)J

eΘ∞ − +

=

+∑ ...(6.10.3)

where 2

2 IkΘ = is rotational characteristic temperature.

(i) When I or T is very small, we can express the partition function as

2 / T 6 / T1 3 5 ...... smaller termsz e e− Θ − Θ= + + + ...(6.10.4)

At low temperature the thermal energy kT is not enough to take the molecules to higher energylevels. The lower rotational energy levels will be heavily populated.

(ii) At higher temperature, the rotational energy levels form continuum and hence Er may beassumed to be continuous. The summation sign in Eqn. (6.10.2) or Eqn. (6.10.3) may be replacedby integral sign. Putting J(J +1) = x, we can write Eqn. (6.10.2) as

2J(J 1)

2I

0

(2J 1) Jz e d∞ β− +

= +∫

2

2I

0

xe dx

∞ β−= ∫

= 2 2

2I 2I Tk=β

=T

Θ...(6.10.5)

The general expression for z taking symmetry into consideration comes out to be

2

2I Tkz =

σ ...(6.10.6)

where = 1 for heteronuclear (asymmetric) molecule and = 2 for symmetric linear molecule.

At higher temperature T > , for an ideal diatomic gas

NN T

Z z = = σΘ

...(6.10.7)

Helmholtz free energy

T

F N T lnk= −σΘ ...(6.10.8)


Mean energy

2 ln Z

E T N TT

k k∂= =∂ ...(6.10.9)

Heat capacity

V

EC N

Tv k∂ = = ∂ ...(6.10.10)

Entropy

E

S ln ZT

k= +

= 2

I T 2N ln ln 1

kk + + σ

...(6.10.11)

6.11 VIBRATIONAL PARTITION FUNCTION

A diatomic molecule made up of two atoms of masses m1 and m2 joined by ‘spring’ of force constant

k, acts like a harmonic oscillator. The classical frequency of the oscillator is ω=µk

, where is

the reduced mass of the molecule. The vibrational energy levels of the molecule are given by

ε = + ω

1

2n n , n = 0, 1, 2, … ...(6.11.1)

The energy 0 = 1/2 ω is called the zero-point energy. The vibrational partition function ofthe molecule is

∞ ∞−β ω + − β ω −β ω

= =

= =∑ ∑

1 12 2

0 0

nn

n n

z e e e ...(6.11.2)

The expression ∞

−β ω

=∑

0

n

n

e can be in an alternative form as

∞

−β ω

=∑

0

n

n

e = −β ω − β ω+ + + 21 ...........e e = −β ω−

1

1 e

Therefore

− β ω

−β ω=−

1

2 1.

1z e

e...(6.11.3)


= / 2T

/ T

1

1e

e−Θ

−Θ−...(6.11.4)

where ωΘ =

k is vibrational characteristic temperature. For a system consisting of N distinguishable

particles the partition function is

1 NNN 2

NN / 2T

/ T

1Z

1

1

1

z ee

ee

− β ω

−β ω

− Θ−Θ

= = −

= −

...(6.11.5)

and

( )1ln Z N N ln 1

2e−β ω= − β ω − −

` ...(6.11.6)

Helmholtz function

( )1F Tln Z N N T ln 1

2k k e−β ω= − = β ω+ −

...(6.11.7)

Mean energy of the system

lnZ

E∂= −∂β

= 1 1

N2 1eβ ω

ω +

−

...(6.11.8)

Heat capacity

Cv = ( )2

2V

EN

T T 1

ek

k e

β ω

β ω

∂ ω = ∂ −

...(6.11.9)

=

( )2 / T

2/ TN

T 1

ek

e

Θ

Θ

Θ −

...(6.11.10)

Entropy

E

S lnZT

k= +


1 1 N

N ln NT T1 1

k kke e−β ω −β ωω ω = + + − −

...(6.11.11)

(i) At high temperature 1,Tk

ωβ ω = << thermal energy kT is much greater than the spacing

of energy levels. In this case the energy of the system is

( )1 1

E N2 1 ..... 1

= ω +

+ β ω + −

= 1 1

N2

ω + β ω

= N

N Tk=β

Therefore Cv = Nk.

(ii) At low temperature Tk

ωβ ω = >> 1. The energy of the system is

1 1

E N N2 2

e−β ω = ω + ⇒ ω

Almost all the molecules will be in the ground state and the energy is temperatureindependent. Therefore Cv will tend to zero.

6.12 GRAND CANONICAL ENSEMBLE AND GRAND PARTITION FUNCTION

In a grand canonical ensemble, each system is enclosed in a container whose walls are both heatconducting and permeable to the passage of particles (molecules). The number of molecules in asystem, therefore, can range over all possible values i.e., each system is open with respect to thetransport of matter. We construct a grand canonical ensemble by placing a collection of such systemsin a large heat bath at temperature T and a large reservoir of molecules. After equilibrium is reached,the entire ensemble is isolated from its surroundings. Since the entire ensemble is at equilibriumwith respect to the transport of heat and matter, each system is specified by volume V, temperature Tand chemical potential .


Fig. 6.12.1 Grand canonical ensemble. Each system has fixed volume V and temperature T butis open with respect to molecular transport

Let the system A under study be in contact with a heat reservoir A'. The system A and A'constitute a composite system A*. Let the system A be in its energy level Er and has number ofparticles Nr. The corresponding values for heat reservoir are E' and N' and that of composite systemare E* and N*. Since the composite system is insulated with respect to energy and passage of particleswe must have

Nr + N' = N* ...(6.12.1)

Er + E' = E* ...(6.12.2)

The probability PNr that the system A in the ensemble is in the state r and contains N particlesis proportional to the number * of states accessible to the composite system A*, which is just equalto the number ' ( E', N' ) or ' (E* – Er, N

* – Nr) of states accessible to the reservoir. Thus

N,P (E ,N ) (E ,N )r r r ′ ′ ′∝ Ω

or * *N,P (E ,N ) (E E ,N N )r r r r r′∝ Ω − − ...(6.12.3)

Since A is very small compared to A' , Er << E* and Nr << N*, we can expand ln ' in Taylorseries as follows.

* *

* * * *

E E N N

ln lnln (E E ,N N ) ln (E ,N ) E N

E Nr r r r′ ′= =

′ ′∂ Ω ∂ Ω ′ ′Ω − − = Ω − − ′ ′∂ ∂ Denoting

* *E E N N

ln lnand

NE ′ ′= =

′ ′∂ Ω ∂ Ω β = −βµ = ′ ′∂ ∂ ...(6.12.4)


we can write

* * * *ln (E E ,N N ) ln (E ,N ) E Nr r r r′ ′Ω − − = Ω −β +βµ

( N E )* * * *(E E ,N N ) (E ,N ) r rr r e β µ −′ ′Ω − − = Ω ...(6.12.5)

The probability that any randomly chosen system contains N particles and be in r th energylevel with energy Er is

( N E )N,P C r r

r eβ µ −= ...(6.12.6)

where C is a constant. Using the condition for normalization of probability viz., PN, r = 1 we have

( N E )

N,

1C

r r

r

e β µ −=∑ ...(6.12.7)

Therefore,

( N E )

( N E )

N,

N,

Pr r

r rr

r

e

eβ µ −

β µ −=∑ ...(6.12.8)

The quantity in the denominator of right hand side of Eqn. (6.12.7) is called Grand PartitionFunction and denoted by ZG. Thus, the grand partition function is

( N E )G

N 0 0

Z r r

r

e∞ ∞

β µ −

= =

= ∑ ∑ ...(6.12.9)

= E N

0 N 0

r r

r

e e∞ ∞

−β βµ

= =

∑ ∑

= N

N 0

Z re∞

βµ

=∑ ...(6.12.10)

where Z is canonical partition function.

Let < nN, r > denote the number of systems in the ensemble that contains Nr (variable) particlesand are in the state r. Then

( N E )

N,N,

G

PM Z

r rrr

n eβ µ −= = ...(6.12.11)

where M is total number of systems in the ensemble. The most probable distribution is then given by

( N E )

N,G

M

Zr r

rn eβ µ −= ...(6.12.12)


6.13 STATISTICAL PROPERTIES OF A THERMODYNAMIC SYSTEM IN TERMS OF GRAND PARTITION FUNCTION

The grand partition function plays a central role in statistical thermodynamics because all the propertiesof system with a variable number of particles can be expressed in terms of it. In order to express ourresults in terms of ZG we shall first derive the entropy S, which is defined as

N, N,N,

S lnr rr

k p p= − ∑

= N, N, GN,

N E lnZr r rr

k p − βµ −β − ∑

= N, N, N, GN, N,

N E ln Zr r r rr r

k p p − µβ −β − ∑ ∑

= GN E ln Zk

− µβ −β −

= GE N

lnZT T

kµ− + ...(6.13.1a)

or –kT ln ZG = E – TS – N ...(6.13.1b)

where E and N are the mean energy and mean number of particles of the system. In a macroscopicsystem the fluctuations are usually negligibly small, so the mean values are just the actual values ofthese quantities. Accordingly we frequently omit bars from such quantities.

Other thermodynamic functions are expressed in terms of grand partition function are as follows.

Total energy E or U G

V,

lnZ

µ

∂ = − ∂β

...(6.13.2)

Mean number of particles GlnZN , where eβµ

∂ = η η = ∂η

...(6.13.3)

Mean pressure of the system G

,

lnZ1

V β µ

∂ Π = β ∂ ...(6.13.4)

6.14 GRAND POTENTIAL

We define grand potential as

G(T,V, ) T ln Z (T,V, ) E TS NkΦ = Φ µ = − µ = − − µ ...(6.14.1)

[This quantity is similar to the Helmholtz function F, which is related to canonical partitionfunction Z (T, V, N) as F = – kT ln Z (T, V, N) = E – TS.]


The other thermodynamic functions are related to grand potential as given below.

V, V, T

S ; NT µ

∂Φ ∂Φ = − = − ∂ ∂µ ...(6.14.2)

6.15 IDEAL GAS FROM GRAND PARTITION FUNCTION

The canonical partition function Z (T, V, N) for a gas of N particles contained in a volume V and attemperature T is given by

3N/2NN

2

1 V 2 TZ(T,V,N) (T,V)

N! N!

mkz

h

π = = ...(6.15.1)

where Z is single particle canonical partition function.The grand partition function for the gas is

N

NG G

N 0 N 0

Z Z (T,V, ) Z(T,V,N)N!

zee

βµ∞ ∞βµ

= =

= µ = =∑ ∑ ...(6.15.2)

Eqn. (6.15.2) may be simplified using the property of exponential function.

∞

=

= + + + =∑2

0

1 .......1! 2! !

nx

n

x x xe

n...(6.15.3)

In view of Eqn.(6.15.3) we can write Eqn. (6.15.2) as

GZ exp zeβµ = ...(6.15.4)

Let βµη = e . Therefore

GZ exp z= η ...(6.15.5)

Now grand potential for the gas is

GT ln ZkΦ = −

= Tk z− η

= 3/ 2

2

2 TT V

mkk e

hβµπ −

...(6.15.6)

The entropy of the gas is

3 / 2

3,

(2 ) 5

( ) 2V

mkTS V e

kT hβµ

µ

∂Φ π = − = −βµ ∂ ...(6.15.7)

This is Sakur-Tetrode equation.The mean number of particles in the system is

3 / 2

3V, T

(2 T)N V

T

mke

khβµ ∂Φ π Φ= − = = − ∂µ


From this equation we can get chemical potential

3 / 2

G3

V (2 T)Tln Tln

N N

zmkk k

h

πµ = − = −

...(6.15.8)

Thus, the chemical potential increases with increase in concentration of particles.The pressure of the gas is given by

T,

NT

V Vk

µ

∂Φ Π = − = ∂ ...(6.15.9)

This gives

V N TkΠ =

which is the perfect gas law.

6.16 OCCUPATION NUMBER OF AN ENERGY STATE FROM GRAND PARTITION FUNCTION: FERMI-DIRAC AND BOSE-EINSTEIN DISTRIBUTION

The state of a system is specified by a set of occupation number n1, n2, …..nr, ….of single particlestates with energies ε ≤ ε ≤ ε ≤1 2 ...... .....r Consider a state of the system in which it contains N-particlesand total energy ENr, which are given by

N,N and Er r r rr r

n n= = ε∑ ∑Of course, in grand canonical ensemble, N is variable and free to take on values

N = 0, 1, 2, ….The grand canonical partition function of the system is

N,

1 2

(N)EN

GN 0 , ,...

Z r

n n

e e∞

−ββµ

=

= ∑ ∑

= 1 2 1 1 2 2

1 2

(N)( .....) ( ....)

N 0 , ...

n n n n

n n

e e∞

βµ + + −β ε + ε +

=∑ ∑ ...(6.16.1)

The superscript N over the summation sign means that the occupation numbers obeys thecondition n1 + n2 +….= N. In the above summation first sum over all the values of n1, n2….for afixed value of N and then sum over all the values of N from N = 0 to . This way of summation isequivalent to summation over all values of n1, n2, …… independently of each other. So the aboveexpression can be expressed as

1 2 1 1 2 2

1 2

( ....) ( ....)GZ n n n n

n n

e eβµ + + −β ε + ε +=∑∑


= ( )( )1 1 2 2

1 2

( ) ( )

0 0

...... .........n n

n n

e e∞ ∞

β µ−ε β µ−ε

= =

∑ ∑

= ( )r r

r

n

rn

e∞

β µ−ε Π ∑ ...(6.16.2)

G(Z )rr

= Π ...(6.16.3)

where (ZG)r is given by

( ) ( )

GZ r r

r

nr

n

eβ µ−ε=∑ ...(6.16.4)

If the system consists of fermions, then nr = 0 or 1 and therefore

( ) ( )( ) ( )G

0, or1

Z 1r r

r

nr

n

e eβ µ−ε β µ−ε

=

= = +∑ ...(6.16.5)

Hence the grand partition function for fermions is

( )β µ−ε= Π + ( )GZ 1 r

re (fermions) ...(6.16.6)

The grand potential for a system of fermions is

GTlnZfermions kΦ = −

= ( )( )T ln 1 r

r

k eβ µ−ε− +∑

= ( )( )T ln 1 r

r

k eβ µ−ε− +∑ ...(6.16.7)

= φ∑ rr

...(6.16.8)

The mean occupation number of rth energy state is

∂φ

= −∂µ

rrn (6.16.9)

= ( )( )T ln 1 rk eβ µ−ε∂ − − + ∂µ

= β µ−ε

β µ−ε+

( )

( )1

r

r

e

e

= β ε −µ +( )

1

1re...(6.16.10)


This expression for the mean occupation number for state of energy r is called Fermi-Diracdistribution for particles with half-integral spin (fermions).

If the system consists of bosons, then nr = 0, 1, 2, 3, …. and therefore the grand partitionfunction for particles with integral spin (bosons) is

( )

G0, 1, 2, 3,.....

Z r r

r

n

rn

eβ µ−ε

=

=Π ∑

= ( )( ) 2 ( )1 ............r r

re eβ µ−ε β µ−εΠ + + +

= β µ−ε µ <ε − ( )

1

1 rr

e...(6.16.11)

The grand potential E TS NΦ = − −µ for a system of bosons is

GT ln Zbosons kΦ = −

= ( ) 1( )T ln 1 r

r

k e−β µ−ε− −∑

= ( )( )T ln 1 r

r

k eβ µ−ε−∑ = ( )( )T ln 1 r

r

k eβ µ−ε−∑ ...(6.16.12)

= φ∑ rr

...(6.16.13)

The mean occupation number of state of energy r is

( )( )T ln 1 rrrn k eβ µ−ε

∂φ ∂ = − = − − ∂µ ∂µ

β µ−ε

β µ−ε=−

( )

( )1

r

r

e

e

or β ε −µ=−( )

1

1rrn

e...(6.16.14)

This is the Bose-Einstein distribution function for a system of particles with integral spin(bosons).

For a system of bosons (such as photons, phonons, etc.) whose number is not conserved, thechemical potential µ is zero. The Bose-Einstein distribution for such particles is

βε= ε = ω−

1

,1r

r r rne

(for photons and phonons) ...(6.16.15)

CHAPTER

APPLICATION OF PARTITION FUNCTION

7.1 SPECIFIC HEAT OF SOLIDS

7.1.1 Einstein Model

In 1907 Einstein made an attempt to explain the temperature-dependence of the heat capacity ofsolids on the basis of quantum theory, and made the following assumptions.

1. The atoms in a solid vibrate about their fixed lattice sites. Their vibrations are independentof each other and the frequency of vibration, called Einstein frequency, is the same forall atoms, and is a characteristic constant of the solid. The vibration of an atom can beresolved into three mutually independent equations say along the three Cartesian coordinateaxes x, y and z. Thus, a solid containing N0 atoms is equivalent to 3 N0 harmonic oscillators.

2. An oscillator of frequency can have only discrete values of energy* given by

ε = + ω

1

2n where n = 0, 1, 2 3, ….. ...(7.1.1)

The energy ω1

2 is called the zero-point energy.

The partition function of one oscillator is

1 12 2

0 0 0

rr

r

r r r

z e e e e

∞ ∞ ∞−β ω + − β ω −β ε −β ω

= = =

= = =∑ ∑ ∑

...(7.1.2)

Let T

xk

ω= β ω = . The expression for partition function becomes

∞

− −

=

= ∑/ 2

0

x x r

r

z e e


= − − − + + + / 2 21 .......x x xe e e

= −−

−

/ 2 1

1x

xe

e...(7.1.3)

The Helmholtz free energy for an oscillator is

( )

( )

1F Tln

T ln 12

1 1ln 1

2

x

k z

xk e

e

−

−β ω

= −

= − − − −

= ω + − β

...(7.1.4)

The average energy of an oscillator is

( )11

FlnF

T

z

k

∂ ∂ ∂ ε = − = − − = β ∂β ∂β ∂β

= ( )−β ω∂ β ω + − ∂β

1

ln 12

e

= β ωωω +

−

1

2 1e...(7.1.5)

The energy of crystal containing N atoms is

1

E 3N 3N2 1eβ ω

ω= ε = ω + −

...(7.1.6)

The specific heat at constant volume is

E

CTv

v

∂ = = ∂

E

T

∂ ∂β ∂β ∂

=

( )2

2 2

3N( )

T 1

e

k e

β ω

β ωω

−

= 2

3NT

kk

ω

( )β ω

β ω −

2

1

e

e...(7.1.7)

Application of Partition Function 361

Let us express this result in terms of Einstein temperature E defined by Ekω = Θ .

( )E

E

/ TE

2/ TC 3N

T 1v

ek

e

Θ

Θ

Θ = − ...(7.1.8)

(i) At high temperature T >> E,

E / T E1 ...........T

eΘ Θ= + +

The expression for Cv simplifies to

( )

2E E E

2E

1 / TC 3N 3N 1

T T/ Tv k k

Θ + Θ Θ = = + Θ

...(7.1.9)

As T , Cv = 3Nk = 3R

Thus, at high temperature the heat capacity of solid is independent of temperature and isequal to 3R which is the Dulong-Petits law.

(ii) At low temperature E/T is large and so is exp(E/T). The expression for heat capacitymay be simplified as follows.

( )E

EE

2 2/ TE E

2 / T/ T

1C 3N 3N

T Tve

k kee

Θ

ΘΘ

Θ Θ = =

= 2 2 3

E E E

E

13N

T 1 11 .......

T 2! T 3! T

k Θ Θ Θ + + + + Θ

= 2

E

E E

13N

T T 1 1....

2! 3! T

k Θ+ + + + Θ Θ

...(7.1.10)

As T 0, Cv 0. The heat capacities computed on the basis of Einstein’s model at differenttemperatures are, in general, in good agreement with the experimental results. However, careful anddetailed comparison of the predictions from Einstein’s theory with experimental results shows thatthe agreement is only approximate. It has also been shown by the experiments that at low temperature,the heat capacity varies as T3 whereas Einstein’s theory predicts exponential variation of specificheat with temperature. This difference between theory and the experiment is not very surprisingbecause the model assumed by Einstein is over simplified.


For example, Einstein has assumed that the atoms in the solid are vibrating independent ofeach other, which cannot be realistic, as atoms are strongly coupled with each other and therefore ifan atom is disturbed it must effect the other atoms. Consequently one must consider the vibrationsof the groups of atom rather than the vibration of each atom independently. Moreover, Einstein hasalso assumed that all the atoms vibrate with a single frequency , which is again not correct. Inorder to develop a more appropriate theory, which can provide a realistic explanation of the observedexperimental facts, Debye assumed a more realistic model of solid and provided a better theory ofspecific heat of solid.

7.1.2 Debye Model

The atoms in a solid cannot be assumed to be independent. We must take into account their cooperativeinteractions. Taking this interaction into consideration, we arrive at a theory of heat capacity that isin agreement with experimental results. At 0 K, the system of atoms comprising a solid is in theground state having the minimum energy. After receiving energy from outside, an atom moves in acertain direction from its equilibrium position. A force striving to return it to the equilibrium positiondevelops. Therefore while leaving the equilibrium position, the atom acts a certain force onneighbouring atoms, which in turn are to leave their equilibrium positions, as a result of which themotion becomes cooperative. This cooperative motion, in which the displacement of one atom istransferred to the neighbouring atom and then to the next neighbour and so on, is an acoustic wavein the solid.

Taking into account the interaction between atoms, a system of atoms must be considered as aset of coupled oscillators. In this case, any motion of the system of atoms can be represented as asuperposition of normal oscillations or normal modes of the system. Each normal mode is characterizedby its frequency and the energy of this mode is given by ε = ω .

A solid can support longitudinal and transverse waves both with different velocities. Transversemode may have two different directions of polarization. There is a standard method of calculatingthe number of modes for each polarization in an isotropic solid. The number of modes of acousticoscillations in the frequency range d at frequency , in a solid of volume V is given by

ω ω = ω ωπ

22 3

( )2

Vg d d

v...(7.1.11)

where v is velocity of the wave. In a isotropic solid the velocities of the two transverse waves areequal. Taking the longitudinal and transverse mode both, the number of modes in the frequency intervald about is given by

ω ω = + ω ω π

22 3 3

1 2( )

2 l t

Vg d d

v v...(7.1.12)

The factor 2 multiplying in 1/vt2 is due to the fact that there are two independent transverse

directions of polarization. We define a mean velocity v by

= +3 3 3

3 1 2

l tv v v...(7.1.13)


According to Debye, a crystal with N atoms possesses 3N modes in all. The maximum frequencyD is such that there are 3N modes altogether i.e.,

D

0

( ) 3Ng dω

ω ω =∫ ...(7.1.14)

The cut-off frequency D occurs because at sufficiently high frequency i.e., short wavelength,we cannot ignore the atomic nature of the solid. That is at short wavelengths, the solid is no longeracts as a continuum. A crystal with inter-atomic spacing a cannot propagate waves with wavelengthsless than min = 2a. In this case neighbouring atoms vibrate in anti-phase.

Let us introduce a characteristic temperature D related to Debye cut-off frequency D definedby

D D Dorkk

ωω = Θ Θ = ...(7.1.15)

So Debye condition (7.1.14) in terms of energy ε = ω becomes

D

0

( ) 3Nk

g dΘ

ε ε =∫

D2

2 3 30

3V3N

2

k

dv

Θ

ε ε =π ∫

2 3 3 3D

3V 9N

2 ( )v k=

π Θ...(7.1.16)

Fig. 7.1.2 Wave of shortest wavelength that can propagate in a one-dimensional crystal

With the help of Eqns. (7.1.12), (7.1.13) and (7.1.14) we can write

( )2

3D

9N( )g d d

kε ε = ε ε

Θ ...(7.1.17)

The average number of normal modes (oscillators) with energy , are given by

/ T

1( )

1kn

eεε =− ...(7.1.18)


The energy of the crystal is

D

0

E ( ) ( )k

g n dΘ

= ε ε ε ε∫

= ( )

D 3

3 / TD 0

9N

1

k

k

d

ek

Θ

εε ε

−Θ ∫

=

3

D

T9N Tk

Θ

D / T 3

0

,1x

x dx

e

Θ

−∫

where D DD,

T T T Tx x

k k k

ω Θω ε= = = =

...(7.1.19)

(i) At high temperature (T >> D)

3 32 as is small.

1 1 ..... 11

x

x xx x

xe= ≈

− + + − Therefore, Eqn. (7.1.19) simplifies to

D / T4 3

3D 0

9N TE 3N T

3k x

k

Θ = = Θ

...(7.1.20)

The molar heat capacity at constant volume is

E

C 3N 3RTv

v

k∂ = = = ∂ ...(7.1.21)

(ii) At low temperature, T << D, x and the integral

D 3 3 4

0 0151 1

x

x x

x xdx dx

e e

∞π= =

− −∫ ∫ (see Appendix A-3)

The energy of the system is

4 4

3D

9N TE

15

k π= Θ ...(7.1.22)

The heat capacity

4

33D

E 9NC T

T 15vv

k ∂ π = = ∂ Θ ...(7.1.23)

Thus, the heat capacity of solid at low temperature varies as T3. This law is known as Debye T3 law.


7.2 PHONON CONCEPT

The energy ε = ω corresponding to a mode of acoustic vibration with frequency in a solid suggeststhat a mode should be treated as a quasiparticle. Such a quasiparticle associated with the modes ofacoustic oscillations is called a phonon. The introduction of the phonon concept is a fruitful approach,which considerably simplifies the reasoning. The thermal vibrations of a lattice are equivalent to anaggregate of phonons and the latter may be treated as an ideal Bose Gas. The phonon is a quanta of

acoustic wave with energy ε = ω and momentum ε=pv

, where v is mean velocity of acoustic wave

in solid and is given by = +3 3 3

3 1 2

l tv v v, vl and vt are the velocities of the longitudinal and transverse

waves.The density of states for a gas of phonons is given by

2

3

3V( ) 4g p dp p dp

h= π ...(7.2.1)

The factor 3 takes into account of three possible polarizations of phonon. V is volume of thesolid. Making use of the relation = vp, the expression for the density of states can be transformedin terms of energy.

2

2 3 3

3V( )

2g d d

vε ε = ε ε

π ...(7.2.2)

Let the maximum frequency of the acoustic oscillations be D and the energy of the

corresponding phonon be Dω . We define a characteristic temperature D such that D Dkω = Θ

D. ., .i ek

ωΘ = The total number of modes of oscillation in the solid containing N atoms is 3N.

Thus

D

0

( ) 3Nk

g dΘ

ε ε =∫

D2

2 3 30

3V3N

2

k

dv

Θ

ε ε =π ∫

( )2

3D

9N( )g

kε = ε

Θ ...(7.2.3)

The phonons are bosons and their number is not conserved so the occupation number of phononsin the state with energy is given by


/ T

1( )

1kn

eεε =−

...(7.2.4)

The energy of phonons in the solid is

D

0

E . ( ). ( ).k

g n dΘ

= ε ε ε ε∫

= ( )

D 3

3 / TD 0

9N

1

k

kd

ek

Θ

εε ε

−Θ ∫

= D3 / T 3

D 0

T9N T

1x

xk dx

e

Θ Θ − ∫ , ...(7.2.5)

where T T

xk k

ω ε= =.

At high temperature x is small and ≅−

32

1x

xx

e. So Eqn. (7.2.5) simplifies to

E 3N Tk=

and hence Cv = 3Nk = 3R (Dulong-Petits Law)

At low temperature, x is very large, and the integral in Eqn. (7.2.5) is given by

D / T 3 3 4

0 0151 1x x

x xdx dx

e e

Θ ∞ π= =− −∫ ∫

Therefore 4 4

3D

9N TE

15

k π= Θ ...(7.2.6)

Heat capacity 4

33

D

3NC T

5v

k π= Θ ...(7.2.7)

Thus, the heat capacity of a solid varies as T3 at low temperature. This is Debye T3 law.


7.3 PLANCK’S RADIATION LAW: PARTITION FUNCTION METHOD

Let us denote the single particle states of photon gas by 1, 2, 3, ….., r,… with energies1, 2, 3,……, r ,…..and occupation numbers n1, n2, n3, ……, nr,….For photons nr = 0, 1, 2, ….forall r. i.e., each of the occupation number assumes all possible values 0, 1, 2, ….independent of thevalues of the other occupation numbers. The partition function of the photon gas is

∞

=∞ −β ε

=∑

∑ 1

1 2, ,....

r rr

n

n n

Z e ...(7.3.1)

where ∑1 2, ...n n

stands for summation over all sets of occupation numbers. Eqn. (7.3.1)) can be written

as

Z = ∞ ∞

−β ε + ε +∑∑ 1 1 2 2

1 2

( ....)......... n n

n n

e ...(7.3.2)

= ( )( )( )∞ ∞

−β ε −β ε ∑ ∑1 1 2 2

1 2

....... ....... ........n n

n n

e e

= ( )( )−βε −β ε − − 1 2

1 1...... ........

1 1e e

= ∞

−β ε=

Π − 1

1

1 rr e...(7.3.3)

Therefore

( )1

lnZ ln 1 r

r

e∞

−βε

== − −∑ ...(7.3.4)

where ε = ω .r

The mean occupation number of photons in the state r is

1 ln Z 1

( )1r

rr

neβ ε

∂ω = − =β ∂ε −

= β ω −

1

1e...(7.3.5)

The degeneracy of energy levels r is given by

2

2 3

V( ) 2

2g d d

c

ω ω = ω ω π ...(7.3.6)


The factor 2 accounts for the two independent directions of polarization of photon.Therefore

( )22 3

VlnZ ln 1 r

r

ec

−β ω = − ω − π ∑

= 2

2 30

Vln(1 )e d

c

∞−β ω− ω − ω

π ∫

...(7.3.7)

= 3 3

2 3 2 30 0

V Vln(1 )

3 3 1

ee d

c c e

∞ ∞ −β ω−β ω

−β ω

ω β ω− − + ω π π −

∫

= 0 +3

2 30

V

3 1d

c e

∞

β ωβ ω ω

π −∫

= 3

2 3 30

V

3 ( ) 1x

xdx

c e

∞

π β −∫

where β ω = x

= 32V T

45

k

c

π

...(7.3.8)

Average energy

2 42 4

3 3

lnZ ln Z VE T T

T 15

kk

c

∂ ∂ π= − = = ∂β ∂

...(7.3.9)

Mean pressure

2 4 4

3 3

1 ln Z TP

V 45

k

c

∂ π= =β ∂

...(7.3.10)

1

PV E3

= ...(7.3.11)

Entropy S [ln Z E]k= + β

= 2 4 3

3 3

4V T

45

k

c

π

...(7.3.12)


Radiation density

( ,T) ( ). ( ) ( )u d g n dω ω = ω ω ω ω

β ω

ω ω = ω π −

2

2 3

1( )

1

d

c e

= β ωω ω

π −

3

2 3

1

1d

c e...(7.3.13)

This is the Planck radiation law.


1. With suitable examples, explain macrostates, microstates of a system. What are accessible states?

2. Explain the concept of ensemble in statistical mechanics. State the fundamental postulates of statistical mechanics.

3. What do you mean by micro-canonical, canonical and grand-canonical ensemble? Which one is more suitable forquantum particles and why?

4. What do you mean by phase space? How do you divide the phase space into cells? What is the minimum size ofa cell according to classical and quantum mechanics?

5. Distinguish between µ-space and -space. Find an expression for density states of a system using the concept ofphase space.

6. Derive an expression for the density of states using the concept of discrete energy levels and quantum states of aparticle restricted to move in a three dimensional box.

7. Four distinguishable particles are distributed in two boxes with equal weights. State clearly (i) various possiblemicrostates (ii) number of macrostates (iii) the probabilities of most probable and least probable states.

8. What do you mean by canonical system? Deduce the Boltzmann canonical distribution law and determine theexpression for probability for a molecule having energy i

−ε

−εε =

∑/ T

/ TP( ) .

i

i

k

i k

i

e

e

9. Define entropy and thermodynamic probability (statistical weight) and establish a relation between them.

10. Give statistical definition of entropy and prove that S k ln .

11. For a single particle of mass m enclosed in volume V, show that the number of accessible states in the energyrange E to E + dE is given by

3/ 2 1/ 2

3

2 V(E) E (2 ) E E.d m d

h

πΩ =

12. Derive the classical Maxwell-Boltzmann velocity distribution law. Discuss the effect of temperature on thedistribution function.

13. Write down Maxwell velocity distribution function for an ideal gas and calculate average velocity, root meansquare velocity and most probable velocity.


14. Write the basic assumptions of Maxell-Boltzmann statistics and derive expression for Boltzmann distributionfunction.

15. What is difference between the classical and quantum statistics? Obtain the condition under which quantum statisticsreduces to classical statistics.

16. Write the basic assumptions which form the basis of Bose-Einstein statistics. Derive B-E statistics.

17. State the conditions of Fermi-Dirac statistics. Derive F-D statistics. In what respect a F-D system differs from aB-E system?

18. Derive Planck’s radiation law from B-E statistics.

19. What do mean by degenerate and non-degenerate system. Discuss the degeneration of Fermi gas and Bose gas.

20. What do you mean by B-E condensation? In what respect does it differ from conventional condensation ofvapour into liquid.

21. Obtain expression for energy, pressure and entropy of a Fermi gas in degenerate and non-degenerate state.

22. Obtain expression for energy, pressure and entropy of a Bose gas in degenerate and non-degenerate state.

23. What do you mean by Fermi energy and Fermi temperature of an electron gas? Obtain an expression for Fermi

energy at T = 0 and T ≠ 0.

24. What do you mean by partition function of a particle? Show that the partition function of a monatomic gas isgiven by

3/ 2

3

VZ (2 T)mk

h= π .

25. Give Einstein theory of specific heat of solids. Give comments about the draw back of the theory.

26. Giving the basic assumptions of Debye model for the specific heat of solids, derive Debye T3 law.

27. What do you mean by grand canonical ensemble? Obtain F-D and B-E statistics from this formulation.

28. Give a comparative study of classical and quantum statistics. Under what conditions quantum statistics merge

into classical statistics?

APPENDIX-A

A–1 Evaluation of integral ∞

−

−∞∫

2xe dx .

Let

2I xe dx

∞−

−∞

= ∫ ...(1)

Also

2I ye dy

∞−

−∞

= ∫ ...(2)


Multiplying (1) and (2)

( )2 2

2Ix y

e dx dy∞ ∞

− +

−∞ −∞

= ∫ ∫ ...(3)

The range of integration is over the entire x-y plane. The integral (3) can be evaluated in termsof polar coordinates (r, ).

x = r cos , y = r sin , x2 + y2 = r2 and dx dy = r dr d

Fig. A-1 Relation between Cartesian and polar coordinates

The range of r is from 0 to and that of is from 0 to 2. Therfore

2

22

0 0

I rr e dr d∞ π

−= θ∫ ∫

= ∞ π

− θ∫ ∫2

2

0 0

rr e dr d

= ∞ ∞

− −π = π∫ ∫2

0 0

2 r ur e dr e du = ( r2 = u, 2 r d r = du )

I∴ = π

Therefore ∞

−

−∞

= π∫2xe dx ...(4)

and ∞

− π=∫2

02

xe dx ...(5)


A-2. Evaluation of Integral 2

0

I ( ) n axn a x e dx

∞−= ∫

2

0

I ( ) n axn a x e dx

∞−= ∫ ...(1)

These integrals are found by calculating I0 (a) and I1 (a).

2 2 2 2

0

0 0

1 1I ( ) , ( , )ax ya e dx e dy ax y dx dy

a a

∞ ∞− −= = = =∫ ∫

I0(a) = π1

2 a...(2)

2 2

1

0 0

1 1I ( ) , where ,

2 2ax ya xe dx e dy ax y x dx dy

a a

∞ ∞− −= = = =∫ ∫

= ∞− − = 0

1 1

2 2ye

a a

11

I ( )2

aa

= ...(3)

2 21

0 0

1I ( ) ( )

2n ax n ax

n a x e dx x d ea

∞ ∞− − −= = −∫ ∫

= ∞∞

− − − − − − −

∫2 21 2

00

1( 1)

2n ax n axx e n x e dx

a

= ∞

− −−+ ∫22

0

10

2n axn

x e dxa

= 21

I ( )2 n

na

a −−

∴ 21

I ( ) I ( )2n n

na a

a −−= ...(4)


From (4)

2 3 42 2

1 1 3I ( ) , I ( ) , I ( ) etc.

4 2 8a a a

a a aa a

π π= = = ...(5)

A-3. Evaluation of 3

0

I .1x

xdx

e

∞

=−∫

This integral can be evaluated by expanding the integrand in a series. Since ex < 1, throughoutthe range of integration, we can write

−

− − −−

= = + + + − −

3 33 2. 1 ........

1 1

xx x x

x x

x e xe x e e

e e

= ∞

−

=∑ 3

1

nx

n

e x

Hence

3

1 0

I .nx

n

e x dx∞∞

−

== ∑∫

= 3

41 0

1, where ,y

n

e y dy nx yn

∞∞−

=

=∑ ∫

= 4 4

4 4 41 1 1

1 1 14 (3! ) 6 6 .

90 15n n nn n n

∞ ∞ ∞

= = =

π πΓ = = = = ∑ ∑ ∑

A-4. Evaluation of

( )4

20

I .1

x

x

x edx

e

∞

=−

∫

34 4

00 0

1 1I . 4

1 1 1x x x

xx d x dx

e e e

∞ ∞∞ = − = − − − − − ∫ ∫

= ∞

−∫3

0

41x

xdx

e

= 4

444 .

15 15

π = π


A-5. Riemann Zeta Function (x).Riemann zeta function is defined as

ζ = + + + +1 1 1( ) 1 .....

2 3 4x x xx ...(1)

∞

=

πζ = =∑2

21

1(2)

6n n

= 1.645

∞

=

πζ = = =∑4

41

1(4) 1.082

90n n

1 2

21

( 1).

12

n

n n

∞ −

=

− π=∑

A-6. Stirling’s Approximation.For large n,

ln n! = n ln n – n = n ln (n/e) ...(1)

By definition n! = 1.2.3………n

ln n! = ln 1 + ln 2 + ln 3 + …… + ln n

= =

∑1

lnm

m

m

This sum is exactly equal to the area under the stepcurve shown by broken line in the figure between n = 1and n = n. This area may be approximated with the areaunder the smooth curve y = ln n between the same limits.For small values of n, the step curve differs appreciablyfrom the smooth curve but the smooth curve becomes moreand more nearly parallel to the n-axis as n increases.

( )11 1

1ln ! ln ln . .

n nn

n x dx x x x dxx

= = −∫ ∫ = − +ln 1n n n

= −lnn n n ...(2)

(n >> 1, we can neglect 1)An exact analysis gives the following series for n !.

Fig. A-6


= π + + − + 2 3

1 1 139! 2 1 ....

12 288 51840

nn

n ne n n n

Retaining the first term only we obtain

= π + −1ln ! ln(2 ) ln

2n n n n ...(3)

n ln n! n ln – n ½ ln 2n

5 4.8 3.0 1.825 58.0 55.5 2.5

100 363.7 360.5 3.2

A-7. Gamma Function.Gamma function is defined as

∞

− −Γ = >∫ 1

0

, 0.n xn x e dx n ...(1)

Now

∞ ∞− − − −Γ = = −∫ ∫1 1

0 0

( ) ( )n x n xn x e dx x d e

= ∞∞− − −

− − − −

∫1 2

00

( 1) ( )n x n xx e n x e dx

= ∞

− −− ∫ 2

0

( 1) n xn x e dx

= − Γ −( 1) ( 1)n n

So Γ = − Γ −( ) ( 1) ( 1)n n n ...(2)

For n = 1/2 we have

∞ ∞

− − − π Γ = = = = π ∫ ∫21/ 2

0 0

12 2

2 2x ux e dx e du ...(3)

( )∞ ∞− −Γ = = − =∫ 00

(1) 1x xe dx e ...(4)


If n is a positive integer then

Γ = − Γ − = − − Γ −( ) ( 1) ( 1) ( 1)( 2) ( 2)n n n n n n

= − − − Γ( 1)( 2)( 3)........ (1)n n n

= (n – 1) (n – 2) (n – 3) …………….1

= (n – 1)! ...(5)

A-8. An useful integral associated with F-D distribution.

( )∞ ∞

+ +−

=

ε ε = ε + − ζ ε + ε − ε + ε ∑∫

21 12

F F2 1 2F F10

1 12( T) 1 (2 )

1 exp( ) / T 1 2

p np pn

n nn

d dk n

k p d...(1)

For p =1/2 , n = 1, 2, 3, ……., πζ = =

2

(2) 1.645.6

A-9. An useful integral associated with B-E distribution.When working with B-E distribution function we often come across an integral of type

1

10

1

1

s

x

xdx

s e

∞ −

−Γ η −∫ . This integral can be expressed as infinite series in the following form:

1

10

1

1

s

x

xdx

s e

∞ −

−Γ η −∫ = 2 3 4

F ( ) ..........2 3 4

s s s s

η η ηη = η + + + + ...(1)

For = 1,

1

0

1 1 1 1F (1) 1 .......... ( )

1 2 3 4

s

s x s s s

xdx s

s e

∞ −= = + + + + = ζ

Γ −∫ ...(2)

For s = 3/2, = 1,

1/ 2

3 / 2 3 / 2 3 / 2 3 / 20

1 1 1 1F (1) 1 ..... (3/ 2) 2.612

(3/ 2) 2 3 4x

xdx

e

∞

= = + + + + = ζ =Γ ∫ ...(3)

For s = 5/2, = 1,

3 / 2

5 / 2 5 / 2 5 / 2 5 / 20

1 1 1 1F (1) 1 ..... (5/ 2) 1.342

(5/ 2) 2 3 4x

xdx

e

∞

= = + + + + = ζ =Γ ∫ ...(4)

ATOMIC SPECTRA

UNIT


blank

CHAPTER

ATOMIC SPECTRA-I

1.1 INTRODUCTION

The history of development of the understanding of atomic structure is of special significance becauseit was the first systematic attempt through which, the relationship between the macroscopic propertiesof matter and its microscopic structure was investigated. By 19th century, it was firmly establishedthat the matter was composed of atoms and molecules. The kinetic theory of gases provided directevidence and realistic information regarding mass and size of atoms and molecules. The kinetic theorywas based on the application of ordinary laws of mechanics to the motion of molecules in gases andprovided relationship between some structural properties of its molecules and the properties of gases.The discovery of electron by J.J. Thomson (1996) gave indication that the atom had inner structure.This led physicists to speculate about the internal structure of atom. Attempts made in this directionmanifested in terms of various atomic models.

1.2 THOMSON’S MODEL

After the discovery of negatively charged electrons, it was realized that the electrons were theconstituent particles of atom. Since atom is electrically neutral, Thomson proposed that the atommight be regarded as a sphere of positive charge in which negatively charged electrons were embeddedin it. The magnitude of positive charge in the sphere was equal to the total charge carried by electrons.This model of atom, called plum-pudding model, received serious set back because it could not explainthe experimental observations made in the famous alpha-particle scattering experiment conducted byGeiger and Marsden under the guidance of Lord Rutherford.

Figure 1.2.1 shows the schematic diagram of the experimental set up used by Rutherford. Acollimated beam of -particles from a radioactive substance was allowed to be fall on a thin goldfoil. The scattered -particles were detected by a zinc sulphide screen placed behind the foil. When-particles strike the screen, they give off visible flash of light.

It was found that majority of the particles suffered small deflection from their original directionbut some of them suffered a deflection of 90° or more. One in ten thousand -particles came off inbackward direction after being scattered by gold foil. Let us see what does the Thomson’s model


predict? Suppose that an alpha-particle is incident on the thin metallic film, which consists of manylayers of atoms. If an alpha-particle while passing through the foil is outside the atom, it shouldsuffer no deviation from its original path. If it penetrates inside the atom and interacts with electron,it should suffer small deviation because the electron is very light in comparison to the alpha-particle.On a foil there are many layers of atoms, the alpha-particle may be scattered in different waysdepending upon its interaction with various atoms.

The problem of finding the deflection of alpha-particles after emerging from the foil is astatistical problem, which is similar to the random walk problem. In accordance with the predictionof this theory, the probability of scattering of alpha-particle by 90° or more is about 1 in every103500 alpha-particles whereas the experiment shows 1 in 104. Therefore, the Thomson’s model wassubjected to serious objections. In an attempt to explain the results of alpha-particle scatteringexperiment, Rutherford suggested another model that is named after him.

Fig. 1.2.1 Schematic diagram of Rutherford experiment

Fig. 1.2.2 Deflection of -particles by atoms of the target foil

Atomic Spectra-I 381

1.3 RUTHERFORD ATOMIC MODEL

On the basis of the results obtained from the scattering experiments, Rutherford suggested an atomicmodel, according to which the entire positive charge in the atom is concentrated to a very smallregion, called nucleus. The entire mass of the atom is due to this nucleus. To explain the stability ofthe atom against the falling of electrons into the nucleus under the electrostatic attraction, he postulatedthat like solar system, electrons revolve around the nucleus in circular orbits; the centripetal forcerequired for the circular motion is obtained from the electrostatic attraction of the nucleus on theelectrons. This model is known as the planetary atomic model.

Let us now consider the dynamics of the simplest atom, the hydrogen atom, consisting of asingle electron revolving round the nucleus (proton). In accordance with the laws of classicalmechanics, the equation of motion of the electron is

2 2

0

1

4

e mv

r r=

πε...(1.3.1)

Fig. 1.3.1 Hydrogen atom according to Rutherford model

From this equation the kinetic energy of electron comes out to be

2

2

0

1 1 1K

2 2 4

emv

r= =

πε ...(1.3.2)

The total energy E of electron moving around the nucleus in circular orbit consists of twoparts: kinetic and potential.

2 2 2

0 0 0

1 1 1 1E K U

2 4 4 2 4

e e e

r r r

−= + = + = −πε πε πε ...(1.3.3)

The negative total energy means that the electron in the atom is bound to the nucleus.The motion of electron in the force field of a nucleus is governed by the two well-established

laws of classical physics namely the Newton’s law and the Coulomb’s law. The electron moves in acircular orbit, which is an accelerated motion. According to classical electrodynamics an acceleratedcharge must radiate energy in the form of electromagnetic waves. The rate at which a particle having


charge e and moving with acceleration a radiates electromagnetic energy is given by

2 2

30

2 1P

3 4

e a

c=

πε ...(1.3.4)

Since the acceleration of electron in a circular orbit is a = v2/r, the rate at which energy isradiated is given by

3 6

3 2 40

2 1P

3 4

e

c m r

= πε

...(1.3.5)

For hydrogen atom r = 5.3 × 10–11m, the value of P comes out to be

P = 4.6 × 10–9 J/s = 2.9 × 1010 eV/s

Since the electron loses energy though emission of radiation,the total energy E become more and more negative, which impliesthat the radius become progressively shorter. It is apparent fromEqn. (1.3.5) that the rate of emission of energy is proportional to1/r4, hence the rate of emission rapidly increases as the orbit becomesshorter and shorter, and if this process continues the electron mustultimately fall into the nucleus. This prediction of classical physicsis in direct contradiction to the fact that the hydrogen atom is stable. Moreover, classicalelectrodynamics predicts that energy is continuously radiated and therefore the resulting spectrum ofthe emitted radiation must be continuous. This is again contradicts the observations. These puzzlingresults led physicists to think that the classical laws of physics, which are valid in macroscopicworld, do not apply to the microscopic world.

1.4 ATOMIC (LINE) SPECTRUM

To obtain line spectrum of a substance, it is transformed into gaseous (atomic) state and is then excitedby an electric discharge. The atoms then emit light that contains only certain wavelengths. To observespectrum we often allow the emitted light to pass through a fine rectangular aperture, called slit, andthen through a dispersive device, such as prism or diffraction grating. The emergent radiation isreceived on a photographic film or can be seen through a telescope. The various wavelengths in thelight appear as well-defined fine lines, which are the images of the slit. Each line corresponds to adefinite wavelength present in the light. These lines taken together constitute what we call atomic orline spectrum. By 1823, the line spectrum of each element was found. The line spectrum is acharacteristic property of the element. In other words, each element can be identified by its linespectrum.

By the middle of the 19th century the study of atomic spectra had held the interest of scientistsbecause of their mysterious varieties. Using improved techniques and spectrographs of high resolvingpower, more and more finer details of atomic spectra were recorded and a wealth of fascinating butunexplained data on the spectral lines of various atoms was collected. The real theoretical work in

Fig. 1.3.2 Spiral path of electron


spectroscopy dates back from 1885 when a Swiss high school mathematics teacher Johann Balmerdiscovered an empirical rule governing the wavelengths of various spectral lines in the visible partof the spectrum of hydrogen atom. Balmer found that the wavelengths of spectral lines of hydrogenatom could be represented by formula

Å2 2

2 23645.6

4 4

m mb

m mλ = = ×

− −

Substituting for m = 3, 4, 5, 6……, we get the wavelength of the first, second, third, fourthlines beginning at the red end. Customarily the lines are denoted by wave number, which is reciprocalof wavelength. In terms of wave number the Balmer formula is represented as

1

2 2 2 2

1 1 1 1 1109678 cm R

2 2m m− ν = = − = − λ

where R = 109678 cm–1 is a constant now known as Rydberg constant.A group of lines, whose wave numbers are represented by giving appropriate values to the

variable integer m, constitute a spectral series. Since the pioneer work of Balmer, a large volume ofwork has been done on the analysis of atomic spectra. In the hydrogen spectrum many spectral serieswere discovered, which were named after their discoverer.

Fig. 1.4.1 Balmer lines of hydrogen spectrum

Spectral Series of Hydrogen Atom:

Lyman series (1906): It is found in ultraviolet region.

2 2

1 1R ; 2, 3, 4,........

1m

m

ν = − =

2

1 1R R

1∞

ν = − = ∞

Å Å.1 1216 , 912∞λ = λ =

Balmer series (1885): Four lines (H, H, H, H) are found in the visible region.

2 2

1 1R ; 3, 4, 5.....

2m

m

ν = − =


2

1 1 RR

42∞

ν = − = ∞

1 = 6563 Å, = 3640 Å.

Paschen series (1908): It is found in infrared region.

2 2

1 1R ; 4, 5, 6,.......

3m

m

ν = − =

2

1 1 RR

93∞

ν = − = ∞

1 = 18760Å, = 8210 Å.

Bracket series (1922): It is found in infrared region.

2 2

1 1R ; 5, 6, .........

4m

m

ν = − =

2

1 1 RR

164∞

ν = − = ∞

1 = 40530Å, = 14590 Å.

Pfund series: It is found in far infrared region.

2 2

1 1R ; 6, 7, ....

5m

m

ν = − =

2

1 1 RR

255∞

ν = − = ∞

1 = 74620 Å, = 22800 Å.

The problem to which the physicists in the second decade of the 20th century were confrontedwas the problem of finding the possible mechanism responsible for the origin of discrete spectrallines. The only atomic model available at that time was that of Rutherford but that too was unstableaccording to the classical concepts of electrodynamics. The observed facts viz the existence of discretespectral lines and the stability of the atom, were inexplicable in terms of classical physics. The reignof confusion was spread over the scientific world. No solution within the realm of classical physicswas seen. At the same time, the departure from the concepts of classical physics was much too adaring step. On this cloudy scene a brilliant young man appeared like an angel who made a bolddeparture from the classical physics and cleared up the mystifying clouds—the man was Niels HendrickDavid Bohr.


NIELS BOHR

Born in 1885 at Copenhagen in Denmark, Bohr studied at Copenhagenwhen Rutherford was performing his epoch making scattering experimentat the university of Manchester. Bohr received a fellowship to work atCambridge and Manchester. There he became acquainted with theRutherford’s atomic model. After returning to his home city Copenhagenin the summer of 1913, he published his celebrated paper on the atomicstructure in the Philosophical Magazine. As mentioned earlier theRutherford’s atomic model was not consistent with the laws of classicalphysics, Bohr’s solution to this contradiction between the conclusions ofthe conventional laws of physics and the facts of nature wasstraightforward and bold. Since nature cannot be wrong, the conventional laws of physics must bewrong at least when applied to the dynamics of electron within the atom. In making his revolutionarystatement regarding the motion of electron within the atom, he took a bold step in applying Planck-Einstein quantum hypothesis to the atomic system. Thus his theory of atomic structure was a hybridof classical and quantum ideas. The way he proposed was too odd and unconventional that he keptthe manuscript locked in his desk for almost two years before he decided to send it for publication.When this epoch making paper finally appeared (1913) it sent out a shock wave of amazement throughthe world of contemporary physics. For this outstanding work Bohr was awarded Nobel Prize in 1922.

In 1918, Bohr became professor of theoretical physics at the university of Copenhagen and in1921 he established an institute in Copenhagen, which became an international center for theoreticalwork in quantum physics. At this institute, world’s outstanding physicists spent some time. WolfgangPauli, P.A.M. Dirac, Werner Heisenberg, Landau, Bloch, Teller, Gamow, Heitler were all alumni ofBohr’s institute, their names and accomplishments tell a large part of what happened in quantummechanics during the crucial decade of the 1920’s. Robert Oppenheimer wrote of this and Bohr’sindispensable role in it. It was a heroic time. It was not the doing of one man: it involved thecollaboration of scores of scientists from many different lands, though from the first to last the deepcreative and critical spirit of Niels Bohr guided, restrained, deepened and finally transmuted theenterprise.

1.5 BOHR’S THEORY OF HYDROGENIC ATOMS (H, He+, Li++)

In 1913, Bohr proposed an atomic model, which explained with amazing accuracy the main featuresof the spectra of hydrogenic atoms. His model was based on the following postulates:

(i) The electron in hydrogen atom moves in circular orbit around the nucleus. The dynamicsof the electron is governed by the Newtonian mechanics i.e., the centripetal force requiredfor circular motion is provided by Coulomb attraction of nucleus on the electron.

2

22

0

1 Z

4

em r

r= ω

πε ...(1.5.1)


where r = Radius of the orbit, = Angular velocity of electron, m = Electronic mass,Z = Atomic number of atom.

(ii) In contrast to classical physics where the radius of electronic orbit can assume anymagnitude, Bohr asserted that only those orbits are allowed in which the angular momentumof electron is integral multiple of (= h/2).

2mr nω = ...(1.5.2)

where n is an integer, called principal quantum number. n = 1, 2, 3….. label the first,second, third ….. orbits of the electron.

(iii) Since the revolving electron around the nucleus is not a stable system under the laws ofclassical electrodynamics. Bohr assumed that the classical laws do not apply, at least, tothe atomic phenomena. That is, the electron revolving in any one of the allowed orbitsdoes not radiate. These non-radiating orbits are called stationary orbits. However, whilemaking transition from a stationary orbit of higher energy to that of lower energy it doesradiate. The electron may also go over from orbit of lower energy to that of higher energyby absorbing energy. If Ei and Ef are the energies of electron in the initial and final orbit,the frequency (or angular frequency ) is given by

E Ei f h− = ν = ω ...(1.5.3)

Let us calculate the radius of electron orbit, orbital frequency of revolution, energy ofelectron and the frequency (wavelength) of the radiation emitted in electronic transition.

Fig. 1.5.1 Hydrogen atom

Radius of orbit: Eliminating from Eqns. (1.5.1) and (1.5.3) and solving the resulting equationfor r , we have

2 2

0 24

Z

nr

me= πε

...(1.5.4)

For hydrogen atom Z = 1, the radius of the first orbit (n = 1), called Bohr orbit (a0) comesout to be

2

1 0 0 24r a

me= = πε

...(1.5.5)


Substituting e = 1.6 × 10–19 C, m = 9.1 × 10–31 kg, = 1.054 × 10–34 J s and(1/ 40) = 9 × 109 m/F, we obtain

a0 = 0.529 × 10–10 m = 0.529 Å

In terms of Bohr radius a0, the radius of the nth orbit is given by

2

0 Znn

r a= ...(1.5.6)

Rotational frequency of electron: Eliminating r from Eqns. (1.5.1) and (1.5.2) we obtain thefrequency of rotation of electron in nth orbit.

2 4 2 2

03 3 30

1 Z Z

4nme

n n

ω = = ω πε

...(1.5.7)

where 0 = 4.14 × 1015 radian/s.Linear velocity of electron: The linear velocity of electron in nth orbit is given by

2

00

1 Z Z

4n n ne

v r vn n

= ω = =πε

...(1.5.8)

where v0 = 2.19 × 106 m/s.

The ratio of velocity of electron in the first orbit of hydrogen atom to the speed of light c iscalled the fine structure constant and is given by

2

1

0

1 1

4 137

v e

c cα == = =

πε ...(1.5.9)

Energy of electron: The kinetic energy of electron is

22 2 2

0

1 1 1 ZK

2 2 4 2

emv mr

r= = ω =

πε ...(1.5.10)

Potential energy of electron is

2

0

1 ZU

4

e

r= −

πε ...(1.5.11)

The total energy of electron is

2

0

1 ZE K U

4 2

e

r= + = −

πε ...(1.5.12)


Substituting the value of r from (1.5.4), we have

2 4 2 2 218

2 2 2 20

1 Z Z ZE (2.176 10 J) (13.6eV)

4 2n

me

n n n−

= − = − × = − πε ...(1.5.13)

Equation (1.5.13) can be rearranged as

2 4 2 2 2

3 2 2 20

1 Z Z ZE (2 ) (2 R) R

4 4n

mec c hc

c n n n

= − π = − π = − πε π

...(1.5.14)

where 2 4

7 13

0

1R 1.097 10 m

4 4

me

c−

= = × πε π

is Rydberg constant.

From Eqns. (1.5.10), (1.5.11) and (1.5.12), we see that

K E and U 2En n= =

Eqation (1.5.13) gives the possible energy levels of electron in hydrogen atom. The electronin the first orbit (n = 1) of hydrogen atom (Z = 1) has energy equal to

E1 = – 13.6 eV

The negative energy means that the electron is bound to the nucleus with this much energy. Inorder to remove the electron from the force field of nucleus, minimum energy equal to 13.6 eVmust be imparted to it and therefore the first ionization energy of hydrogen atom is 13.6 eV.

Frequency of emitted radiation: If the electron makes transition from an orbit of quantumnumber ni to the orbit of quantum number nf, the frequency of the emitted radiation is given by

2

2 2

1 1E E (2 R)Zi f

f i

cn n

ω = − = π −

The wavelength ( = 2c/) of the radiation is

2

2 2

1 1 1RZ

f in n

= − λ

...(1.5.15)

This formula is known as the Balmer formula. It is remarkable to note that the value of Rydbergconstant R calculated from fundamental constants comes out to be the same as that obtained fromspectroscopic measurements. This gives the dramatic confirmation of Bohr’s theory.


1.6 ORIGIN OF SPECTRAL SERIES

Lyman series: When the electron jumps from the energy levels labeled by ni = 2, 3, 4,…… to theenergy levels nf = 1, a series of spectral lines are emitted. These lines constitute Lyman series afterthe name of its discoverer. The wavelengths of these lines are given by

2

2 2

1 1 1RZ ; 2, 3, 4..... (ultraviolet).

1i

i

nn

= ν = − = λ

Balmer series: When the electron makes transition from energy levels ni = 3, 4, 5,…..tonf = 2, the spectral lines of Balmer series are emitted. The wavelengths of these lines are representedby formula

2

2 2

1 1 1RZ ; 3, 4, 5....

2i

i

nn

= ν = − = λ

The first four lines of this series lie in the visible region and are denoted by H, H, H, andH.

Paschen series: The lines of Paschen series are emitted when the electron jumps fromni = 4, 5, 6,….. to nf = 3 and their wavelengths are given by the formula

2

2 2

1 1 1RZ ; 4, 5, 6........

3i

i

nn

= ν = − = λ

(infrared)

Bracket series: The wavelength of Bracket series are given by

2

2 2

1 1 1RZ ; 5, 6, 7....

4i

i

nn

= ν = − = λ

(far-infrared).

Pfund series: The wavelength of lines of this series are represented by formula

2

2 2

1 1 1RZ ; 6, 7, 8,...

5i

i

nn

= ν = − = ∞ λ

The criterion for the success of any new theory in physics is not only that it should give acorrect interpretation of the previous observations but it should also provide new predictions, whichcan later be confirmed. At the time when Bohr presented his theory, the only spectral series knownwas Balmer series. Bohr’s theory not only explained the origin of this series but also predicted theexistence of other spectral series. In fact Lyman and other series were discovered much after theBohr’s theory was presented.

Energy level diagram : The energy level diagram of hydrogen atom corresponding to differentvalues of principal quantum number n are shown in the Figure (1.6.1). The energy level for n = 1 iscalled the ground state and has energy equal to –13.6 eV. All other energy states are called the excited


states. The electronic transitions leading to various spectral series are shown on the diagram by verticallines.

Fig. 1.6.1 Spectral series of hydrogen atom

Criticism of Bohr’s theory: Although Bohr’s theory met with spectacular success in explainingthe hydrogen spectrum, nevertheless it was too revolutionary to get a warm reception. In order tosave the stability of the atom from catastrophe Bohr threw away the only classical picture of mechanism


of emission of radiation by an accelerated charge. His postulates viz., only those orbits are allowedin which angular momentum of electron is integral multiple of and the frequency of emitted radiationduring electronic transition is given by = (Ei – Ef)/ , were taken from the quantum theory, whichitself was questioned. Bohr’s theory has nothing to say about the intensity of spectral lines. All attemptsto construct a theory of helium atom (and other multi-electron atoms) failed. The weakest point ofBohr’s theory was its internal logical contradiction. It was neither consistent classical theory nor aconsistent quantum theory. It was a hybrid theory based on classical and quantum concepts both andso it was felt by many physicists to be unsatisfactory. In spite of all its deficiencies Bohr’s theorybrought a new light and hope for spectroscopy and for quantum theory as well. This theory may beregarded as a transition step on the path to the creation of a consistent theory of atomic phenomena.It provided the foundation on which theoretical physicists erected a vast structure of atomic andmolecular physics.

1.7 CORRECTION FOR NUCLEAR MOTION

In the development of Bohr’s theory the nucleus of the atom was assumed to stationary. In fact, themotion of electron and nucleus under their mutual interaction is a two body problem. In hydrogenatom both the nucleus and the electron rotate with the same angular velocity, say about an axispassing through their center of mass and perpendicular to the line joining them. Let r be distancebetween the nucleus and the electron and r1 and r2 be their distances from the center of mass. If Mand m are the masses of the nucleus and the electron then

Fig. 1.7.1 Rotation of nucleus and electron about their center of mass

Mr1 = mr2 and r1 + r2 = r

From these two equations, we find

1 2M

andM M

mr r r r

m m= =

+ +

The total angular momentum of the system is

2 2 2 2

1 2M

L MM

mr mr r r

m= ω + ω = ω = µ ω

+


where is called the reduced mass of nucleus and electron and is given by =M

M

m

m + or

1 1 1

M m= +

µ.

The above result shows that the angular momentum of electron–nucleus system rotating about theircenter of mass is the same as that of a fictitious particle of mass µ revolving in a circular orbit ofradius r. Thus, the effect of taking the nuclear motion into consideration is equivalent to replacingelectronic mass m by the reduced mass . In the special case: when M , m. Thus, theassumption that the nucleus is stationary is equivalent to assuming the nucleus to be infinitely heavy.When nuclear motion is taken into consideration, the formulas for orbital radius, angular frequencyof rotation, energy and Rydberg constant become

2 2 2 2 2

0 0 02 2

M M4 4

Z M Z M Znn m n m n

r ae me

+ + = πε = πε = µ

...(1.7.1)

2 24 2 4 2

3 3 3 30 0

1 Z 1 M Z

4 4 Mne me

mn n

µ ω = = πε πε + ...(1.7.2)

2 4 2

2 20

1 M ZE

4 M2

me

m n

= − πε + ...(1.7.3)

2 4

30

1 M M 1R R R

4 M M4 1M

memm mc

∞ ∞

= = = πε + +π +

...(1.7.4)

where R is the value of Rydberg constant when nucleus is assumed to infinitely heavy. The Balmerformula now becomes

2 2

2 2 2 2

1 1 1 M 1 1RZ R Z

Mf i f i

mn n n n∞

= − = − λ + ...(1.7.5)

Positronium atom: When an electron and positron come together, they form a short-lived atompositronium in which both the particles revolve about their center of mass. Since both the particlesare equally massive, the center of mass lies mid-way between them. For positronium atom, reducedmass = m/2 and (M + m)/ M = 2. The radius of the circular path of either particle is

Å0 0M

2 1.06Mn

mr a a

+ = = =

The orbital frequency is

2 415

20

1 M(4.14 10 rad/s)(2)

4 Mnme

m

ω = = × πε +


= 2.07 × 1015 rad/s

The energy of the atom is

2 4 2 2

2 2 20

1 M Z M ZE (13.6 eV)

4 M M2n

me

m mn n

= − = − πε + +

= – 6.79 eV

Rydberg’s constant

M 1

R R RM 2m∞ ∞

= = +

The wavelength of spectral line

2 2

1 1 1 1R

2 f in n∞

= − λ

Muonic (mesic) atom: In this atom a muon which has charge equal to that of electron andmass equal to 207 times the electronic mass, revolves around a proton. The reduced mass of thesystem is

(207 )(1836 )M

186M 207 1836

m mmm

m m m

′µ = = =

′ + +

The orbital radius

2 2 2 2 2

00 02 2

14 4

Z Z 186 186 Znan n n

re me

= πε = πε =µ

For n = 1 and Z = 1,

o30

1 2.84 10 A186

ar −= = ×

Orbital frequency

2 24 2 4 2

3 3 3 30 0

1 Z 1 Z(186)

4 4ne me

n n

µω = = πε πε

2

153

Z(4.14 10 rad/s)(186)

n= ×

For n = 1 and Z =1,

1= 7.70 × 1017 rad/s

Energy of atom 2 24 2 4 2

3 2 3 20 0

1 Z 1 ZE (186)

4 42 2n

e me

n n

µ= − = − πε πε


For 1 and Z 1n = =

1E (13.6 eV) (186) 2530 eV= − = −

Rydberg’g constant R =4 4

7 13 3

0 0

1 1(186) 186 R 204.04 10 m

4 44 4

e m e

c c

−∞

µ = = = ×πε πεπ π

Wavelength of spectral line.

2

2 2

1 1 1RZ

f in n

= − λ

The first line of Lyman series has wavelength = 6.5Å (x-ray region.)

1.8 DETERMINATION OF ELECTRON-PROTON MASS RATIO (m/MH)

The Rydberg constant for hydrogen atom is

HH

1R R

1 ( / M )m∞=+ ...(1.8.1)

and for helium atom is

HeHe

1R R

1 ( / M )m∞=+ ...(1.8.2)

where MHe = 4 MH


HeH H

He H H

1 ( / M )R 1 ( / 4M )

R 1 ( / M ) 1 ( / M )

m m

m m

+ += =

+ +

Or He H

H H He

R R

M R (1/ 4)R

m −=

− ...(1.8.3)

Substituting the experimental value of RH = 1096758 m–1 and RHe = 10972226 m–1 we findthat m/MH = 1/1848 which agrees with other measurements.

1.9 ISOTOPIC SHIFT: DISCOVERY OF DEUTERIUM

The Rydberg’s constant for an atom depends on nuclear mass hence it will have different values fordifferent isotopes. The Rydberg’s constants for ordinary hydrogen (1H1) and deuterium (1H2) are

1 21 2

H HH H

1 1R R R R

1 / M 1 / Mm m∞ ∞= =+ +


The wavelengths of the first Balmer lines emitted by these two isotopes are given by

1

2

1 2H 2 2

1

2 2H 2 2

1

1 1 1(H ) R Z

2 3

1 1 1(H ) R Z

2 3

α

α

= − λ = − ′λ

Substituting the values of RH1 and RH2, we get

1(H) = 6562.79Å and 1(H

) = 6561.00Å.

Thus, the two Balmer lines should be separated by 1.79Å. Using a concave grating spectrographUrey, Brickwedde and Murphy observed that the H lines emitted by these two isotopes were separatedby this amount. In fact, the isotope deuterium was discovered in this way.

1.10 ATOMIC EXCITATION

According Bohr’s theory, atomic energy levels are quantized. Normally the atom resides in its lowestenergy state (ground state). By imparting energy to it from external agency, it can be raised to oneof its excited states. There are many ways to cause excitation in the atom. One way is to make itcollide with another particle possessing appropriate energy. If the kinetic energy of the collidingparticle is less than the energy difference between the ground state and the first excited state of theatom, the collision is elastic and the particle bounces off (because of its much smaller mass incomparison to that of the atom). If the energy of the colliding particle is large, the atom may absorbsome of its energy during collision and make a transition from ground state to one of its excitedstate. In such collisions, the kinetic energy is not conserved. Such types of collisions are called inelasticcollisions. The excited atom returns to its ground state in an average time of 10–8 second by emittingone or more photons. These atomic processes can be realized in a discharge tube containing a gas atlow pressure. Electrons and ions produced due to discharge are accelerated under the intense electricfield produced by the applied voltage between the electrodes. The electrons acquire sufficient energyand are capable of causing excitation in the atoms, which come their way.

The atom can also make a transition from the ground state to one of its excited state by absorbinga photon whose energy is exactly equal to the energy difference between the ground state and theexcited state. Absorption spectra have their origin because of this type of excitation.

At the time Bohr published his theory of atomic structure, two scientists James Franck andGustav Hertz were performing experiments on the excitation of atoms. In 1914, they submitted theirresults, which gave striking evidence in the favour of quantization of atomic energy states. It isremarkable to note that Franck and Hertz were not aware of Bohr theory. If they had read it beforecollecting their results they would not have believed it which is evident from the Franck’s candidremark we had a colloquium at that time in Berlin at which all important papers were discussed.Nobody discussed Bohr’s paper. Why not? The reason is that fifty years ago one was so convincedthat nobody would, with the state of knowledge we had at that time, understand spectral lines emission,so that if somebody published a paper about it, one assumed, probably it is not right. So we did notknow it (this statement of Franck was given in an interview in 1961).


1.11 FRANCK-HERTZ EXPERIMENT

Franck and Hertz experiment provides strong and conclusive evidence in support of the existence ofdiscrete energy states of atoms. The apparatus, as shown in the Fig. (1.11.1), consists of a glass tubefilled with mercury vapor, a filament with heating arrangement, an anode plate for receiving electronsand a grid near the anode. On heating the filament electrons are emitted, which are accelerated towardsthe anode. When they pass through the grid, the retarding potential V0 prevents them from reachingthe plate. Thus, electrons having very small kinetic energy will not be able to reach the anode. Anammeter measures the current due to electrons reaching the plate. When the accelerating voltage isincreased, the current increases. For a particular value of the accelerating voltage the current suddenlydrops and then again increases with increasing voltage. At certain voltage again the current drops. Itis observed that the current drops at equal interval of accelerating voltage. The results obtained areshown in the Fig. (1.11.2).

Interpretation of observed results: When the accelerating voltage is increased from its zero value,the kinetic energy of the electrons increases and therefore more and more electrons reach the anodeovercoming the retarding potential thus the current increases. When the accelerating voltage becomes4.9 volts, the electrons acquire kinetic energy equal to 4.9 eV on reaching the grid. In front of thegrid, they suffer inelastic collisions with the mercury atoms and lose most of their kinetic energyand are unable to reach the anode because of the retarding potential. This explains the drop in currentwith increasing voltage. During the collisions with electrons the mercury atoms are raised to theirfirst excited state. When the accelerating voltage is further increased above 4.9 volts, electrons acquireso much energy that even after suffering inelastic collision they are left with sufficient energy toovercome the retarding potential and thus reach the anode. This explains the reason for the increasein current after the first drop of current. Again when electrons acquire energy equal to 9.8 eV theysuffer inelastic collisions with mercury atoms in their trip from cathode to anode and therefore currentfalls.

This simple experiment shows that the energy required to raise mercury atoms from their groundto their first excited state is 4.9 eV. Electrons of energy less than 4.9 eV do not excite the mercuryatoms. Thus, the mercury atoms can exist in the ground state or in the first excited state which it hasenergy 4.9 eV relative to the ground state.

Fig. 1.11.1 Schematic sketch of Franck-Hertz experiment


Fig. 1.11.2 Results of Franck-Hertz experiment

When an excited mercury atom returns to its ground state a photon of energy 4.9 eV is emitted.In order to find the energy of the emitted photon, Hertz observed the emission spectrum of mercuryvapor filled in the tube. To his surprise, when the accelerating voltage was less than 4.9 volts, nospectral line appeared but when it was 4.9 volts a spectral line of wavelength 2536 Å was observed.The energy of this photon is

E = Å

Å

12400 eV.4.89 eV

2536

ch = =λ

which is in excellent agreement with the experiment.

1.12 BOHR’S CORRESPONDENCE PRINCIPLE

In 1923, Bohr pointed out that quantum and classical theories yield identical result in the region ofhigh quantum numbers. This requirement is called the Correspondence principle. In the earlydevelopment of the quantum theory, this principle played an important role in checking the formulaeobtained from quantum principles. Let us verify this principle by taking hydrogen atom as an example.

According to classical electromagnetic theory; an electron revolving in circular orbit radiateselectromagnetic radiation of frequency equal to the orbital frequency and to the harmonics of theorbital frequency. The orbital frequency of electron in the hydrogen atom is

2 4 2 2

3 3 30

1 Z4 R (orbital)

4nme Z

cn n

ω = = π πε

...(1.12.1)

The frequency of the emitted radiation when electron jumps from (n + p)th orbit to nth orbitis

22 2

2 22

2 2

2 1 12 RZ

( )

( )2 RZ

( )

cc

n n p

n p nc

n n p

πω = = π − λ + + −= π +


2

2 2

(2 )2 RZ

( )

n p pc

n n p

+= π +

In the limit of large quantum numbers p << n, we have

2

3

22 RZ

pc

nω = π ...(1.12.2)

For p = 1, this frequency coincides with the orbital frequency of electron. Harmonics are obtainedby letting p = 2, 3, 4,…. This illustrates that: In the limit of large quantum numbers, classical andquantum physics provide identical results.

1.13 SOMMERFELD THEORY OF HYDROGEN ATOM

Bohr’s theory of hydrogen atom in its simplest form met with spectacular success in predicting thecorrect positions of the spectral lines in the hydrogen atom. Spectrographs of higher resolving powerrevealed that the spectral lines which were thought to be single, actually consisted of a group oflines very close together. This means that the energy levels corresponding to a principal quantumnumber possess fine structure i.e., the energy level consists of a number of energy levels lying veryclose together. Michelson with his interferometer found that the H and H lines of Balmer serieswere close doublet with separation of only 0.14 Å and 0.48 Å respectively.

In an attempt to explain the existence of fine structure, Wilson and Sommerfeld proposed ageneral rule for quantum conditions, known as Wilson-Sommerfeld quantization rule. The Planck’squantum condition for a harmonic oscillator: energy of a harmonic oscillator is integral multiple of and Bohr’s condition: angular momentum of an electron moving in circular orbit is integral multipleof are the particular cases of this general quantum condition.

For a harmonic oscillator with momentum p and position q, the Wilson-Sommerfeld quantumcondition states that

pdq nh=∫ ...(1.13.1)

where the integration is to be carried out over the complete cycle. The integer n is called the principalquantum number. The energy of a one-dimensional harmonic oscillator is

2

2 21E K U

2 2

pm q

m= + = + ω

Above equation can be written as

2 2

21

2 E 2E /

p q

m m+ =

ω...(1.13.2)

The state of harmonic oscillator is described by momentum coordinate p and position coordinateq. If we plot the instantaneous values of q and p on q – p plane for one cycle we get an ellipse with


semi-major axis 2

2Ea

m=

ω and semi-minor axis 2 E.b m= Each point on the ellipse represents

some state of the oscillator. Such a two-dimensional space with position and momentum as its axesis called phase space. As the oscillator completes its one cycle, its representative point completesellipse in phase space. The actual motion of the oscillator should not be confused with the motion ofits representative point in phase space.

Fig. 1.13.1 Area between any two successive ellipses is h

In Wilson-Sommerfeld quantum condition ,p dq nh=∫ the integral on the left hand side

represents the area of the phase trajectory and is equal to ab. Thus, Eqn. (1.13.1) can be written as

ab nhπ =


2

2E2 Em nh

mπ =

ω

E2

hn n= ω = ω

π ...(1.13.3)

which is the Planck’s quantization rule for harmonic oscillator. The states of oscillator with energiesE = , 2, 3……are described by a series of ellipses such that the area between two successiveellipses is equal to Planck’s constant h.

The Wilson-Sommerfeld condition for electron moving in circular orbit around the nucleus isobtained by replacing linear momentum p with angular momentum L and position coordinate q withangular position . So the quantum condition in this case reduces to

L d nhθ =∫ ...(1.13.4)

In Coulomb force (which is a central force) the angular momentum remains constant andtherefore

L d nhθ =∫ L.2 nhπ =

L2

hn n= =

π ...(1.13.5)

which is the Bohr quantum condition.

Sommerfeld Theory of Hydrogen Atom

In 1916, Arnold Sommerfeld presented a theory of hydrogen atom according to which the electronin hydrogen atom revolves round the nucleus in an elliptical orbit with nucleus at one of its foci.This system requires two coordinates for the description of motion. In polar coordinates it is specifiedby radial distance r and angular position . The Wilson-Sommerfeld quantum conditions in this caseare

r rp dr n h=∫ ...(1.13.6)

p d n hθ θθ =∫ ...(1.13.7)

where pr and p are radial and angular momentum respectively. nr and n are radial and azimuthalquantum numbers.

In central force angular momentum of electron is a constant and therefore equation (1.13.7)simplifies to

2

hp n nθ θ θ= =

π ...(1.13.8)



( )2 2

2 2 2

0 0

1 1E K U ( )

2 4 2 4

e emv m r r

r r

−= + = + = + θ −πε πε

22 2

202 42

r pp e

m rmrθ= + −

πε2( , )p mr p mrθ= = θ

Solving for pr we get

22

20

2 12 E

4rpme

p mr r

θ= + −πε ...(1.13.9)

Substituting the expression for pr in (1.13.7), we have

2 24

20

1 22 E

4 rnme

m n hr r

θ+ − =πε∫

This integral on simplification gives

4 4

2 2 2 2 2 2 2 20 0

1 1E

32 ( ) 32r

me me

n n nθ= − = −

π ε + π ε

where n = nr + n is called the total principal quantum number.In polar coordinates, the equation of ellipse is

1 cosl

er

= + θ

Fig. 1.13.2 Elliptical trajectory of electron

Taking logarithm and then differentiating, we get

1 sin

1 cos

dr e

r d e

θ=θ + θ

...(1.13.10)


Now

2

rdr dr d dr dr d

p dr m dr m d m ddt d dt d d dt

θ θ = = θ = θ θ θ θ

2

1 drp d

r d θ = θ θ

2 dp mr

dtθθ =

2 2

2

sin

(1 cos )

ep d

eθ

θ= θ+ θ

...(1.13.11)

In view of Eqn. (1.13.11), the quantum condition (1.13.7) becomes

2 2

2

sin

(1 cos )r r

ep dr p d n h

eθ

θ= θ =+ θ∫ ∫ ...(1.13.12)

or 2

12

1r rp dr p n h

eθ

= π = −

∫ ...(1.13.13)

Since p = n, Eqn. (1.13.13) may be written as

2

22

1( )r

ne

n nθ

θ

− =+

...(1.13.14)

For ellipse, we know that

2

22

1b

ea

− = ...(1.13.15)

where a and b are the semi-major and semi-minor axes of the ellipse. From Eqns. (1.13.14) and(1.13.15), we have

2 22

22 2 2

1( )r

n nbe

a n n nθ θ

θ

− = = =+

...(1.13.16)

Since nr and n are integers, the total quantum number n is also an integer. In ellipse b < a, son < n. When b = a, the ellipse becomes circle and n = n. So the maximum value of azimuthalquantum number n can be n. When n = 0, b = 0 and the ellipse a straight line. Physically thismeans that the electron would move along a straight line passing through the nucleus, which is notpossible. So n cannot be zero. Thus, allowed values n are the integers between 1 and n, both valuesinclusive. For n = 4, n can assume values 1, 2, 3, 4 and the radial quantum number nr takes 3, 2, 1and 0. Corresponding to these four values of n we have four orbits with different eccentricities.The orbits with their usual notations are given in the table.

n n nr Orbit notation (nn)

4 1,2,3,4 3,2,1,0 41, 42, 43, 44


Fig. 1.13.3 Sommerfeld elliptical orbits

Since the total energy of the atom depends on the total quantum number n, which has the samevalue for all elliptical orbits, the introduction of elliptical orbit does not lead to the prediction ofnew energy levels for hydrogen atom. The orbits corresponding to a given total quantum number nwith the same energy are said to be degenerate. Thus, the Sommerfeld theory as such is in no waysuperior to Bohr’s theory.

1.14 SOMMERFELD’S RELATIVISTIC THEORY OF HYDROGEN ATOM

The revolving electron in an elliptical orbit around the nucleus of hydrogen atom has greater velocitywhen it is near the nucleus and has smaller velocity when it is relatively far away from the nucleus.According to the special theory of relativity, the mass of a moving body varies with velocity. Whenthis result is applied to the motion of electron, its energy levels, except the ground level, are foundto split into a number of closely spaced components called the fine structure, a term for the firsttime used by Sommerfeld. The orbit of electron now becomes a complicated curve — a precessingellipse — similar to the orbit of the planet mercury about the sun.

According to Wilson-Sommerfeld, each degree of freedom of electron is quantized. So, wehave two quantum conditions.

p d n hθ θθ =∫ ...(1.14.1)

r rp dr n h=∫ ...(1.14.2)

The first condition reduces to

p nθ θ= ...(1.14.3)


E = K + U = 2

20 2 0

1

41

em c

r

− πε− β

...(1.14.4)


Making use of Eqns. (1.14.3) and (1.14.4) the total energy of electron can be found. The actualcalculation is some what tedious. The final expression for the energy comes out to be

4 2

2 2 2 2 40

2 R 1 3E

432

c me n

nn n θ

π α= − − − π ε

...(1.14.5)

where 2

04

e

cα =

πε is fine structure constant. The expression for E may be symbolically represented

as

0E E E ( , )n c n nθ= − − ....(1.14.6)

where E0n stands for the first term on the right hand side of Eqn. (1.14.5) and represents the energyof electron as obtained from Bohr’s theory. Ec is the relativistic correction term that depends on nand n. Allowed values of n are 1, 2, 3, .... and of n are 1, 2, ……n. The expression ofEqn. (1.14.6) is sufficient to explain the fine structure of H line of hydrogen atom.

The H line in the spectrum of hydrogen atom results from the transition from energy leveln = 3 to the energy level n = 2. For n = 3, n = 1, 2, 3. And for n = 2, n = 1, 2. The energy levelscorresponding to n = 3 are given by

E3 = – E03 – Ec (3, 1)

E3 = – E03 – Ec (3, 2)

E3 = – E03 – Ec (3, 3)

The energy levels corresponding to the total quantum number n = 2 are: E2 = – E02 – Ec (2, 1)

E2 = – E02 – Ec (2, 2)

Fig. 1.14.1 Fine structure of H line. Transitions marked x are forbidden

These energy levels and the possible transitions are shown in the Figure (1.14.1). Theory showsthat there should be six transitions. Experimentally H line is found to be a doublet. In order toexplain the existence of only two lines out of six possibilities it was further assumed that certain


selection rules are obeyed in such transitions. Selection rules are statements that allow certain transitionsand forbid others. In the present case only those transitions are allowed for which the quantum numbern changes by ± 1.

n = ±1 (allowed transitions)

Thus, the selection rule permits only three lines. The spectral lines resulting from the transitions31 22 and 33 22 are so close together, that they cannot be resolved at normal temperature. Thetriplet character of H line was later observed using heavy hydrogen at low temperature.

The agreement between the prediction of Sommerfeld’s relativistic theory and the observedfine structure of H line provided another remarkable confirmation of special theory of relativityand reinforces conviction of its universal validity. In fact one can visualize the slight change in energyfor electrons of different azimuthal quantum number in the following way: It is apparent that thevelocity of electron in an atom is comparable to the velocity of light and therefore relativistic effectsare plausible. Since the electrons with the same principal quantum number n but different azimuthalquantum numbers have orbits of elliptical shape with varying eccentricity, electron possesses higherspeed in the vicinity of nucleus for more eccentric orbit and therefore it can have different effectivemass, different n and hence different total energy.

SOLVED EXAMPLES

Ex. 1. Calculate for He+ (i) radius of the first Bohr orbit (ii) velocity of electron moving in the firstorbit (iii) orbital frequency in the first orbit (iv) kinetic energy and binding energy of electron in theground state (v) ionization potential and the first excitation potential (vi) wavelength of the resonanceline emitted in the transition n = 2 n = 1.

Sol. (i) For helium atom Z = 2.

Radius of the first orbit

ÅÅ

2

0

01

( 1)Z

0.5290.264

2 2

nn

r a n

ar

= =

= = =

(ii) Velocity of electron 2

6

0

1 Z Z(2.19 10 m/s)

4ne

vn n

= = ×πε

6 6

12

(2.19 10 m/s) 4.38 10 m/s1

v = × = ×

(iii) Orbital frequency

2 4 2 215

3 3 30

1 Z Z(4.14 10 )

4nme

n n

ω = = × πε

151 16.56 10 rad/sω = ×


(iv) Kinetic energy Kn = 2

2

ZE (13.6 eV)n

n=

K1 = 54.32 eV

Binding energy = 54.32 eV

(v) Ionization potential = 54.32 eV.

First excitation energy is equal to the energy required to raise the electron from groundstate (n = 1) to the first excited state (n = 2).

E = E2 – E1 = (13.58 eV)Z2 2 2

1 1

1 2

− = 40.74 eV (Z = 2)

(vi) Wavelength of the resonance line

2 2

2 2 2 2

1 1 1 1 1 3R Z R R Z

41 2f in n∞ ∞ ∞

= − = − = λ

Å.2 7 1

4 4303.8

3R Z 3 (1.097 10 m ) 4−∞

λ = = =× × ×

Ex. 2. A stationary hydrogen atom emits photon corresponding to the first line of Lyman series.Calculate (i) recoil velocity of the atom, (ii) recoil kinetic energy of the atom and (iii) energy of emittedphoton.

Sol. (i) When the electron makes transition from n = 2 to n = 1, the energy of transition isshared by photon and the atom. The energy emitted in the transition is given by

2 2

2 1 2 2

1 1 3E E E 2 RZ RZ

21 2c c

∆ = − = π − = π ...(1)

The conservation of momentum requires that: Momentum of atom = Momentum of photon

P = /c ...(2)

Energy of recoil atom E = 2 2

2

( )

2M 2M

p

c

ω= ...(3)

The energy of transition is equal to the sum of energy of emitted photon and the recoil energyof atom.

2

2

( )E

2Mc

ω∆ = ω + ...(4)

whence

2

2 E

1 2 E / Mc

∆ω =+ ∆

...(5)


For hydrogen atom, E/Mc2 << 1 hence

23

E RZ2

cω = ∆ = π ...(6)

The velocity of recoil of the atom is given by

23 RZ

M M 2M

pv

c

ω π= = = = 3.26 m/s ...(7)

(ii) Recoil kinetic energy of atom

2 2 2 2 2 4

82

( ) 9 R ZE 5.5 10 eV

2M 8 M2Mr

p

c−ω π= = = = ×

(iii) Energy of emitted photon E = E – Er (Er << E)

= E = 10.20 eV.

Ex. 3. A stationary He+ ion emits a photon corresponding to the first line of Lyman series. Thisphoton, when strikes a stationary hydrogen atom in the ground state liberates the electron from the latter.Find the kinetic energy of the photoelectron.

Sol. The energy of the emitted photon

22 1

3E E E RZ 6 R

2c cω = ∆ = − = π = π (Z = 2)

Ionization energy of hydrogen atom

0E 2 Rc∆ = π

The excess energy of the photon will appear as the kinetic energy of photoelectron. The kineticenergy of photoelectron

K = E – E0 = 6cR – 2cR = 4cR = 27.2 eV.

Ex. 4. What element has a hydrogen like spectrum whose lines have wavelengths four times shorterthan those of atomic hydrogen?

Sol. The reciprocal of wavelength (1/) emitted by hydrogen like atoms is proportional to Z2.Let 1 and 2 be the wavelengths of hydrogen atom and of unknown atom respectively. Then

21 2

22 1

Z

Z

λ= ⇒

λ Z2 = 2 (helium).

Ex. 5. Find the quantum number n corresponding to the excited state of He+ if on transition to theground state that ion emits two photons in succession with wavelengths 1.85 and 30.4 nm.

Sol. Let the n0 and n1 be the quantum number of the ground and intermediate state respectivelyand 1 and 2 be the wavelengths corresponding to transitions n n1 and n1 n0.


2

2 21 1

1 1 1RZ ,

n n

= − λ

2

2 22 0 1

1 1 1RZ

n n

= − λ

Adding these two equations, we have

2

02 21 2 0

1 1 1 1RZ 1n

n n

+ = − = λ λ

Substituting 1 = 108.5 × 10–9 m, 2 = 30.4 × 10–9 m, R = 1.097 × 107 m–1, Z = 2, n0 = 1we find n = 5.

Ex. 6. Calculate the Rydberg’s constant R if He+ions are known to have the wavelength differencebetween the first (longest wavelength) lines of the Balmer and Lyman series equal to = 133.7 nm.

Sol. Wavelength of the first line of Balmer series

2

2 2

1 1 1 9RZ

5R2 3

= − ⇒ λ = λ

Wavelength of the first line of Lyman series

2

2 2

1 1 1 1RZ

3R1 2

′= − ⇒ λ = ′λ Therefore

22

15R′∆λ = λ − λ =

7 122

R 1.097 10 m15

−= = ×∆λ


Ex. 7. What hydrogen—like ion has the wavelength difference between the first lines of the Balmerand Lyman series equal to 59.3 nm?

Sol. 2 2 2

36 4 88

5RZ 3RZ 15RZ∆λ = − =

Whence

7 1 9

88 88Z 3.

15R 15 1.097 10 m 59.3 10 m− −= = =∆λ × × ×

Ex. 8. Determine the separation of the first line of the Balmer series in a spectrum of mixture ofordinary hydrogen and tritium. R = 1.097 × 107 m–1.

Sol. Rydberg’s constant of ordinary hydrogen is

HR

R1 / Mm

∞=+

and that of tritium is

TR

R1 / 3Mm

∞=+

Let 1 and 2 be the wavelengths of the first lines of Balmer series of the two isotopes. Then

H H2 21

1 1 1 5R R

362 3

= − = λ

1H

36

5R⇒ λ =

T T 22 22 T

1 1 1 5 36R R

36 5R2 3

= − = ⇒λ = λ

The wavelength difference

1 2H T

36 1 1 361 1

5 R R 5R M 3M

m m

∞

∆λ = λ − λ = − = + − +

Å10

7 1

36 2 36 2 12.4 10 m 2.4 .

5R 3 M 18365 1.0973 10 m 3

m −−

∞

× = = = × = × × ×


1. Describe Rutherford scattering experiment of alpha particles. What were the conclusions drawnthe results of this experiment? How did Rutherford calculate the nuclear dimensions? DescribeRutherford atomic model. What were the objections raised against this model?

2. Describe Bohr’s theory of hydrogen atom. How does this theory explain the various spectralseries observed in the spectrum of hydrogen atom? What are the shortcomings of Bohr’s theory?


3. Find expression for the Rydberg constant for hydrogen atom taking nuclear motion intoconsideration. How will you find the ratio of electron to proton mass?

4. Describe an experiment gives experimental evidence for the existence of discrete energy levelsof atomic system. State and explain Bohr correspondence principle.

5. Describe non-relativistic Sommerfeld theory of hydrogen atom. Why does this theory notexplain the fine structure of H line?

6. Give an outline of relativistic Sommerfeld’s theory of hydrogen atom. How does this theoryexplain the fine structure of Ha line? Mention the shortcomings of this theory.

7. Calculate for doubly ionized lithium (Li++)

(i) radius of the first Bohr orbit.

(ii) velocity of electron in the first orbit.

(iii) orbital frequency of electron in the first orbit.

(iv) ground state energy of electron.

(v) first ionization potential.

(vi) the wavelengths of first line of Lyman and Balmer series.

8. Calculate the longest and the shortest wavelength in Lyman and Balmer series of

(i) hydrogen atom.

(ii) singly ionized helium atom.

9. What minimum energy must an electron have for all lines of all the series of the hydrogenspectrum to appear when the hydrogen atoms are excited by impacts of the electrons? What isthe minimum velocity of these electrons? [Ans. 13.58 eV, 2.2 × 106 m/s]

10. Within what limits (in eV) should the energy of bombarding electrons be for the hydrogenspectrum to have only one spectral line when hydrogen atoms are excited by impacts of theseelectrons?

[Hint: First excitation energy is 10.2 eV, second excitation energy is 12.1 eV. To obtain only onespectral line the bombarding electrons must have energy E such that 10.2 E 1212.1 eV.].

11. How many revolutions does an electron in the state n = 2 of hydrogen atom make beforedropping to n = 1? The average life-time of excited atom is 10–8 s.

[Hint: 15(4.14×10 rad/s)(1/8)w = ]

[Ans. Required number of revolutions N = /2 = 8.2 × 105 rev.]

12. How many spectral lines are emitted when hydrogen atom are excited to the fourth energylevel?

[Hint: Number of spectral lines N = nC 2 = 4C2 = n(n – 1)/2 = 6]

13. Calculate for He+ and Li++ the following quantities:

(i) Radius of the first orbit.

(ii) Frequency of revolution of electron in the first Bohr orbit.

(iii) Velocity of the electron in the first orbit.

(iv) Kinetic energy, potential energy and total energy of electron.


(v) Ionization energy in the ground state.

(vi) First three excitation potentials.

(vii) Wavelengths of spectral lines when electron jumps from n = 3 to n = 2.

14. For positronium atom calculate

(i) the radius of the first Bohr orbit.

(ii) ground state energy.

(iii) Rydberg constant.

(iv) wavelengths of the first line of Lyman and Balmer series.

15. Calculate the following for mesic atom:

(i) Radius of the first Bohr orbit.

(ii) Ground state energy of the atom.

(iii) Binding energy in the ground state.

(iv) Wavelengths of the first lines of Lyman and Balmer series.

(v) In what region of electromagnetic spectrum do these radiations lie?

16. For atom of light and heavy hydrogen find the difference between

(i) the binding energies of their electrons in the ground state.

(ii) the wavelengths of the first lines of the Lyman series.

[Ans. (i) ED – EH = 3.7 × 10–3eV , (ii) D – H = 0.33 Å ]

CHAPTER

ATOMIC SPECTRA-II

2.1 ELECTRON SPIN

In an attempt to explain the doublet character of spectral lines emitted by alkali atoms and thephenomenon of anomalous Zeeman effect, two Dutch physicists Samuel Goudsmit and GeorgeUhlenbeck in 1925 postulated that electron might be rotating about its own axis. The name ‘spin’was given to this kind of motion of the electron. The angular momentum associated with the spinmotion of the electron is called intrinsic spin angular momentum. In classical picture electron isregarded as a charged sphere, which rotates about its own axis. The motion of the electron in anatom may be compared with that of earth’s motion. The angular momentum of the earth due to itsrotation about its own axis corresponds to the intrinsic spin angular momentum. The hypothesis ofspinning electron was proposed before the discovery of Schrodinger equation and had no theoreticalbasis. It was merely an ad-hoc hypothesis introduced to explain experimental observations. The conceptof electron spin was missing in Schrodinger theory. Later in 1928, English physicist P.A.M. Dirac,showed that in relativistic formulation of Schrodinger equation for hydrogen atom the intrinsic angularmomentum of electron appeared in a natural way and the concept of electron spin got theoreticalbasis. In quantum picture, the spin is regarded as an intrinsic property characterizing an electron inthe same way as its charge and mass do.

2.2 QUANTUM NUMBERS AND THE STATE OF AN ELECTRON IN AN ATOM

When Schrodinger equation is applied to the motion of electron in an atom, it is found that thequantum state or the wave function of an electron is characterized by four numbers, called quantumnumbers. They are: principal quantum number n, orbital quantum number l, magnetic quantumnumber ml and spin quantum number ms. The solution of Schrodinger wave equation, calledwave function, gives all kind of information about the electron in the atom. The importantcharacteristics and significance of these quantum numbers are as follows.

Principal Quantum Number (n): This quantum number determines the total energy of electronin the atom and the average distance of electron from nucleus. It can take integral values 1, 2, 3, ….The greater the value of n; greater is the energy of electron.

Atomic Spectra-II 413

Orbital (azimuthal) quantum number (l): This quantum number determines the orbitalangular momentum of electron. The magnitude of orbital angular momentum of electron is given by

( 1)l l= + l ...(2.2.1)

where l is a number, called orbital quantum number. For a given value of principal quantum numbern, the orbital quantum number can take integral values 0, 1, 2, ……(n – 1). The quantum number lalso gives the shape of probability distribution curve. The electrons with l = 0, 1, 2, 3…. are calleds, p, d, f electrons respectively.

Magnetic Quantum Number (ml ): The angular momentum vector l cannot take all orientationsin space; only certain directions are allowed. This feature of vector l is called space quantization.The allowed orientations of vector l are such that its components along any fixed direction, sayz-axis, are given by

z lm= l ...(2.2.2)

where ml is an integer called magnetic quantum number. For a given value of l, the quantum numberml can take integrally spaced values from – l to + l.

The other components of vector l are uncertain which is in accord with the uncertainty principle.This means that the vector l traces out a cone in space about z-axis such that its projection ontoz-axis is ml . The average values of x and y-components of l turn out to be zero.

Spin quantum number (ms): Relativistic quantum mechanics shows that electron possessesan intrinsic angular momentum S whose magnitude is given by

( 1)s s s= + ...(2.2.3)

where s is spin quantum number. It assumes only one value 1/2. The vector s can have only twodirections. The projection of vector s onto any fixed axis, say z-axis, are given by

1

2z sm= = ± s ...(2.2.4)

where ms = ± 1/2 is called the magnetic spin quantum number.

Fig. 2.2.1 The allowed values of quantum number ms are ± 1/2


Thus, the state of an electron, in an atom is described by four quantum numbers n, l, ml and ms

Now we shall find the number of quantum states corresponding to various values of principalquantum number n.

Corresponding to the principal quantum number n = 1, we have l = 0, ml = 0. ms = ± 1/2.Thus, for n = 1, there are 2 states defined by the quantum numbers

n = 1, l = 0, ml = 0, ms = +1

2

n = 1, l = 0, ml = 0, ms = – 1

2

According Pauli principle, each state is occupied by a single electron. The quantum states havingthe same value of principal quantum number n are said to constitute a shell. Shells are designatedaccording to the following scheme

n 1 2 3 4 5Shell K L M N OThus, K shell contains two quantum states and hence two electrons. The quantum states, which

have the same value of l are said to constitute a sub-shell. The above two states have the same valueof l (= 0) and therefore form a sub-shell. The sub-shells are designated according to the followingscheme:

Azimuthal 0 1 2 3 4 5 ...quantum number lSub-shell s p d f g h ………The K shell contains only one sub-shell denoted by s.For n = 2, l = 0, 1. For l = 0, the allowed value of ml is 0. For l = 1, the allowed values ml are

– 1, 0, 1. For each value of ml, ms = ± 1/2. Thus, the quantum states for n = 2 are as follows.

n l m l ms Quantum states

2 0 0 + 1/2 (2, 0, 0, 1/2)

– 1/2 (2, 0, 0, – 1/2)

2 1 –1 + 1/2 (2, 1, –1, 1/2)

– 1/2 (2, 1, –1, – 1/2)

0 1/2 (2, 1, 0, 1/2)

– 1/2 (2, 1, 0, – 1/2)

1 1/2 (2, 1, 1, 1/2)

– 1/2 (2, 1, 1, – 1/2)

Thus, the L shell (n = 2) contains one s sub-shell and three p sub-shells. In all there are eightquantum states. The pair of quantum states of a sub-shell differing in spin quantum numbers only,are called orbital. The s sub-shell contains one orbital and p sub-shell contains three orbitals, usuallydesignated as px, py, pz. Each orbital can accommodate two electrons with opposite spins.

The quantum states corresponding to principal quantum number n = 3 are shown in the tablegiven below.


The M shell (n = 3) contains one s sub-shell, three p sub-shells and five d sub-shells and in alleighteen quantum states. Thus, it can accommodate 18 electrons.

The number of electrons that can be accommodated in shell can be calculated as follows.Consider a shell characterized by a principal quantum number n. For this value of n, the orbital(azimuthal) quantum number l can take integral values from 0 to n – 1. For each value of l, magneticquantum number ml assumes integrally spaced values from – l to + l i.e., in all 2l +1 values. Foreach value of ml, the spin quantum number takes two values +½ and –½. Thus, the total number ofquantum states is given by

1

0

2(2 1) 2[1 3 5 ........... (2 1)]n

l

l n−

=+ = + + + + −∑

2 (1 (2 1)2

nn

= + − = 2n2

n l ml ms Quantum states

3 0 0 +1/2 (3, 0, 0, 1/2)– 1/2 (3, 0, 0, –1/2)

1 –1 1/2 (3, 1, – 1, 1/2)– 1/2 (3, 1, –1, –1/2)

0 1/2 (3, 1, 0, 1/2)–1/2 (3, 1, 0, –1/2)

1 1/2 (3, 1, 1, 1/2)–1/2 (3, 1, 1, –1/2)

2 – 2 1/2 (3, 2, –2, 1/2)– 1/2 (3, 2, –2, –1/2)

– 1 1/2 (3, 2, –1, 1/2)– 1/2 (3, 2, –1, –1/2)

0 1/2 (3, 2, 0, 1/2)– 1/2 (3, 2, 0, 1/2)

1 1/2 (3, 2, 1, 1/2)– 1/2 (3, 2, 1, –1/2)

2 1/2 (3, 2, 2, 1/2)– 1/2 (3, 2, 2,–1/2)

2.3 ELECTRONIC CONFIGURATION OF ATOMS

The electronic configuration of atoms are governed by following rules:Aufbau’s principle: The word Aufbau means build up. According to this principle the first

electron in an atom occupies the quantum state with lowest possible energy and then the second electron


goes to the next quantum state having higher energy. The sequence of energy levels in increasingorder of energy is

1s < 2s < 2p < 3s <3p <4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d.Pauli’s exclusion principle: In 1925, A German Physicist Wolfgang Pauli enunciated this

fundamental principle, which governs the electronic configuration of complex atoms. This principle,known as Pauli’s Exclusion Principle states that “no two electrons in an atom can exist in the samequantum state or each quantum state is occupied by a single electron.

Hund’s rule: The filling of electron in various orbitals of a sub-shell take place according toHund’s rule. This rule states that electrons prefer to occupy separate orbitals so that they have parallelspins. In other words, the pairing of electrons will occur in any orbital of a given sub-shell when allthe available orbitals have one electron each. According to this rule the electronic configuration ofcarbon, nitrogen and oxygen atom will be as follows:

6C ↑↓ ↑↓ ↑ ↑ 1s 2s 2px 2py 2pz

7N 1 2 2 2 2x y zs s p p p

↑↓ ↑↓ ↑ ↑ ↑

8O 1 2 2 2 2x y zs s p p p

↑↓ ↑↓ ↑↓ ↑ ↑

2.4 MAGNETIC MOMENT OF ATOM

When a charged particle moves along a closed path or rotates about its own axis, an electric currentis associated with it. This current loop has magnetic moment. The magnetic moments of electrondue to orbital and spin motions are related to their corresponding angular momenta.

Consider an electron moving with velocity v in a circular orbit of radius r. The orbital currentassociated with this motion is

I2

ev

r= −

π ...(2.4.1)

The magnetic moment associated with orbital motion is

( ) ( )2IA2 2 2

ev e er mvr

r m m µ = = − π = − = − π

L ...(2.4.2)

where L = mvr is the orbital angular momentum of electron. Eqn. (2.4.2) expresses the fact that

magnetic moment is associated with angular momentum of micro-particle. The minus sign indicatesthat the direction of the magnetic moment is opposite to that of the angular momentum. It is aremarkable fact that this classical result is also valid in quantum mechanics. The ratio of magnetic

Reddyvari Venugopal

Highlight


moment to the angular momentum L 2

e

m

µ = is called the gyromagnetic ratio.The magnetic moment

associated with orbital motion of electron can be written as

L| |

| | ( 1) ( 1)2 le

l l g l lm β βµ = − = −µ + = − µ +

L...(2.4.3)

where gl = 1, called orbital g-factor and

24 59.27 10 J/T 5.79 10 eV/T

2

e

m− −

βµ = = × = ×

is Bohr Magneton, a unit for measuring magnetic moment of atom.For purely quantum mechanical reason, the magnetic moment associated with spin motion is

related to its intrinsic (spin) angular momentum. The relation between them is

| |

2 ( 1)2 S

e eg s s

m m β µ = − = − µ +

ss ...(2.4.4)

where gS = 2, spin g-factor. Notice that the gyromagnetic ratio of spin motion is not (–e/2m) buttwice of it. For this reason the spin is said to have double magnetism.

2.5 LARMOR THEOREM

Consider an electron moving in a circular orbit. The orbital angular momentum l and correspondingmagnetic moment µ are mutually related through the relation

= – 2

e

m

l ...(2.5.1)

Let a magnetic field B be applied to the electron. In the magnetic field the electron experiencesa torque = B sin , where is angle between vector l and B. This torque cause the vector l and µto precess about the direction of the magnetic field B. The angular velocity of precession of l or µabout B is called Larmor frequency (l).

Let the torque cause a change in angularmomentum l by amount dl in time dt. Then

dl = dt

...(2.5.2)

Now refer to the Fig. (2.5.1). The magnitudeof change dl is given by

dl = l sin d

or dt = l sin d

or B sin dt = l sin d

whence Bd

dt l

θ µ= Fig. 2.5.1 Precession of vector l about B


B

2

d e

dt m

θ =

B

2le

mω = ...(2.5.3)

The result Eqn. (2.5.3) is known as Larmor theorem.

2.6 THE MAGNETIC MOMENT AND LANDE g-FACTOR FOR ONEVALENCE ELECTRON ATOM

Electrons in an atom are distributed in various sub-shells according to Pauli’s principle and Hund’srule. Every electron has orbital angular momentum and spin angular momentum. In a closed sub-shell, every electron is matched by another electron with opposite orbital and spin angular momenta.So the resultant angular momentum of a closed sub-shell is zero. Only the valence electrons outsidethe closed sub-shell contribute to the total angular momentum and the magnetic moment of the atom.The optical properties of atoms are all due to valence electrons only.

The magnetic moment of an atom with one valence electron is due to the single valence electron.Magnetic moments are associated with orbital and spin motion both and are given by

µl = – 2

e

m

l ...(2.6.1)

µs = – 2

e

m

s ...(2.6.2)

A schematic vector diagram ofmagnetic moments l and s andvectors l and s is shown in the figure.Due to double spin magnetism theresultant magnetic moment ls is notcollinear with the resultant j. The spin-orbit interaction causes vectors l and sto combine to form resultant vector j.The vectors l and s precess about thedirection of their resultant j. Thefrequency of precession is equal to theLarmor frequency.

The consequence of precession isthat only the component of ls parallelto j contributes to the magneticmoment of atom. The perpendicularcomponent averages out to zero Fig. 2.6.1. Vector addition of l, s, and ml , ms


Therefore µj = l cos + s cos

= cos 2 . cos2 2

e e

m m − θ + − ϕ

l s

= – ( 1) cos 2 ( 1) cos2

el l s s

m + θ + + ϕ

...(2.6.3)

The cosine formula for angles and are

2 2 2( 1) ( 1) ( 1)

cos2 2 ( 1) ( 1)

j j l l s s

j j l l

+ − + + + − +θ = =+ +

j l s

j l...(2.6.4)

2 2 2( 1) ( 1) ( 1)

cos2 2 ( 1) ( 1)

j j s s l l

j j s s

+ − + + + − +ϕ = =+ +

j s l

j s ...(2.6.5)

Substituting the expressions of cos and cos in Eqn. (2.6.3), we have

J( 1) ( 1) ( 1) ( 1) ( 1) ( 1)

22 2 ( 1) 2 ( 1)

j j l l s s j j s s l le

m j j j j

+ + + − + + + + − +µ = − + + +

3 ( 1) ( 1) ( 1)

( 1)2 2 ( 1)

e j j s s l lj j

m j j

+ + + − += − + +

( 1) ( 1) ( 1)

1 ( 1)2 2 ( 1)

e j j s s l lj j

m j j

+ + + − += − + + +

( 1)2

eg j j

m= − +

( 1)g j jβ= −µ + ...(2.6.6)

where ( 1) ( 1) ( 1)

12 ( 1)

j j s s l lg

j j

+ + + − += ++

...(2.6.7)

Thus the magnetic moment of an atom can be written as

J ( 1) ( 1)2 2

e eg g j j g j j

m m βµ = − = − + = −µ +j ...(2.6.8)


The projection of j onto z-direction is given by

( )J 2 2z j jz

e eg g m g m

m m βµ = − = − = −µ

j ...(2.6.9)

where mj = 0, ±1, ±2, ±3……i.e., mj can take on integrally spaced values from – j to + j. mj is

called the magnetic quantum number of the atom.

2.7 VECTOR MODEL OF ATOM

The quantum mechanical theory of atom is capable of providing satisfactory explanation of all atomicphenomena but the application of this theory to the interpretation of puzzling features of atomicspectra presents great mathematical complications. Before the advent of quantum mechanics, Landedeveloped an atomic model, called Vector Model, to explain the experimental observations made aboutatomic spectra and the behaviour of atom in magnetic field. This model is based on many conceptsand empirical rules, which had no theoretical justification at that time, but later after the advent ofquantum mechanics they got theoretical basis. This model not only provided the satisfactory explanationof many features of atomic spectra with amazing accuracy but also predicted many phenomena thatwere discovered later.

Electrons in an atom are distributed in various sub-shells according to Pauli’s principle andHund’s rule. Every electron has orbital angular momentum and spin angular momentum. In a closedsub-shell, every electron is matched by another electron with opposite orbital and spin angularmomenta. So the resultant angular momentum of closed sub-shell is zero. Only the valence electronsoutside the closed sub-shell contribute to the total angular momentum and the magnetic moment ofthe atom. Only valence electrons are responsible for the optical properties of atoms.

In this model, the orbital angular momenta and spin angular momenta and associated magneticmoments of valence electrons are treated as vectors. These angular momenta combine under theinfluence of two kinds of interactions to from resultant angular momentum for the atom as a whole.Since the interactions responsible for the coupling of angular momenta are of two types thereforethere are two types of coupling between the angular momenta.

Russell-Saunders Coupling or L-S Coupling

When the mutual repulsion between electrons due to their electrical charge is treated quantummechanically an unexpected result is found. The energy of the system contains two terms: onecorresponding to classical Coulomb interaction and the other is known as exchange interaction. Theexchange forces have no classical analogue but play an important role in atomic theory. One effectof exchange forces is to couple together the various spin vectors si to form a resultant S vector. Theorbital momenta li couple under the influence of electrostatic forces to from a resultant L. This methodof coupling of angular momenta is known as Russell-Saunders or L-S coupling. Finally, under theaction of spin-orbit interaction, the resultant orbital angular momentum L and resultant spin vectorS couple to form resultant angular momentum J for the atom as a whole. The L-S coupling may besummarized as follows.


(i) Coupling of orbital angular momenta

In light atoms containing many valence electrons, the electrostatic interaction between theelectrons is large and by virtue of this interaction the individual orbital angular momenta of valenceelectrons add up to form a resultant L. l1 + l2 + l3 +….= li = L ...(2.7.1)

where li are the orbital angular momenta of valence electrons and are given by

1 1 1 2 2 2| | ( 1) , | | ( 1) etc.l l l l= + = + l l

The magnitude of vector L is quantized and is given by

L(L 1)= + L

where L is total orbital quantum number and is determined by

1 2 3L ....l l l= ⊕ ⊕ ⊕ ...(2.7.2)

Here l1, l2, l3 stand for orbital quantum number of valence electrons and ⊕ for quantizedvector addition. For example, consider an atom with two valence electrons both in p sub-shell i.e.,l1 = 1, l2 = 1. Then

1 2L 1 1 0, 1, 2l l= ⊕ = ⊕ =

Here ( 1 2l l⊕ ) takes on all integrally spaced values from | l1 – l2 | to (l1+ l2). The allowed

values of magnitude of the total orbital angular momentum L of these two valence electrons are:

0(0 1) 0

1(1 1) 2

2(2 1) 6

= + =

= + =

= + =

L

L

L

The geometrical addition of orbital angular momenta of the two electrons with l1 = 1 andl2 = 1 is shown in the Fig. (2.7.1).

Fig. 2.7.1 Addition of orbital angular momenta of two electrons with l1 =1 and l2 = 1. Symbols with star assuperscript denote magnitude of corresponding vectors in units of


Fig. 2.7.2 Addition of orbital angular momenta of two electrons with l1 = 1 and l2 = 2

If one of the valence electrons is in p sub-shell and the other is in d sub-shell i.e., l1 = 1,l2 = 2 then

1 2 1 2 1, 2, 3

L(L 1) 2 , 6 , 12

l l= ⊕ = ⊕ =

= + =

L

L

The geometrical addition of angular momenta are shown in the Fig. (2.7.2).The total orbital angular momentum vector L can have only certain orientations in space. This

implies that its projection along any fixed direction (z-axis) can have only discrete values given by

Z LM= L ...(2.7.3)

where ML, called total orbital magnetic quantum number, can take on integrally spaced values from– L to L. In all ML can take (2L + 1) values.

(ii) Coupling of Spin Angular Momenta

Each electron has spin angular momentum. Because of strong quantum mechanical effect, knownas exchange interaction, which has no classical analogue, the spin angular momenta of valence electronsare coupled to form a resultant spin angular momentum vector S.

S = s1 + s2 + s3 +….. ...(2.7.4)

The magnitude of vector S is quantized and is given by

S(S 1)= + S

where S is total spin quantum number and is obtained from the following quantum sum:

1 2 3S .....s s s= ⊕ ⊕ ⊕

1 1 1

.....2 2 2

= ⊕ ⊕ ⊕ ...(2.7.5)


The direction of vector S is quantized and its projection along any fixed direction has discretevalues given by

Z SM= S ...(2.7.6)

where MS, called total magnetic spin quantum number, can take integrally spaced values from –S toS.

Let there be five valence electrons in an atom. The possible orientations of spin and thecorresponding values of total spin quantum number S are shown below.

1 1 1 1 1 5S

2 2 2 2 2 21 1 1 1 1 3

S2 2 2 2 2 21 1 1 1 1 1

S2 2 2 2 2 2

↑↑↑↑↑ = + + + + =

↑↑↑↑↓ = + + + − =

↑↑↑↓↓ = + + − − =

Corresponding to S = 5/2, the allowed values of MS are –5/2, –3/2, –1/2, 1/2, 3/2, 5/2. Similarlythe values of MS for S = 3/2 and S = 1/2 can be written.

(iii) Coupling of L and S

Now the total orbital angular momentum vector L and total spin angular momentum vector Sinteract magnetically through their associated magnetic moments and form a resultant J called thetotal angular momentum vector of the atom. This coupling is called spin-orbit coupling.

L + S = J ...(2.7.7)

The magnitude of vector J is quantized and is specified by

J(J 1)= + J

where J is total angular momentum quantum number of the atom. The allowed values of J are givenby

J L S= ⊕ ...(2.7.8)

i.e., J can take on integrally spaced values from L + S down to | L – S |.The direction of vector J is quantized. Its projection onto any axis is given by

Z JM= J ...(2.7.9)

where MJ called total magnetic quantum number of the atom. It can take integrally spaced valuesfrom –J to +J. In absence of external magnetic field, the total angular momentum J is conserved inmagnitude and direction. The effect of the internal torques can be only produce precession of L andS around the direction of their resultant J. However, in presence of external magnetic field B, theresultant vector J precesses about B while L and S continue to precess about their resultant J. If wecalculate the component of J along a specified direction, say z-direction, then z-component has awell-defined value but the other components of J viz. Jx and Jy are uncertain. This means that vectorsL and S precess about their resultant J as shown in the figure.


If L 2, and S 1 then J L S 2 1 1, 2, 3.

1 3 5 7If L 2, S 3 / 2 then J L S 2 3 / 2 , , , .

2 2 2 2

= = = ⊕ = ⊕ =

= = = ⊕ = ⊕ =

* *1 1 1 2 2 2( 1), ( 1)l l l l l l= + = + * *

1 1 1 2 2 2( 1), ( 1)s s s s s s= + = +

* ( 1)L L L= + * ( 1)S s s= +

Fig. 2.7.3 Vectors l1 and l2 precess about their resultant L. Vectors s1 and s2 precess about theirresultant S

Fig. 2.7.4 Vectors L and S precess about their resultant J. Vector J precesses about external magneticfield B


j-j Coupling

As the nuclear charge increases the magnetic spin orbit forces become stronger which dominatethe electrostatic interactions. The L-S coupling now breaks down. Under the influence of large spinorbit interaction, the orbital and the spin angular momenta of each electron couple to form a resultantangular momentum j. The resultant angular momentum j of each electron combine to form a resultantJ called total angular momentum vector of the atom. The spin-orbit coupling is primarily magneticin origin and arises from the magnetic moments of the orbital and spin motion of electrons. Thiscoupling is known as j-j coupling and is summarized below li + si = ji

ji + j2 + ….. = ji = JFor illustration of j-j coupling consider the electron configuration pd. For p electron, l1 = 1,

s1 = 1/2, therefore j = 1/2, 3/2. Similarly, for d electron, l2 = 2, s2 = ½, and j = 3/2, 5/2. Now j1and j2 may combine in four ways as describe below.

(i) j1 = 1/2, j2 = 3/2, J = j1 j2 = 1, 2.

(ii) j1 = 1/2, j2 = 5/2, J = j1 j2 = 2, 3.

(iii) j1 = 3/2, j2 = 3/2, J = j1 j2 = 0, 1, 2, 3.

(iv) j1 = 3/2, j2 = 5/2, J = j1 j2 = 1, 2, 3, 4.

It should be noted that L-S and j-j coupling both give the same number of terms and the sameJ values. The spectral term arising from j-j coupling are designated as ( j1, j2).

Fig. 2.7.5 Precession of j1 and j2 about their resultant J

For j-j coupling, L and S lose their meaning. The selection rules that operate in transitionsbetween terms arising from this type of coupling are:

j = 0, ± 1, j = 0, ± 1.


2.8 ATOMIC STATE OR SPECTRAL TERM SYMBOL

The state of an atom is characterized by quantum numbers L, S and J and is represented by a symbolaccording to the following scheme

L 0 1 2 3 4 5 Symbol S P D F G HHere S is not to be confused with total spin quantum number. The value of J is written as post

subscript and the multiplicity r (= 2S + 1 if S L and = 2L + 1 if L < S) as pre-superscript. Forexample if an atom is characterized by L = 2, S = 3/2 and J = 5/2 then it is designated as 4D5/2 ( r = 2S + 1= 3 + 1 = 4)

2.9 GROUND STATE OF ATOMS WITH ONE VALENCE ELECTRON(HYDROGEN AND ALKALI ATOMS)

The ground state configuration of valence electron is n s1. The valence shell has only one electronwith l = 0.

L = l = 0

S = s = 1/2

J L S 0 1/ 2 1/ 2= ⊕ = ⊕ =

Multiplicity r = 2L + 1 = 1

The multiplicity is equal to the number of sub-levels differing in their values of J. The groundstate symbol should be 1S1/2 but it is written as 2S1/2 because this term belongs to a system, which isdoublet. (In the excited state the valence goes to p or some other sub-shell. In all excited statesL > S and the multiplicity r = 2S + 1 = 2. To specify this fact that these terms belong the systemwhose excited states are doublet (r = 2) we write the ground state as 2S1/2 .

Excited states of alkali atoms: In the excited state the valence electron in an alkali atom goesfrom s to p state or other higher states. If it is in p state then

L = l1 = 1

S = s1 =1/2

r = 2S + 1 = 2

J1 3 1

1 ,2 2 2

= ⊕ =

Spectroscopic symbol 2P1/2, 2P3/2.


If the valence electron is promoted to d state then

L = 2, S = 1/2, J = 2 1 5 3

,2 2 2

⊕ = .

Multiplicity r = 2S + 1 = 2

Spectroscopic symbol 2D3/2, 2D5/2 .

Multiplet

Due to spin-orbit interaction each level characterized by an L value splits into a group ofsub-levels called fine-structure levels. The collection of fine-structure levels belonging to a givenL value is called a multiplet.

2.10 SPECTRAL TERMS OF TWO VALENCE ELECTRONS SYSTEMS (HELIUM AND ALKALINE-EARTHS)

I. L-S Coupling

Two non-equivalent electrons (n, l different)

In such atoms all the four quantum numbers of the two electrons are not identical. The methodof writing the spectral terms will be illustrated with examples.

(1) sp configuration

In this case l1 = 0, l2 = 1, L = l1 l2 = 0 1 = 1,

s1 = 1/2, s2 = 1/2 , S = s1 s2 = 1/2 1/2 = 0, 1

Singlet states (S = 0): J = L S = 1 0 = 1. Spectral term is 1P1.

Triplet states (S = 1): J = L S = 1 1 = 0, 1, 2. Spectral terms are 3P0, 1, 2

Fig. 2.10.1 Fine structure levels of sp configuration

Breit’s scheme: The same spectral terms can be obtained with the help of Breit’s schemeas illustrated below.

For the first electron, l1 = 0, ml1 = 0. ms1

= 1/2 , – 1/2 .


For the second electron, l2 = 1, ml2 = 1, 0, –1. ms2 = 1/2 , – 1/2 .

Write the values of ml of the two electrons in row and in column as shown in the table.In the similar way write the values of ms of the two electrons in row and in column.

From the table on the left we see that ML = 1, 0, –1. This implies that L = 1. TheL-shaped dotted line in the right table separates the two sets of values of MS, viz. MS = 0and MS = 1, 0, –1. From the values of MS we obtain the value of S which are S = 0 andS = 1.

Thus, we have L = 1 and S = 0 and 1. Therefore,

Singlet states (S = 0): J = L S = 1 0 = 1. Spectral term is 1P1.

Triplet states (S = 1): J = L S = 1 1 = 0, 1, 2. Spectral terms are 3P0, 1, 2

(2) pd configuration

In this case l1 = 1, l2 = 2, L = 1 2 = 1, 2, 3. (P, D, F terms)

s1 = 1/2 , s2 = 1/2 , S = 1/2 1/2 = 0, 1

Singlet states (S = 0):

(i) J = L S = 1 0 = 1

(ii) J = L S = 2 0 = 2

(iii) J = L S = 3 0 = 3

The spectral terms are: 1P1, 1D2, 1F3.

Triplet states (S = 1): (i) J = L S = 1 1 = 0, 1, 2. The spectral terms are: 3P0, 1, 2

(ii) J = 2 1 = 1, 2, 3. The spectral terms are: 3D1, 2, 3

(iii) J = 3 1 = 2, 3, 4. The spectral terms are: 3F2, 3, 4.

Breit’s scheme

l1 = 1, ml1 = 1, 0, –1 and l2 = 2, ml2 = 2, 1, 0, –1, –2. ms1

= 1/2 , –1/2 , ms2 = 1/2 , –1/2

.


In table on the left, the L-shaped dotted line separates the values of ML. From the allowedvalues of ML we find L = 1, 2, 3. Similarly from the table on the right, we get S = 0, 1.From the values of L and S we can easily find the J values and the spectral terms 1P1,1D2, 1F3, 3P0, 1, 2, 3D1, 2, 3 and 3F2, 3, 4.

Fig. 2.10 .2 Spectral terms of pd configuration

(3) np n' p configuration (non-equivalent electron)

In this case the principal quantum number n is different for the two electrons hence theyare non-equivalent electrons.

Here l1 = 1, l2 = 1, s1 = 1/2 , s2 = 1/2.


Allowed values of L = 0, 1, 2. Allowed values of S = 0, 1.

Singlet states (S = 0):

(i) J = 0 0 = 0

(ii) J = 1 0 = 1

(iii) J = 2 0 = 2

Singlet states are: 1S0, 1P1, 1D2.

Triplet states (S = 1):

(i) J = 0 1 = 1

(ii) J = 1 1 = 0, 1, 2

(iii) J = 2 1 = 1, 2, 3.

The spectral terms are: 3S1, 3P0, 1, 2, 3D1, 2, 3.

Fig. 2.10.3 Spectral terms of non-equivalent p-p configuration


Two equivalent electrons (n, l same)

p-p configuration: Two electrons having the same n and l quantum numbers are called equivalentelectrons. In the ground state of carbon atom (6C = 1s2 2s22p2), the two p-electrons are equivalentelectrons. In this case

l1 = 1, ml1 = 1, 0, –1, ms1

= 1/2 , – 1/2 .

l2 = 1, ml 2 = 1, 0, –1, ms2

= 1/2 , – 1/2 .

The Breit’s scheme for obtaining spectral terms is shown in the following tables.

According to Pauli’s principle, if (i) ms1 = ms2

then ml1 ≠ ml 2

and if (ii) ml1 = ml2

thenms1

≠ ms2. We consider the two cases one by one.

(i) ms1 = ms2

(parallel spins) and ml1 ≠ ml2

. In the left table, the numbers (2, 0, –2) alongthe diagonal, enclosed in the dotted curve, correspond to the case ml1

= ml2. So they must

be excluded. The remaining values of ML are 1, 0, –1. This gives L = 1(P state). Thetable on the right gives S = 0 and 1. The value S = 0 is excluded because this correspondsto opposite spins of electrons whereas we are considering the case of parallel spins(ms1

= ms2). The allowed values of J are given by

J = L S = 1 1 = 0, 1, 2.

The spectral terms are: 3P0, 1, 2.

(ii) ml1 = ml2

and ms1 ≠ ms2 (opposite spins). Allowed values of ML fall along the diagonal

of the left table and these are 2, 0, – 2. This gives L = 0 , 2. To exclude the values of MS

corresponding to parallel spins we omit the values MS = 1, –1. The remaining value ofMS is zero. This gives S = 0. The values of J are given by

(i) J = L S = 0 0 = 0. The spectral term is 1S0.

(ii) J = 2 0 = 2. The spectral term is 1D2.

So the spectral terms of pp configuration of equivalent electrons are1S0, 1D2, 3P0, 1, 2.


II. Spectral Terms in J-J Coupling

(1) p-d configuration of non-equivalent electrons

In this case

l1 = 1, s1 = 1/2 , j1 = l1 s1 = 1 1/2 = 1 3

,2 2

l2 = 2, s2 = 1/2, j2 = l2 s2 = 2 1/2 = 3 5

,2 2

(i) J = j1 j2 = 1 3

1, 22 2

⊕ =

(ii) J = 1 5

2, 32 2

⊕ =

(iii) J = 3 30, 1, 2, 3.

2 2⊕ =

(iv) J = 3 5

1, 2, 3, 4.2 2

⊕ =

Spectral terms are:

1, 2 2, 3 0, 1, 2, 3 1, 2, 3, 4

1 3 1 5 3 3 3 5, , , , , , ,

2 2 2 2 2 2 2 2 . There are 12 states.

(2) p-p configuration of non-equivalent electrons

l1 = 1, s1 = 1/2, j1 = 1 1/2 = 1/2, 3/2.

l2 = 1, s2 = 1/2, j2 = 1 1/2 = 1/2, 3/2.

(i) J = j1 j2 = 1/2 1/2 = 0, 1

(ii) J = 1/2 3/2 = 1, 2

(iii) J = 3/2 1/2 = 1, 2

(iv) J = 3/2 3/2 = 0, 1, 2, 3

Spectral states are:

0, 1 1, 2 1, 2 0, 1, 2, 3

1 1 1 3 3 1 3 3, , , , , , ,

2 2 2 2 2 2 2 2


(3) p-p configuration of equivalent electrons

l1 = 1, s1 = 1/2, j1 = l1 s1 = 1 1/2 = 1/2, 3/2.

l2 = 1, s2 = 1/2, j2 = 1 1/2 = 1/2, 3/2.

In j-j coupling the state of an electron is described by four quantum numbers n, l, j and mj.According to Pauli’s principle, at least one of the four quantum numbers of the two electrons mustbe different. The values of mj (ml1

ms1) may be obtained from Breit’s scheme.

From the left table mj1 = 1/2 , –1/2 and mj1

= 3/2, 1/2 , – 1/2 , –3/2. These values of mj1 give

j1 = 1/2 and 3/2. Similarly, the right table gives j2 = 1/2 and 3/2.

Possible combinations of j1 and j2 are

1 1 1 3 3 1 3 3, , , , , , ,

2 2 2 2 2 2 2 2

Of the four combinations, the two in the middle give identical terms and one of these must beomitted. So we are left with only three combinations

1 1 1 3 3 3, , , , ,

2 2 2 2 2 2

.

Determination of J for state 1 1

,2 2

with j1 = j2 and mj1 ≠ mj2

. (Pauli’s principle)

Diagonal terms must be excluded.

Spectral term is 0

1 1,

2 2

.


Determination of J for the state 1 3

,2 2

with j1 ≠ j2 (no need of Pauli’s principle)

MJ = 2, 1, 0, –1, –2 gives J = 2. MJ = 1, 0, –1, gives J = 1.

The spectral term is 1,2

1 3,

2 2

.

Determination of J for 3 3

,2 2

when j1 = j2 and mj1 ≠ mj2

(Pauli’s principle)

Diagonal terms must be excluded. MJ = 0, gives J = 0 and MJ = 2, 1, 0, –1, –2 gives J = 2

Spectral term is 0, 2

3 3,

2 2

.

2.11 HUND’S RULE FOR DETERMINING THE GROUND STATE OF AN ATOM

(i) Of the terms belonging to a given electronic configuration, the term with the greatest possiblevalue of S (highest multiplicity) and greatest possible value of L at this S will have the lowest energy.


(ii) The multiplet formed by equivalent electrons are normal i.e., the energy of the state growswith increase in the value of J if not more than half of the sub-shell is filled.

(iii) The multiplet formed by equivalent electrons are inverted i.e., the energy diminishes withan increase in J if more than half of the sub-shell is filled.

In other words when not more than half of a sub-shell is filled, the component of the multipletwith having minimum value of J has the lowest energy.

Equivalent Electrons with Closed Sub-shell

When the valence sub-shell of an atom is closed the atom has only one term symbol for which S = 0,L = 0, J = 0. The spectral term is 1S0.

For a closed sub-shell, 0, 0l sm m= =∑ ∑ . ( Example: for p6 , ml = 1, 1, 0, 0, –1, –1 and

ms = 1/2, –1/2, 1/2 –1/2, 1/2, –1/2.). That is, ML = 0, and MS = 0. This implies that L = 0, S = 0and J = 0.

The spectral terms of p4, and p5 are same as those of p6–4 = p2 and p6–5 = p1. Similarly, termsof d6, d7, d8 are same as those of d10–6 = d4, d10 – 7 = d3, d10 – 8 = d2.

2.12 LANDE g-FACTOR IN L-S COUPLING

In light element containing more than one valence electron, the electrostatic interaction betweenelectrons is larger than the spin orbit interaction. As a result of electrostatic interaction the individualorbital angular momenta of electrons combine vectorially to give a resultant L. The electrons in theclosed shells do not contribute towards the resultant L. So only valence electrons need to be considered.

L = l1 + l2 + …… ...(1.12.1)

Where |l1| = [l1(l1 + 1)]1/2 , |l2| = [l2 (l2 + 1)]1/2,…l1, l2 are the orbital angular quantum

numbers of valence electrons. The magnitude of resultant orbital angular momentum of atom is givenby

| | L(L 1)= + L ...(2.12.2)

where L = 1 2 .....l l⊕ ⊕ The symbol stands for quantized sum.

Similarly, the spin angular momenta of valence electrons add up vectorially to form a resultant S. S = s1 + s2 +…... ...(2.12.3)

where |s1| = [s1(s1 + 1)]1/2, |s2| = [s2(s2 + 1)]1/2

, s1 = 1/2, s2 = 1/2. The magnitude of S is givenby

|S| = [S(S + 1)]1/2 ...(2.12.4)

The total spin quantum number S is given by

1 11 2 2 2

S ...... .....s s= ⊕ = ⊕ ⊕ ...(2.12.5)

Now total orbital angular momentum L and total spin angular momentum S combine to form aresultant angular momentum J.


L S = J ...(2.12.6)

The magnitude of J is given by

| | J(J 1)= + J ...(2.12.7)

where the total angular quantum number J is given by

J L S= ⊕ ...(2.12.8)

This type of coupling of angular moment is called L-S or Russell-Saunders coupling. Like Land S, the associated magnetic moments also add up to form a resultant. The magnetic moment ofan atom is given by

( ) ( )L S 22 2

e e

m mµ = µ + µ = − + = − +

L S J S ...(2.12.9)

The projection of µ onto J is

( )

J

+ . +

2 2

e e

m m

⋅ µ = = − = −

J S JJ J. J J.SJ J J

µ...(2.12.10)

Now L . L = (J – S) . (J – S) = J . J + S . S – 2 J . S

J . S = J . J + S . S L . L

2

−

( )

2Je

m

− −µµ = = −

1J. J J . J + S . S L . L. J 2

J J

2 2 2 21J(J 1) J(J 1) S(S 1) L(L 1)

22 J(J 1)

e

m

+ − + + + − += −

+

J(J 1) S(S 1) L(L 1)1 J(J 1)

2 2J(J 1)

e

m

+ + + − += − + + +

J(J 1)2

eg

m= − +

J (J 1)2

eg

m= − +

J(J 1)gβ= −µ + ...(2.12.11)


where

J(J 1) S(S 1) L(L 1)

12J(J 1)

g+ + + − += +

+ ...(2.12.12)

g is called Lande g-factor or spectroscopic splitting factor.The g-factor depends on the atomic state (i.e., on L, S, J). For pure orbital motion S = 0,

L = J and hence g = 1. For pure spin motion L = 0, S = J and hence g = 2. The g-factor can also becalculated as follows.

Aliter

The relation between orbital angular momentum vector L, spin angular momentum vector Sand their resultant total angular momentum vector J is depicted by vector diagram as shown in thefigure. Also shown are the associated magnetic moments on the same diagram. Because of doublemagnetism of spin motion, the resultant of L and S, which has been depicted as µatom, is notcollinear with J. The projection of atom onto the direction of J is J. Let LJ and SJ be the anglesdefined in the figure. From the geometry of the figure we have

J L LJ S SJcos cosµ = µ θ + µ ϕ

LJ SJcos 2 cos2 2

e e

m m= − θ − ϕL S

2

e

m= −

LJ SJL(L 1) cos 2 S(S 1) cos + θ + + ϕ ...(2.12.13)

The cosine formula for angles LJ and SJ are

2 2 2

LJJ (J 1) L(L 1) S(S 1)

cos2 2 J(J 1) L(L 1)

+ − + + + − +θ = =+ +

J L S

J L ...(2.12.14)

2 2 2

SJJ (J 1) S(S 1) L(L 1)

cos2 2 J(J 1) S(S 1)

+ − + + + − +ϕ = =+ +

J S L

J S ...(2.12.15)

Substituting the expressions of cos LJ and cos SJ in Eqn. (2.12.13), we have

JJ(J 1) L(L 1) S(S 1) J(J 1) S(S 1) L(L 1)

22 2 J(J 1) 2 J(J 1)

e

m

+ + + − + + + + − +µ = − + + +

3J(J 1) S(S 1) L(L 1)

J(J 1)2 2J(J 1)

e

m

+ + + − + = − + +


J(J 1) S(S 1) L(L 1)

1 J(J 1)2 2J(J 1)

e

m

+ + + − += − + + +

J(J 1)2

eg

m= − +

J(J 1)gβ= −µ + ...(2.12.16)

where J(J 1) S(S 1) L(L 1)

12J(J 1)

g+ + + − += +

+

Thus, the magnetic moment of an atom can be written as

J J J(J 1) J(J 1)2 2

e eg g g

m m βµ = − = − + = −µ +...(2.12.17)

Fig. 2.12.1 Addition of µL and µS to form µatom

The projection of µ J onto z-direction is given by

( )J J JM M2 2zz

e eg g g

m m βµ = − = − = −µ

J (2.12.18)

where MJ = 0, ±1, ±2, ±3……i.e., MJ can take on integrally spaced values from –J to +J. MJ is

called the magnetic quantum number of the atom.


2.13 LANDE g-FACTOR IN J-J COUPLING

In heavy atoms, the spin-orbit interaction between magnetic moments associated with orbital andspin motion of an electron is greater than the electrostatic interaction between orbital magneticmoments of valence electrons and that between spin magnetic moments. As a result of this the orbitalangular momentum l and spin angular momentum s of the same electron combine to form a resultantj. Now these j’s of valence electrons combine to form a resultant J. This type of coupling of angularmomenta is called j-j coupling.

1 1 1 2 2 2 .l + s = j ,l + s = j ..... ...(2.13.1)

1 2 ......+ + = Jj j ...(2.13.2)

Consider two valence electrons with angular momentum j1 and j2 and associated magneticmoments µ1 and µ2. The resultant magnetic moment in the direction of J is equal to the sum ofcomponents of µ1 and µ2 parallel to J.

J 1 1 2 2cos( , J) cos( , J)j jµ = µ + µ

= 1 2 2| | cos( , J) | | cos( , J)2

eg g j

m+ 1 1 2j j j

= 12 2 2

eg

m

− −

2 2 2 2 2 21 2 2 1

1 2 21 2

+ +| | + | |

| || | | || |

J j j J j jj g j

j J j J

= 12 2 2

eg

m

− −

2 2 2 2 2 21 2 2 1

2+ +

+| | | |

J j j J j jg

J J

= 1 | J |2 2 2

eg

m

− −

2 2 2 2 2 21 2 2 1

22 2

+ ++

| | | |

J j j J j jg

J J

= 1 J(J 1)2 2 2

eg

m

− −+

2 2 2 2 2 21 2 2 1

22 2

+ ++

| | | |

J j j J j jg

J J

= 1 J(J 1)2 2 2

eg

m

− −+

2 2 2 2 2 2

1 2 2 122 2

+ ++

| | | |

J j j J j jg

J J

= 1 J(J 1)2 2

gβ − −

µ +

2 2 2 2 2 21 2 2 1

22 2

+ ++

| | | |

J j j J j jg

J J


= J(J 1)gβµ + ...(2.13.3)

where

g = 1 22 2

g g− −

+2 2 2 2 2 2

1 2 1 22 2

+ +

| | | |

J j j J j j

J J...(2.13.4)

where g1 and g2 are the Lande g-factors of individual electrons and are given by

1 1 1 1 1 11

1 1

( 1) ( 1) ( 1)1

2 ( 1)

j j s s l lg

j j

+ + + − += +

+.

2 2 2 2 2 22

2 2

( 1) ( 1) ( 1)1

2 ( 1)

j j s s l lg

j j

+ + + − += +

+.

2.14 ENERGY OF AN ATOM IN MAGNETIC FIELD

When an atom possessing magnetic moment is placed in a uniform magnetic field, it experiences atorque equal to µ . B. Referred to a zero of potential energy when µ and B are perpendicular to eachother, the potential energy in an arbitrary orientation is given by – µ . B. Thus, in the magnetic fieldan atom acquires an extra energy – µ . B. If E0 is the energy of an atom in absence of magneticfield, the energy in presence of magnetic field is

0E = E − ⋅Bµ ...(2.14.1)

If z-axis is chosen in the direction of the magnetic field i.e., B = B z , then the energy of theatom can be written as

0

0

E E B

E J B2

z

ze

gm

= − µ

= − −

0 JE M B2

eg

m = +

0 JE BM2

eg

m= +

0 JE BMg β= + µ ...(2.14.2)

where MJ = J, J – 1, ……0 …….–(J – 1), – J. Since MJ can take on 2J + 1 values, an atomic energylevel is splits into 2J + 1 equally spaced sub-levels as shown in the Fig. (2.14.1). The splitting of anenergy level results from the interaction of magnetic field with the magnetic moment of the atom. Itis evident from Eqn. (2.14.2) that an atomic level with g = 0 does not split at all. For example, the


state 4D1/2 has L = 0, S = 3/2, J = 1/2 and g = 0. Similarly for a state with S = 0 (called singlet), nosplitting occurs. 1S state is an example of this case. In magnetic field the separation of two adjacentsub-levels is gµB and the total splitting is given by

E 2 BJg β∆ = µ ...(2.14.3)

Fig. 2.14.1 Splitting of an energy level in magnetic field

An energy level characterized by L = 0, S = 1/2 , J = 1/2 splits into two levels, one with isparallel and the other with anti-parallel to field B. The level with µ parallel to B has minimumenergy and lies deepest.

2.15 STERN AND GERLACH EXPERIMENT (SPACE QUANTIZATION):EXPERIMENTAL CONFIRMATION FOR ELECTRON SPIN CONCEPT

The confirming evidence of space quantization of angular momentum (and hence of magnetic moment)came from the celebrated atomic beam experiment of Stern and Gerlach (1922), which was originallydevised to measure the magnetic moment of individual silver atoms. This experiment also providesan evidence for the spin hypothesis of electron. A well-defined beam of silver atoms was obtainedby evaporating silver in a hot oven and letting the atoms through a series of holes as shown in theFig. (2.15.1). The beam of silver atoms was allowed to pass through an inhomogeneous magneticfield B, which was produced between a sharp edged and a flat faced pole piece of a magnet. Theemergent beam was received on photographic plate. The geometry of the experimental set up is shownin the figure. The magnetic field acts in z-direction and the atomic beam enters the field alongx-axis.

In silver atom, the angular momentum and hence the magnetic moment comes from the spin ofvalence electron. Let µ be the magnetic moment of silver atom. In an inhomogeneous magnetic fieldhaving gradient in z-direction a magnetic dipole with magnetic moment µ experiences a translationalforce Fz in z-direction.

B B

F cosz z z z

∂ ∂= µ = µ θ ⋅∂ ∂ ...(2.15.1)


where is the angle that magnetic moment vector makes with the field B. Classically the magneticmoment µ can take all possible orientations and hence is a continuous variable. Atoms for whichcos is positive, will be pulled up and those for which cos is negative, will be pulled downward.Atoms whose magnetic moments are perpendicular to the magnetic field will be subjected to no forceand hence they will go straight. Atoms with µ parallel to B will suffer maximum upward deflectionand those with anti-parallel to B will suffer maximum downward deflection. Thus, the beam ofsilver atoms, after emerging the field B will spread out; the spreading of atoms will be proportionalto the z-component of µ. Thus, the classical physics predicts a smeared out pattern in vertical directionon the photographic plate. Stern and Gerlach, however, observed that the beam of silver atoms wassplit into two distinct parts giving two separate lines arranged symmetrically with respect to the traceof the beam obtained in absence of magnetic field.

A beam of ions cannot be used because the ions will be acted upon by Lorentz force in magneticfield and hence the beam will suffer deflection due to this force also and unnecessary complicationwill arise.

Quantum Mechanical Explanation

As stated the entire magnetic moment of silver atom comes from the spin of its valence electron.The spin of silver atom is 1/2. In magnetic field, the angular momentum and hence the magneticmoment can have only two orientations, parallel and anti-parallel to B. Atoms with parallel to Bare deflected upward and those with µ anti-parallel to B are deflected downward and hence the beamgets split into two parts. Thus, only two traces are expected. This is observed in the experiment. Thenumber of traces on the photographic plate depends on the angular momentum (spin) of the atom. Inexperiments with beams of aluminium, copper and alkali metals two traces were obtained. Vanadium,nitrogen, halogens gave four, oxygen five, manganese six, iron nine and cobalt 10 traces. Mercuryand magnesium gave a single trace at the central position. This means that these atoms have no magneticmoments.

Fig. 2.15.1 Stern-Gerlach experiment


2.16 SPIN ORBIT INTERACTION ENERGY

The classical theory of atom assumes that the electron moves in the Coulomb field of the stationarynucleus. In electron’s rest frame, the nucleus moves around the electron. The circulation of nucleusaround the electron is equivalent to a current loop, which produces a magnetic field B at the site ofelectron. The intrinsic magnetic moment (µ = – (e/m) s) of electron interacts with the magnetic fieldB. This interaction is known as the spin-orbit interaction and leads to a structure in energy spectrum,known as fine structure. The change in energy of atom due to this interaction is called spin orbitinteraction energy. We shall now derive an expression for this energy.

The spin-orbit interaction causes the orbital angular momentum l and spin angular momentums of electron to combine to form a resultant j and this type of coupling is called L-S coupling. Weshall see that spin orbit interaction causes splitting of energy levels, and hence the splitting of spectrallines.

Consider an electron moving in a circular orbit of radius r around the nucleus with velocity v.In electron’s rest frame, the nucleus is moving in a circular orbit of radius r in the direction oppositeto that of the electron with the velocity –v. The current associated with the motion of nucleus producesmagnetic field at the location of electron. This magnetic field B is given by

0 0 2( )

c

×= −µ ε × = − EB E

vv ...(2.16.1)

where E is electric field of nucleus, v is velocity of electron. Making use of the result

1 V

r r

∂= −∂

E r , 0

ZV

4

e

r=

πε

We can write

( )2 2

1 V 1 V

r rc r mc r

∂ ∂= − × = −∂ ∂

B r v l ...(2.16.2)

where l = r × mv is orbital angular momentum of electron. The electron possesses spin magneticmoment given by

e

m= −

sµ ...(2.16.3)

The spin magnetic moment interacts with magnetic field B, the corresponding magnetic energyis

E .ls = − Bµ

We call this energy spin orbit interaction energy. Substituting the values of B and , we have

( )2

2 2 30

1 Z 1E

4lse

m c r

= ⋅ πε l s ...(2.16.4)


where we have put 0

ZV

4

e

r=

πε.

So far our calculation was non-relativistic. The relativistic correction imparts the electron aprecessional motion about the nucleus. The effect of this precession, called Thomas Precession, isthat the magnetic field seen by electron is only half as large as the one assumed in the forgoingderivation. With this correction, the spin orbit interaction energy becomes

2

2 2 30

1 Z 1 1E ( )

2 4lse

m c r

∆ = ⋅ πε l s ...(2.16.5)

The magnitude of spin-orbit energy correction is very small in comparison with the total energyof the electron. It may be regarded as a small perturbation. The Hamiltonian operator correspondingto this spin-orbit correction is

2

2 2 30

ˆ ˆ1 1 ZH . . .

2 4

e

m c r=

πεl .s

If the wave function of electron for the state characterized by quantum numbers n, l, j is n, l, j

then the average value of spin-orbit energy is given by

2

*2 2 3

0

ˆ ˆ1 ZE .

2(4 )ls n l j n l je

dm c r

∆ = ψ ψ τπε ∫ l .s

...(2.16.6)

Now (l + s)2 = j . j

| l |2 + | s |2 + 2. l . s = | j |2

The average values of l2, s2 and j2 are l (l + 1) 2, s( s + 1 ) 2 and j ( j + 1 ) 2. Therefore

2( 1) ( 1) ( 1)

2 2

j j l l s s− − + − + − += =

2 2 2| j | | l | | s |l . s ...(2.16.7)

The average value of 1/r3 in a state characterized by quantum numbers n, l, j is given by

3

3 3 3 3 10 2

1 1 Z( )

( )( 1)n l j d

r r a n l l l∗= ψ ψ τ =

+ +∫ ...(2.16.8)

In view of results Eqns. (2.16.7) and (2.16.8) the expression for spin orbit interaction energybecomes

2 2 4

. 2 2 3 3 10 0 2

1 Z ( 1) ( 1) ( 1)E

4 4 ( )( 1)ls

e j j l l s s

m c a n l l l

+ − + − +∆ =πε + +

2 43 1

2

1 ( 1) ( 1) ( 1)R Z

2 ( )( 1)

j j l l s sch

n l l l

+ − + − += α+ +

, l ≠ 0, ...(2.16.9)


The corresponding change in term value is

( )2 4

3 12

E 1 R ZT ( 1) ( 1) ( 1)

2 ( 1)ls

ls j j l l s shc n l l l

∆ α∆ = − = − + − + − + + + ...(2.16.10)

or T ( 1) ( 1) ( 1)2lsa

j j l l s s∆ = − + − + − + ...(2.16.11)

where

2 4

3 1( )( 1)

2

R Za

n l l l

α=+ + ...(2.16.12)

The term value of an energy level, taking spin orbit energy into consideration, is T = T0 + Tls

where T0 is the term value of some reference level (or hypothetical level). If Tls is positive, theshift of level is downward and if Tls is negative, the shift is upward with respect to the referencelevel. The splitting of states with the same n is called fine structure.

We shall illustrate it with the help of examples.

1. Fine Structure of Doublet 2P1/2 and 2P3/2

For the first spectral term, l = 1, s = 1/2 , j = 1/2.

( )1 3 1 3T . 1(1 1) .

2 2 2 2 2

aa

∆ = − − + − = ↓ For the second spectral term, l = 1, s = 1/2 , j = 3/2.

( )3 5 1 3T . 1(1 1) .

2 2 2 2 2 2

a a ∆ = − − + − = − ↑

Fig. 2.16.1

2. Fine Structure of Doublet 2D3/2 and 2D5/2

For state 2D3/2, s = 1/2 , l = 2, j = 3/2.

Therefore T = 3a/2 ( )

For the state 2D5/2 , s = 1/2 , l = 2, j = 5/2.

Therefore T = – a ()

Fig. 2.16.2


3. Fine Structure of Doublet 2F5/2 and 2F7/2

For the state 2F5/2 s = 1/2 , l = 3, j = 5/2.Therefore T = 2a ()For state 2F7/2 s = 1/2 , l = 3, j = 7/2.Therefore T = – 3a/2 ()

2.17 FINE STRUCTURE OF ENERGY LEVELS IN HYDROGEN ATOM

The spin orbit interaction energy is

2 43 1

2

( 1) ( 1) ( 1)1E R Z , 0

2 ( )( 1)ls

j j l l s sch l

n l l l

+ − + − +∆ = α ≠+ +

...(2.17.1)

For one electron atom, s = 1/2, j = l s = l 1/2 = l + 1/2 and l – 1/2. Therefore,

2 4

, 1/ 2 3 12

1 1E R Z

2 ( )( 1)ls j l ch

n l l= +∆ = α

+ + ...(2.17.2)

and 2 4

, 1/ 2 3 12

1 1E R Z

2 ( )ls j l ch

n l l= −∆ =− α

+ ...(2.17.3)

The spin orbit interaction is not the only effect that contributes to the fine structure. Two otherfactors which add to the spin orbit energy are: relativistic effect and self-energy effect. The relativisticeffect arises due to increase in electron mass and the self-energy effect due to interaction of electronwith its own electromagnetic field. The change in energy due to latter effect is called Lamb shift.

The relativistic increase in electron mass gives rise to the following expression for the changein energy.

2 4

3 12

R Z 1 3E

4rch

nln

α∆ = − − +

...(2.17.4)

n and l are the principal and orbital quantum number of electron.The total energy shift due to spin orbit interaction and relativistic effect for j = l + 1/2 is

E E Els r∆ = ∆ + ∆

2 4

3 1 12 2

R Z 1 1 3

42( )( 1)

ch

nl l ln

α = − − + + + 2 4

3 12

R Z 1 3

4

ch

njn

α= − − +

...(2.17.5)

The expression for the total energy shift for state j = l – ½ also comes out to be the same.The corresponding term value is

2 4

3 12

R Z 1 3T

4njn

α∆ = − +

...(2.17.6)

where j = 1/2, 3/2, 5/2, ….. n – 1/2.

Fig. 2.16.3


Equation (2.17.6) shows that for a given value of n, the total correction term depends on j andeach level with l > 0 is split into two levels, the level of higher j having the higher energy. Therefore,the states having the same value of j but different l values have the same energy. For example, thestates 2 2S1/2 and 2 2P1/2 are degenerate. Other degenerate pairs are (3 2S1/2, 3 2P1/2) and(3 2P3/2, 3 2D3/2). The spin orbit energy shift and relativistic corrections add up in such a way thatfinally the states 2S1/2 and 2P1/2 are degenerate for a given n = 2, 3, 4, …

In 1947, Lamb and Rutherford observed a very small splitting of 0.033 cm–1 between the energiesof 2S1/2 and 2P1/2 in hydrogen atom. This shift is called Lamb shift. The cause of this shift is theinteraction of the electron with its own electromagnetic field (self energy).

Since R2 = 1.097 × 105 cm–1 × (1/137)2 = 5.84 cm–1, the shift of energy level n = 1,(l = 0, s = 1/2) 1 2S1/2 is given by

T = 3 1

2

5.84 1 3

4njn

−

+ = 1.46 cm –1

For n = 2, l = 0, 1, s = 1/2 . For l = 0, j = 1/2, and for l = 1, j = 1/2 and 3/2. Therefore, thereare three states 2 2S1/2, 2 2P1/2, 2 2P3/2. The states 2 2S1/2 and 2 2P1/2 shift by equal amount givenby

1

1 12 2

5.84 1 3T 0.456 cm

8 8−

∆ = − = +

The state 2 2P3/2 shifts by amount

1

3 12 2

5.84 1 3T 0.09125 cm

8 8−

∆ = − = +

For n = 3, l = 0, 1, 2 and s = 1/2. There are five states 3 2S1/2, 3 2P1/2, 3 2P3/2, 3 2D3/2 and3 2D5/2. The states 3 2S1/2 and 3 2P1/2 shift by equal amount given by

1

1/ 2 1 12 2

5.84 1 3T 0.1622 cm

27 12j−

=

∆ = − = +

Without Lamb shift these states are degenerate. The states 3 2P3/2 and 3 2D3/2 also shift byequal amount and hence are degenerate without Lamb shift. The energy shift is given by

13/2 3 1

2 2

5.84 1 3T 0.054 cm

27 12j−

=

∆ = − = +

Similarly, the state 3 2D5/2 shifts by amount

T = – 0.018 cm–1

The fine structure splitting of energy levels n = 1, n = 2 and n = 3 for hydrogen atom areshown in the figure.


The cumulative effect of these corrections is to split the energy level n = 2 into three components(2 2P1/2, 2 2S1/2, 2 2P3/2).

Fine structure of energy level n = 3(Relativistic + Spin orbit + Lamb shift)

The cumulative effect of these corrections is to split the energy level n = 3 into five components (3 2S1/2, 3 2P1/2,

3 2P3/2, 3 2D3/2, 3 2D5/2).

Fig. 2.17.1 Fine structure of first three energy levels of hydrogen atom


2.18 FINE STRUCTURE OF H LINE

The H line of hydrogen spectrum results from the transition of electron from the energy levelcorresponding to n = 3 to the energy level n = 2. The entire state of the atom is determined by itssingle valence electron.

Corresponding to n = 2, there are two sub-levels, s sub-level (l = 0) and p sub-level (l =1).When electron is in s sub-level (l = 0).

L = l = 0S = s = 1/2

J L S 0 1/2 1/2= ⊕ = ⊕ =This state is represented by 2S1/2.When the electron is in p sub-level (l = 1)

L = l = 1S = s = 1/2

J L S 1 1/2 3/2,1/2= ⊕ = ⊕ =This state is represented by 2P3/2, 2P1/2.Corresponding to n = 3 there are three sub-levels, s, p and d sub-levels. When the electron is in

s sub-level (l = 0)L = l = 0S = s = 1/2

J L S 0 1/2 1/2= ⊕ = ⊕ =The corresponding state is 2S1/2.When the electron is in p sub-level (l = 1)

L = l = 1S = s = 1/2

J L S 1 1/2 3/2, 1/2= ⊕ = ⊕ =

and the corresponding states are 2P3/2 and 2P1/2.When the electron is in d sub-level (l = 2)

L = l = 2S = s = 1/2

J L S 2 1/2 5/2,3/2.= ⊕ = ⊕ =The corresponding states are 2D5/2, 2D3/2.The energy levels corresponding to n = 3 and n = 2 are shown in the Fig. (2.18.1). It can be

shown that a state with lower value of J has smaller energy than the state with higher value of J. Inall fifteen transitions are possible but selection rules limit their number. Allowed transitions are thosein which L changes by ±1 or J changes by 0 or ±1, i.e.,

L = ± 1, J = 0, ± 1 (allowed)The selection rules permit only seven transitions. The allowed transitions are:2D5/2 2P3/2, 2D3/2 2P1/2,

2P1/2 2S1/2, 2D3/2 2P3/2,

2P3/2 2S1/2, 2S1/2 2P3/2, 2S1/2 2P1/2.


Fig. 2.18.1 Fine Structure of H line (6563 Å). Lines marked 4, 4 are coincident. This is also true for lines marked 5, 5.

It was Lamb who in collaboration with Rutherford in 1947, discovered that in hydrogen-likeatom, for a given value of n, the levels with the same value of J but L values differing by unity arenon-degenerate. In fact, he detected a small difference of 0.0353 cm–1 between the levels 22S½ and22 P½. This shift in energy is called Lamb Shift. Lamb shift of amount 0.0105 cm–1 is also observedbetween levels 32 S1/2 and 32 P1/2. Because of very small magnitude of Lamb shift, the resultingsplitting of spectral lines is normally not observed.

2.19 FINE STRUCTURE OF SODIUM D LINES

The D lines of sodium spectrum result from the transitions of electron from 3p to 3s level. In theground state of Na atom the valence electron lies in 3s level. In this state

L = l = 0S = s = 1/2

J L S 0 1/2 1/2= ⊕ = ⊕ =The ground state is denoted by 2S1/2. When the valence electron is excited to 3p level

L = l = 1S = s = 1/2

J L S 1 1/2 3/2,1/2= ⊕ = ⊕ =This state is denoted by 2P3/2, 2P1/2.The energy level diagram of 3s and 3p levels are shown in the Fig. (2.19.1). Three transitions

from upper level to lower level are possible. Selection rules L = ± 1, J = 0, ± 1 allow only twotransitions. The D1 line ( = 5896 Å) originates from the transition 2P1/2 2S1/2 and D2 line( = 5890 Å) from the transition 2P3/2 2S1/2.

Fig. 2.19.1 Fine Structure of sodium spectrum (origin of D1 and D2 lines)


2.20 INTERACTION ENERGY IN L-S COUPLING IN ATOM WITH TWO VALENCE ELECTRONS

In an atom with two valence electrons there are four angular momenta l1, s1, l2 and s2 and hencethere are six terms for interaction energy corresponding to six combinations of these momenta viz.(s1, s2), (l1, l2), (l1, s1), (l2, s2), (l1, s2) and (l2, s1). The general expression for interaction energy in l-s coupling is

T | || | cos( , ) [ ( 1) ( 1) ( 1)]2 2ls lsa a

a l s j j l l s sΓ = −∆ = = − − = + − + − +2 2 2I s j l s ...(2.20.1)

where | | ( 1) ,| | ( 1) ,| | ( 1)l l s s j j= + = + = + l s j

and

2 2

3.

1( )( 1)

2

R Za

n l l l

α=+ + ...(2.20.2)

For convenience we shall make a change in notation as given below.

* * *( 1), ( 1), ( 1)l l l s s s j j j= + = + = +

* * *L L(L 1), S S(S 1), J J(J 1)= + = + = +Thus, the symbols l*, s* and j* represent the magnitude of orbital, spin and total angular momenta

in unit of and so on. In terms of new symbols the general expression for interaction energy is

( )* * * * *2 *2 *21T cos( , )

2ls a l s l s a j l sΓ = −∆ = = − − ...(2.20.3)

Now the interaction energies for two valence electrons can be expressed as

1Γ = * * * *1 1 2 1 2cos( )a s s s s = ( )*2 *2 *21

1 2S2

as s− − ...(2.20.4)

2Γ = * * * *2 1 2 1 2cos( )a l l l l = ( )*2 *2 *22

1 2L2

al l− − ...(2.20.5)

3Γ = * * * *3 1 1 1 1cos( )a l s l s = ( )*2 *2 *23

1 1 12

aj l s− − ..(2.20.6)

4Γ = * * * *4 2 2 2 2cos( , )a l s l s = ( )*2 *2 *24

2 2 22

aj l s− − ...(2.20.7)

5Γ = * * * *5 1 2 1 2cos( , )a l s l s = ( )*2 *2 *25

12 1 22

aj l s− − ...(2.20.8)

6Γ = * * * *6 2 1 2 1cos( , )a l s l s = ( )*2 *2 *26

21 2 12

aj l s− − ...(2.20.9)

The terms 5 6andΓ Γ are negligibly small and will be omitted. The spin-spin and orbital-orbital interactions are electrostatic in nature whereas spin-orbit interaction is magnetic in origin.


In L-S coupling the quantum mechanical exchange interactions between spin vectors s1 and s2

and between orbital vectors l1 and l2 predominate over the spin-orbit interactions between vectors land s. A consequence of this is that orbital vectors l1 and l2 precess more rapidly about their resultantL. This result also holds for spin vectors s1 and s2. Due to weaker spin orbit interaction, vectors Land S precess slowly about their resultant J. This means that 1 and 2 are greater than 3 and 4.The interaction energies 1 and 2 can be calculated from Eqns. (2.20.4) and (2.20.5).

To calculate 3 and 4, let us transform Eqns. (2.20.6) and (2.20.7) into a convenient form.The angles between l1 and s1 and between l2 and s2 continuously change. The average values ofcosine terms are then given by

* * * * * * * *1 1 1 1cos( , ) cos( ,L )cos(L ,S )cos(S , )l s l s= ...(2.20.10)

* * * * * * * *2 2 2 2cos( , ) cos( ,L )cos(L ,S )cos(S , )l s l s= ...(2.20.11)

Fig. 2.20.1

* * * * * * * * * * * *3 3 1 1 1 1 3 1 1 1 1cos( , ) cos( , L )cos(L , S )cos(S , )a l s l s a l s l sΓ = =

= *2 *2 *2 *2 *2 *2*2 *2 *2

* * 1 2 1 23 1 1 * * * * * *

1 1

L SJ L S

2 L 2L S 2 S

l l s sa l s

l s

+ − + −− −

= ( )*2 *2 *2 *2 *2 *2

*2 *2 *23 1 2 1 2*2 *2

L SJ L S

8 L S

a l l s s + − + −− −


Similarly,

( )*2 *2 *2 *2 *2 *2

*2 *2 *24 2 1 2 14 *2 *2

L SJ L S

8 L S

a l l s s + − + −Γ = − −

( )

*2 *2 *2 *2 *2 *21 2 1 2

3 *2 *2*2 *2 *2

3 4 *2 *2 *2 *2 *2 *22 1 2 1

4 *2 *2

S L

2S 2L1J L S

2 S L

2S 2L

s s l la

s s l la

+ − + − + Γ + Γ = − −

+ − + −

= *2 *2 *2

3 3 4 41

(J L S )2

a aα + α − −

= ( )*2 *2 *2AJ L S

2− − ...(2.20.12)

where A = 3 3 4 4a aα + α ...(2.20.13)

and 3α =*2 *2 *2 *2 *2 *2

1 2 1 2*2 *2

S L

2S 2L

s s l l + − + −

...(2.20.14)

4α =*2 *2 *2 *2 *2 *2

2 1 2 1*2 *2

S L

2S 2L

s s l l + − + −

...(2.20.15)

The term value of the state is

0 1 2 3 4T T= + Γ + Γ + Γ + Γ

= ( ) ( ) ( )*2 *2 *2 *2 *2 *2 *2 *2 *21 20 1 2 1 2

AT S L J L S

2 2 2

a as s l l+ − − + − − + − −

...(2.20.16)where T0 is the hypothetical (reference) level from which the shift of energy levels are measured.

Heisenberg, on the basis of quantum mechanical analysis, showed that a1 and a2 are negative,a3 and a4 are positive.

Splitting of sp Configuration in L-S Coupling

For this configurationl1 = 0, l2 = 1, s1 = 1/2 , s2 = 1/2 .L = 0 1 = 1,S = 1/2 1/2 = 0, 1,J = L S


For singlet state (S = 0), J = 1 0 = 1. 1P1.

For triplet state (S = 1), J = 1 1 = 0, 1, 2. 3P0, 1, 2.

(i) For singlet state S = 0

Calculation of 1 + 2 : (S = 0, l1 = 0, l2 = 1, s1 = 1/2, s2 = 1/2)

( )*2 *2 *11 1 2S

2

as sΓ = − −

1 133 31 1(0.1 . . )2 2 2 22 4

a a= − − = −

and ( ) ( )*2 *2 *22 22 1 2L 1. 2 0 .1 1. 2 0

2 2

a al lΓ = − − = − − =

It is worth to notice that for any configuration involving s-electron (l = 0) the interaction energy2 always comes out be zero.

General observations show that the singlet level lies above the corresponding triplet level. Thisindicates that the coefficient a1 in 1 should be negative. This was also shown by Heisenberg byquantum mechanical calculations. Therefore,

11 2

3

4

aΓ + Γ = −

So the singlet level (S = 0) shifts upward by 13

4

a− from the hypothetical reference level T0.

This is also supported by Hund’s rule: A term with highest multiplicity (hence highest value of S)will lie deepest, and of these the term with largest value of L will lie deepest.

Calculation of 3 + 4 : (S = 0, l1 = 0, l2 = 1, s1 = 1/2, s2 = 1/2.)

The coefficients 3 4andα α are given by

*2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *21 2 1 2 2 1 2 1

3 4*2 *2 *2 *2

S L S L. , .

2S 2L 2S 2L

s s l l s s l l+ − + − + − + −α = α =

Obviously, 3 = 4 = 0Therefore, for the singlet state 3 + 4 = 0. Hence this state does not split into components

due to spin-orbit interaction.

(ii) For triplet state S = 1

1 11 2 1 2, 0. ,

4 4

a aΓ = Γ = ∴Γ + Γ = a1 is negative.

The triplet states shift downward by amount 1 .4

a


Fig. 2.20.2

Now we shall calculate the splitting of this state due to spin-orbit interaction.(i) J = 0, L = 1, S = 1, 3P0

( ) ( )*2 *2 *23 4 4

A AJ L S 0 .1 1. 2 1. 2 2A

2 2aΓ + Γ = − − = − − = − = −

(ii) J = 1, L = 1, S =1, 3P1

( )*2 *2 *2 43 4

A AJ L S (1. 2 1 . 2 1. 2) A

2 2 2

aΓ + Γ = − − = − − = − = −

(iii) J = 2, L = 1, S = 1, 3P2

43 4

A(2 . 3 1. 2 1. 2) A

2 2

aΓ + Γ = − − = = , a3 and a4 are positive.

Splitting of triplet level is shown in Fig. (2.20.2).

2.21 INTERACTION ENERGY IN J-J COUPLING IN ATOM WITH TWOVALENCE ELECTRONS

For j-j coupling the interaction energies are given by

* * * *1 1 1 2 1 2cos( )a s s s sΓ = ...(2.21.1)

* * * *2 2 1 2 1 2cos( )a l l l lΓ = ...(2.21.2)

* * * *3 3 1 1 1 1cos( , )a l s l sΓ = ...(2.21.3)

* * * *4 4 2 2 2 2cos( , )a l s l sΓ = ...(2.21.4)

* * * *5 5 1 2 1 2cos( , )a l s l sΓ = ...(2.21.5)

* * * *6 6 2 1 2 1cos( , )a l s l sΓ = ...(2.21.6)

The terms 5 and 6 are negligibly small and will be omitted. In j-j coupling the spin-orbitinteraction are much stronger than spin-spin and orbit-orbit interaction and therefore the interactionenergies 3 and 4 predominate over the terms 1 and 2. The angle between l1 and s1 and between


l2 and s2 are fixed but the angle between s1 and s2 and between l1 and l2 continuously change. Sothe average values of the changing angles should be taken.

1, 2 1 1 1 2 2 2cos( ) cos( , )cos( , )cos( , )s s s j j j j s= ...(2.21.7)

1 2 1 1 1 2 2 2cos( , ) cos( , ) cos( , ) cos( , )l l l j j j j l= ...(2.21.8)

Now,

*2 *2 *2 *2 *2 *2* *1 1 1 2 2 21 2* * * * * *

1 1 1 2 1 1 2 2

cos( )2 2

j s l j s lj j

a s s s j s j

+ − + − Γ =

*2 *2 *2 *2 *2 *2

* * * *1 1 1 2 2 22 2 1 2 1 2* * * *

1 1 2 2

cos( )2 2

j l s j l sa l l j j

l j l j

+ − + −Γ =

1 2Γ + Γ =*2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2

* *1 1 1 2 2 2 1 1 1 2 2 21 2 1 2* * * *

1 2 1 2

cos( )2 2 2 2

j s l j s l j l s j l sa a j j

j j j j

+ − + − + − + −+

=

*2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *21 1 1 2 2 2 1 1 1 2 2 2 1 2

1 2*2 *2 *2 *21 2 1 2

J

22 2 2 2

j s l j s l j l s j l s j ja a

j j j j

+ − + − + − + − − −+

or *2 *2 *2

1 2 1 2B

J2

j j Γ + Γ = − − ...(2.21.9)

where

*2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *21 1 1 2 2 2 1 1 1 2 2 2

1 2*2 *2 *2 *21 2 1 2

B .2 2 2 2

j s l j s l j l s j l sa a

j j j j

+ − + − + − + −= +

...(2.21.10)

or 1 1 2 2B a a= β + β ...(2.21.11)

where 1 = *2 *2 *2 *2 *2 *21 1 1 2 2 2

*2 *21 2

.2 2

j s l j s l

j j

+ − + −

...(2.21.12)

and 2 = *2 *2 *2 *2 *2 *21 1 1 2 2 2

*2 *21 22 2

j l s j l s

j j

+ − + −

...(2.21.13)


The interaction energies 3 and 4 are given by

( )* * * * *2 *2 *233 3 1 1 1 1 1 1 1cos( )

2

aa l s l s j l sΓ = = − − ...(2.21.14)

( )* * * * *2 *2 *244 4 2 2 2 2 2 2 2cos( )

2

aa l s l s j l sΓ = = − − ...(2.21.15)

The total shift is

( )( )*2 *2 *21 1 2 2 1 2

1T J

2a a j j∆ = β + β − − + ( )*2 *2 *23

1 1 12

aj l s− − ( )*2 *2 *24

2 2 22

aj l s+ − −

...(2.21.16)

Interaction Energy in sp Configuration (j-j coupling)

l1 = 0, s1 = 1/2 , j1 = 0 1/2 = 1/2 .l2 = 1, s2 = 1/2 , j2 = 1 1/2 = 1/2 , 3/2.J = j1 j2

(i) J = 1/2 1/2 = 0, 1(ii) J = 1/2 3/2 = 1, 2.

The spectral terms are 0, 1 1, 2

1 1 1 3, and ,

2 2 2 2

For * 2 *2 *2 *2 *23 4

3 4 1 1 1 2 2 21 1

, , ( ) ( )2 2 2 2

a aj l s j l s

Γ + Γ = − − + − −

= ( ) ( )3 44

3 3 3 31 1 1 10 1.22 2 2 2 2 2 2 22 2

a aa− − + − − = −

For 43 4

1 3, ,

2 2 2

a Γ + Γ =

(i) Calculation of 1 + 2 for state ( )0

1 1,2 2

s1 = 1/2 , s2 = 1/2 , l1 = 0, l2 = 1, j1 = 1/2 , j2 = 1/2 , J = 0

1 2Γ + Γ = *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *2 *21 1 1 2 2 2 1 1 1 2 2 2 1 2

1 2*2 *2 *2 *21 2 1 2

J

22 2 2 2

j s l j s l j l s j l s j ja a

j j j j

+ − + − + − + − − −+

= 1

4

a

(ii) Calculation of 1 + 2 for state 1

1 1,

2 2

s1 = 1/2 , s2 = 1/2 , l1 = 0, l2 = 1, j1 = 1/2 , j2 = 1/2 , J = 1

1 + 2 = 1

12

a− , a1 is negative


(iii) Calculation of 1 + 2 for state 1

1 3,

2 2

s1 = 1/2, s2 = 1/2, l1 = 0, l2 = 1, j1 = 1/2, j2 = 3/2, J = 1

1 + 2 = 15

12

a− , a1 is negative.

(iv) Calculation of 1 + 2 for state 2

1 3,

2 2

s1 = 1/2, s2 = 1/2 , l1 = 0, l2 = 1, j1 = 1/2, j2 = 3/2, J = 2

1 + 2 = 1

4

a, a1 is negative.

Fig. 2.21.1

2.22 LANDE INTERVAL RULE

Spin-orbit interaction energy is given by

E . . ,ls a∆ = L S where a is a constant.

Now, L . S is given by

21 1

J(J 1) L(L 1) S(S 1)2 2

= − − = + − + − + 2 2 2L . S J L S

Therefore interaction energy becomes

Els = 2J(J 1) L(L 1) S(S 1) A J(J 1) L(L 1) S(S 1)2

a + − + − + = + − + − +

where A is a constant. The fine structure levels are characterized by the same value of L and S but


differ in J values. The energy difference between two fine structure levels corresponding to two valuesof J viz. J and J + 1 is

J 1 JE E A (J 1)(J 2) J(J 1) 2A J 1+ − = + + − + = + This is the mathematical statement of Lande interval rule. It can be stated as follows:The energy interval (spacing) between two fine structure levels of a multiplet characterized by

J and J + 1 is proportional to the larger of the two J-values of the levels.Let us find the ratio of intervals between the fine structure levels 3P0, 3P1, 3P2. The interval

between the first pair of levels is proportional to larger J-value 1 and the interval between the lastpair of levels is proportional to 2. Thus, the ratio of energy interval is 1 : 2.

Similarly, the ratio of intervals between the levels 3D1, 3D2, 3D3 is 2 : 3 and that between thelevels 4D1/2, 4D3/2, 4D5/2, 4D7/2 is 3 : 5 : 7.

SOLVED EXAMPLES

Ex. 1. Write down the spectral designations of the terms of the hydrogen atom whose electron is inthe state with principal quantum number n = 3.

Sol. Hydrogen atom has a single electron. For n = 3, l = 0, 1, 2 and s = 1/2. ThereforeL = l = 0, 1, 2.S = 1/2,

J = L ⊕ S= (0 1/2), (1 1/2), (2 1/2)= 1/2, (3/2, 1/2), (5/2, 3/2).

Multiplicity, r = 2S + 1 = 2 (except for L = 0)

Spectral terms 2S1/2, 2P1/2, 2P3/2 2D3/2, 2D5/2.

Ex. 2. Write the spectroscopic notation of the following states: (a) L = 0, S = 0, J = 0 (b) L = 2, S = 0, J = 5/2, (c) L = 3, S = 1/2, J = 5/2, (d) L = 4,

S = 1, J = 5.Sol. (a) 1S0, (b) 1D2, (c) 2F5/2 (d) 3G5.

Ex. 3. Find the values of S, L and J in the following states. 1S0 , 3P2 , 2D3/2 , 5F5 , 6H5/2.Sol.

State L S = (r – 1)/2 J1S0 0 0 03P2 1 1 22D3/2 2 1/2 3/25F5 3 2 56H5/2 5 5/2 5/2


Ex. 4. Find the allowed values of total angular momenta of electron shells of atom in the states 4Pand 5D.

Sol.

State r S = ( r – 1)/2 L J = L S ( )J J J 1= +

4P 4 3/2 1 5/2, 3/2, 1/235 15 3

, ,2 2 2

5D 5 2 2 4, 3, 2, 1, 020, 12, 6,

2,0

Ex. 5. Write the spectral terms of atoms possessing besides filled shells(a) two electrons, one in s and the other in p

(b) two electrons one in p and the other in d.

Sol. (a) for s electron l1 = 0, and for p electron l2 = 1.

L = l1 l2 S = s1 s2 r = 2S + 1 J = L S Spectral Terms

0 1=1 1/2 1/2 1, 3 1 0 = 1 1P1

= 0, 1 1 1 = 2, 1, 0 3P2, 3P1, 3P0

(b) For p electron l1 = 1 and for d electron l2 = 2

L = 1 2 = 1, 2, 3S = 1/2 1/2 = 0, 1J = L S

(i) J = 1 0 = 1, r = 2S + 1 = 1, 1P1

(ii) J = 2 0 = 2, r = 2S + 1 = 1, 1D2

(iii) J = 3 0 = 3, r = 2S + 1 = 1, 1F3

(iv) J = 1 1 = 2, 1, 0, r = 2S + 1 = 3, 3P0, 3P1, 3P2

(v) J = 2 1 = 3, 2, 1, r = 2S + 1 = 3, 3D1, 3D2, 3D3

(vi) J = 3 1 = 4, 3, 2, r = 2S + 1 = 3, 3F2, 3F3, 3F4.

Ex. 6. How many different types of terms can a two electron system consisting of d and f electronspossess?

Sol. For d electron l1 = 2, and that for f electron l2 = 3.L = l1 l2 = 2 3 = 5, 4, 3, 2, 1.S = s1 s2 = 1/2 1/2 = 0, 1r = 2S + 1 = 1, 3.J = L S


(i) J = 5 0 = 5

(ii) J = 4 0 = 4

(iii) J = 3 0 = 3

(iv) J = 2 0 = 2

(v) J = 1 0 = 1. These are singlet terms.

(vi) J = 5 1 = 6, 5, 4.

(vii) J = 4 1 = 5, 4, 3.

(viii) J = 3 1 = 4, 3, 2.

(ix) J = 2 1 = 3, 2, 1.

(x) J = 1 1 = 2, 1, 0.

These are triplet terms.

Ex. 7. What is the ground state of atoms composed of filled sub-shells?Sol. The z-component of total orbital angular momentum of an atom is given by

LMz = L

where ML = ml, summation is carried over all electrons. Since the values of ml lie in the range – l,(– l + 1), ….– 1, 0, 1, …..(l – 1) , l,

therefore ml = 0. Thus L 0z = . This implies that L = 0.

Similarly, the z-component of total spin angular momentum is given by

SMz = S

where MS = ms . In a closed sub-shell, electron are paired withopposite spins hence MS = ms = 0. This means that S = 0.Thus, for a closed sub-shell, L = 0, S = 0 and hence J = 0. So, the ground state of the atom is

1S0.

Ex. 8. Obtain L . S in terms of L, S, J. Calculate the possible values of L . S for L = 1 and S = 1/2.

Sol.J = L + S

2 2 2J = (L + S) . (L + S) = L + 2L . S + S

( )2 2 21

2= − −L . S J L S

21J(J 1) L(L 1) S(S 1)

2= + − + − +

For L = 1, S = 1/2, J = L S = 1 1/2 = 3/2, 1/2.


For J = 3/2, L . S = 21

2

For J = 1/2, L . S = – 2

Ex. 9. Find the angle between vectors l and s in 2P3/2 state of one electron atom.

Sol. *2 *2 *2

* *

( 1) ( 1) ( 1)cos

2 ( 1) ( 1)2

j l s j j l l s s

l l s sl s

− − + − + − +θ = =+ +

Given that l = 1, s = 1/2, j = 3/2.

3 5 31. 1.2 .2 2 2 2 1cos .

3 612 1.2. .2 2

− −θ = =

Ex. 10. An atom in the state 2P3/2 is located in an external magnetic field of 1.0 kg. Find g-factor,difference of energies of adjacent levels after splitting and frequency of Larmor precession.

Sol. For the state 2P3/2, L = 0, S = 1/2, = 3/2, g = 4/3. Possible values of Mj are – 3/2, – 1/2,1/2 , 3/2. The level splits into 4 sub-levels. The spacing of adjacent sub-levels is

2

B B B4

E B (0.1 Wb/m ) 0.1333

g∆ = µ = × µ = µ

where B = 9.27 × 10 – 24 J/T.

Larmor frequency19 2

9L 31

B 1.6 10 C 0.1 Wb/m1.4 10 Hz.

4 4 3.14 9.1 10 kg

e

m

−

−× ×ν = = = ×

π × × ×

Ex.11. Determine the maximum separation of a beam of hydrogen atoms that moves a distance of20 cm with a speed of 2 × 10 5 m/s perpendicular to a magnetic field whose gradient is 2 × 10 2 T/m.Neglect the magnetic moment of proton.

Sol. For hydrogen atom sz = ms .Resolved part of magnetic moment in the direction of magnetic field

1

B B22 2 ( )

2 2z s s se e

g m mm m

µ = − = − = − µ = −µ

Force on the atom 24 2B

B B| F | (9.27 10 J/T)(2 10 T/m)z z

d d

dz dz−= µ = µ = × ×

= 1.85 × 10 –21 N

Displacement of beam

22 F1 1

2 2z

zl

z a tm v

∆ = ± = ±

= 221

27 5

1 1.85 10 N 0.20 m

2 1.67 10 kg 2 10 m/s

−

−

× ± × × = 5.54 × 10 –7 m

Total separation = 2 61.11 10 mz −∆ = × .


Ex. 12. Calculate the magnitudes of orbital, spin and total angular momenta and also the anglebetween l and s for a p electron in a one-electron atom.

Sol. For p electron l = 1, s = 1/2.

3

| | ( 1) 2 , | | ( 1)4

l l s s= + = = + =

l s

1 1 3

1 ,2 2 2

j l s= ⊕ = ⊕ =

| | ( 1)j j= + j = 3 15

,4 4

Angle between l and s

2 2 2| | | | | | ( 1) ( 1) ( 1)cos

2 | || | 2 ( 1) ( 1)

j j l l s s

l l s s

− − + − + − +θ = =+ +

j l sl s

For l = 1, s = 1/2, j = 3/2 . 1

cos , 666

θ = θ =

For l = 1, s = 1/2, j = 1/2 1

cos 2 , 145 .6

θ = − θ =

Ex. 13. Calculate the two possible orientations of spin vector of an electron in a magnetic field of0.5 T. Also calculate the separation of the energy levels.

Sol. Magnitude of spin vector 31 1| | ( 1) ( 1)2 2 4

s s= + = + = s

Projection of s onto the magnetic field

12

coszs s= θ = ±

112cos , 54.7 and 125.3

| | 3 34

zs ±θ = = = ± ⇒ θ = θ =

s

In magnetic field, the energy level of electron is split into two components with separation

E 2 B 2 2 (0.5) 2g β β β∆ = µ = × × µ = µ .

Ex. 14. Show that for hydrogen atom (or one electron atom) the term separation of spin-orbitdoublet is given by

4

3

ZT 5.84

( 1)n l l∆ =

+Sol. The term value of spin-orbit interaction energy is given by

2 4*2 *2 *2

3 12

E R ZT

2 ( )( 1)ls

ls j l shc n l l l

−∆ α ∆ = = − − − + +


For one electron atom, s = 1/2, j = l s = l ± 1/2.

For j = l + 1/2, j*2 – l*2 –s*2 = l

For j = l – 1/2, j*2 – l*2 – s*2 = – (l + 1)

2 4

31/2 12

R ZT

2 ( )( 1)ls j l n l l= +

α∆ = − + +

And

2 4

31/2 12

R ZT

2 ( )ls j l n l l= −

α∆ = +

The separation of the doublet is

2 4 4 41 1

3 3 3

R Z Z Z( T ) 584 m 5.84 cm

( 1) ( 1) ( 1)ls

n l l n l l n l l− −αδ ∆ = = =

+ + +Thus, the level splitting for one electron atom (H and alkali atoms)

(i) increases with increasing atomic number Z

(ii) decreases with increasing principal quantum number

(iii) decreases with increasing l value

The splitting is zero for l = 0 (s-state).

Ex. 15. If the doublet splitting of the first excited state, 2 23/ 2 1/ 22 2P P− of He+ is 5.84 cm–1.

Calculate the corresponding separation for hydrogen atom.Sol. The term separation (doublet splitting) of a state in one electron atom due to spin-orbit

interaction is given by

2 4

3

R Z( T)

( 1)n l l

αδ ∆ =+

Since 4( T) Z we haveδ ∆ ∝

41 1HH H

H4 4H H

( T)( T) Z 1 5.84( T) cm 0.365cm .

( T) 16 16Z 2e

e e+ +

− −δ ∆δ ∆= = ⇒ δ ∆ = = =

δ ∆

Ex.16. Calculate the spin-orbit interaction splitting of a level corresponding to n = 2, l = 1 of thehydrogen atom.

Sol. 4

1 1 13

Z 5.84 1( T) 5.84 cm cm 0.365cm .

8 1(1 1)( 1)n l l− − −×δ ∆ = = =

× ++Ex. 17. Write down the spectral terms of carbon atom in the normal and first excited state. Indicate

the allowed transitions.

Sol. In the normal state 6 2 2 2( C 1 2 2 )s s p the carbon atom has two equivalent optical electrons.

The spectral terms are 3 3 3 1 10 1 2 2 0P , P , P , D , S . In the excited (2p 3s) the spectral terms are

3 3 3 10 1 2 1P , P , P , P . The allowed transitions are shown in the Fig. E-17.


Fig. E-17

Ex. 18. In an atom obeying L-S coupling, the components of a normal triplet state have separations20 cm–1 and 40 cm–1 between adjacent components. There is a higher state for which the separations are22 cm–1 and 33 cm–1 respectively. Determine the terms for the two states and show with the help of anenergy level diagram the allowed transitions and the pattern of the spectrum.

Sol. Calculation of L, S, J for the lower triplet state.Let the J values of the lower triplet state are J, J + 1 and J + 2. According to Lande interval

rule,

J 1 20

J 2 40

+ =+

From this we get J = 0. Therefore, the J-values of the lower triplet state are J = 0, 1, 2. Nowthe values of J are given by

| L – S |, | L – S | + 1, ………………………(L + S)This implies that | L – S | = 0 and L + S = 2.

(i) Let L > S, then L – S = 0 and L + S = 2. This gives L = 1 and S = 1 and multiplicityr = 2S + 1 = 3 (triplet). The states are 3P0, 1, 2.

(ii) Let S > L, then S – L = 0 and L + S = 2. This gives L = 1, and S = 1.


Calculation of L, S, J for the Higher State

According to Lande rule J 1 22

J 2 33

+ =+

. This gives J = 1. Therefore, the J-values of the successive

states of higher level are J = 1, 2, 3.Since J-values are given by

| L – S |, ………………………L + SIf L > S then |L – S| = 1 and L + S = 3. From these equations, we get L = 2 and S = 1.The states are 3D1, 2, 3.If S > L then S – L = 1 and L + S = 3. From these equations, we get S = 2, and L = 1. Given

that the higher state is triplet (S = 1). So this state does not exist.Allowed transitions and pattern of spectrum are shown in the Fig. E-18.

Fig. E-18

Ex. 19. Assuming j-j coupling derive the spectral terms of 4p4d configuration.Sol. For p electron: l1 = 1, s1 = 1/2 , j1 = 1/2 , 3/2.

For d electron: l2 = 2, s2 = 1/2 , j2 = 3/2, 5/2.Possible combination of j1 and j2 are ( 1/2, 3/2 ), ( 1/2, 5/2 ), ( 3/2, 3/2 ), ( 3/2, 5/2 ).Of these, (1/2, 3/2) lies lowest and (3/2, 5/2) highest. Each of these four levels is further splits

by residual electrostatic interaction and spin-spin correlation into a number of J-levels. The J-valuesof the four levels are given below.

(1/2, 3/2) J = 1, 2(1/2, 5/2) J = 2, 3(3/2, 3/2) J = 0, 1, 2, 3.(3/2, 5/2) J = 1, 2, 3, 4.


Fig. E-19

Ex. 20. Derive the spectral terms of oxygen atom in normal state.Sol. O8 : 1s2 2s2 2p4

The terms of p4 configuration are the same as those of p2 configuration.The terms are 1S0, 1D2 and 3P0, 1, 2.According to Hunds rule, the terms of highest multiplicity lie lowest. These are 3P0, 1, 2.

Since the valence sub-shell is more than half filled, the terms of the triplet will be inverted.

Fig. E-20


1. Calculate Lande g-factor for the energy levels 1S0, 1P1, 2S1/2, 2P3/2.

2. Discuss L-S coupling scheme for a two valence electron system and find the expression for the interactionenergy. Show schematically the fine structure for the electron configuration sp.

3. Write the normal electronic configuration of carbon atom and obtain spectral terms arising from it. Alsowrite the first excited configuration of C atom and obtain the resulting spectral terms.


4. Find the angle between l and s vectors in 2P3/2 , 2D5/2 states of one electron system.

5. Find out the different angular momentum states for a d-s electron configuration in L-S and j-j coupling inground state.

6. The ground state of Cl is 2P3/2. Find its magnetic moment.

7. Deduce an expression for the change in energy due to spin-orbit interaction.

Explain that the relative splitting of 2P level is greater than 2D level.

8. (a) Explain what is meant by L-S and j-j coupling in atoms having more than one electron.

(b) The quantum numbers of two electrons in a two-valence electron atom are n1 = 6, l1 = 3, s1 = 1/2and n2 = 5, l2 = 1, s2 = 1/2. Assuming (i) L-S coupling find the possible values of L and J. (ii) j-j

coupling find the possible values of J.

9. Calculate the values of l, s, j for d electron of sodium atom. What are the spectroscopic symbols ford electron?

10. Deduce an expression for spin-orbit interaction energy for D lines of sodium atom.

11. Describe the different types of coupling schemes found in the spectra of two valence electron system. Findexpressions for their interaction energies.

12. Show that for p-s configuration the total 3P separation is the same in both L-S and j-j coupling schemes.

13. Deduce the spectral terms for a 3p4p configuration in both L-S and j-j couplings. Show that the samenumber of states is obtained under both coupling schemes.

14. Calculate the spectral terms arising from p2 configuration in L-S coupling.

15. (a) In a lithium atom the valence electron is in the state n = 3. What is the maximum value of angularmomentum in this state?

(b) For a p2 configuration in L-S coupling, 1S0, 1D2, and 3P0, 1, 2 states are obtained. Which one is

ground state?

(c) In a certain state, the angular momentum of the atom is 2 , spin is 2 and the Lande g-factor is zero.

Find the term symbol.

16. Calculate the energy states for p2 configuration in j-j coupling.

17. (a) In a lithium atom the valence electron is in the state n = 3. What is the angular momentum of electron

in this state? [35

4h

πAns. ]

(b) For a p2 configuration in L-S coupling, 1S0, 1D2, and 3P0, 1, 2 states are obtained. Which one is the

ground state?

(c) Amongst the following transitions, which one is forbidden?2S1/2 2P1/2, 1S0 1D2, 1P1 1D2, 3S1 3P0

[Ans. 1S0 1D2 (Selection rules: L 1, J 0, 1∆ = ± ∆ = ± )]

(d) In a state the angular momentum of the atom is 2 , spin is 2 and Lande’s g-factor is zero. Write

down the state. [Ans. 5F1]

18. Calculate the energy states for p2 configuration in j-j coupling.

19. From a two electron configuration 3F4 state is obtained. What is the magnetic moment of the atom in this

state? [Ans. 5 5 24, 9.27 10 J/TB B

2 2

e

m

−µ µ = = ×

]


20. Using Breit scheme derive the spectral terms arising from p2 configuration in L-S coupling.

21. In sodium atom the first member of principal series consisting of two lines of wavelengths 5890 and5896Å arises from 3p 3s transition. This happens due to splitting of 3p level into two levels 32P3/2 and32P1/2 due to spin orbit interaction. Find the value of spin orbit coupling coefficient in cm–1.

22. Write out the allowed terms for the atoms possessing, in addition to filled sub-shells, two p electrons withdifferent principal quantum numbers. [Ans. 1D2, 1P1, 1S0, 3D1 , 2 , 3, 3P0, 1 , 2 , 3S1]

23. D term consists of five components. What are the multiplicities of this term?

[Ans. 5, 6, 7…]

24. Find the possible multiplicities of the terms S0, P2, D3/2, F1/2.

[Ans. (1), (3, 5, 7), (2, 4, 6, 8), (6, 8)]

25. Write out the possible terms of atoms with following configurations:

(i) 2s2 (ii) 2p3s [Ans. (i) 1S0 (ii) 1P1, 3P2, 3P1, 3P0]

26. What is the minimum possible value of total angular momentum of lithium atom whose valence electron is

in a state n = 3. Write the symbol of the corresponding state.35 2, D5/ 22

Ans.

27. An atom is in a state whose multiplicity is 4 and has angular momentum 63

2= J . What values can the

quantum number L of this state have? [Ans. L = 2, 3, 4, 5]

28. Find the Lande g-factor in the following states: 4D1/2, 5F1, 7H2. [Ans. g = 0 in all the states]

29. Obtain the magnetic moment of the atom in the following states: 3S1, 1P0, 1P1, 4D1/2, 5F1, 7H2

[Ans. 2 ,0,2 2 , 0, 0, 0µ µβ β ]

30. Calculate the magnetic moment of hydrogen atom in the ground state. Hint: L = 0, S = 1/2, J = 1/2, g = 2.

[Ans. 3µβ ]

31. Find µ and the allowed values of the projection of an atom in the state 1F and 2D3/2.

[Hint: In the first case L = 3, S = 0, J = 3, g = 1, 2 3µ = µβ

In the second case L = 2, S =1/2, J = 3/2, g = 4/5, 3

25

µ = µβ ].

32. Write the spectral terms of atom with (i) S = 1/2, J = 5/2, g = 6/7

(ii) S = 1, L = 2, g = 4/3 [Ans. (i) 2F5/2 (ii) 3D3]

33. Calculate the Lande g-factor for atom (i) with one valence electron in S, P, D states (ii) with one electron in3P state (iii) with one electron in S state (iv) in the singlet state.

[Ans. (i) 2, (2/3, 4/3), (4/5, 6/5) (ii) 0/0 in 3P0 state, 3/2 in 3P1 and 3P3/2 state

(iii) 2 except in the singlet state for which g = 0/0 (iv) 1]

34. The valence electron of a sodium atom is in a state with n = 4. The values of other quantum numbers of the

electron are such that the atom has maximum possible value of the total angular momentum J . Determine

the magnetic moment of the atom.

[Hint: n = 4, l = 0, 1, 2, 3. Lmax= 3, Smax = 1/2, Jmax = 7/2, g = 8/7. Hence 4 63

7 βµ = µ ].


35. A carbon atom with electronic configuration 1s2 2s22p 3d has maximum possible total angular momentumat such a configuration. What is the magnetic moment of the atom in this state?

[Hint: L = 1 2 = 3, 2, 1. Smax = 1, Lmax = 3, Jmax = 4. Hence 5 5

2µ = µβ ]

36. Into how many components will the following terms split in magnetic field? 1S, 1P, 1D, 2D5/2.

[Ans. 3, 3, 5, 6]

37. A magnetic field of 1.0 T is switched on an atom in the following states: (i) 1P (ii) 2D5/2. Find the totalsplitting in electron volt.

[Hint: (i) J = 1, g = 1. This state splits into three components.

E = 2 g B = 1.6 × 10–4eV

(ii) J = 5/2, g = 6/5. This state splits into six components.

E = 5 g B = 3.47 × 104 eV].

38. Derive the Russel-Saunders terms for the configuration 2s 2p. [Ans. 3P2, 3P1, 3P0, 1P1 ]

39. Derive the spectral terms for the configuration 2p3p.[Ans. (3D3, 3D2, 3D1), (1D2), (3P2, 3P1, 3P0), (1P1), (1S0)]

40. The quantum numbers of two electrons in a two valence electron atom are

n1 = 6, l1 = 3, s1 =1/2

n2 = 5, l2 = 1, s2 = 1/2

Find the possible values of J (i) assuming L-S coupling scheme (ii) assuming j-j coupling scheme.

[Hint: (i) L = 2, 3, 4. S = 0, 1. For S = 0, J = 2, 3, 4 and for S = 1, J = 1, 2, 3; 2, 3, 4; 3, 4, 5.

(ii) j1 = 5/2, 7/2, j2 = 1/2, 3/2.

j1 + j2 = (2, 3), (3, 4), (1, 2, 3, 4), (2, 3, 4, 5)].

41. Write down the electronic configuration of an atom with Z = 21. Determine the values of l, s, j, L, S and Jfor the electron in the ground state.

[Hint: Sc (21) = 1s2 2s2 2p6 3s2 3p6 4s2 3d1

For 3d electron n =3, l = 2, s = ± 1/2. Hence j = 2 ± 1/2 = 5/2, 3/2.

L 5 / 2, 3 / 2, S 1/ 2, J L S 3 / 2 1/ 2and 5 / 2 1/ 2l s= ⊕ = = = ⊕ = ⊕ ⊕ ]

42. Derive the spectral terms of carbon atom in the normal state.

[Hint: The spectral terms are: 3P0, 3P1, 3P2, 1D2, 1S0. Since the outermost sub-shell is less than half filled,the triplet in normal. The terms of maximum multiplicity lie lowest. The ground state is 3P0.]

CHAPTER

ATOMIC SPECTRA-III

3.1 SPECTRA OF ALKALI METALS

Like the spectrum of hydrogen atom the emission spectrum of alkali metals consists of discrete lines asshown in the figure. The analysis of spectra requires diligent and patient study of a large mass ofwavelength data. Spectroscopists of 19th century were able to isolate four kinds of spectral series inthe spectrum of alkali metals. These are: Principal Series, Sharp Series, Diffuse Series and Fundamental(Bergmann) Series.

3.2 ENERGY LEVELS OF ALKALI METALS

A comparison of energy levels of an alkali metal with those of hydrogen atom shows that the energystates of former with higher values of l (viz, d and f ) are nearly equal to those of their equivalentsin hydrogen atom but there are considerable discrepancies at the lower values of l (such as S and Pstates). This can be seen in the energy level diagram. This discrepancy can be explained by usingGauss’s law and radial probability of valence electron by taking sodium as an example.

The electronic configuration of sodium is [1s2 2s2 2p6]3s1. The electric field at the location ofvalence electron due to nucleus (11 protons) and the remaining 10 electrons in the inner closed sub-shells is given by E = (1/40) (qeff /r

2) where qeff is the charge inside the Gaussian surface enclosing11 protons and 10 core electrons. Obviously qeff = 11e – 10e = + e. Thus, the valence electron ofsodium atom experiences the electric field of effective nuclear charge +e. In other words, we cansay that 10 electrons of the closed sub-shells screen the 10 protons of nucleus, leaving an effectivenet charge of +e, so Zeff = 1. In this model, the ionization potential of sodium would be

2

2 2

R Z (13.6eV) 1I

3

effch

n

×= = = 1.5 eV.

This value of ionization potential is much lower than the observed value 5.1 eV. In order toremove this large discrepancy and to retain the idea of principal quantum number, a new term, ,called quantum defect, is introduced in the expression for energy of electron. That is, the concept


of quantum defect is used to explain the increased binding energy of the electron. In terms ofquantum defect the energy of valence electron in alkali metals is written as

2 2 2 2

R R R RE or T

( ) ( )eff eff

ch ch

n n n n= − = − = =

− ∆ − ∆ ...(3.2.1)

The value of depends on the value of l of valence electron and is greatest for s-state (l = 0).The quantum defect is a measure of the penetration of electron into the sub-shells of inner electrons.For a given value of n, the value of decreases rapidly with increasing value of l. For this reasonthe state with larger value of l approaches to the corresponding state of hydrogen. The quantumdefect also depends on n but its variation with n is very slow. All these facts can be understood byfollowing arguments.

The variation of radial probability density P(r) of valence electron with distance r from nucleusfor electrons with different l is shown in the figure. We see that for l = n – 1, the maximum possiblevalue of l, P(r) has one peak, which gives the most probable distance of valence electron from thenucleus. For the electron with l = n – 2, P(r) has two peaks, and for the electron with l = n – 3,P (r) has three peaks and so on. In sodium atom P(r) has one peak for d electron (l = 2), two peaks,for p electron (l = 1) and three peaks for s electron (l = 0).

Fig. 3.2.1 Line spectrum of sodium : p principal series, s sharp series, d diffuse series. Dotted line indicates the position of series limit

Fig. 3.2.2 Radial probability density of valence electron in sodium atom. The shaded area represents theprobability for the ten core electrons. The 3d electron spends almost 100% of its time outside the core

atom. A 3p electron spends less time outside and an s electron even less.

Atomic Spectra-III 473

Fig. 3.2.3 Penetrating and non-penetrating orbits of valence electron

Thus, for a given value of n, the smaller the value of l, the valence electron has higher theprobability of finding itself closer to the nucleus. That is the probability of penetrating the core ofinner electrons is maximum when the valence electron is in s state. The valence electron in p statehas less chance of penetrating the inner core of electrons and d electron has lesser chance to do so.In other words, the s electron finds itself most often within the inner electron sub-shells. Therefore selectron is least shielded from nucleus and hence experiences highest effective nuclear charge and ismost tightly bound to the nucleus. This makes the energy of s electron most negative i.e., least. Forthis reason the s state in alkali atom is displaced downward by large amount from its equivalent inhydrogen atom. The p electron spends some of its time within the inner closed sub-shells and isshielded less and its energy is less negative. The d electron spends least of its time inside the core ofinner electrons so it is most shielded and hence experiences least effective nuclear charge. This iswhy as l increases, the downward shift of energy levels relative to their equivalents in hydrogendecreases.

We can arrive at the same conclusion by making use of the classical picture of electron orbits.The s electron moves in most eccentric elliptical orbit and penetrates all the inner orbits and it findsitself most often in the vicinity of the nucleus and hence it is most tightly bound. This causes itsenergy most negative. The p electron moves in less eccentric elliptical orbit and finds itself less oftenclose to the nucleus. So it is less tightly bound and hence its energy is less negative. The energy of pstate is greater than that of s state. The d electron moves in almost circular orbit and it finds itselfleast often near the nucleus and is least tightly bound. The energy of d electron is least negative i.e.,greater than that of p electron.

In the last decade of 19th century Rydberg showed that the wave number (1/) of a spectralline of a spectral series can be written as the difference of two spectral terms, of which one is fixedand the other is variable (running). [A term T and energy E of an atom are mutually related throughthe relation T = – E /ch]. In terms of principal quantum number n of valence electron, the termvalue is expressed by Eqn.(3.2.1).


Table 3.2.1: The values of quantum defect for sodium atom

State n = 3 n = 4 n = 5 n = 6

S 1.373 1.357 1.352 1.350

P 0.883 0.867 0.862 0.859

D 0.010 0.011 0.013 0.011

F ——— 0.000 0.001 0.008

Since the value of depends on the value of l, we shall denote it values by symbols S, P,D, F respectively.

3.3 SPECTRAL SERIES OF ALKALI ATOMS

The spectral lines of an alkali atom can be classified into four groups: Principal series, sharp series,diffuse series and fundamental series.

Principal series: The lines of principal series are observed in emission and absorption bothand are the brightest lines in the spectrum. The lines of this series are emitted then the optical electronmakes transition from P states to the ground state 3 2S1/2 . The wave numbers of lines of this seriesare given by

2 2

R R

(3 ) ( )p

s pnν = −

− ∆ − ∆, n = 3, 4, 5 ,……… ...(3.3.1)

The wave number of the series limit is

2

R

(3 )p

s∞ν =

− ∆ ...(3.3.2)

Therefore 2

R.

( )p p

pn∞ν = ν −

− ∆...(3.3.3)

Sharp series: The spectral lines of this series are very sharp in physical appearance and hencetheir name. These lines are emitted when the valence electron makes transition from higher S statesto the lowest P state. Their wave numbers are given by

2 2

R R,

(3 ) ( )s

p snν = −

− ∆ − ∆n = 4, 5, 6,….. ...(3.3.4)

2

R

(3 )s

p∞ν =

− ∆ ...(3.3.5)

2

R

( )s s

sn∞ν = ν −

− ∆ ...(3.3.6)


Fig. 3.3.1(a) Spectral series of sodium


Fig. 3.3.1(b) Principal series doublets in alkali atoms

Diffuse series: The lines of this series are blurred in comparison to the lines of other series.These are emitted when the valence electron undergoes transition from higher D states to 3P states.Their wave numbers are given by

2 2

R R,

(3 ) ( )d

p dnν = −

− ∆ − ∆ n = 3, 4, 5, ……. ...(3.3.7)

2

R

(3 )d

p∞ν =

− ∆ ...(3.3.8)

2

R

( )d d

dn∞ν = ν −

− ∆ ...(3.3.9)

Each line of this series consists of three components but under low resolution it appears asdoublet. It is in fact called compound doublet. The weakest component is called satellite.

Fundamental series: The lines of this series are very much like those of hydrogen spectrumand hence called fundamental. The wave numbers of these lines are given by

2 2

R R

(3 ) ( )f

d fnν = −

− ∆ − ∆, n = 4, 5, 6…… ...(3.3.10)

2

R

(3 )f

d∞ν =

− ∆ ...(3.3.11)

2

R

( )f f

fn∞ν = ν −

− ∆. ...(3.3.12)

Selection rules: Out of many possible transitions, only those are allowed which obey the selectionrules.

L 1, S 0, J 0, 1, (J 0 J 0 is forbidden)∆ = ± ∆ = ∆ = ± = → =


3.4 SALIENT FEATURES OF SPECTRA OF ALKALI ATOMS

Some of the important features of alkali spectra are as follows:(i) The separation of spectral line of a particular series regularly decreases and ultimately the

lines converge.

(ii) The sharp and diffuse series have a common limit 2

R( )

(3 )s d

p∞ ∞ν = ν =

− ∆. This common

limit is equal to the first running term of the principal series.

(iii) The wave number difference between the limit of principal series and the common limitof sharp or diffuse series is equal to the wave number of first line of principal series.

,1pp s d

∞ ∞ν − ν = ν

This is called Rydberg-Schuster law.

(iv) The wave number difference of series limit of diffuse series and fundamental series isequal to the wave number of first line of diffuse series.

∞ ∞ν − ν = ν1d f d

This is called Runge’s law.

(v) When the spectra of alkali atoms are examined with instruments of high resolving power,it is found that each line of principal and sharp series of a particular atom consists of anarrow doublet and each line of diffuse series consists of a group of three lines (triplet).For example, the sodium D line, which a member of principal series, is actually a doublet(two closely spaced lines) with wavelength 5890 and 5896 Å. The spectral line is said tohave a fine-structure. The complex lines consisting of several components are calledmultiplets.

(vi) On going from Li to Cs, the wave number separation of principal doublet increases.

(vii) The wave number separation of principal doublets decreases with increasing principalquantum number n.

(viii) The wave number separation of sharp doublet remains constant (=17.2 cm–1).

(ix) The wave number separation of diffuse series doublet remains constant (=17.2 cm–1).

3.5 ELECTRON SPIN AND FINE STRUCTURE OF SPECTRAL LINES

The concept of electron spin plays fundamental role in determining the state of an atom. The spin isquantum as well as relativistic property of electron. Its theoretical justification came in a naturalway from relativistic formulation of Schrodinger wave equation by Dirac. According to this theoryelectron possesses intrinsic angular momentum s whose magnitude is given by

| s | = + =( 1) , 1/ 2s s s ...(3.5.1)


s is called spin quantum number and has value ½. The magnetic moment associated with spin angularmomentum is

µ =2s

eg

m

−

s ...(3.5.2)

| µs | =| |

( 1)s sg g s sβ β− µ = − µ +

s...(3.5.3)

where gs is spin g factor and has value gs = 2.00230. Roughly it is taken equal to 2.Alkali metals have a single electron outside closed sub-shells. The angular momentum of closed

sub-shell is zero. So the angular momentum of an alkali atom is due to its single valence electron.The angular momentum of valence electron has two components: the orbital angular momentum Land spin angular momentum S. The resultant of these two angular momenta gives the total angularmomentum J. The magnitude of L, S and J are given by

| | L(L 1)

| | S(S 1)

| | J(J 1)

= +

= +

= +

L

S

J

where L = l = total orbital angular quantum number, S = s = total spin quantum number andJ = total angular momentum quantum number, and is given by

= ⊕ = + + − −J L S L S, L S 1,............ | L S |

i.e., J can take integrally spaced values. The orbital magnetic moment (µL ) and spin magnetic moment

(µ S ) interact with each other like magnetic dipoles. This interaction is called spin-orbit interactionand the energy of interaction depends on the mutual orientation of the magnetic moments.

The spectral terms of sodium atom in ground and excited states are given below:Ground state: 3s1.Here n = 3, L = 0, S = 1/2, J = 1/2.Spectral term: 3 2 S1/2.

Excited states: 3s 3p1 3 2P1/2, 3/2

3s 3d1 3 2D3/2, 5/2

3s 4s1 4 2S1/2

3s 4p1 4 2P1/2, 3/2

3s 4d1 4 2D3/2, 5/2

3s 4f1 4 2F5/2, 7/2

In the ground state of alkali atom L = 0, S = 1/2, J = 1/2. The term symbol is 2S1/2. Thenumber of J value is one. So the ground state is singlet i.e., it has no fine structure. If the valenceelectron is excited to p state, then L = 1, S = 1/2 , J = 1/2, 3/2. The term symbols are 2P1/2, and2P3/2. The spin-orbit interaction splits the P level into two components i.e., P state has doubletstructure. In this way, we can show D level splits into two sub-levels 2D3/2 and 2D5/2, F level splitsinto two sub-levels 2F5/2 and 2F7/2. Thus each level, except S, splits into two sub-levels i.e., all the


excited states of alkali atoms have doublet structure. The multiple splitting of D and F terms forsodium is very small and therefore the sub-levels of D and F differing in their J values are shown ascoincident lines. As one goes from Li to Cs the multiplet splitting increases.

Fine Structure of Spectral Lines

In alkali metals, due to spin-orbit interaction the s vector of valence electron combines with l vectorof the same electron to form a resultant vector j. In one-electron atom vectors s, l and j represent therespective quantities for the atom as a whole hence we represent them by S, L and J. The effect ofspin-orbit interaction is to the energy levels and hence the spectral terms into two components (exceptthe S term) one with J = L + 1/2 and other with J = L – 1/2. The S term does not split, the P term(L = 1) splits into P1/2 and P3/2, D term (L = 2) splits into D3/2 and D5/2, F term (L = 3) splits intoF5/2 and F7/2 etc.

Doublet Structure of Principal Series in Sodium

The transitions leading to the emission of first doublet of principal series of sodium viz., the well-

known sodium D-lines is shown in the figure. Let 1 2andν ν are the wave numbers of the lines of a

doublet emitted in transitions 3 2S1/2 n 2P1/2 and 3 2S1/2 n 2P3/2 where n = 3, 4, 5, …

Table 3.5.1: The first doublet separation of the principal series in spectra of alkali metals

Metal Atomic numberWave number

separation Åcm–1

Li 3 0.34 0.15

Na 11 17.2 6

K 19 58 34

Rb 37 238 147

Cs 55 554 422

Fig. 3.5.1 Principal series doublets


Sharp series doublet in sodium: The doublets of this series are emitted in the transitions

3P1/2 nS1/2 , and 3P3/2 nS1/2

where n = 4 for first doublet, n = 5 for second doublet and so on.Diffuse series doublet: The lines of the first doublet of diffuse series in sodium are emitted in

the transitions 3P1/2 nD3/2, 3P3/2 nD3/2, 3P3/2 nD5/2

where n = 3, 4, 5, ……(The transition 3P1/2 3D5/2 is forbidden by selection rule J = ± 1).The fine structure of diffuse series consists of three lines instead of two. The transitions leading

to these lines in cesium atom are shown below:6 2P3/2 5 2D3/2 = 36127 Å

6 2P3/2 5 2D5/2 = 34892 Å

6 2P1/2 5 2D3/2 = 30100 Å

Fig. 3.5.2 First two doublets of sharp series

Fig. 3.5.3 Compound doublet of diffuse series


One of the three lines is very weak and is not observed under low resolution. Such a group oflines is called not a triplet but a compound doublet.

Fundamental series doublet: The transitions leading to the origin of compound doublet areshown in the Fig. (3.5.4).

Fig. 3.5.4 First doublet of fundamental series (compound doublet)

It can be seen that the separation ∆ν of the doublets increases rapidly with increasing atomicnumber Z. In the following table the approximate separation of the first doublet of principal seriesof alkali metals are shown:

Li Na K Rb Cs

Z 3 11 19 37 55∆ν (cm–1) 0.34 17 58 238 554

3.6 INTENSITY OF SPECTRAL LINES

The qualitative idea about the relative intensities of spectral lines can be obtained from the followingrules:

1. In any doublet, the spectral line resulting from transition in which L and J both eitherincrease or both decrease (i.e., both change in the same way) is strongest.

If there are more than one line satisfy this condition, the strongest is one which involveslargest J value.

A spectral line resulting from the transition in which L and J both change in opposite ways is notallowed. For example, the transition accompanied by L = – 1, J = +1 and that with L = +1,J = –1 are not allowed.

Consider the principle series doublet 2S1/2 2P1/2 and 2S1/2 2P3/2. For the first lineL = –1, J = 0 and for the second line L = –1, J = –1. For the second line L and J both decreasehence it stronger than the first line.


Fig. 3.6.1

Consider the diffuse series compound doublet.

Fig. 3.6.2 The line arising from transition D5/2 P3/2 involves largest J and hence is strongest

2. The relative intensities of lines of a compound doublet may be calculated from the Burger-Dorgello-Ornstein sum rule.

(i) The sum of the intensities of all the lines of a multiplet, which arise from the transitionoriginating from same initial state is proportional to the statistical weight 2J + 1 ofthe initial state.

(ii) The sum of the intensities of all the lines of a multiplet, which arise from the transitionending on same final state is proportional to the statistical weight 2J + 1 of the finalstate.

Now consider the compound doublet of diffuse series.

Fig. 3.6.3


Let , , , and denote the intensities of the lines as shown in the Fig. (3.6.3). The intensityof forbidden line = 0. In the following table the spectral terms with their statistical weights areshown.

2P3/2 [4] 2P1/2 [2]

2D5/2 [6] = 02D3/2 [4]

Now, + is the sum of intensities of lines with the same initial state 2D5/2 [2J + 1 = 6].Similarly, + is the sum of intensities of lines with the same initial state 2D3/2 [2J + 1 = 4].Therefore,

α + δ α

= ⇒ =β + γ β + γ

6 3

4 2...(i)

+ is the sum of intensities of lines with the same final state 2P3/2 [2J + 1 = 4] and + isthe sum of intensities of lines with the same final state 2P1/2 [2J + 1 = 2]. Therefore,

α + β α + β

= ⇒ =δ + γ γ

4 2

2 1...(ii)

From Eqns. (i) and (ii), we have 9 and 5 .α = β γ = β Hence

α β γ =: : 9 :1: 5

Consider the compound doublet of fundamental series. The possible transitions are shown inthe Fig. (3.6.4).

Fig. 3.6.4

According to the intensity rule,

α + δ α

= ⇒ =β + γ β + γ

8 4

6 3...(i)

α + β α + β

= ⇒ =δ + γ γ

6 3

4 2...(ii)

From Eqns. (i) and (ii), 20 , 14 .α = β γ = β

α β γ =: : 20 :1:14 .


SOLVED EXAMPLES

Ex. 1. If the valence electron in sodium is excited to the 42D state, what are the different routesopen for electron in returning to the normal state?

Sol. Subject to the selection rules L = ± 1, the following are the possible routes for electronictransitions.

1. 2 2 2 2 24 D 4 P 3 D 3 P 3 S→ → → →

2. 2 2 24 D 4 P 3 S→ →3. 2 2 24 D 3 P 3 S→ →4. → → → →2 2 2 2 24 D 4 P 4 S 3 P 3 S.

Ex. 2. Obtain an expression for the doublet separation caused by spin orbit interaction in alkaliatoms. Interpret the results so obtained.

Sol. In alkali atoms the splitting of levels and hence the splitting of spectral lines due to spinorbit interaction is more important than that due to relativistic effects. All the energy levels of opticalelectron except l = 0 (s-state) are split into two components. One level corresponds to J = l + 1/2and the other to J = l – 1/2. The change in term value due to spin orbit interaction is

2 43 1

2

E 1 [ ( 1) ( 1) ( 1)]T R Z , 0

2 ( )( 1)

lsls

j j l l s sl

hc n l l l

∆ + − + − +∆ = − = − α ≠

+ +

For one electron atom, s = 1/2, j = l s = l 1/2 = l + 1/2 and l – 1/2 . Therefore,

= +∆

′∆ = − = − α+ +

, 1/2 2 43 1

2

E 1 1T R Z

2 ( )( 1)

ls j lls l

hc n l l l

and = −∆

′′∆ = − = α ++ +

, 1/ 2 2 43 1

2

E 1 1T R Z ( 1)

2 ( )( 1)

ls j lls l

hc n l l l

Doublet separation

2 4 41

3 3

R Z ZT T T 584 .

( 1) ( 1)ls ls ls m

n l l n l l−α

′′ ′δ = ∆ − ∆ = =+ +

Thus, the doublet separation is (i) proportional to Z4 (ii) inversely proportional to n3. Withincreasing value of l, the doublet separation decreases.

Ex. 3. Show that the doublet separation in sharp series of alkali atoms is constant.Sol. Wave numbers of the lines of nth doublet of sharp series are:

1 2 1/ 2 1 1/ 2 2 2 3 / 2 1 1/ 2T ( P ) T (3S ), T ( P ) T (3S )v n v n= − = −Wave number separation of sharp series doublet

∆ = − = − =1 2 2 3/ 2 2 1/ 2T (3P ) T (3P ) independent of = constant.v v v n


Ex. 4. The principal and sharp series for Li atom converge to continuum at 43487 and 28583 cm–1

respectively. Calculate the quantum defect for the common term in each series.(R = 109729 cm–1)Sol. For common term T = 28583 cm–1.

Now2

RT

( )n=

− ∆

R 1097293.83896 1.9593

T 28583n∴ − ∆ = = = =

∆ = − = − =1.9593 2 1.9593 0.0407n .

Ex. 5. The effective quantum number for the ground state of rubidium is 1.805. Determine theionization potential of the atom. R = 109737 cm–1.

Sol. 12 2

R 109737 33682T 33682cm eV 4.176eV

8065(1.805)effn−= = = = = (1 eV = 8065 cm–1).

Ex. 6. The ionization potential of hydrogen is 2.5 times the ionization potential of sodium. Calculatethe effective atomic number of sodium.

Sol. Energy of atom2

2

RZE

eff

n= −

Ionization potential I is equal to | E |. Therefore,

( )

= =

22 2Na Na

2 2 2H

Na H

ZZI 1.

I 1Z 3

effeff

eff

n

n

( ) ( )= ⇒ = =

2

Na

Na

Z1Z 3.6 1.89.

2.5 9

eff

eff

Ex. 7. The first member of principal series of sodium has a wavelength of 5890 Å. The first excitedS-state of sodium lies 3.18 eV above the ground state. Find the wavelength of the first member of sharpseries.

Sol. The separation of 3S and 3P levels E = 12400eV. Å

2.10eV5890 Å

hc= =

λSeparation of 4S and 3P level

E 3.18 2.10 1.08eV∆ = − =Wavelength of the first line of sharp series

12400eVÅ

11481Å.E 1.08eV

hcλ = = =

∆


Ex. 8. Calculate the doublet separation of the 3p state of sodium atom. The wavelengths of theprincipal series doublet are 1 = 5890 Å and 2 = 5896 Å.

Sol. 72

1 2

1 6Å 1| | 1.73 10

5890Å 5896Å Å−∆λ ∆λ

ν = ⇒ ∆ν = = = = ×λ λ λ ×λ

−−

−

×∆ν = =

71

8

1.73 1017.3 cm

10 cm

Separation of corresponding energy levels in wave number units is

− − − −∆ = ∆ν = × × × × = ×34 8 1 1 22E 6.63 10 Js 3 10 ms 1730 m 3.43 10 Jhc

E = 2.14 × 10–3 eV.

Ex. 9. The sodium yellow line 5893Å arises from the transition 3p 3s. The p-level is split by spinorbit interaction into two components separated by 2.1 × 10–3 eV. Evaluate the wavelength separationbetween the two components of the yellow line.

Sol. 3T 2.1 10 eV−∆ν = ∆ = × ( )3 1cm2.1 10 eV 8065 16.9cm

eV− −

= × =

Now 2 8 2 1 8

2(5893 10 m) (16.9cm ) 5.87 10 cm.− − −∆λ

∆ν = ∴∆λ = λ ∆ν = × = ×λ

Ex. 10. The mean position of the levels giving the firstpair of principal series of sodium atom is 16960 cm–1. Theconvergence limit of sharp series is 24490 cm–1. Calculate theionization potential of sodium atom.

Sol. Ionization energy of sodium atom

I = (16960 + 24490 ) cm–1 = 41450 cm–1

= 41450

eV 5.1395 eV.8065

=

Ex. 11. The principal and sharp series for sodium atomconverge to continuum at 41450 and 24477 cm–1 respectively.Calculate the ionization potential of sodium atom.

Sol. Ionization energy

I = 41450 cm–1

= 41450

eV 5.139 eV.8065

=


Ex. 12. The mean position of first pair of lines of theprincipal series of Li is 14904 cm–1. If the convergence limit ofthe sharp series is at 28583 cm–1, calculate the ionizationpotential of Li.

Sol. Ionization energyI = (14904 + 28583 ) cm–1 = 43487 cm–1

= 5.39 eV.

Ex. 13. Calculate the quantum defect for 3p configuration of sodium. The term value for this stateis 24477 cm–1. (R = 109734 cm–1).

Sol. 2 2

2 2

RZ RZT , Z 1, 3

( )

eff effeff

eff

nn n

= = = =− ∆

R 109734

2.117T 24477

n − ∆ = = =

= 3 – 2.117 = 0.883.Ex. 14. If the doublet splitting of the first excited state 2 2P state in an atom with Z = 2 is

5.84 cm–1. Calculate the corresponding separation in hydrogen atom.

Sol. 4( T) Zδ ∆ ∝

= −

=

δ ∆= ⇒ = ⇒ δ ∆ =

δ ∆ δ ∆

4 42 2 1

4 41 1

( T) Z 5.84 2( T) 0.365cm .

( T) ( T)Z 1

Z

Z

3.7 SPECTRA OF ALKALINE EARTHS

The elements of Group II of periodic table viz., Be, Mg, Ca, Sr, Cd, Ba, Hg are called alkalineearths. The atoms of these elements have two valence electrons outside a closed shell. The heliumatom also has two electrons in its outermost shell. Hence, the spectra of alkaline earths resemblewith that of helium atom. The spectrum of an element of the group II consists of twosystems—singlet and triplet, and each system can be grouped into four kinds of series, sharp, principal,diffuse and fundamental. The valence electrons determine the entire optical properties of theseelements. When one or both electrons go to higher energy levels, the atom is said to be in excitedstate. It is found that the chief series of spectral lines result from the electronic transitions in whichonly one of the valence electrons is involved, the other electron remains in the ground state. Whenboth electrons are excited and participate in electronic transitions, the resulting spectrum is complexin nature.

An alkaline earth atom has two valence electrons and hence there are two orbital angular momental1 and l2 and two spin angular momenta s1 and s2. These angular momenta can combine in twoways: (i) L-S coupling and (ii) j-j coupling. In light atoms electrostatic interaction (electron-electronrepulsion) and spin-spin exchange interaction predominate over the spin-orbit interaction. As aconsequence of this, vectors l1 and l2 combine to form a resultant L. Similarly, vectors s1 and s2


combine to form a resultant S. Next, the spin orbit interaction causes the vectors L and S to combineto form a resultant J. These couplings may be summarized as follows:

l1+ l2 = L (electrostatic interaction)

| l | = 1 1 2 2 2 1 2( 1) , | | ( 1) , , 0,1, 2, 3,.....l l l l l l+ = + = l

| L | = + = ⊕ = − + 1 2 1 2 1 2L(L 1) , L .....( )l l l l l l

s1 + s2 = S (exchange interaction)

| s1| 1 1 2 2 2 1 2( 1) , | | ( 1) , 1/ 2, 1/2s s s s s s= + = + = = s

| S | = + = ⊕ = ⊕ = 1 2( 1) , S 1/ 2 1/ 2 0,1S S s s

L + S = J, = + = ⊕ = −| J | J(J 1) , J L S | L S | .. integrally spaced values

.. L + S.

Explanation of Essential Features of Spectra of Alkaline-Earths

In an alkaline earth atom there are two valence electrons and hence there are four angular momental1, l2, s1 and s2, which may couple in two ways to form their resultant.

(i) L–S Coupling (ii) j-j CouplingIn light atoms, the electron-electron electrostatic interaction and spin-spin correlation predominate

over the weak spin-orbit interaction. Owing to strong electrostatic interaction, the orbital angularmomenta l1 and l2 couple to form their resultant L. Similarly, spin angular momenta s1 and s2

couple to form their resultant S. Then the spin-orbit interaction causes the vectors L and S to coupleto form their resultant J. The magnitudes of the various angular momenta are determined by theirrespective quantum numbers. This type of coupling may be summarized as follows:

l1 + l2 = L, s1 + s2 = S, L + S = J

1l 1 1 2 2 2( 1) , ( 1) , L (L 1) , S(S 1)l l l l= + = + = + = + L Sl

J (J + 1)= J

Where ⊕ = ⊕ = ⊕1 2 1 2L = , S , J L Sl l s s

The spin-orbit interaction splits each level characterized by L-value into a number of components(fine structure) each characterized by J-value. The group of fine structure levels constitutes amultiplet.

Ground state: In the normal state (ns2) of alkaline earths both electrons have the same orbitalquantum numbers as well as the same spin quantum numbers (l1 = 0, l2 = 0, s1 = 1/2, s2 = 1/2). Ifboth electrons have parallel spins (), their all the four quantum numbers will be identical. Thisstate is not allowed by Pauli’s exclusion principle. Thus, the state in which valence electrons withparallel spin that leads to S = 1, is not permitted. Consequently the state 3S1 does not exist. Therefore,the ground state of alkaline earth atom is one in which the valence electrons have opposite spins( ). In this state L = 0, S = 0, J = L S = 0. This state is singlet and is designated as 1S0.

Excited states: Assuming that only one electron is promoted to higher energy state, the termsymbol and their multiplet nature of excited states can be deduced as explained in the followingtable.


Thus, we see that the spectral terms of elements of group II of periodic table consists of singletand triplet energy states corresponding to one value of J and three values of J respectively for eachvalue of L. The singlet and triplet energy states of calcium are shown in the Fig. (3.7.1a).

l1 l2 L s1 s2 S J Symbol

ss 0 0 0 1/2 1/2 1 1 3S1 does not exist

ss 0 0 0 1/2 1/2 0 0 1S0

sp 0 1 1 1/2 1/2 1 2, 1, 0 3P0, 1, 2

sp 0 1 1 1/2 1/2 0 1 1P1

sd 0 2 2 1/2 1/2 1 3, 2, 1 3D1, 2, 3

sd 0 2 2 1/2 1/2 0 2 1D2

sf 0 3 3 1/2 1/2 1 4, 3, 2 3F2, 3, 4

sf 0 3 3 1/2 1/2 0 3 1F3

Fig. 3.7.1(a) Energy level diagram of calcium


Ca(20) 1s2 2s2p6 3s2p6 4s2

Fig. 3.7.1 (b)


Chief Series Resulting from Transitions Between Singlet Levels

The allowed transitions between the singlet levels give rise to chief series of spectral lines. No splittingof spectral lines occurs in these transitions.

Principal series: In calcium, the spectral lines of this series originate from the transitions4s 4s 4s np, n = 4, 5, …….that is, from n1P levels to the ground state 1S0.

Sharp series: The lines of this series are emitted from electronic transitions 4s4p 4s ns,n = 5, 6,…..that is, from 1S levels to the lowest 1P level.

Singlet sharp and singlet diffuse series have a common convergence limit.

∞ ∞ν = νs d

Diffuse series: The lines of this series appear in electronic transitions 4s 4p 4s nd,n = 4, 5…… that is, from 1D levels to the lowest 1P level.

Fundamental Series: The electronic transitions 4s 3d 4s nf, n = 4, 5,….. that is, from 1Flevels to the lowest 1D level give rise to the lines of this series.

The wave numbers difference between the common limit of singlet sharp and diffuse seriesand the limit of singlet principal series is equal to the wave number of the first line of principalseries. This is also observed in triplet series.

∞ ∞ ∞ν ν ν = ν 1(or ) ~ ps d p (in singlet and triplet both)

This result is called Rydberg-Schuster law.The wave number difference between the common limit of singlet sharp and diffuse series and

the limit of singlet fundamental series is equal to the wave number of the first line of diffuse series.This is also found in triplet series.

∞ ∞ ∞ν ν − ν = ν1(or ) fs d f (in singlet and triplet both)

This result is known as Runge’s law.Chief Series Resulting from Transitions between Triplet LevelsPrincipal series: Lines of this series are emitted from the transitions from 3P levels to the

lowest 3S level. Each line of this series consists of three components with decreasing separation. Thelines approach a single series limit.

Sharp series: The transitions from 3S levels to the lowest 3P levels give rise to the lines ofsharp series. The lines of this series also consist of three components but with constant separation.The triplet of principal and sharp series are called simple triplet. The triplet sharp and triplet diffusehave a common limit.

Diffuse and fundamental series: The diffuse series originates from transitions from 3D levelsto lowest 3P level and fundamental series from transitions from 3F to the lowest 3D level. The linesof both series have six components — three strong and three satellites. The multiplet of lines is calledcompound triplet.

The intervals between 3P0, 1, 2 states are greater than those of the 3D1, 2, 3 states. Similarly,3D1, 2, 3 intervals are greater than those of 3F2, 3, 4. These intervals obey the Lande interval rule.The order of these levels is regular (normal) that is, the energy of a level increases with increasing J


value. According to Lande interval rule, the intervals 3P2 – 3P1 and 3P1 – 3P0 are in the ratio 2 : 1.The intervals 3D3 – 3D2 and – 3D2 – 3D1 are in the ratio 3 : 2.

Those components of the triplet fine structure are more intense for which L and J change inthe same way and of these, the one involving larger value of J is strongest.

Fig. 3.7.2


3.8 TRANSITIONS BETWEEN TRIPLET ENERGY STATES

Series Number of components Transitions

Sharp 3 43P2 n3S1

43P1 n3S1

43P0 n3S1

Principal 3 53S1 n3P2

53S1 n3P1

53S1 n3P0

Diffuse 6 43P2 n3D3

43P2 n3D2

43P2 n3D1

43P1 n3D2

43P1 n3D1

43P0 n3D1

Fundamental 3 33D3 n3F4, 3, 2

33D2 n3F4, 3, 2

33D1 n3F4, 3, 2

The allowed transitions amongst the triplet levels also produce the four chief series—sharp,principal, diffuse and fundamental. The transitions leading to these series and the number of componentsof each line are given in the table.

These transitions obey the selection rules:S = 0, L = 0, ± 1, J = 0, ± 1, (0 0 excluded)

There is no restriction on the change in value of n.In some cases transitions between singlet and triplet levels are also observed. These lines are

called inter-combination lines.

3.9 INTENSITY RULES

The lines resulting from the transitions, in which changes in L and J are in the same sense, are stronger.Among these, the transition involving by largest values of L and J gives rise to the strongest line.

3.10 THE GREAT CALCIUM TRIADS

Calcium, strontium and barium emit three prominent groups of lines which do not belong to thechief series of singlet and triplets. These three groups of lines are called the great calcium triads. Incalcium, these lines arise from the transitions which start from the three triplet terms 3P0, 1, 2,


3D1, 2, 3 and 3F2, 3, 4 and end to the triplet term 3D1, 2, 3. The upper levels arise when both electronsare excited, one to 4p and the other to 3d level. The lower terms arise from the state 4s 3d.

Each spectral line of the triad consists of three strong lines and three or four faint lines. Thespectral lines of the triads are shown in the Figure 3.10.1.

Fig. 3.10.1 The great triads of calcium

3.11 SPECTRUM OF HELIUM ATOM

Helium atom has two electrons in first shell. In many respects its spectrum resembles with that of analkaline earth metal. Assuming L-S coupling to be in operation, we will first write the ground stateof helium atom. In the normal state, the atom has configuration 1s 1s. The helium atom whose boththe electrons have opposite spins ( ) has total spin quantum number S = s1 s2 = 0 and is calledparahelium. In the ground state (1s 1s) of parahelium, l1 = 0, l2 = 0, L = 0, s1 = 1/2 (),s2 = 1/2 ( ), S = 0, J = L S = 0. Its term symbol is 11S0 (singlet). The term symbols for excitedstate of parahelium can be deduced as follows:


Configuration l1 l2 L s1 s2 S J Symbol

1s 2s 0 0 0 1/2 1/2 0 0 2 1S0

1s 2p 0 1 1 1/2 1/2 0 1 2 1P1

1s 3s 0 0 0 1/2 1/2 0 0 3 1S0

1s 3p 0 1 1 1/2 1/2 0 1 3 1P1

1s 3d 0 2 2 1/2 1/2 0 2 3 1D2

1s 4s 0 0 0 1/2 1/2 0 0 4 1S0

So the states of parahelium are singlets i.e., J has only one value for each value of L.The state 1s 1s of helium atom in which the two electrons have parallel spins () has total

spin quantum number S = 1 and is called orthohelium. In this configuration all the four quantumnumbers of both the electrons are identical, which violates the Pauli’s exclusion principle. This state,which has L = 0, S = 1, J = 1, r = 3, is designated by 3S1. Pauli’s principle forbids the existence ofthis state (3S1). Therefore, the lowest state 1s1s of orthohelium does not exists. Its first excited stateis 1s 2s.

The spectral terms of excited states of orthohelium can be deduced as explained below.

Configuration l1 l2 L s1 s2 S J Symbol

1s 2s 0 0 0 1/2 1/2 1 1 2 3S1(lowest excited state)

1s 2p 0 1 1 1/2 1/2 1 0, 1, 2 2 3P0, 2 3P1, 2 3P2

1s 3s 0 0 0 1/2 1/2 1 1 3 3S1

1s 3p 0 1 1 1/2 1/2 1 0, 1, 2 3 3P0, 3 3P1, 3 3P2

1s 3d 0 2 2 1/2 1/2 1 1, 2, 3 3 3D1, 3 3D2, 3 3D3

Fig. 3.11.1 Energy levels of helium atom


The excited states of orthohelium are triplet. Thus, the energy states of helium divide into singlet(parahelium) and triplet (orthohelium) states. Parahelium has additional energy level correspondingto the configuration 1s 1s. The corresponding level does not exist in orthohelium.

The four chief series (sharp, principal, diffuse and fundamental) are observed in both singletand triplet systems. Both the principal series in the two systems lie in visible and near and far ultravioletregions. In parahelium the principal series arises as a result of transitions from higher P states to theground state.

The energy level diagram and the transitions obeying selection rulesL = 0, ± 1, S = 0,J = 0, ± 1 (0 0 excluded)

are depicted in the figure. Because of selection rule S = 0, no transitions between singlet states andtriplet states can occur and hence no inter-combination lines are observed in helium spectrum. Thehelium spectrum results from the transitions taking place in one set or the other.

PARAHELIUM (S = 0 ) ORTHOHELIUM (S = 1 )

Fig. 3.11.2 Spectrum of helium atom

The S rule forbids the triplet states (S = 1) from decaying to the ground state (S = 0). Thesetransitions can thus only occur by violating these selection rules and since that is a very unlikelyevent, these transitions occur with very low probability. Energy levels that have a low probability ofdecay must live for a long time before they decay. Such states are known as metastable states.

An orthohelium atom () can lose its excitation energy by collision to become a paraheliumatom and a parahelium ( ) atom can be excited by collision to become an orthohelium atom. Ordinaryhelium is mixture of the two forms.


Fig. 3.11.3 Origin of helium spectrum


1. Draw an energy level diagram of sodium atom and using this discuss the salient features of the spectra ofalkali metals.

2. (a) Why does one get only principal series of lines in the absorption spectrum of alkali metals, while all

the four series are observed in their emission spectra.

(b) Describe the doublet fine structure in the spectra of alkali elements and interpret on the basis of spinorbit interaction model.

3. If the valence electron in sodium atom is excited to 42D state, what are the different routes open for theelectron in returning to the normal 3 2S1/2 state? Draw energy level diagram to show the transitions.

4. Doublet structure of spectral lines is a characteristic feature of the spectra of all alkali metals. How is itexplained by introducing the concept of electron spin?

5. Write down the expression for the energy levels of alkali atoms and explain the reason of introducingquantum defect term in it.


Name the different series of sodium atom have quantum defect 1.37 and 0.88 respectively. Obtain thewavelength of the spectral line for the transition 3p-3s.

6. What is meant by fine structure of spectral lines? Describe how does electron spin coupled with orbitalmotion explain the fine structure of alkali spectra?

7. Differentiate between penetrating and non-penetrating orbit. Prove that the quantum defect for an atomdepends upon azimuthal quantum number and is independent of principal quantum number.

8. Discuss the salient features of the spectra of alkaline earth elements.

9. What are calcium triads? What are their origins? Using a tentative energy level diagram, draw the transitionsfor the three calcium triads appearing in the visible region around 6500, 5600 and 5270 Å.

10. Explain the singlet and triplet series in two valence electron system.

11. Explain diagrammatically the singlet and triplet series of lines in the spectra of helium atom.

12. Draw the energy level diagram of calcium atom on the basis of one electron excitation. Using this diagramoutline the essential features of the spectra of alkaline earth element.

CHAPTER

MAGNETO-OPTIC AND ELECTRO-OPTIC

4.1 ZEEMAN EFFECT

In 1896 Peter Zeeman discovered that when a light source is placed in a magnetic field, the spectrallines emitted by the atoms split into a number of components. This phenomenon is called the ZeemanEffect. Suppose that a source emits a spectral line of frequency 0 in absence of magnetic field.When a magnetic field is switched on and the light emitted by the source is viewed transverse to thefield, three equally spaced spectral lines of frequencies 0 – , 0, and 0 + are observed. Thechange in frequency of emitted light is called Zeeman shift.

Fig. 4.1.1 Zeeman effect

In transverse view, the central line is designated as component and the outer lines as

components. The plane of vibration (electric vector) of the central line ( component) is parallel to

PHENOMENA


the magnetic field B and that of the outer lines ( components) is perpendicular to the magneticfield. When viewed parallel to the magnetic field (longitudinal view) only two spectral lines withfrequencies 0 – and 0 + are observed. Both of the lines are circularly polarized, the line ofhigher frequency shows left hand polarization and that with lower frequency shows right handpolarization.

Quantum Mechanical Explanation

Normal Zeeman effect occurs in atoms having zero spin (S = 0). In an atom having even number ofelectrons, the spins can pair off and cancel, so that the atom behaves like a spin-less particle. So thenormal Zeeman effect is observed only in atoms with even number of valence electrons. In otherwords, the condition for normal Zeeman effect to occur is that the pair of energy levels involved inthe transitions leading to emission of spectral lines must be singlet i.e., S = 0 and g = 1 for initialand final levels both.

When an atom possessing magnetic moment is subjected to a magnetic field, its energy levelsare split into a number of components, called magnetic sub-levels (or Zeeman levels). Obviously,the transitions between the two sets of sub-levels will result in splitting of spectral lines. Supposethat the atom is placed in a magnetic field B acting in z-direction. The change in energy of an energylevel of the atom due to magnetic field is

β

∆ = −µ = −µ = − − = = µ

Z Z J JE .B B J B BM BM

2 2

e eg g g

m m ...(4.1.1)

where J is total angular quantum number and MJ is total magnetic quantum number of atom. MJ cantake on integrally spaced values from J, J – 1, …….0 ………– J. Thus, the energy level characterizedby quantum number J is split into 2J + 1 equally spaced sub-levels. The amount of splitting dependson the Lande g-factor (i.e., L, S, J). In absence of magnetic field, the sub-levels with different MJ

have the same energy i.e., they were degenerate. Application of the magnetic field removes thedegeneracy and the sub-levels with different MJ now possess different energies.

For illustration we first consider the splitting of spectral line resulting from the transition1P1 1S0 (These levels are involved in emission of lines of principal series in alkaline earth atom).For lower level (1S0), L = 0, S = 0, g = 1, MJ = 0. This level does not split in the magnetic field.For the upper level (1P1), L = 1, S = 0, J = 1, g = 1, MJ = –1, 0, 1. This level splits into three sub-levels. The allowed transitions are shown in the Fig. (4.1.2). It is evident from the figure that theoriginal line will split into three components.

Let E1 and E2 be the energies of the lower and upper level in absence of magnetic field. Thewave number of the spectral line emitted from the transition between these energies levels is

2 1

0

E E

hc

−ν = ...(4.1.2)

In presence of external magnetic field B, the lower level does not split whereas the upper levelsplits into three sub-levels with different energies. The energies of the upper and lower levels aregiven by

2 2 2 JE E E E BM ,β′ = + ∆ = + µ MJ = – 1, 0, 1

1 1E E′ =

Magneto-optic and Electro-optic Phenomena 501

The wave numbers of the lines resulting from transitions between these levels are given by

J2 1 2 1

J

BME E E E, M 1, 0,1

hc hc hcβµ′ ′− −

ν = = + = −

0 J

BM

hcβµ

ν = ν +

...(4.1.3)

The number 0

B BL

4

e

hc mcβµ

= =π

= 46.7 B m–1 is called Lorentz number. In terms of Lorentz

number we can write Eqn. (4.1.3) as

0 0 JL Mν = ν + ...(4.1.4)

Putting MJ = –1, 0, 1 we obtain three wave numbers 0 0 0 0 0L , , Lν − ν ν + of the spectral

lines into which the original line is split under the action of magnetic field. The Zeeman shift inwave number units is

0 0

B~ L

4

e

m c∆ν = ν ν = =

π ...(4.1.5)

The corresponding shift in wavelength ( )2| | .∆λ = λ ∆ν is

2B

4

e

mc

λ∆λ =

π ...(4.1.6)

Thus, the normal Zeeman pattern consists of three lines of wave numbers 0 0 0, ,ν − ∆ν ν ν + ∆ν

or wavelengths , and .λ + ∆λ λ λ − ∆λPolarization: Quantum mechanical analysis shows that the spectral line with polarization

results from the transitions in which MJ = ± 1 and polarization from the transitions in whichMJ = 0 (excluding 0 0 transitions).

Fig. 4.1.2 Normal Zeeman effect


For another example let us consider the normal Zeeman splitting of the spectral line of diffuseseries in cadmium resulting from the transition 1D2 1P1 between singlet levels.

For the lower level L = 1, S = 0, J = 1, g = 1, MJ = –1, 0, 1. This level splits into three sub-levels.

For the upper level, L = 2, S = 0, J = 2, g = 1, MJ = – 2, – 1, 0, 1, 2. This level splits into fivecomponents.

The selection rules for allowed transitions are MJ = 0, ± 1

The transitions for which MJ = 0 lead to emission line of a single line with wave number ν0 ;

that for which MJ = 1 leads to a single of wave number ν − ∆ν0 and that for which MJ = –1

corresponds to wave number ν + ∆ν0 . These transitions with their frequencies are shown in the Fig.(4.1.3)

Fig. 4.1.3 Normal Zeeman splitting of a spectral line

The transitions marked 1 on the diagram are accompanied by equal amount of energy changeand hence give rise to a single spectral line. Its wave number is given by

2 1

0

BE E

hc hcβµ−

ν = − = ν − ∆ν ...(4.1.7)


Where ∆ν represents the Zeeman shift and is given by

0

B BL is Lorentz number

4

e

hc mcβµ

∆ν = = =π

or B

4

e

mc∆ν =

π ...(4.1.8)

The transitions marked 2 on the diagram are accompanied by a change in energy E2 – E1,which give rise to a line of wave number

2 1

0

E E

hc

−ν =

Similarly, the transitions marked 3 on the diagram lead to a single line of wave number

ν = ν + ∆ν0

The corresponding shift in wavelength is given by

2B

4

e

mc

λ∆λ =

π ...(4.1.9)

Polarization Rules

Viewed perpendicular to B: The spectral line originating from the transition obeying the selectionrule MJ = 0, have the plane of vibration parallel to the magnetic field B and that obeying the selectionrule MJ = ± 1, have the plane of vibration perpendicular to B.Viewed parallel to B: The transition obeying MJ = ± 1 gives spectral lines with circular polarization.Transitions for which MJ = 0 are forbidden.

4.2 ANOMALOUS ZEEMAN EFFECT

When an atom is placed in an external magnetic field, the behaviour of the angular momentum vectorin general will be very complicated. The reason is that each of the magnetic moments associatedwith orbital and spin motion interact with each other as well as with the external field. No simpledescription of the motion is possible when these fields are of the order of the same magnitude.However, when one is much larger than the other an approximate treatment is possible.

Weak field approximation: When the external magnetic field is small compared with the fielddue to the spin-orbit coupling, the coupling between L and S remains intact. The resultant of L andS viz. J has physical significance. The vectors L and S precess around J at much faster rate than Jprecesses around external field B. The space quantization of vector J allows it to have only discretevalues of its projection onto the direction of magnetic field B given by

J JM , M , 1,............. .z j j j= = − − + +J

The magnetic interaction energy is given by

β∆ = −µ = µ

JE .B BMg ...(4.2.1)


[In case of one valence electron atom, we replace the vectors L, S, J by l, s, j and quantumnumbers L, S, J, and MJ by l, s, j and mj.]

In 1907 Runge observed that when a spectral line was split into more than three components,the magnitude of splitting was a rational fraction of the normal Zeeman splitting.

normal

p

q

∆ν = ∆ν

, p and q being integers. (Runge’s Law) ...(4.2.2)

We shall explain the anomalous Zeeman effect with two examples; the splitting of sodium Dlines. The D1 line 5896 Å arises from the transition 2P1/2 2S1/2 and the D2 line 5890 Å from thetransition 2P3/2 2S1/2. The values of L, S, J, g, MJ and gMj for the levels involved in thesetransitions are given in the following table:

Table

Term L S J g MJ gMJ

2P3/2 1 1/2 3/2 4/3 3/2, 1/2 2, 2/3

–1/2, – 3/2 – 2/3, – 22P1/2 1 1/2 1/2 2/3 1/2, – 1/2 1/3, – 1/32S1/2 0 1/2 1/2 2 1/2, – 1/2 1, – 1

The allowed transitions obey the selection rules: MJ = 0, ± 1.The level 2S1/2 (for which L = 0, S = 1/2, J = 1/2 , g' = 2) splits into two sub-levels. The level

2P1/2 (for which L = 1, S = 1/2, J = 1/2 . g'' = 2/3.) splits into two sub-levels.The level 2P3/2 splits into four Zeeman pevels.Let E1, E2 and E3 be the energies of the levels 2S1/2, 2P1/2, and 2P3/2 in absence of magnetic

field. The corresponding energies in presence of magnetic field will be denoted by prime. The wavenumbers of lines in absence of B are

2 1

0 1

E E, D line

hc

−ν = −

3 1

0 2

E E, D line

hc

−ν = −

The energies of the levels in presence of magnetic field are

1 1 JE E B Mgβ′ ′ ′= + µ

2 2 JE E B Mgβ′ ′′ ′′= + µ

3 3 JE E B Mgβ′ ′′′ ′′′= + µ

The wave numbers of lines arising from transitions 2P1/2 2S1/2 are given by

2 1 2 1 J JE E E E ( M M )

Bg g

hc hc hc β′′ ′ ′′ ′′ ′ ′− − −

ν = = + µ

′′ ′′ ′ ′= ν + −0 J J 0( M M ) Lg g ...(4.2.3)


The value of (g'' M''J – g' M'J) for each line is given beside the vertical line showing the transitionsin the Fig. (4.2.1). See that the original line disappears when the magnetic field is switched on.Substituting the values of g, and MJ we get the wave numbers of the four lines resulting fromtransitions between the two levels.

1 0 0 2 0 0 3 0 0 4 0 0

4 2 2 4L , L , L , L

3 3 3 3ν = ν − ν = ν − ν = ν + ν = ν +

Thus

∆ν = − = − ∆ν − ∆ν ∆ν ∆ν0 normal normal normal normal

4 4 2 2 4L , , , .

3 3 3 3 3

This is Runge Law.The wave numbers of the lines originating from the transitions 2P3/2 2S1/2 are given by

β′ ′ ′′′ ′′′ ′− − − ′

ν = = + µ3 1 2 1 J JE E E E ( M )MB

g g

hc hc hc...(4.2.4)

0 J J 0( M M ) Lg g′′′ ′′′ ′ ′= ν + −Substituting the values of g, and MJ, we get the wave numbers of the six lines resulting from

transitions between the two levels.

1 0 0 2 0 0 3 0 0

5 3 1L , L , L

3 3 3ν = ν − ν = ν − ν = ν −

4 0 0 5 0 0 6 0 0

1 3 5L , L , L

3 3 3ν = ν + ν = ν + ν = ν +

The wave number shift can be expressed as

∆ν = − − − = − − − ∆ν

0 normal

5 3 1 1 3 5 5 3 1 1 3 5, , , , , L , , , , ,

3 3 3 3 3 3 3 3 3 3 3 3

This is Runge law.Polarization Rules: If viewed perpendicular to B, the selection rule MJ = 0 leads to

polarization and MJ = ± 1 to polarization.If viewed parallel to B, the selection rule MJ = ± 1 leads to circular polarization of spectral

line.

Intensity Rules

The transition J J, MJ = ± 1, gives the intensity of line given by

I = I0 (J ± MJ + 1)(J MJ) ...(4.2.5)And MJ = 0 gives the intensity

I = 4I0 MJ2 ...(4.2.6)

The transition J J + 1, MJ = ± 1, gives the intensity of line given byI = I0 (J ± MJ + 1)(J ± MJ + 2) ...(4.2.7)

And MJ = 0 gives the intensityI = 4I0 (J + MJ + 1)(J – MJ + 1) ...(4.2.8)


Fig. 4.2.1 Anomalous Zeeman effect with principal series doublet (Sodium D1 and D2 lines) Dots represents the position of normal triplet

4.3 PASCHEN-BACK EFFECT

(i) For one valence electron atom

When a very strong magnetic field is applied to an atom, such thatthe interaction energy of the external field with the atom becomeslarger than the spin-orbit interaction, the coupling between vectorsl and s breaks down and the individual vectors l and s precess aroundthe magnetic field independent of each other. In this condition thecomplicated anomalous Zeeman pattern transform into a simpleZeeman pattern. This phenomenon is called thePaschen-Back effect.

Fig. 4.3.1


In the strong magnetic field, the total interaction energy of the atom is equal to the sum ofthree parts.

(i) The energy due to the precession of vector l around B(ii) The energy due to the precession of vector s around B

(iii) The energy due to interaction between l and s.

Thus

B BE E E El s ls∆ = ∆ + ∆ + ∆

* * * *E B B cos( )l l s sg m g m al s l sβ β∆ = µ + µ + ...(4.3.1)

where ml and ms are orbital and spin magnetic quantum numbers.In absence of magnetic field B, the angle between l* and s* is constant but in presence of B it

changes continually. So its average value must be calculated.

Now, * * * *cos( ) cos( B)cos( B)l s l s=

* * * * * * * *cos( ) cos( B). cos( B)al s l s al l s s=

= l sam m ...(4.3.2)

The total energy of atom in magnetic field can be expressed as

E B( 2 ) , 1, 2l s l s l sm m am m g gβ∆ = µ + + = = ...(4.3.3)

As an example we consider the Paschen-Back effect with principal series doublet 2S1/2 2P1/2, 3/2.

The following table displays the computation of magnetic energies in strong magnetic field for thelevels involved in the transitions. Fig. (4.3.2) shows the allowed transitions and resulting spectrallines.

The selection rules for the allowed transitions areML = 0, ± 1, and MS = 0.

Term ml ms ml + 2ms a ml ms

2P3/2 1 1/2 2 a/2

0 1/2 1 0

–1 1/2 0 –a/2

1 – 1/2 0 –a/2

2P1/2 0 – 1/2 – 1 0

–1 – 1/2 – 2 a/22S1/2 0 1/2 1 0

0 – 1/2 –1 0


0 -components, 1 -componentsl lm m∆ = π ∆ = ± σ

Fig. 4.3.2 Magnetic energy levels and Paschen-Back pattern for a principal series doublet

If we neglect the spin-orbit interaction energy term aml ms the expression for total energybecomes

E B( 2 )l sm mβ∆ = µ + ... (4.3.4)

The quantity (ml + 2ms) is an integer and thus the splitting of energy levels is integral multiplesof the normal splitting (0 = µB). In terms of frequency the shift is

0 0 upper lower( 2 ) ( 2 ) ( 2 )l s l s l sm m m m m m ∆ω = ∆ω ∆ + = ∆ω + − + ...(4.3.5)

The splitting of principal series doublet terms 2P0, 1, 2 and 2S1/2 in strong field are shown inFig. (4.3.3). Since energy shift E depend on (ml + 2ms), the terms having the same value of(ml + 2ms) now coincide. Thus, there are five P-states and two s-states. The selection rules for allowedtransitions are

∆ + = ±( 2 ) 0, 1l sm m


Fig. 4.3.3 Paschen-Back effect without L-S interaction

In the Fig. (4.3.3) the transition marked a,d correspond to ∆ + =( 2 ) 1l sm m and hence

have the same frequency 0 – 0, the transitions marked b,e correspond to ∆ + =( 2 )l sm m 0 and

have the frequency 0 and those marked c, f correspond to ∆ + = −( 2 ) 1l sm m and have the frequency

0 + 0. Thus, we have only three spectral lines in the strong magnetic field.Let us calculate the Zeeman shift d for the line = 1210 Å (in hydrogen) emitted in the

transition 2p 1s in a magnetic field of 5 Wb/m2. The frequency shift in normal Zeeman effect is

0 = Bβµ

The frequency shift, from Eqn. (4.3.5) is found to be

0

B B

2

e

mβµ

∆ω = ∆ω = =

The corresponding shift in wavelength

Åλ

∆λ = =π

2B0.034

4

e

mc

The wavelengths of the Zeeman triplet are 1210.034 Å, 1210 Å and 1209.966 Å.


(ii) Two valence electron atom

All the arguments made in case of one-electron system are also valid for two-electron system. Intwo-electron system vectors L and S play the role of l and s, the appropriate quantum numbers inthis case are L, S, ML and MS. The total energy of atom in strong magnetic field is given by

L S L SE B(M 2M ) AM Mβ∆ = µ + + ...(4.3.6)

As an example we consider the splitting of triplet terms 3S1, 3P0, 1, 2 in strong magnetic field.The computation of magnetic levels is shown the table given below. The Fig. (4.3.4) shows the splittingof terms and the transitions leading to the Paschen-Back pattern of the principal series triplet. Theselection rules for the strong field are

L

S

M 0 for -components

for -components

M 0

∆ = π= ± σ

∆ =...(4.3.7)

It should be noted that if the small interaction between L and S viz., the term aml ms is omitted,the Paschen-Back pattern will transform into a normal triplet shown by thick lines in the figure.

Table 4.3.1: Computation of magnetic energies in strong magnetic field.

Terms L ML MS ML + 2MS ML + 2MS + AMLMS

3P2 1 1 1 3 3 + A

1 0 1 2 2 + 0

1 – 1 1 1 1 + 0

1 1 0 1 1 – A

1 0 0 0 0 + 03P1 1 –1 0 –1 –1 + 0

1 1 –1 –1 –1 – A

1 0 –1 –2 –2 + 03P0 1 –1 –1 –3 –3 + A

Paschen-Back effect has been observed only for very narrow multiplet. In Li, this multiplet(doublet) has a separation of 0.34 cm–1 and hence this effect can be observed. In sodium doubletseparation is about 17 cm–1. To observe Paschen-Back effect a very large magnetic field is required,which is a difficult task. In general, Paschen-Back effect can observed if the magnetic splitting(Zeemansplitting) exceeds the fine structure splitting due to spin orbit interaction.


Fig. 4.3.4 Paschen-Back effect in a principal series triplet. Thick linesshow the positions of normal triplet


Meaning of weak and strong magnetic field

A good Zeeman pattern is produced when the magnetic field of the order of 0.1 T is applied on theatom. A field is said to be weak when the total spread of Zeeman pattern of each line is small incomparison to the line separation cause by spin-orbit interaction. The line separation of D1 and D2

lines of sodium is about 17.2 cm–1. A magnetic field of 3T produces Zeeman spread of3.76 cm–1 in line D1 and 4.70 cm–1 in D2 line. The Zeeman spread is small compared with the lineseparation. Therefore, a magnetic field of 3T is weak in this case.

For Li, the doublet separation is about 0.34 cm–1. A field of 3T produces a Zeeman patternwith spread of 1.4 cm–1. In this case Zeeman spread is large compared with the line separation.Therefore, the magnetic field of 3T is a strong field in this case.

Fig. 4.3.5

4.4 STARK EFFECT

The splitting of spectral lines under the action of electric field is called Stark effect. It was Stark,who in 1913 demonstrated the splitting of Balmer lines of hydrogen atom in an electric field. Aschematic diagram of the apparatus used for the study of transverse Stark effect is shown in thefigure. A perforated cathode is placed at a distance of about 3 mm from the plate P. The pressure ofthe gas in the discharge tube is kept at a value such that the Crookes dark space is several cm long.Under this condition the energy is less than that required to produce ionization. A very high potentialgradient is applied between the plate P and the cathode.

Under the action of applied electric field , the initially degenerate level with quantum numbern, splits into 2n – 1 components with different energies. An important difference between the Zeemanand Stark effect is that each pair of levels with MJ = + J and MJ = – J arising from a given level hasthe exactly the same energy in electric field.

For a weak field , the Stark effect in hydrogen lines produces a symmetrical pattern and theline separations are proportional to the strength of the applied electric field . This effect is knownas linear or first order Stark Effect.


Fig. 4.4.1 Experimental set up to observe Stark effect

When exceeds 107 V/m, the line separations are proportional to 2. This called quadratic orsecond order Stark Effect.

Let examine the linear Stark effect in Na D-lines, which are emitted in the transitions32P1/2 – 32S1/2 and 32P3/2 – 32S1/2

Fig. 4.4.2 (a) Field free transitions (b) Transitions in presence of field

The level with J = 3/2 splits into two sub-levels with MJ = ± 3/2 and MJ = ± 1/2. The levels2P1/2 and 2S1/2 do not split. In presence of electric field three transitions will take place.

The quadratic Stark effect can be explained as follows. The electric field produces a relativeshift of center of negative and positive charge in the atom. Thus, the electric field induces an electricdipole moment in the atom. This action of field is called electric polarization. The magnitude of


electric dipole moment is proportional to the strength of the field. The dipole moment interacts withthe applied field . The electric field exerts a torque, which makes the angular momentum J of theatom precess about the direction of the electric field such that the component of J along is a constant.An increase in electric field causes increase in precessional velocity. The energy shift is given by

E = µ , µ is dipole moment.Since µ is proportional to , the splitting of energy level is proportional to the square of the

electric field . For potassium doublet, ( = 4044 Å, 52P3/2 – 42S1/2 and = 4047 Å, 52P1/2 – 52S1/2)the wavelength shift is plotted against 2, a straight line is observed.

Fig. 4.4.3 Quadratic Stark effect in potassium doublet 4044 Å 4047 Å.

Fig. 4.4.4 Quadratic Stark effect

SOLVED EXAMPLES

Ex. 1. Calculate the normal Zeeman splitting of the line 6438 Å in a magnetic field of 0.5 T.

Sol. Zeeman shift 2B

4

ed

mc

λλ =

π.


Substituting e = 1.6 × 10–19 C, B = 0.5 T, = 6438 × 10–10 m, m = 9.1 × 10–31 kg,c = 3 × 108 m/s, we find

d = 0.097 × 10–10 m = 0. 097 Å.

Ex. 2. Compute the separation in Angstrom units of the outer two lines of a normal Zeeman patternfor spectral line of wavelength 612 nm in a magnetic field of 10 kg (1 g = 10–4 T).

Sol. Normal Zeeman shift

2B

4

ed

mc

λλ =

πSubstituting the given values we find

d = 0.175 ÅSeparation of outer lines = 2 d = 0.35 Å.

Ex. 3. An atom in the state 2P3/2 is located in an external magnetic field of 1.0 kg. In terms ofvector model, find the angular velocity of precession of the total angular momentum of the atom.

Sol. Lande g-factor of the state 2P3/2

J(J 1) S(S 1) L(L 1) 41

2J(J 1) 3g

+ + + − += + =

+

The angular velocity of precession

24 110

34

B 4 (9.27 10 J/T)(1 10 T)1.2 10 rad/s

3 (1.054 10 Js)

g − −β

−

µ × × ×ω = = = ×

× ×

Ex. 4. Into what number of sub-shells are the following terms split in a weak magnetic field?

(i) 3P0 (ii) 2F5/2 (iii) 4D1/2.

Sol. (i) For the state 3P0 , J = 0, MJ = 0. This energy level does not split.(ii) For the state 2F5/2 , g 0, J = 5/2, MJ has 2J + 1 = 6 values. Hence, it splits into six sub-

levels.(iii) For the state 4D1/2, g = 0, it does not split at all.

Ex. 5. An atom is located in a magnetic field of 2.5 kg. Find the value of total splitting of thefollowing terms (i) 1D (ii) 3F4.

Sol. (i) For the state 1D, we have S = 0, L = 2, J = 2, g = 1. The total splitting is given byE = 4 g m B

= 4 × (5.79 × 10 –5 eV/T) (0.25T)= 57. 9 × 10–6 eV.

(ii) For the state 3F4, S = 1, L = 3, J = 4, g = 5/4. This level splits into 9 sub-levels. Thetotal splitting is given by

E = 8 (g µB B)

= 55 eV8 5.79 10 0.25T

4 T−× × × ×

= 144.8 5 10–6 eV.


Ex. 6. What kind of Zeeman effect, normal or anomalous is observed in a weak magnetic field inthe case spectral lines caused by the following transitions:

(i) 1P 1S (ii) 2D5/2 2P3/2 (iii) 3D1 3P0 (iv) 5I5 5H4.Sol. (i) 1P 1S. For the lower level L = 0, S = 0, J = 0, g = 1. It does not split.For the upper level L = 1, S = 0, J = 1, g = 1, MJ = –1, 0, 1. It splits into three levels.Normal Zeeman effect is observed.(ii) 2D5/2 2P3/2. For the lower level L = 1, S = 1/2, J = 3/2, g = 4/3. It splits into four sub-

levels.For the upper level L = 2, S = 1/2, J = 5/2, g = 6/5. It splits into six sub-levels.Anomalous Zeeman pattern is observed.(iii) 3D1 3P0. Lower level: L = 1, S = 1, J = 0, g = 1. It does not split.Upper level : L = 2, S = 1, J = 1, g = 1/2. It splits into three sub-levels.Normal Zeeman pattern is observed.(iv) 5I5 5H4. Lower level: L = 5, S = 2, J = 4, g = 19/20. Upper level:L = 6, S = 2, J = 5, g = 19/20. Since, the two levels have identical Lande g-factor, normal

Zeeman pattern is observed.

Ex. 7. Draw the diagram of the permitted transitions between the terms 2P3/2 and 2S1/2 in a weakmagnetic field. Find the displacements of the Zeeman components of that line in a magnetic field of0.45 µb/m2.

Sol. Zeeman shift is given by

( )βµ′ ′∆ω = −

J J

BM Mg g

State L S J g M J2P3/2 1 1/2 3/2 4/3 3/2, 1/2, –1/2, –3/22S1/2 0 1/2 1/2 2 1/2, –1/2

Now 11B3.95 10 rad/sβµ

= ×

and

′ ′ − = − − −J JM M 5 / 3, 3/ 3, 1/ 3, 1/ 3, 3 / 3, 5 / 3.g g

Hence = ± 6.59 × 1010, ± 3.95 × 1010, ± 1.31 × 1010 rad/s.

Fig. E-7


Ex. 8. A beam of electrons enters a uniform magnetic field of flux density 1.2 b/m2. Find theenergy difference in electron volts between electrons whose spins are parallel and anti-parallel to thefield.

Sol. The energy difference

E 2 BJsg β∆ = µ

where = 249.273 10 J/T2

e

m−= ×

= 5.79 × 10–5 eV/T

For a pure spin system J = s = 1/2 and g = gs = 2.

51 1

E 2(2) 2(2)(5.79 10 eV. )(1.2T)2 2 2

es

m−

∆ = = ×

= 1.39 × 10–4 eV.

Ex. 9. The spectral line resulting from the transition 2p 1s ( = 1210 Å) in hydrogen atom issubjected to a huge magnetic field of 5T. Find the wavelengths of the spectral lines in the pattern.

Sol. In strong magnetic field B, the spin-orbit interaction is negligible. The orbital and spinangular momenta are separately quantized. The anomalous Zeeman pattern reverts to normal Zeemanpattern (Paschen-Back effect).

Fig. E-9


The change in magnetic energy of upper level due to interaction of magnetic moment of atomwith applied magnetic field is given by

( )upper L SE B M 2M2

ed

m= +

5L S

eV5.79 10 (5T)(M 2M )

T−

= × +

5L S(28.94 10 eV)(M 2M )−= × +

The values of ML + 2MS for upper levels are 2, 1, 0, 0, – 1, – 2. The states with the samevalues of ML + 2MS coincide. Hence, the upper level splits in 5 sub-levels with energies shown inthe figure. Similarly the lower energy level splits into two sub-levels.

The wavelength of a spectral line is given by

upper lowerE Ech

= −λ

The change in wavelength due to change in energy is

upper lower2E E

chd d d− λ = −

λ

2

upper lower( E E )d d dch

λλ = − −

Å

Å= − −

2

upper lower

(1210 )( E E )

12400 V.d d

e

Åupper lower118 ( E E )

Vd d

e

= − −

The values of (dEupper – dElower) for the transitions marked a, b, c, d, e, f in the figures are

(28.94, 0, – 28.94, 28.94, 0, – 28.94 ) × 10 – 5 eV respectively.The transitions (a, d), (b, e) and (c, f) coincide and hence, we get only three spectral lines.For transitions marked a and d

Åupper lower118 ( E E )

eVd d d

λ = − −

ÅÅ5118 (57.88 28.94) 10 0.034

eV−

= − − × = −

Similarly for the transitions marked b, e, d = 0 and that for c, f, d = 0.034 Å. The wavelengthsof the lines in the pattern are 1210 Å and 1210 ± 0.034 Å.



1. What is Zeeman effect? Give the simple theory of normal Zeeman effect and obtain expression for theZeeman effect. Explain the difference between a normal and anomalous Zeeman effects.

What should be the minimum value of the magnetic field to observe normal Zeeman effect?

Evaluate the magnetic field at which the Zeeman shift at 5400 Å spectral line is 0.1 Å.

2. Explain why does normal Zeeman effect occur only in atoms with even number of electrons? Differentiatebetween the normal Zeeman effect and anomalous Zeeman effect. Find an expression for Zeeman splittingfor a two electron system.

Compute the Zeeman pattern for a diffuse series singlet-singlet transition.

3. What is anomalous Zeeman effect? Discuss the Zeeman splitting pattern of D1 and D2 lines of sodium.

4. Discuss the theory of Paschen-Back effect in one electron system and compute the pattern for the principalseries doublet.

5. Distinguish between Zeeman and Paschen-Back effect. Outline the theory of Paschen-Back effect in oneelectron system and discuss the Paschen-Back pattern of 2P 2S transition.

6. What is Lande g-factor? Calculate its value for the energy levels involved in the transitions 2P3/2 2S1/2

and 1P1 1S0.

Can you get anomalous Zeeman pattern in both these transitions? Explain your answer.

7. Why Paschen-Back effect is called transition effect? Explain.

Obtain an expression for the change in energy in Paschen-Back effect.

Differentiate between normal Zeeman effect and Paschen-Back effect.

What will be the Zeeman shift for 6000 Å line in a magnetic field of 4 Tesla?

8. What is difference between normal Zeeman and Paschen-Back effect? Explain splitting of D-lines of sodiumby Paschen-Back effect?

9. What is Stark effect? Discuss the weak-field Stark effect in H line of hydrogen. What are the main differencebetween Stark effect and Zeeman effect?

10. Calculate the g value for (i) 2G9/2 (ii) 3F2.

[Ans. (i) L = 4, S = 1/2 , J = 9/2, g = 10/9

(ii) L = 3, S = 1, J = 2, g = 2/3]

11. Find the magnetic moment of an atom in state 3P2.

[Ans. J(J 1),g βµ = µ + g = 3/2, µ = βµ3

32

]

12. Find the magnetic moment of an atom in the state 2P3/2 . In how many sub-states will this state split in aweak magnetic field?

[Ans. βµ = µ2

15 .3

Number of sub-states = 2J + 1 = 4]

13. An element emits light of wavelength 4500 Å. It is placed in a magnetic field of strength 0.3 Tesla. Howfar apart are the Zeeman components? e/m = 1.76 × 1011 C/kg, c = 3 × 108 m/s.

[Ans. Å1 2 14B14.0 m , 283.5 10 m 0.02835

4

e

mc− −∆ν = = ∆λ = λ ∆ν = × =

π]

CHAPTER

X-RAYS AND X-RAY SPECTRA

5.1 INTRODUCTION

The German physicist Roentgen, while studying the properties of cathode rays in 1885 found that avery penetrating radiation was coming from the discharge tube. Since the nature of radiation washitherto unknown, it was called X-rays. Subsequently, the properties of this radiation were thoroughlyinvestigated and established but they are still known as X-rays. As we have already shown in theprevious chapter that the characteristic spectrum in the visible region is correlated with the motionof electrons in atom, it is natural to seek a relationship between the structure of matter and characteristicX-ray spectrum. However, before turning to find such correlation, it is necessary to follow theexperimental steps through which its basic nature was established.

The discovery of new and unknown radiation stimulated the imagination of many workers whomade serious attempts to establish its nature. On the basis of several experiments, it was recognizedthat most of the properties of this radiation are similar to those of electromagnetic waves of veryshort wavelengths. Many attempts were made to observe the diffraction of X-rays by passing themthrough narrow slits. In 1906, Walter and Pohl carried out such an experiment and recorded thebroadening of the image. Arnold Sommerfeld (1912) calculated the wavelength of X-rays and foundthat it is greater than 10 Å. Later on, ruled gratings were used to observe X-ray diffraction and itwas found that the wavelength of this radiation was of the order of 1 Å.

5.2 LAUE PHOTOGRAPH

Although several methods of accurate measurements of the wavelength of X-rays were beingdeveloped, a German scientist, Von Laue made a remarkable discovery, which proved to be oftremendous importance for future development of physics, chemistry, metallurgy and medical science.Laue’s ingenious suggestion was as follows. Since the wavelength of X-rays is of the order of 1 Å,therefore, it is possible to observe the diffraction pattern if they are scattered by a crystal, which is athree-dimensional grating with inter atomic spacing of equal magnitude. On performing the experimenthe actually observed a series of regularly spaced spots, called Laue’s spots. Laue definitely establishedthat these spots could only be due to diffraction of waves from atoms of the crystal. This experiment

X-Rays and X-Ray Spectra 521

proved two assumptions simultaneously, first: X-rays are electromagnetic waves of short wavelengths,second: atoms are regularly arranged in the crystal. Laue’s experiment heralded a new era in physicsby providing a very powerful tool for the determination crystal structure.

Fig. 5.2.1 Schematic diagram arrangement for getting Laue’s photographs

5.3 CONTINUOUS AND CHARACTERISTIC X-RAYS

X-rays are produced when high-energy electrons, accelerated under high potential difference, strikea heavy metallic target. The distribution of X-rays intensity among various wavelengths at differentaccelerating potentials with molybdenum and tungsten as targets is shown in the Fig. (5.3.1). Theimportant features of the curves may be stated as follows.

Fig. 5.3.1 Continuous and characteristic X-rays

(i) For each accelerating potential there exists a short wavelength limit min below which noradiation is produced. This short-wavelength limit depends only on the magnitude of theaccelerating potential and not on the target material. Since the energy is continuouslydistributed among the various wavelengths like that in white radiation therefore the X-rayradiation is called continuous or white radiation. In 1915 Duane and Hunt observed thatmin is proportional to the accelerating voltage.

(ii) With increasing voltage the amount of radiation and intensity of radiation increases. Foreach target when the voltage is increased beyond a certain value, the intensity-wavelength


curve shows several peaks. In case of M0, peaks are observed at 35 kV. The wavelengthsat which these peaks are observed are the characteristic of the target and are calledcharacteristic radiation. The characteristic radiation is superposed on the continuousradiation.

Mechanism of Production of X-rays (Quantum Theory)

The classical theory provides no explanation for the existence of the short wavelength limit and thecharacteristic radiation. On the other hand, quantum theory provides straightforward explanation ofboth the experimental observations. As the highly energetic electron beam passes through the targetmaterial, it collides with the nuclei and the electrons of the target atoms. The electron-electroncollisions, which are slow deceleration process, are not responsible for the production of X-rays.

X-rays are actually produced from less frequent but more catastrophic encounters of the incomingelectrons with the target nuclei. In Fig. (5.3.2), an electron with initial kinetic energy T is shownpassing nearby a nucleus. It interacts with the nucleus via Coulomb field, transferring momentum tothe nucleus. In the process a part of energy of electron is converted into photon. The target nucleusis so heavy that it does not recoil. After encounter the electron moves with remaining energy T'. Thefrequency or the wavelength of the emitted radiation is given by

h = ch/ = T – T' ...(5.3.1)

The incoming electron suffers many encounters with target nuclei before coming to rest andloses different amount of energy in such collisions. The emitted radiation, therefore, forms acontinuous spectrum.

The X-ray photon of shortest wavelength (highest frequency) is emitted when the incident electronloses all of its kinetic energy in a single encounter. In this case T' = 0 and hence

maxmin

Tch

hν = =λ where T = eV

Fig. 5.3.2 Emission of continuous X-rays

V is the accelerating voltage. From above relation, we have

min

eVch

=λ

whence

Åmin

12400

eV V(volt)

chλ = = ...(5.3.2)


In expression (5.3.2) V represents the numerical value of applied voltage in volt. The aboveequation is known as Duane and Hunt law. Thus, we see that quantum theory provides an easy andconvincing explanation for the existence of short wavelength limit. If we set h = 0 we get min

= 0.This means that the existence of min is a quantum mechanical phenomenon. The emission of X-rayfrom decelerating electron is called Bremsstrahlung process. This process is sometimes called inversephotoelectric effect. In a photoelectric process, photon is absorbed; its energy and momentum aretransferred to electron. In bremsstrahlung process, a photon is created; its energy and momentum arederived from the electron-nuclear collision.

5.4 X-RAY ENERGY LEVELS AND CHARACTERISTIC X-RAYS

When a high-energetic electron of cathode ray strikes the target (anticathode) of the X-ray tube, itpenetrates the target atom and knocks out one of the electron from the inner shell such as K or Lshell. Thus, a vacancy or hole is created in the atom. If the vacancy or hole is created in K shell, itmay be filled by electron from the L or M or N shell etc. If the vacancy is created in L shell, it maybe filled by electron from L or M shell or higher shells. If the vacancy of K shell is filled up byelectron from L shell, the vacancy moves from K to L shell. Thus, the electronic transition is oppositeto the vacancy (hole) transition.

Electronic transition L KHole transition K L

Fig. 5.4.1 Electron and hole transition

When the electrons make transition from L, M, N… shells to K shell, the spectral lines K series(K, K, ….) are emitted. Similarly, the electronic transitions from M, N, ….. levels to L level giverise to spectral lines of L series (L, L….). The observations, made by high resolving powerinstruments, indicate that the individual lines of K series are not single but possess fine structure.This is true for the lines of L series too. Thus, the X-ray spectral lines show fine structure. Theexistence of fine structure can be explained as follows.

X-ray energy levels are specified by four quantum numbers: principal quantum number (n),orbital angular momentum quantum number (l) and total angular momentum quantum number ( j)

and magnetic quantum number mj. For K shell, n = 1, l = 0 and j = ⊕ = ⊕ =0 1/ 2 1/ 2.l s Thus wehave a single K level, which can accommodate two electrons in the states corresponding to mj = – 1/2

and 1/2. Similarly for L shell, n = 2, l = 0, 1, j = l 0 1/ 2 and 1 1/ 2 . ., 1/ 2,1/ 2, 3/ 2.s i e j⊕ = ⊕ ⊕ =Thus, there are three levels in L shell designated by LI, LII, LIII. Of these the first level accommodates


two electrons corresponding to mj = – 1/2, 1/2, the second level two electrons corresponding tomj = – 1/2, 1/2, and the third level four electrons corresponding to mj = − −3/ 2, 1/ 2, 1/ 2, 3/ 2. Thus,the L shell can contain 8 electrons. Similarly for M shell, n = 3, we have five levels designated asMI, MII, MIII, MIV, MV. The N shell contains seven levels, which are denoted by NI, NII, NIII, NIV,NV, NVI, NVII. The spectral terms corresponding to these levels are:

K, LI, LII, LIII, MI, MII, MIII, MIV, MV,12S1/2, 22S1/2, 22P1/2, 22P3/2, 32S1/2, 32P1/2, 32P3/2, 32D3/2, 32D5/2.

NI, NII, NIII, NIV, NV, NVI, NVII

42S1/2, 42P1/2, 42P3/2, 42D3/2, 42D5/2, 42F5/2, 42F7/2

The filling of electrons in these shells is shown in the table.In X-ray spectroscopy a neutral atom in the ground state is assigned zero energy and that having

vacancy (hole) in the inner level is assigned positive energy. For instance, if K electron is missing, ithas a vacancy there and it is assumed to have positive energy equal to the energy required to removethe K electron from the atom. Obviously the energy of the atom having vacancy in K shell has highestenergy. For tungsten (74W) the energy levels are shown in the Fig. 5.4.2.

Table 5.4.1

MaximumShell l j mj no. of Level

electrons notation

K 0 1/2 1/2 , – 1/2 2 K(12S1/2)

(n = 1)

L 0 1/2 1/2 , – 1/2 2 LI (22S1/2)

(n = 2) 1 1/2 1/2 , – 1/2 2 LII(22P1/2)

1 3/2 3/2 , 1/2 , – 1/2 , – 3/2 4 LIII(22P3/2)

M 0 1/2 1/2 , – 1/2 2 MI (32S1/2)

(n = 3) 1 1/2 1/2 , – 1/2 2 MII(32P1/2)

1 3/2 3/2 , 1/2 , – 1/2 , – 3/2 4 MIII(32P3/2)

2 3/2 3/2 , 1/2 , – 1/2 , –3/2 4 MIV(32D3/2)

2 5/2 5/2, 3/2 , 1/2 , – 1/2 , –3/2, –5/2 6 MV(32D5/2)

0 1/2 1/2 , – 1/2 2 NI (42S1/2)

N 1 1/2 1/2 , – 1/2 2 NII (42P1/2)1 3/2 3/2, 1/2, – 1/2 , –3/2 4 NIII (42P3/2)

n = 4 2 3/2 3/2, 1/2 , –1/2 , –3/2 4 NIV (42D3/2)

2 5/2 5/2, 3/2, 1/2 , –1/2 , 3/2, 5/2 6


The splitting of L level into three components (LI, LII, LIII), M level into five components(MI, MII, MIII, MIV, MV), N level into seven components (NI, NII, NIII, NIV, NV, NVI, NVII) etc., iscalled fine structure.

When a hole makes transition from K to L, M, N … levels etc., the X-ray lines emitted aresaid to constitute K-series. Similarly, when a hole moves from L to M, N, ......etc., level, the linesof L series originate. The allowed transitions are those, which obey the selection rules

l = ± 1, j = 0, ± 1

A few lines emitted from tungsten are shown in the Fig. (5.4.2).

Fig. 5.4.2 X-ray energy levels of tungsten


In hydrogen atom the electron experiences the attractive force due to full nuclear charge ofproton and the energy of electron is given by En = – RchZ2/n2 or the term value by Tn = RZ2/n2. Ina heavy atom with filled K, L, M…. shells, one of the K electron screens the nucleus as a result ofwhich the other electron does not experience the full attractive force of the nucleus. In other words,each of the K electron ‘sees’ the nuclear charge reduced by one unit. This amounts to say that thescreening constant of K electron is 1. Similarly, the each electron in L shell doesn’t ‘see’ the fullnuclear charge due to screening of K electrons (called internal screening) and due to the remainingelectrons of L shell as well as, to some extent, due to outer electrons (called external screening).These two kinds of screening combine and reduce the nuclear charge for L electron. The screeningconstant L electron is nearly 2. For electrons of M shell, two K electrons and eight L electrons screenthe nucleus and hence the screening constant for M shell is close to 10.

Taking spin-orbit interaction and relativity correction into consideration Dirac theory of electrongives the following expression for the term value of an energy level with principal quantumnumber n.

2 2 4

2 4 12

R(Z ) R (Z ) 3T

4n

n

jn n

− σ α − σ = + − +

3 24 6

6 1 1 12 2 2

R (Z ) 1 3 3 5

4 4 2 8

n n n

j j jn

α − σ + + − + + + + ...(5.4.1)

5.5 MOSELEY’S LAW

In 1913, a young British physicist Henry Moseley (1887–1915) was undertaking a systematic studyof the characteristic spectra of a large number of elements. During his investigation he found andthat the frequencies of the emitted lines were the characteristic properties of the elements. He establishedthat square root of the frequencies of the spectral lines, for example K line, were proportional tothe atomic number Z of the element.

Fig. 5.5.1 Moseley’s law


(Z )a bν = − ...(5.5.1)

where a and b are constants for a given series.Moseley’s discovery is of extreme importance because it established that atomic number is morefundamental quantity than atomic weight. This fact formed the basis of deciding the correct order ofelements in the periodic table.

Moseley discovery came after Bohr’s theory of hydrogen atom was published. It is remarkableto note that Bohr’s theory, although proposed for one electron atom, was capable of explaining theMoseley law. According to Bohr’s theory the wavelength of a spectral line is given by

2

2 2

1 1 1RZ

f in n

= − λ

For K line nf = 1, ni = 2 and therefore

2

K

1 3RZ

4α=

λ ...(5.5.2)

In Bohr’s theory of one electron atom, the energy of electron in a shell is derived by assumingthat the electron “sees” full nuclear charge. This assumption is valid for one electron only. In manyelectron atoms this assumption requires modification. In K shell there are two electrons; each electronshields or screens the nucleus for the other as a result of which an electron in the atom does notexperience the attraction of the full nuclear charge. Similarly, an electron in the L shell is shieldedby K electrons and to some extent by the other electrons of the L shell itself. Therefore each electronin the L shell does not see the full nuclear charge. This lessening effect of nuclear charge bysurrounding electrons is known as shielding effect or screening effect. If the shielding effect is takeninto account, the nuclear charge number Z in Eqn. (5.5.2) must be replaced by (Z – b) where b is ameasure of shielding effect. The formula (5.5.1) then becomes

2

K

1 3R(Z )

4b

α= −

λ ...(5.5.3)

For the lines of K series, b = 1 and those of L series, b = 7.4. Hence

2

K

1 3R(Z 1)

4α= −

λ...(5.5.4)

or αν = − 2K

3 R(Z 1)

4

c

5.6 SPIN-RELATIVITY DOUBLET OR REGULAR DOUBLET

A pair of energy levels having the same n, S, L values but different J values is called spin-relativity(regular) doublet. Such pairs are (LII, LIII), (MII, MIII), (MIV, MV), (NII, NIII), (NIV, NV),


(NVI, NVII) or (22P1/2, 22P3/2), (32P1/2, 32P3/2), (32D3/2, 32D5/2), (42P1/2, 42P3/2), (42D3/2,42D5/2), (42F5/2, 42F7/2).

It is observed that the difference of wave numbers of spin relativity doublets is approximatelyproportional to the fourth power of effective atomic number. This fact is known as spin-relativity(regular) doublet law. This law can be obtained from the Dirac result for the term value. Let uscalculate the wave number separation for the screening doublet LII and LIII.

For LII, n = 2, L = 1, S = 1/2 , J = 1/2, and for LIII, n = 2, L = 2, S = 1/2, J = 3/2. Thus

T2(LII) = 2 2 4

342 4

R(Z ) R (Z )2

2 2

− σ α − σ+ −

+

3 24 6

6

R (Z ) 1 2 3 2 3 2 5

4 1 4 1 2 1 82

α − σ + − +

...(5.6.1)

T2(LIII) =

2 2 4

2 4

R(Z ) R (Z ) 31

42 2

− σ α − σ + −

+

4 6

6

R (Z ) 1 3 3 5

4 4 2 82

α − σ + − +

...(5.6.2)

T2(LII) – T2(LIII) =

2 42R (Z ) 5

1 (Z )16 8

α − σ + − σ

Neglecting the second term in the curly bracket we find that

α − σ

∆ν = − = σ =2 4

2 II 2 III

R (Z )T (L ) T (L ) , 3.5

16

For a given value of L, = constant, hence 4(Z ) .∆ν ∝ − σSimilarly, for MII and MIII, we find

2 42 2

II IIIR (Z ) 3 31

(M , M ) (Z ) , 8.581 2 32

α − σ ∆ν = + α − σ σ =

...(5.6.3)

5.7 SCREENING (IRREGULAR) DOUBLET

A pair of levels having the same values of n, S and J but different values of L is called screeningdoublet. Examples of such pairs are:

(LI, LII), (MI, MII), (MIII, MIV), (NI, NII), (NIII, NIV), (NV, NVI) or (22S1/2, 22P1/2),

(32S1/2, 32P1/2), (32P3/2, 32D3/2), (42S1/2, 42P1/2), (42P3/2, 42D3/2), (42D5/2, 42F5/2).


It is observed that the difference in term values of pair of screening doublets is proportional tothe difference in the values of screening constants of the two components of the doublet. This law isknown as the screening (irregular) doublet law.

Confining ourselves to the first term in the expression for term value, we have

2

2

R(Z )Tn

n

− σ= ...(5.7.1)

R

T (Z )n n= − σ ...(5.7.2)

For the screening doublet LI and LII, n = 2, we have

= − σ2 I I

RT (L ) (Z )

2...(5.7.3)

and 2 II II

RT (L ) (Z )

2= − σ ...(5.7.4)

Therefore

[ ]− = σ − σ2 I 2 II II I

RT (L ) T (L )

2...(5.7.5)

or ( ) II IT const.( )∆ = σ − σ

which is independent of Z.

5.8 ABSORPTION OF X-RAYS

If a collimated beam of X-rays is made to pass through a medium, the intensity of beam decreases.Let I0 be the intensity of incident beam, I the intensity after traversing a distance x , – dI the reductionin intensity after traversing an infinitesimal distance dx. It is found that fractional loss of intensity isproportional to dx

I

I

ddx− ∝

− = µI

I

ddx ...(5.8.1)

where is a constant called the linear absorption coefficient and has the dimensions of m –1. Thevalue of depends on the wavelength of X-rays and the absorbing material.

Integrating Eqn. (5.8.1), we have

0

I

I

I

I

d∫ =

0

x

dx−µ∫I = I0 e – x ...(5.8.2)


Fig. 5.8.1 Absorption of X-rays

Thus, the intensity of X-rays falls exponentially with the distance x traversed in the absorbingmaterial. Eqn. (5.8.1) is usually expressed in terms of mass absorption (attenuation) coefficient m

which defined by m = µ/, being the density of the material. The linear absorption coefficient ismeasure of the probability per unit length for removal of X-ray photons while passing through amaterial medium.

Mechanism of Absorption

The primary interaction processes responsible for the absorption of electromagnetic radiation are(i) photoelectric effect, (ii) compton scattering and (iii) pair production. Because X-ray photons haveenergies in the range 1–100 KeV, they cannot produce pair production (for which threshold energy =1.02 MeV). Therefore, the reduction in intensity X-rays is caused by only first two factors, thephotoelectric effect being the dominant mechanism.

Fig. 5.8.2


Photoelectric Effect

The photoelectric effect is the term applied to a process in which an atom absorbs a photon andemits an electron. A peculiar characteristic of photoelectric effect is that a free electron cannot absorbor emit a photon because of the combined effect of the laws of conservation of energy and momentum.The probability of photoelectric effect is maximum if the energy of photon is comparable to thebinding energies of electron. An increase in energy of photon should be accompanied by a drasticfall in the absorption coefficient because the atomic electrons become progressively more like freeelectrons. The binding energies of K electrons vary from 1.56 KeV for Al to 88.10 KeV for Pb,which lies in the X-ray range. Binding energies of L are less than those of K electrons. Hencephotoelectric effect is more likely to take place with K and L electrons. Now we can explain thedependence of absorption coefficient on the energy (or wavelength) of X-ray photon. This dependenceis shown in the Fig. (5.8.2). At very high energies (small wavelengths) of X-ray photons, the atomicelectrons of the absorbing material behave like free electrons and hence the probability of photoelectricabsorption is small. With decrease in energy (or increase in wavelength) of X-ray photons, theprobability of photoelectric absorption increases and becomes maximum when the energy of photonbecomes equal to the binding energy of K electrons. At this energy maximum number of K electronsare ejected from the absorbing material. The wavelength () of the X-ray corresponding to maximumabsorption is called K absorption edge. Obviously the wavelength of K absorption edge givesthe binding energy E of K electron.

KK

Ech

=λ ...(5.8.3)

Fig. 5.8.3 K and L absorption edges of Pb

At energy lower than EK or at wavelength greater than k, the absorption suddenly drops becauseX-ray photons are unable to eject K electrons. Although this energy is sufficient to eject L electronsbut probability of this ejection is small. With further decrease in energy (or increase in wavelength)the photoelectric absorption (associated with the ejection of L electrons) increases and again becomesmaximum when the energy of photon is just equal to the binding energy of L electron. Since, thereare three L levels, there are three absorption edges LI, LII and LIII. With further increase in energywe observe five absorption edges MI, MII, MIII, MIV, MV corresponding to five M levels.


Except at absorption edges, the dependence of photoelectric absorption on atomic number Z ofthe absorbing material and energy (E) of the photon is approximately described by the formula

4 3

photo CZ E−µ =

4 3C Z′= λ ...(5.8.4)

Absorption due to Compton Scattering

The scattering of X-ray photons by weakly bound atomic electrons is called the Compton effect. Asa result of Compton scattering the X-ray photons are deflected from their original direction and hencedon’t reach the detector. The absorption coefficient corresponding to Compton effect is given by

Compton

ZC , C = constant

Eµ = ...(5.8.5)

Fig. 5.8.4 Absorption coefficient vs energy of X-ray photon

In lead the Compton effect supersedes the photoelectric effect at energy E > 0.5 MeV. Thecontributions of photoelectric absorption, Compton scattering and pair production to the total absorptionas function of energy is shown in the Fig. (5.8.4).

5.9 BRAGG’S LAW

A simplified way of looking at the process of X-ray diffraction by a crystal was proposed by W.L.Bragg. He suggested that through any crystal a set of equidistant parallel planes might be imaginedthrough all the atoms of the crystal. In Fig. (5.9.1) some typical systems of planes with their spacingare shown. These planes are called Bragg planes and their spacing Bragg spacing.


Fig. 5.9.1 Atoms in a crystal are arranged in a regular way in three dimensions. Here a hypothetical two-dimensional arrangements of atoms is shown. Atoms in a crystal are very close together. In the figure

they are shown far apart. Some families of Bragg planes with their spacing are shown

Consider a set of parallel planes of atoms in a crystal, two of which are represented by thelines AA and BB. The actual planes are perpendicular to the plane of the paper. Suppose that a beamof monochromatic X-rays is incident at these planes. Let the incident rays make angle with theplanes. This angle is called the glancing angle. The incident rays 1 and 2 will be scattered by atomsof the upper plane. These scattered waves reinforce in the direction ' = , which is the condition ofspecular reflection. Thus, the atomic planes act as a mirror. Now consider the condition ofreinforcement of waves reflected from successive planes that are parallel to AA. The requirement tobe satisfied for the constructive interference is that the path difference for rays reflected from successiveplanes be equal to integral number of wavelength.

Fig. 5.9.2 Reflection of X-rays from Bragg planes

From the Fig. (5.9.2), we can see that the path difference between rays reflected from thesuccessive planes having inter-planar distance d is 2d sin . Hence the condition for constructiveinterference is

2d sin = n , n = 1, 2, 3,….. ...(5.9.1)

This equation is known as Bragg’s law.


Bragg’s Spectrometer

This spectrometer was designed by W. H. Bragg and his son W. L. Bragg to determine the wavelengthof X-rays. It consists of (i) a crystal (usually of rock-salt or calcite or mica) mounted on a table,which can be rotated about a vertical axis (ii) a device for detecting reflected X-rays from the crystal.The detector, which is usually an ionization chamber, is mounted on a arm capable of rotating aboutthe same vertical axis. Making use of slits, a narrow beam of X-rays is allowed to fall on the crystalat angle, say and the detector rotated to receive those rays, which are reflected at angle . Fromthe very setting of the detector, it is clear that the X-rays reaching it obey the Bragg’s condition. Byvarying the angle (by rotating the crystal and the detector both) reflections of different orders canbe recorded. If 1, 2, 3,… are the Bragg’s angles corresponding to the first, second, third ordersthen we have

2d sin 1 = 2d sin 2 = 2

If the Bragg’s spacing d is known, the wavelength can be calculated.From the following data let us calculate the inter-planar distance in sodium chloride crystal,

which is a cubic lattice.Molecular weight = 58.5 kg/k molDensity = 2.16 × 10 3 kg/m3

Avogadro’s number = 6.02 × 1026 molecules/k mol

Molar volume M

Vm =ρ

Volume available to a single molecule = A A

V M

N Nm =

ρ

Volume available to a single ion = A

M

2 Nρ

If d is the inter-atomic distance then 3

A

M

2 Nd =

ρ

Fig. 5.9.3 Schematic diagram of Bragg’s spectrometer


1/3 1/ 3

3 26A

M 58.5

2 N 2 2.16 10 6.02 10d

= = ρ × × × ×

102.82 10 m 2.82 Å.−= × =

SOLVED EXAMPLES

Ex. 1. What is the minimum voltage across an X-ray tube that will produce an X-ray having (i) theCompton wavelength, (ii) a wavelength of 1 Å and (iii) a wavelength to be capable of pair production.

Sol. (i) Compton wavelength = min = 0.024 Å

Å

min

12400 eV

eVxλ = where x is the numerical value of applied voltage in volt.

Å Å

Å3

min

12400eV. 12400 eV511 10 volt

0.024x = = = ×

λ

(ii) Å Å

Åmin

12400eV. 12400eV.12.4 KV.

1x = = =

λ(iii) For pair production, the minimum energy of X-ray photon is 1.02 MeV. The electrons

striking the target in X-ray tube must have at least this much energy. The tube, therefore,must be operated at 1.02 million volt.

Ex. 2. A voltage applied to an X-ray being increased n times, the short wave limit of X-raycontinuous spectrum shifts by d = 26 pm. If n = 3/2, find the initial voltage applied to the tube.

Sol. At initial voltage

Vm

ch

eλ =

At the final voltage

Vm

ch

e′λ =

′

1 1 1 1

V V V Vm m

ch chd

e e n

′λ = λ − λ = − = − ′

Å

Å

1 1 12400eV.V 15.9 KV

. 3 .(0.26 )

n ch

n e d e

− = = = λ

Ex. 3. The wavelength of K line of an element is 1.54 Å. Determine the atomic number of thetarget element.

Sol. For K line, 21 3 4R(Z 1) whence Z 1

4 3 R= − = +

λ λ = + ≈Z 1 28.2 29.


Ex. 4. Find the wavelength of K line in Copper (Z = 29) if the wavelength of K line in iron isknown to be 193 pm.

Sol. For copper the wavelength of K line is

2

1

1 3R (Z 1)

4= −

λand that for iron is

2

2

1 3R (Z 1)

4= −

′λ

221

22

(Z 1) 28

27(Z 1)

′λ − = = λ −

228

(193 pm) 154 pm.27

′λ = =

Ex. 5. Proceeding from Moseley’s law find the wavelength of K line in Al and Co.

Sol. Moseley’s law: 21 3

R(Z 1)4

= −λ

2 7 2

4 4

3R(Z 1) 3 1.097 10 (13 1)λ = =

− × × −= 844 × 10 –12 m = 844 pm

For cobalt (Z = 27), = 180 pm.

Ex. 6. How many elements are there in a row between those whose wavelengths of K line are equalto 250 pm and 179 pm?

Sol. For K line: 21 3 4R(Z 1) Z 1

4 3 R= − ⇒ = +

λ λ

Putting 1 = 250 pm and 2 = 179 pm in above equation, we get

Z1 = 23 and Z2 = 27

The required elements are: Z = 24, 25 and 26.

Ex. 7. Calculate the binding energy of a K electron in vanadium (Z = 23) whose L absorption edgehas a wavelength L = 2.4 nm.

Sol. Binding energy of L electron

LL

Ech

=λ

Å

Å

12400 eV.516 KeV

24= =


Wavelength of K line is given by

2

K LK

3E E R (Z 1)

4

chch

α− = = −

λ , (Rch = 13.6 eV)

23 13.6 V (23 1)4.937KeV

4

e× × −= =

Binding energy of K electron EK = 0.516 KeV + 4.937 KeV

= 5.55 KeV.

Ex. 8. The K absorption edge of tungsten is 0.178 Å and the wavelength of K line is 0.210 Å.Determine the wavelength of L absorption edge.

Sol. The wavelength of absorption edge is measure of the binding energy of the correspondingelectron. The binding energy of K electron is

Å

ÅKK

12.4KeV.E 69.67 KeV

0.178

ch= = =

λ

The wavelength of K line is given by

K LK

E Ech

α− =

λ

12.4 KeV.Å59.04 KeV

0.210 Å= =

EL = (69.67 – 59.04) KeV = 10.63 KeVThe wavelength of L absorption edge is given by

Å

Å.L LL L

12.4KeV.E 1.17

E 10.63KeV

ch ch= ⇒ λ = = =

λ

Ex. 9. For tungsten the K absorption edge is 0.18 Å. It is irradiated with X-rays of wavelength0.10 Å. What is the maximum kinetic energy of photoelectrons that are emitted from K shell?

Sol. Binding of K electron Å

ÅKK

12.4 KeV.E 68.89 KeV

0.18

ch= = =

λ

Energy of incident photon Å

Å

12.4 KeVE 124 KeV

0.10

ch= = =

λ

The maximum kinetic energy of ejected electron

K = (124 – 68.89) KeV = 55.11 KeV.


Ex. 10. Find the kinetic energy of the photoelectrons liberated by K radiation of zinc from the Kshell of iron whose K band absorption edge wavelength K = 174 pm.

Sol. Binding energy of K electron in iron

Å

ÅKK

12.4KeV.E 7.126KeV

1.74

ch= = =

λ

Energy of photon of K radiation

E = 2

K

3R (Z 1) 8.578 KeV

4

chch

α= − =

λ

Kinetic energy of photoelectrons liberated from iron

K = KE E (8.578 7.126) KeV 1.452 KeV− = − = .


1. What do you mean by continuous and characteristic X-rays? Describe the mechanism of their production.

2. What is Moseley’s law? How can it be derived from Bohr’s theory?

3. Derive Bragg’s law. How is the wavelength of X-rays determined?

4. Giving energy level diagram, explain the origin of various series of characteristic X-rays.

5. The Duane-Hunt limit of a continuous spectrum, when an X-ray tube is operated at 50 kV is 0.249 × 10–10 m.Calculate the value of the Planck’s constant.

6. If X-rays of wavelength 0.5 Å are detected at an angle of 5º in the first order, what is the spacing betweenthe adjacent planes of the crystal. At what angle will the second order maximum occur?

[Ans. d = 2.86 Å, = 18º3' ]

7. X-rays of wavelength 1.6 Å are diffracted by X-ray spectrograph at an angle of 30º in the second order.Calculate the interatomic spacing.

8. Calculate the longest wavelength that can be analyzed by a rock salt crystal of spacing d = 2.82 Å (i) in thefirst order and (ii) in the second order.

[Ans. 5.64 Å, 2.82 Å]

MOLECULAR SPECTRA OFDIATOMIC MOLECULES

UNIT


blank

CHAPTER

ROTATIONAL SPECTRA OF DIATOMIC

1.1 INTRODUCTION

We know that an atom has its own characteristic discrete energy levels. These energy levels arise dueto different electronic configurations of the atom. When an electron in an atom makes transitionfrom a higher energy state Ei to a lower energy state Ef, a photon of frequency = (Ei – Eƒ)/h isemitted. On the other hand, when an atom absorbs a photon of frequency , it is raised from a lowerenergy state to a higher energy state such that the difference of energy in the final and initial state isequal to the energy of photon h. The electronic transitions from higher energy states to lower energystates give rise to emission spectra whereas those from lower energy states to higher energy statesgive rise to absorption spectra. The emission spectra consist of bright lines and absorption spectraconsist of dark lines. A spectral line is characterized by its frequency and intensity. Like an atom, amolecule has also its own characteristic discrete energy levels. The total energy E of a molecule ismade up of three parts: electronic energy Ee, vibrational energy Ev and rotational energy Er.

E = Ee + Ev + Er

The electronic energy of a molecule arises due to electronic configuration of electrons bondingthe constituent atoms. Different electronic configurations give rise to different electronic energy levels.The difference between two electronic states of a molecule is of the order of 2–10 eV. An electronictransition with a change of energy 5 eV is accompanied by emission or absorption of radiation ofwavelength given by

Å

Å12400 eV.

2480E 5 eV

chλ = = =∆

The corresponding wave number is

6 11

4.03 10 m−ν = = ×λ

This radiation lies in ultraviolet part of the electromagnetic spectrum. In general, electronicspectra of molecules lie in visible and ultraviolet regions.

The vibrational energy of a molecule is due to vibration of its constituent atoms. A simplestdiatomic molecule (such as H2, O2, CO, HCl) may be considered as two point masses m1 and m2

MOLECULES


connected by a spring-like force. Such a molecule can vibrate along the line joining the atoms andmay be treated as a two-body oscillator whose frequency of vibration is given by

kω =µ

or 1

2

kν =π µ

where k = force constant and = 1 2

1 2

m m

m m+ is the reduced mass of the point masses. For O2 molecule,

k = 500 N/m and = 1.4 × 10 –26 kg, the frequency of vibration comes out to be

= 6 × 10 13 rad/s.

The quantum mechanical treatment of harmonic oscillator with potential V = 12

k2, where is

displacement from equilibrium position, shows that the energy of oscillator is quantized and is givenby

( )12

E , 0, 1, 2, 3,.....= ν + ω ν =

where is called vibrational quantum number. The separation of adjacent energy levels is ω i.e.,

34 13 21E (1.06 10 Js)(6 10 s) 6 10 J 0.014 eV.− −∆ = ω = × × = × =

The transition between two adjacent energy levels gives rise to a spectral line of wavelength given by

Å

Å12400 eV

885714.28E 0.014 eV

chλ = = =∆

4 11.13 10 m−ν = ×The radiation of this wavelength lies in infrared region.The rotational energy of a molecule arises due to its rotation about one of its axes. For example,

a diatomic molecule can rotate about an axis passing through its center of mass and perpendicular tothe line joining the atoms. Quantum—mechanical analysis rotational motion of this type of moleculesshows that the energy of molecule is quantized and is given by

2

E J(J 1) ,2Ir = +

J = 0, 1, 2, 3,…….

where J is called angular momentum quantum number. I is moment of inertia of molecule and isgiven by I = r2, µ is reduced mass of molecule and r is distance between atoms.

The spacing of levels with J = 0 and J = 1 is

2

J 1 JE E EI+∆ = − =

For O2 molecule, I = 1.9 × 10 –46 kg m2.

Therefore 34 2

23 446 2

(1.06 10 Js)E 5.9 10 J 3.69 10 eV.

1.9 10 kgm

−− −

−×∆ = = × = ×

×

Rotational Spectra of Diatomic Molecules 543

The wavelength corresponding to this change in energy is

Å

Å74

12400 eV3.36 10 3.36 mm

E 3.69 10 eV

ch−λ = = = × =

∆ ×

2 12.98 10 m−ν = ×The radiation of this wavelength lies in microwave region.The molecular spectra when observed by an instrument of medium resolving power are seen to

consist of bands. When instruments of high resolving power are used the bands are seen to consist ofa great number of closely spaced lines.

It is worth to notice that a molecule can interact with electromagnetic radiation only if it haspermanent electric dipole moment. Homo-nuclear diatomic molecules, such as H2, O2, N2 do nothave electric dipole moment and hence give no absorption spectra. Hetero-nuclear molecules, suchas HCl, CO, have permanent electric dipole moment and hence give absorption spectra. Vibrationalspectra require a change in electric dipole moment during motion of constituent atoms in the molecule.Homo-nuclear diatomic molecules have no dipole moment and hence they do not interact withradiation. No absorption spectra result from these molecules. Hetero-nuclear diatomic molecules havepermanent electric dipole moments and change in dipole moment always occurs during vibration andhence they give rise to absorption spectra. Electronic spectra are shown by all molecules because achange in electronic configuration in a molecule is always accompanied by a change in dipole moment.

The existence of three kinds of energy levels corresponding to three kinds of motion and couplingof these motions give rise to a very complicated energy level diagram of a molecule. To avoid thecomplication in analysis and interpretation of molecular spectra, we shall limit our discussion to simplestmolecules: the diatomic molecules.

1.2 ROTATIONAL SPECTRA—MOLECULE AS RIGID ROTATOR

Pure rotation spectra of diatomic molecules are observed only when all other kinds of energy transitionsdo not occur. For free rotation, the substance must be in gaseous state. A sufficiently low temperature,the thermal energy is too small to alter vibrational and electronic energy of the molecule. At highertemperature, other forms of motion introduce additional energy levels and make the analysis ofspectrum difficult.

A hetero-nuclear diatomic molecule may be thought of as a system of two point particles ofmasses m1 and m2 rigidly connected with a mass-less rod of length r. The molecule is capable ofrotating about an axis passing through center of mass and perpendicular to the line joining theconstituent atoms. Its moment of inertia I about the axis of rotation is

2 21 2

1 2

Im m

r rm m

= = µ+ ...(1.2.1)

where 1 2

1 2

m m

m mµ =

+is reduced mass of the molecule. The rotational kinetic energy of molecule is

2

21 | |E I

2 2I= ω = J

...(1.2.2)


where J is angular momentum of the molecule. According to quantum mechanics, the angularmomentum of a microscopic system is quantized and its magnitude is given by

| | J(J 1)= + J , J = 0, 1, 2, 3,...... ...(1.2.3)

where J is angular momentum quantum number. In view of Eqn. (1.2.3) the rotational kinetic energycan be expressed as

2

2E J(J 1) J(J 1)

2I 8 I

hhc

c

= + = + π

...(1.2.4)

It is customary to express the energy in terms of rotational constant B, defined by

2B

8 I

h

c=

π...(1.2.5)

In terms of B, the energy E is expressed as

E(J) = Bch J (J + 1) ...(1.2.6)

To indicate that E depends on J we write E as E(J), therefore

E(J) = Bch J (J + 1) ...(1.2.7)

The rotational energy levels are

E0 = 0, E1 = 2Bch, E2 = 6Bch, E3 = 12Bch,……

The rotational term values of a rigid rotator are

E(J)

F(J) BJ(J 1)ch

= = + ...(1.2.8)

In practice, rotational spectra are always observed is absorption. Such spectra result due totransitions of molecules from lower rotational energy states to higher energy states by absorbingphotons from the radiation. It is usual practice to denote the rotational quantum number of higherenergy level by J' and that of lower energy level by J". Not all transitions are permitted. Quantummechanics permits only those transitions which obey the selection rules

J = Jf – Ji = ± 1

+1 for absorption and – 1 for emission. In absorption Ji = J" and Jf = J'.

The wave number 1 E

c ch

ν ∆ ν = = = λ of the absorbed radiation corresponding to the transition

J" J' or J" J" + 1 is given by

F(J ) F(J ) B J (J 1) J (J 1)′ ′′ ′ ′ ′′ ′′ν = − = + − +

B (J 1)(J 2) J (J 1)′′ ′′ ′′ ′′ν = + + − + = 2B(J 1)′′ += 2B(J 1)+ , J = 0, 1, 2, 3, ….. ...(1.2.9)

= 2B, 4B, 6B, 8B,…..where we have put J" = J = rotational quantum number of the lower energy level.


The frequency separation on wave number scale is

22B

4 I

h

c∆ν = =

π. ...(1.2.10)

Thus, the absorption spectrum consists of lines which on wave number scale are equally spacedwith constant separation 2B. Measuring the separation of lines, we can calculate the moment of inertiaI and inter-nuclear distance r.

For HF molecule, ∆ν = 4050 m–1, B = 1/2, ∆ν = 2025 m–1. From Eqn. (1.2.10)

34

2 2 8 1

6.6 10 JsI

8 8(3.14) (3 10 m/s)(2025 m )

h

c

−

−×= =

π ∆ν ×I = 1.38 × 10 –47 kg m2

Since I = r2, Å.47 2

1027

I 1.38 10 kg m0.935 10 m 0.94

1.58 10 kgr

−−

−×= = = × =

µ ×

Fig. 1.2.1 Rotational absorption spectrum

The absorption spectrum of HCl contains wave numbers shown in the table given below. Thedifference of consecutive wave numbers is also given. These results show that wave-number separationis nearly constant. A closer look at the separation indicates that it slightly decreases with increasing Jvalues.


Slightly decreasing trend in wave number separation at higher energies implies decrease inrotational constant B and hence increase in moment of inertia I. This means that our rigid rotatormodel for diatomic molecules needs correction. In fact, with increasing J values or energy the frequencyof rotation of molecule increases. The centrifugal action stretches the bond and the atoms are, therefore,pulled apart. This increases the moment of inertia with increasing J values. When the effect ofcentrifugal distortion is taken into consideration, the energy of molecule comes out to be

Fig. 1.2.2 Centrifugal stretching of bond causes decrease in wave number separationand convergence of lines at higher energies

2 2E B J(J 1) D J (J 1)ch ch= + − + ...(1.2.11)

F(J) = B J ( J + 1 ) – D J2 ( J + 1 )2 ...(1.2.12)where B and D are constants and are related to each other through the relation

3

2

4BD ,

k= ω =µω ...(1.2.13)

K is force constant characterizing the elastic force between atoms. The effect of centrifugaldistortion of the energy levels and wave number of spectral lines is shown in the Fig. (1.2.2).


The frequency of lines in wave numbers is given by

F(J ) F(J )′ ′ν = −

2 2 2 2 2B J (J +1) DJ (J +1) B J (J +1) DJ (J +1) ′ ′ ′ ′ ′′ ′′ ′′ ′′= − − −

= 32B(J 1) 4D(J 1)′′ ′′+ − +

= 32B(J 1) 4D(J 1)+ − + ...(1.2.14)

From this equation it is obvious that as J increases, the separation of lines ∆ν decreases. This isin agreement with the experimental observations.

1.3 ISOTOPIC SHIFT

The isotopic exchange of atoms in a diatomic molecule alters the moment of inertia but not the inter-

nuclear distance. As a result of this the wave number separation 2

2B 28 I

h

c

∆ν = = π slightly decreases

with increasing I. If I1 and I2 are the moments of inertia of molecules corresponding to isotopicmasses m1 and m2, the wave numbers of the spectral lines of these molecules are given by

1, J J 1 12B (J 1)→ +ν = + ...(1.3.1)

2, J J 1 22B (J 1)→ +ν = + ...(1.3.2)

where B1 and B2 are the rotational constants of the two molecules. From Eqns. (1.3.1) and (1.3.2)

1 2 1 22(B B )(J 1)∆ν = ν − ν = − + ...(1.3.3)

Therefore, 2 1 1

1 2 2

B I1 1 1

B I

µ∆ν = − = − = − ν µ ...(1.3.4)

Fig. 1.3.1 Effect of isotopic exchange in CO molecule. Continuous lines representabsorption lines in 12CO and dotted lines in 13CO


From Eqn. (1.3.3), we see that isotopic shift ∆ν increases with increasing J values.Equation (1.3.4) may be used to determine the mass of one isotope if other is known. Effect ofisotopic exchange on energy levels and wave number separation is shown in the Fig. (1.3.1).

1.4 INTENSITIES OF SPECTRAL LINES

The intensity of a spectral line is proportional to the number of molecules in the initial state. Thenumber of molecules in the energy state EJ at temperature T is given by

( )J 0 JN N exp E / Tk= − ...(1.4.1)

where N0 is the number of molecules in the state J = 0. The degeneracy of the state J is (2J + 1).Taking degeneracy of the Jth state into consideration, above formula for population of state J becomes

J 0 JN (2J 1)N exp( E / T)k= + − ...(1.4.2)

The variation of NJ with J is shown in the Fig. (1.4.1). The number NJ is maximum for thevalue of J given by

T 1

J2B 2

k

ch= − ...(1.4.3)

Fig. 1.4.1 Variation of population of an energy level with J

Hence, the intensity of spectral line is maximum for this value of J. For lower and higher valuesof J, the intensity is less.

CHAPTER

VIBRATIONAL SPECTRA OF DIATOMIC

2.1 VIBRATIONAL SPECTRA—MOLECULE AS HARMONIC OSCILLATOR

Pure vibrational spectra are observable when other forms of molecular energies, except vibrationalone, remain unchanged. Such spectra are obtained in liquid because molecular interaction betweenneighbouring molecules suppresses rotational motion. Hetero-nuclear diatomic molecules(HCl, CO, CN) have intrinsic electric dipole moment and are capable of interacting withelectromagnetic radiation. Hence these molecules exhibit vibrational spectra. Homo-nuclear diatomicmolecules do not have dipole moment and hence do not give vibrational spectra.

A diatomic molecule can vibrate along the inter-nuclear axis and may be regarded as a two-body oscillator. Its classical frequency of vibration is given by

1

2osckν =

π µ ,1

2osc k

c c

νω = =

π µ ...(2.1.1)

where k is force constant of the elastic force binding the atoms and µ is the reduced mass of themolecule. For CO molecule, k = 1870 N/m, µ = 1.14 × 10 –26 kg, the frequency of vibration isosc = 2.04 × 1013 Hz. For small amplitude or energy, the motion of atoms is pure harmonic and thepotential energy of the molecule is V = 1/2 k 2, being the displacement of the oscillator. Thispotential energy is called harmonic or parabolic potential energy.

The quantum mechanical treatment of harmonic oscillator with potential V = 1/2 k2 showsthat oscillator energy is quantized and is given by

( )12

1E

2 osch hc = ν + ν = ν + ω

...(2.1.2)

The vibrational term value is

( ) ( )1 12 2

EG( ) osc

ch c

νν = = ν + = ω ν + ...(2.1.3)

where is an integer ( = 0, 1, 2, 3, ...), called vibration quantum number. ω is frequency ofclassical oscillator in wave number units. It is also called vibrational constant. (Note that it is not the

MOLECULES


angular frequency.) Minimum energy of oscillator is 1E for 0.

2osc osch= ν ν = This energy Eosc is

called zero-point energy. Eqn. (2.1.2 or 2.1.3) shows that the energy levels of a harmonic oscillatorare equally spaced with constant separation E .osch∆ = ν When continuous electromagnetic radiationis passed through an assembly of polar molecules, which act as harmonic oscillator, they interactwith radiation and go over to higher energy states by absorbing radiation. Such transitions give riseto absorption spectrum. The allowed vibration transitions are those which obey the selection rules

= ±1 ...(2.1.4)

+ sign indicates absorption and – sign emission. The vibrational quantum numbers of the lower andupper states are denoted by v" and v' respectively. The selection rule then becomes

1 (absorption)′ ′′ν − ν =The frequency of absorption line in wave number units when the molecule makes transition

from " () ' = + 1) is

1 12 2

G( ) G( ) ( ) ( )′ ′′ ′ ′′ν = ν − ν = ω ν + − ω ν +

3 12 2

( ) ( )= ω ν + − ω ν +

osc= ω = ωThus, all the allowed transitions leading to absorption lines have the same frequency equal to

the frequency of the oscillator and the pure vibrational spectrum will consist of a single absorptionline. This is in accordance with the classical electrodynamics.

Fig. 2.1.1 Permitted vibrational transitions leading to a single spectral band

2.2 ANHARMONIC OSCILLATOR

Experimental investigations reveal that the vibration absorption spectrum of HCl shows, in addition

to fundamental frequency 5 11 2.886 10 m−ν = × , lines at frequencies 5 1

2 5.668 10 m−ν = × and

Vibrational Spectra of Diatomic Molecules 551

5 13 8.347 10 m−ν = × . The frequencies 2 3andν ν are called overtones or harmonics and are slightly

less than twice and thrice of the fundamental frequency. The intensities of the overtones are muchsmaller than that of the fundamental line. The existence of overtones indicates that the selection rules = ± 1 are not valid.

At high energy the amplitude of vibration is so large, the oscillations are no longer pureharmonic and the harmonic potential V = 1/2 k 2 does not accurately describe behaviour of thesystem. The large amplitude vibration is called anharmonic vibration and in such vibrations thepotential energy is best approximated by following expression:

2 3 4

1 2 31 1 1

V .....2! 3! 4!

k k k= ξ + ξ + ξ + ...(2.2.1)

The task of finding energy eigenvalues and corresponding eigen functions by solving Schrodingerequation for oscillator with this form of potential is very complicated. P.M. Morse suggested a simpleand more realistic potential function represented by

( ) 2V D 1 exp ( )e ea r r = − − ...(2.2.2)

where De is dissociation energy of the molecule (which is equal to the minimum energy that must beadded to the molecule to bring the atoms at an infinite separation), re is equilibrium separation ofatoms and r is separation of atoms.

Fig. 2.2.1 Morse potential

Schrodinger equation for oscillator with Morse potential gives energy levels given by

2 3

1 1 1E .....

2 2 2e e e e ehc hc x hc y = ν + ω − ν + ω + ν + ω +

...(2.2.3)

where 1

2ek

cω =

π µ...(2.2.4)

is the vibrational frequency in wave number units that the anharmonic oscillator would have classicallyfor an infinitesimal amplitude, xe, ye,.. are anharmonicity constants. As an approximation the third


term can be omitted in Eqn. (2.2.3). With this approximation the energy of an anharmonic oscillatorcan be represented as follows:

E = 2

1 1

2 2e e ehc hc x ν + ω − ν + ω

...(2.2.5)

or term value

21 12 2

G( ) ( ) ( )e e exν = ω ν + − ω ν + ...(2.2.6)

where .e e exω << ωThe wave number of the absorption band arising from the transition " ' is given by

2 21 1

G( ) G( ) ( )2 2e e ex′′ ′ν →ν

′ ′′ ′ ′′ ′ ′′ν = ν − ν = ν − ν ω − ν + − ν + ω

( ) ( 1) ( 1)e e ex′ ′′ ′ ′ ′′ ′′= ν − ν ω − ν ν + − ν ν + ω ...(2.2.7)

The selection rules for transitions between the energy levels of an anharmonic oscillator arefound to be

= ± 1, ± 2, ±3,.......The wave numbers of the bands arising from the transitions 0 , 0 2, and 0 3 can be

found from Eqn. (2.2.7). These are:

0 1 1 (1 2 )e ex→ν = ν = − ω

0 2 2 2(1 3 )e ex→ν = ν = − ω ...(2.2.8)

0 3 3 3(1 4 )e ex→ν = ν = − ω

0 4 4 4(1 5 )e ex→ν = ν = − ω

The first band of wave number 1ν arising from the transition " = 0 ' = 1, is called

fundamental band and is most intense. The band corresponding to the transition " = 0 ' = 2,with = 2 has small intensity and is called first overtone. The band corresponding to the transition" = 0 ' = 3 with = 3, has much smaller intensity and is called second overtone. Only thesethree band have observable intensities.

The energy difference between the first excited state and the ground state is E = h0. For COmolecule this energy is E = h0 = (6.62 × 10–34 Js)(2.04 × 1013) = 13.5 × 10–21 J = 8.4 × 10–2 eV= 0.084 eV. At room temperature, thermal energy kT is about 0.026 eV. Thus, the thermal energy isnot enough to raise the molecules from ground state to excited states. So the majority of the moleculeslie in the ground state at normal temperature. This explains why the fundamental line is most intense.If the temperature of the sample is raised, the population of = 1 state is increased and therefore,the probability of transition from higher states with selection rule = 1 increases. The bandsoriginating from such transitions constitute what we call hot bands.

Vibrational Spectra of Diatomic Molecules 553

From the measurements of frequency of fundamental band and of overtones, the frequency eωand anharmonicity constant xe can be calculated making use of Eqn. (2.2.8). The force constant k isfound from the Eqn. (2.2.4) viz.,

2 2 24 .ek c= π µω ...(2.2.9)

2.3 ISOTOPIC SHIFT OF VIBRATIONAL LEVELS

The harmonic oscillator frequency 1

2osckν =

π µ depends on the reduced mass of the molecule. Since

µ is different for different isotopes so is the oscillator frequency. Let andie eω ω be the frequencies

of the heavier and lighter isotope respectively. Let

(say)ie

ie

ω µ= = ρω µ ...(2.3.1)

The vibrational terms for the two isotopes are

1

G( )2e

ν = ω ν + ...(2.3.2)

1 1

G ( )2 2

i ie e

ν = ω ν + = ρω ν + ...(2.3.3)

The isotopic shift in term values of a vibrational level of quantum number is

1

G ( ) G( ) ( 1)2

ie

ν − ν = ρ − ω ν + ...(2.3.4)

Since < 1, G ( ) G( ).i ν < νThus, the vibrational levels of heavier isotope lie a little deeper than the corresponding levels

of lighter isotope.If anharmonicity is taken into consideration, the vibrational term for lighter isotope is represented

as

( ) ( )21 12 2

G( ) e e exν = ω ν + − ω ν +

From theoretical analysis, the anharmonicity constant x ei is found to be related to xe through

the relation

ie ex x= ρ

Making use of this relation and ie eω = ρω we can find the vibrational term for heavier isotope

( ) ( )221 12 2

G ( )ie e exν = ρω ν + − ρ ω ν +


The isotopic shift of a vibrational level in this case comes out to be

2 21 12 2

G ( ) G( ) ( 1) ( ) ( 1) ( )ie e exν − ν = ρ − ω ν + − ρ − ω ν +

Since is only slightly different from 1, we can assume + 1 = 2. With this approximationwe have

1 12 2

( ) G( ) ( 1)( ) 2 ( )ie e eG x ν − ν = ρ − ν + ω − ω ν + ...(2.3.5)

Since < 1, Gi() – G() is negative. We come to the same conclusion that vibrational levelsof heavier isotope lie deeper than those of lighter isotope. This is shown in the Fig. (2.3.1).

The isotopic shift of energy levels causes doubling of vibrational bands. The band shift is givenby

G ( ) G ( ) G( ) G( )i i i ′ ′′ ′ ′′∆ν = ν − ν = ν − ν − ν − ν

= ( 1) ( ) 1 ( 1) ( 1)e ex′ ′′ ′ ′′ρ − ω ν − ν − ρ + ν + ν + For fundamental band

0 1 ( 1)(1 4 )e ex→∆ν = ρ − − ω ...(2.3.6)

For first and second overtones

0 2 2( 1)(1 6 )e ex→∆ν = ρ − − ω ...(2.3.7)

0 3 3( 1)(1 8 )e ex→∆ν = ρ − − ω ...(2.3.8)

Fig. 2.3.1 Vibrational levels of two isotopes

CHAPTER

VIBRATION-ROTATION SPECTRA OFDIATOMIC MOLECULES

3.1 ENERGY LEVELS OF A DIATOMIC MOLECULE ANDVIBRATION-ROTATION SPECTRA

The spacing of electronic energy levels is nearly ten times that of vibrational energy levels and thespacing of vibrational levels is nearly hundred times that of rotational levels. Each electronic levelhas many equally spaced vibrational levels and each vibrational level has many closely spaced rotationallevels. The energy level diagram of a molecule is schematically shown in the Fig. (3.1.1).

When an electromagnetic radiation of appropriate frequencies is passed through an assembly ofmolecules, the latter after absorbing some characteristic frequencies undergo transitions from lowerenergy states to higher energy states. In such transitions changes in rotational, vibrational and electronicenergies of molecules occur. The resulting absorption spectrum consists of a large number of linescharacteristic of molecules. We first consider the simple case in which the electronic energy states ofmolecules remain unchanged, only rotational and vibrational energies undergo change. To a firstapproximation, we assume that (i) vibrational and rotational energies do not interact, (ii) vibrations ofmolecules are pure harmonic and (iii) centrifugal distortion of molecule is negligibly small. Under thisapproximation the energy of a molecule is sum of its vibrational and rotational energies.

1

E B J(J 1)2 ehc hc

= ν + ω + + ...(3.1.1)

where 2B

8 I

h

c=

πis the rotational constant and

1

2ek

cω =

π µ

is the wave number corresponding to classical frequency of harmonic oscillator.The vibration-rotation term value is

12

T G( ) F(J) ( ) BJ(J 1)e= ν + = ω ν + + + ...(3.1.2)


The selection rules for vibrational transitions are = ± 1 and those for rotational transitionsare J = ± 1. The allowed transitions between the rotational and between the vibrational energylevels give rise to vibrational-rotational spectra. It is customary to denote the quantum numbers ofrotational levels in the upper vibrational level ' and in the lower vibrational level " by J' and J"respectively. It should be noted that energy of molecules in level with higher vibrational quantumnumber ' is always greater than that in level with lower quantum number " irrespective of thevalues of rotational quantum number J corresponding to these vibrational levels.

Fig. 3.1.1 Energy level diagram of a molecule

The wave number of the absorption band originating in the transition " ' and J" J' isgiven by

G( ) BJ (J 1) G( ) BJ (J 1)′ ′ ′ ′′ ′′ ′′ν = ν + + − ν + +

= ( ) B J (J 1) J (J 1)e′ ′′ ′ ′ ′′ ′′ν − ν ω + + − + ...(3.1.3)

For the vibrational transition " = ' = ' + 1, this equation becomes

ν = B J (J 1) J (J 1)e ′ ′ ′′ ′′ω + + − + ...(3.1.4)

These transitions fall in two groups. The collection of absorption lines obeying the selectionrule J = J' – J" = –1 constitute P-branch of the spectrum and the set of lines obeying the selectionrule J = J' – J" = + 1 constitute R-branch of the spectrum. If we write J" = J for the quantum

Vibration-Rotation Spectra of Diatomic Molecules 557

number of the initial state (lower state) then the wave numbers of the lines of P-branch will begiven by

P B (J 1)J J (J 1) 2BJe e′′ ′′ ′′ ′′ν = ω + − − + = ω − , J = 1, 2, 3, … ...(3.1.5)

Similarly, the wave numbers of the lines of R-branch will be given by

R 2B(J 1)eν = ω + + , J = 0, 1, 2, 3…… ...(3.1.6)

Fig. 3.1.2 Rotation-vibration spectrum of diatomic molecule

The wave number of spectral lines both P-and Q-branches can be expressed by a single formulaas follows:

2Be mν = ω ± , m = 1, 2, 3, …. ...(3.1.7)

In P-branch, P eν = ω requires that J = J" = 0, this implies J' = – 1, which is not possiblebecause J cannot be negative. So the line at eω does not appear in the spectrum. Similarly, inR-branch, appearance of line at eω requires J = J" = – 1, which again not possible. The wave numberseparation of successive lines either in P-branch or in R-branch of the spectrum is 2B. Thus, therotation-vibration absorption spectrum of diatomic molecule consists of equally spaced lines on eachside of the band origin eω or the center of the band. The lines of P-branch have lower frequenciesand those of R-branch have higher frequencies. The vibrational part of frequency eω determinesspectral region in which the band is located. The rotational part ± 2Bm determines the fine structureof the band. The region in which the vibration-rotation bands are found extends approximately from8,000 Å to 50,000 Å. That is these band lie in infrared portion of the spectrum.


Fig. 3.1.3 (a) Energy level diagram of rotation-vibration band

(b) Absorption spectrum of HCl


3.2 EFFECT OF INTERACTION (COUPLING) OF VIBRATIONAL ANDROTATIONAL ENERGY ON VIBRATION-ROTATION SPECTRA

The assumption that vibrational and rotational energies of a diatomic molecule are independent ofeach other is far from reality. Certain features of vibration-rotation spectra indicate that there existssome kind of coupling between vibrational and rotational motion. As the vibrational energy of moleculeincreases, the average separation of constituent atom also increases. This causes increase in momentof inertia (I = r2). The rotational constant B is inversely proportional to moment of inertia, andhence it becomes function of vibrational quantum number . B is smaller in state with high vibrationalquantum number. We can represent the dependence of B on in the form

1

B B2eν

= − α ν + ...(3.2.1)

where B and Be refer to the values of B in state with quantum number and in equilibrium staterespectively and is a positive constant. Similarly, the constant D, which describes the non-rigidity ofmolecular bond, may be expressed as

1

D D2eν

= + β ν + ...(3.2.2)

where is a constant and is very small compared to De. The correction term in Dv is very small andmay be ignored.

Ignoring the centrifugal distortion, the energy of anharmonic oscillator is found to be

21 1

J 2 2E E ( ) ( ) B J(J 1)e e ech chx chν ν+ = ν + ω − ν + ω + +

Or term value

21 1

2 2T( ,J) G( ) F( ,J) ( ) ( ) B J(J 1)e e ex νν = ν + ν = ν + ω − ν + ω + +

The wave number of the absorption lines of a particular band (', ") is given by

G( ) F( ,J ) G( ) F( ,J )′ ′ ′ ′′ ′′ ′′ν = ν + ν − ν + ν

= G( ) G( ) F( ,J ) F( , J )′ ′′ ′ ′ ′′ ′′ν − ν + ν − ν

= 0 B J (J 1) B J (J 1)ν ν′ ′ ′ ′′ ′′ ′′ν + + − + ...(3.2.3)

where 0ν is the wave number corresponding to pure vibrational transition between two vibrational

levels with both J' and J" equal to zero. This wave number is the center of the band or band origin.Most of the diatomic molecules have no angular momentum about the inter-nuclear axis and theselection rules J 1∆ = ± are valid. The selection rule J 1∆ = − gives a series of lines called P-branchand J 1∆ = + gives the series of lines called R-branch. For such molecules the transitions with J 0∆ =are forbidden.

For a molecule possessing electronic angular momentum about the inter-nuclear axis, selectionrule is J 0, 1.∆ = ± Here is quantum number for resultant electronic angular momentum. Nitric


oxide has an unpaired electron and hence has electronic angular momentum ( 0).Λ ≠ In this molecule

the transition with J = 0 is allowed. The transitions obeying the selection rule J = 0 give rise toQ-branch of spectrum.

The wave numbers of lines of P-branch are obtained making use of the selection rule

J 1 . ., J J 1i e ′ ′′∆ = − − = − or by replacing J' by J" – 1 in Eqn.(3.2.3).

2P 0 (B B )J (B B )J ,′ ′′ ′ ′′ν ν ν ν′ ′′ ′′ ′ ′′ ′′ν = ν − + + − J" = 1, 2, 3......

Writing J for J" we have

2P 0 (B B )J (B B )J , J 1, 2, 3.....′ ′′ ′ ′′ν ν ν ν′ ′′ ′ ′′ν = ν − + + − = ...(3.2.4)

The lines of P-branch are denoted by P(1), P(2), P(3),…..

Since Bν′ < Bν′′ ( B decreases slightly with increasing . B 1/ r2, r increases with increasing

value of ), the difference Bν′ – Bν′′ is negative. Thus, both the linear and quadratic terms in J are of

the same sign (negative) and hence the lines of P-branch get farther apart as J values go increasing.Similarly, the wave numbers of lines of R-branch making use of selection rule J = + 1 i.e.,

by replacing J' by J" + 1 in Eqn. (3.2.3).

2R 0 2B (3B B )J (B B )J′ ′ ′′ ′ ′′ν ν ν ν ν′ ′ ′′ ′′ ′ ′′ ′′ν = ν + + − + − J" = 0, 1, 2, ….

Writing J for J", we obtain

2R 0 2B (3B B ) J (B B )J ,′ ′ ′′ ′ ′′ν ν ν ν ν′ ′ ′′ ′ ′′ν = ν + + − + −

J = 0, 1, 2,…. ...(3.2.5)

The lines of R-branch are denoted by R(0), R(1), R(2), ….

Since Bν′ < B ,ν′′ the term Bν′ – Bν′′ is negative but has very small value. Therefore, the term

3B Bν ν′ ′′− is positive. The linear and quadratic terms in J are of opposite sign. As a consequence ofthis, the line spacing in R-branch decreases very slowly and the lines ultimately converge and thepoint where lines converge is called band head.

For Q-branch J' = J" and we obtain

2Q 0 (B B )J (B B )Jν ν ν ν′ ′′ ′′ ′ ′′ ′′ν = ν + − + − J" = 0, 1, 2, …

Writing J for J", we get

2Q 0 (B B )J (B B )Jν ν ν ν′ ′′ ′ ′′ν = ν + − + − J = 0, 1, 2, … ...(3.2.6)

It is possible to represent the wave numbers of lines of P-and R-branch by means of a singleformula

2P, R 0 (B B ) (B B )m mν ν ν ν′ ′′ ′ ′′ν = ν + + + − ...(3.2.7)

where m takes the values – 1, –2, –3…..for P-branch and 1, 2, 3, …. for R-branch.


Fig 3.2.1 Absorption curves for fundamental transitions 0 1 in HCl

The absorption curve for the fundamental transition (" = 0 ' = 1) for HCl is shown in theFig. (3.2.1). The lines of P-branch and R-branch can be seen in the curve. No line is observed atthe center of the P- and R-branches. In fact, this is the position of Q-branch, which is not observedin HCl. In the case B' = B", a single line is observed in Q-branch. Actually B' slightly differs fromB" and hence Q-branch consists of a number of lines which are very closely spaced. It is observed

only in those molecules having 0Λ ≠ .The average inter-nuclear distance re increases with increase in vibrational energy and hence

the rotational constant B 22

, I8 I

eh

rc

= = µ π

is smaller in the upper vibrational state than that in the

lower state. Thus, B' < B", from Eqn. (3.2.4) it is evident that the separation of lines in P-branchincreases as one moves towards the lower frequency side of band origin. This means band head (wherethe lines converge) appears on higher frequency side of band origin i.e., band head is formed inR-branch. Such a band is said to be degraded (shaded) towards the red. In vibration-rotation spectrabands are always degraded towards the red only.

Measuring the values of Bv for two or three vibrational levels, Be can be calculated. From thisvalue of Be, we can calculate moment of inertia and hence inter-nuclear distance.

CHAPTER

ELECTRONIC SPECTRA OF DIATOMICMOLECULES

4.1 ELECTRONIC SPECTRA OF DIATOMIC MOLECULES

When atoms combine to form a molecule, the inner electrons in each of the participating atom maybe regarded as constituent part of the parent atom while the outer electrons belong to the moleculeas a whole. A molecule in ground state has a definite electronic configuration and the energy of itselectrons is called ground state electronic energy. The energy of molecule due to its electrons dependsupon the relative positions of the nuclei i.e., electronic energy includes the electrostatic energy ofnuclei of the molecule. The variation electronic energy of a diatomic molecule with inter-nucleardistance in ground state is shown in the Fig. (4.1.1). In excited states the dependence of electronicenergies may be represented by similar curves.

Any change in electron configuration is accompanied by a change in electronic energy. Theelectronic energy levels of a molecule are much more complicated than that in atoms. The outerelectron levels of a molecule have energies with absolute value in the range electron volt. Accordingto Born-Openheimer approximation the total energy of a molecule can be written as sum of threeparts: electronic energy Ee, vibration energy Ev and rotational energy Er.

E E E Ee rν= + +

or in terms of term values

Fig. 4.1.1

Electronic Spectra of Diatomic Molecules 563

T T G( ) F(J)e= + ν +The vibration energy is expressed as

21 12 2

E ( ) ( ) .....e e ech chxν = ν + ω − ν + ω + ...(4.1.1)

or vibration term is

21 12 2

G( ) ( ) ( ) .e e exν = ν + ω − ν + ω + ……. (4.1.2)

where eω is the equilibrium frequency and xe anharmonicity constant in a particular electronic stateand has different values in different electronic state. The inclusion of xe takes into account theinteraction between electronic and vibrational energy states.

The rotational energy is expressed as

2 2E B J(J 1) D J (J 1) ........r ch chν ν= + + + + ...(4.1.3)

The corresponding term value is

2 2F(J) B J(J 1) D J (J 1) ..........ν ν= + + + + ...(4.1.4)

The values of rotational constant B and centrifugal distortion constant D are different indifferent vibrational and electronic states.

A change in electronic energy involves change in all the three kinds of energy. It is worth toobserve that there is always a change in electronic dipole moment in an electronic transition whetherthe molecule is non-polar or polar and therefore all kinds of molecule exhibit electronic band spectra.

The wave numbers of spectroscopic lines originating from the transitions between two electronicstates are given by

(T T ) [G ( ) G ( )] [F (J ) F (J )]e e

e rν

′ ′′ ′ ′ ′′ ′′ ′ ′ ′′ ′′ν = − + ν − ν + −= ν + ν + ν ...(4.1.5)

Single prime refers to higher state and double prime to lower state. The frequency T Te e e′ ′′ν = −

is constant for a given pair of electronic levels. The terms G'(') and G"('') belong to differentelectronic states with different ande e exω ω and it is also possible that G'(') < G''(''). Similarly,F'(J') and F''(J'') also belong to different electronic levels, possibility of F'(J') < F''(J'') is alwaysthere.

Vibrational (Course) Structure of Electronic Spectra

In order to have a general picture of electronic spectra, we shall, for the time being, ignore therotational transitions and shall concentrate on the possible transitions between the different vibrationallevels belonging to different electronic levels. The wave numbers of the lines are given by

ν = e νν + ν

= [G ( ) G ( )]e ′ ′ ′′ ′′ν + ν − ν


2 31 1 12 2 2

2 31 1 12 2 2

( ) ( ) ( )

( ) ( ) ( )

e e e e e e

e e e e e

x y

x y

′ ′ ′ ′ ′ ′ ′ ′= ν + ω ν + − ω ν + + ω ν + − ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ω ν + − ω ν + + ω ν +

...(4.1.6)

Fig. 4.1.2 Vibrational and rotational levels of two electronic states A and B of a molecule (schematic)


For a given pair of electronic levels, Eqn. (4.1.6) gives all possible transitions between differentvibrational levels of the two electronic levels. For electronic transitions, there is no strict selectionrule for the vibration quantum number . Each vibration state of upper electronic level may combinewith each vibration state of lower electronic level. These transitions create numerous lines in thevibration spectrum. When observed with an instrument of high resolving power, each of these linesis found to be composed of many closely spaced lines, called fine structure. So each line resultingfrom vibrational transition is in fact a group of lines and hence called a band. The lines of finestructure of each band result from rotational transitions which we have ignored for a while. All thebands due transitions between a given pair of electronic states, for all possible values of ' and ''are said to form a band system.

If we consider all possible transitions between all electronic levels, the electronic spectrum of amolecule consists of many band system.

The intensity of band drops rapidly with increasing value of | |. At room temperature, mostmolecules occupy the ground state '' = 0 and therefore most intense band is 0 1 and is called

fundamental band. The bands 0 2, 0 3→ → etc. are called overtones. The lines resulting from all the

possible transitions are shown in the Deslandre table.

The set of bands having constant value of ' – '' is called a sequence. The main sequence liesalong the diagonal from upper left corner to the lower right corner. Other sequences consist offrequencies lying along the lines parallel to the above diagonal.

The set of bands arising from the transitions '' ' (or ' '') having a definite value ofeither ' or '' while the other increases by unity constitute a progression. The set of bands with " = 0 and ' = 1, 2, 3,… is called '-progression with '' = 0. This progression consists of bands

00 10 20 30, , , ....ν ν ν ν The bands of this progression lie in the first vertical column of Deslandre table.The set of bands with ' = 0 and '' = 0, 1, 2, 3… is called ''-progression with ' = 0. These bandswith frequencies 00 01 02, , ,...ν ν ν lie along the horizontal row of the Deslandre table.

Eqn. (4.1.6) giving the frequency of bands can be written as

2 3 2 300 0 0 0 0 0 0 0 0 0 0 0( ...) ( ......x y x y′ ′ ′ ′ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ν = ν + ω ν − ω ν + ω ν + − ω ν − ω ν + ω ν + ...(4.1.7)

where ( ) ( ) ( )001 1 1

2 4 8e e e e e e e e e e ex x y y′ ′′ ′ ′ ′′ ′′ ′ ′ ′′ ′′ν = ν + ω − ω − ω − ω + ω − ω ...(4.1.8)

is the term independent of ' and '' i.e., it is the frequency of transition '' = 0 ' = 0 (0-0 band).

or ( ) ( )1 1 1 1 1 100 2 4 8 2 4 8

..... ....e e e e e e e e e e ex y x y′ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ν = ν + ω − ω + ω + − ω − ω + ω + ...(4.1.9)

and 30 4

.... coefficient ofe e e e ex yω = ω − ω + ω '.


Fig. 4.1.3 ''- progressions with ' = constant

Fig. 4.1.4 ' -progressions with '' = constant


Fig. 4.1.5 Deslandre Table

It is interesting to observe that electronic absorption spectrum consists of a single progressiononly with '' = 0. This can be explained as follows. At room temperature, most molecules are in thelowest vibrational state ('' = 0). After absorbing energy from surrounding radiation they are raisedto higher electronic state. Since the molecules can jump to any quantum state of higher quantum

number without restriction, the most probable vibrational transitions are 0 1,0 2, 0 3,.......→ → → etc.

Obviously, these constitute a single progression with '' = 0. If the temperature is raised, an appreciablenumber of molecules are excited to ' = 1 state. Then the probability of transitions 1 0, 1 1,

1 2, 1 3, 1 4......→ → → etc. is enhanced and the progression ' = 1 appears in the absorption spectrum.

Rotational Structure (Fine Structure) of Electronic Bands

So far in the analysis of structure of electronic bands we considered the transitions between vibrationalstates of different electronic levels and ignored the rotational states associated with each vibrationalstate. Now we shall consider the role of rotational states in determining the structure of electronicbands. Examination of electronic spectrum of molecules with instruments of high resolving powerreveals that each vibrational band is composed of a large number of closely spaced lines, called


fine structure of band. These lines originate from transitions between rotational states of differentelectronic levels.

For a given vibrational transition, the wave numbers of lines resulting from rotational transitionsare given by

G ( ) G ( ) F ( , J ) F ( , J )e ′ ′ ′′ ′′ ′ ′ ′ ′′ ′′ ′′ν = ν + ν − ν + ν − ν

= 0 F ( ,J ) F ( ,J )′ ′ ′ ′′ ′′ ′′ν + ν − ν ...(4.1.10)

where 0 eν = ν + G ( ) G ( )′ ′ ′′ ′′ν − ν is the wave number of band origin. F' and F'' are the rotational

terms belonging to upper and lower electronic states respectively. In terms of rotational constantsB and Bν ν′ ′′ of upper and lower electronic states we can express ν as

0 B J (J 1) B J (J 1)ν ν′ ′ ′ ′′ ′′ ′′ν = ν + + − + ...(4.1.11)

If the total electronic angular momentum of the molecule is zero in both upper and lowerelectronic states i.e., both are states, the selection rule is

J 1∆ = ±The set of lines obeying the selection rule J J J 1′ ′′∆ = − = − is called P-branch and the set of

lines obeying the selection rule J J J 1′ ′′∆ = − = + is called R-branch. Putting J' = J '' – 1 in

Eqn. (4.1.11) we get the wave numbers of lines of P-branch.

2P 0 (B B )J (B B )J , J 1, 2, 3,.....ν ν ν ν′ ′′ ′′ ′ ′′ ′′ ′′ν = ν − + + − = ...(4.1.12)

Similarly, putting J' = J'' + 1 in Eqn. (4.1.11), we get the wave numbers of lines of R-branch.

2R 0 2B (3B B )J (B B )J , J 0,1, 2,3....ν ν ν ν ν′ ′ ′′ ′′ ′ ′′ ′′ ′′ν = ν + + − + − = ...(4.1.13)

If the total orbital angular momentum in any of the two electronic states (upper or lower) is

not zero, the selection rule for J is J J J 0, 1′ ′′∆ = − = ± . In this case lines of Q-branch make their

appearance and the wave number of lines of Q-branch are given by

2Q 0 (B B )J (B B )Jν ν ν ν′ ′′ ′′ ′ ′′ ′′ν = ν + − + − , ...(4.1.14)

The wave numbers of lines of P- and R-branches can be represented by a single formula givenby

2PR 0 (B B ) (B B )m mν ν ν ν′ ′′ ′ ′′ν = ν + + + − ...(4.1.15)

where m = – J'' = – 1, – 2, – 3, …for P-branchand m = J'' + 1 = 1, 2, 3, ……for R-branch

Since m 0≠ either in P-branch or in R-branch, a line is missing for m = 0. This missing line

of frequency 0ν = ν is the center of the band and corresponds to the transition

J'' = 0 J'' = 0.


Fig. 4.1.6 P-branch consists of lines P(1), P(2), P(3),….etc. and R-branch consists of lines R(0), R(1),R(2), R(3),…..etc.

Formation of Band Head

The rotational constants B and Bν ν′ ′′ belong to different electronic states as well as, in general, differentvibrational states and therefore they differ considerably. Also the difference B Bν ν′ ′′− may be negativeor positive. [In case of vibration-rotation spectrum, which lies in infrared region, B and Bν ν′ ′′ belongto the vibrational states of the same electronic states, so their difference is quite small.]

Now let us consider the case, B Bν ν′ ′′− = negative i.e., B B .ν ν′ ′′< The wave numbers of lines ofR-branch is given by

2R 0 (B B ) (B B ) , 1, 2, 3.....m m mν ν ν ν′ ′′ ′ ′′ν = ν + + + − = ...(4.1.16)

Since B B ,ν ν′ ′′< the last term containing m2 is negative. Substituting m = 1, 2, 3,….. in

eqn. (4.1.16), we find that ν increases first and after attaining a maximum value it begins to decreasewith further increase in the value of m. This happens because for the smaller values of m, the changein the term containing m dominates over the change in the term containing m2. After the frequencyν attains maximum value, this trend is reversed. Before maxν , the separation between successive linesgradually decreases and therefore the lines begins to crowd. The frequency at which the separationof lines becomes zero is called band head. A further increase in the value of m, causes decrease infrequency but increase in separation between consecutive lines. In other words, the successive linesbegin to turn back upon themselves and this trend continues toward the lower frequency side. Theband head is point where the reversal of frequency change takes place. [In electronic band there is astrong tendency of head formation whereas in vibration-rotation band the tendency of head formationis almost negligible and band head is not observed.]


In P-branch the wave number of lines is given by

2P 0 (B B ) (B B ) , 1, 2, 3,.......m m mν ν ν ν′ ′′ ′ ′′ν = ν + + + − = − − − ...(4.1.17)

The terms containing m and m2 both are negative. As the value of m increases numerically, thefrequencies of lines gradually decreases but the separations of consecutive lines increases i.e., thelines get more and more widely separated. At the same time the intensity of lines diminishes withincreasing value of | m |. This results in gradual shading off the band on the lower frequency side ofband origin.

Fig. 4.1.7 Rotational fine structure of vibration-electronic transition

Thus, if B Bν ν′ ′′< , the band will be degraded (shaded) toward the red end (lower frequency

side or higher wavelength side) of the spectrum. The band head is formed in R-branch i.e., on thehigher frequency side of zero or null line.

If B Bν ν′ ′′> , arguments made above are reversed. The band is formed in P-branch and it is

degraded (shaded) toward violet end.[In electronic spectrum both red and violet degraded bands are observed whereas in vibration-

rotation bands only red degraded bands are observed.]

If B Bν ν′ ′′= , i.e., vibration-rotation interaction is zero, both the branches consist of equally spaced

lines and band would be headless.All the bands resulting from transitions between a given pair of electronic states, for all possible

values of ' and '' are said to form a band system.There are many electronic states of a molecule, therefore, all possible transitions between various

electronic states give rise to many band system in the electronic spectrum of a molecule.


Fortrat Diagram

The dependence of frequency ν of line in electronic spectra on m (or J) is represented by Fortrat

diagram. In this diagram ν is plotted on abscissa and m (or J) on ordinate. The points ( .mν ) arerepresented by small circles. The curve joining these points is a parabola. The vertex of the parabolacorresponds to the band head. The value of m corresponding to band head is obtained by solving theequation

0d

dm

ν =

(B B ) 2(B B ) 0mν ν ν ν′ ′′ ′ ′′+ + − =

B B

2(B B )headm ν ν

ν ν

′ ′′+= −

′ ′′− ...(4.1.18)

If m comes out to be fraction, the nearest whole number will be the value of m.

Fig. 4.1.8 Fortrat diagram

The separation of band head from the band origin is given by

20 (B B ) (B B )head head headm mν ν ν ν′ ′′ ′ ′′ν − ν = + + −

Substituting the value of mhead from Eqn. (4.1.18) in above equation, we find

( )

( )2

0

B B

4 B Bheadν ν

ν ν

′ ′′+ν − ν = −

′ ′′− ...(4.1.19)


For a band degraded toward red, 0( )headν − ν is positive and for that degraded toward violetthis difference is negative. Since B and Bν ν′ ′′ have different values in different vibrational states, thedistance of different bands in a band system are different.

The value of J'' corresponding to Q-head is obtained by setting Q 0.

J

d

d

ν=

′′ From Eqn. (4.1.14),

we have (B B ) 2(B B ) 0ν ν ν ν′ ′′ ′ ′′− + − =

1

J2

′′ = −

Thus, the position of Q-band head does not depend on the values of B and Bν ν′ ′′ .

Fig. 4.1.9 Energy level diagram for a band with P, Q, and R-branches

R(3

)R

(2)

R(1

)R

(0)

Q(1

)

Q(7

)P(

2)

P(3)

P(4)

P(5)

R(3

)R

(2)

R(1

)R

(0)

Q(1

)

Q(7

)P(

2)

P(3)

P(4)

P(5)

v ν0λ

1Σ

02

45

6

7

8

9

1345

6

7

8J'

J'

1ΠΛ = 1

Λ = 0


4.2 FRANCK-CONDON PRINCIPLE: ABSORPTION

Examination of electronic spectra reveals that in some spectra (0, 0) band is most intense, in othersintensity of bands rises up to certain value of ' and then diminishes for higher values of '; yet inothers intensity of band increases with increasing value of ' and finally a continuum is observed.Three typical cases of intensity distribution in absorption band series are shown schematically.

Fig. 4.2.1 Intensity distribution in a band ('-progression with '' = 0)

The intensity distribution in different bands of a band system can be understood in terms ofFranck-Condon principle. According to this principle, the transition from one vibrational level oflower electronic state to a vibrational level of upper electronic state takes place so rapidly (10 –15 s)in comparison to the vibrational motion of nuclei that nuclei before and after the transition havevery nearly the same inter-nuclear distance and velocity. This means that the electronic transitionsleading to the appearance of absorption bands must be represented by vertical lines. Except thetransitions starting from lowest electronic state, the most probable transitions will be those whichstart from the extreme positions (turning points) of the nuclei for any given vibrational level. Themost probable transition will be attended with most intense band.

Case I: e er r′ ′′= : To explain the intensity distribution shown in Fig. (4.2.1a), we consider the

potential energy curves of the two electronic states as sketched in Fig. (4.2.2a). The curves are sodrawn that the minima of two curves lie one above the other. Inter-nuclear distance in the two stateshas the same value. Before absorption the molecule is in the lowest energy level '' = 0. The mostprobable position of finding the vibrating nuclei in this level is the central point of the classicallimits and the kinetic energy of the nuclei (which is equal to the vertical distance from minima ofthe curve to the lowest energy level) is minimum compared with those in the higher levels. It caneasily be seen that the transition AB satisfies the requirements Franck-Condon principle i.e., the position


and velocity be almost unaltered in the transition. Obviously, the (0, 0) transition is most probableand hence the corresponding band will the most intense band.

Transitions from '' = 0 to higher values of ' is accompanied by an appreciable amount ofchange in inter-nuclear distance and velocity. The change in inter-nuclear distance and velocityincreases progressively with increasing value of ' and hence the probabilities of such transitions goon diminishing and hence the intensities of corresponding bands.

Fig. 4.2.2

Case II: e er r′ ′′> : To explain the intensity distribution shown in Fig. (4.2.3b), we consider two

potential energy curves sketched for the lower and upper electronic states. The curves are so drawnthat the minimum of the upper electronic state is some what displaced towards the right (r'e > r'''e).According to the Frank-Condon principle, the most probable transition is AB, where B is verticallyabove A and near the turning point of the vibrational level corresponding to some higher value of '.Obviously, the (0, 0) transition is not the most probable one hence the corresponding line will beless intense. For the transition AB, the internuclear distance has the same value before and aftertransition, the kinetic energy has nearly the same value at A and B, the probability of finding thenuclei at inter-nuclear distance corresponding to these points is maximum. Thus, vibrational level inthe upper electronic state in the neighbourhood of B, will be the upper vibrational level of the mostprobable transition hence the corresponding band will be the most intense band. The probability oftransition starting from '' = 0 to still higher values of ' will go on diminishing. This explains theintensity distribution of Fig. (4.2.1b). Similar intensity distribution is obtained when the minimumof the upper curve is some what displaced towards the left of the minimum of the lower curve.

(a) Vibrational wave functions of harmonic oscillator.

(b) Variation of probability 2

( ) .rψ In the ground state ( = 0), the probability

density is maximum at r = re (mid-point of turning points. In this state

( = 0), kinetic energy is minimum. In the excited states ( > 0), the

probability density 2

( )rψ is maximum near the turning points. The nuclei

are most likely to be found at this value of internuclear distance. Obviously,kinetic energies of nuclei will be minimum near the turning points.


Fig. 4.2.3

Case III: e er r′ ′′>> : Let us explain the intensity distribution of Fig. (4.2.1c). The potential energy

curves needed for this case are shown in Fig. (4.2.3c). In this case the minimum of the curve forupper state is displaced by a greater distance than that in Fig. (4.2.3b). The vertical transition ABstrictly fulfills the requirement of Franck-Condon principle. Notice that the point B lies at the levelof the continuous region of the vibrational term spectrum of the upper state. After such an electronictransition, the atoms will get themselves at infinite distance apart and the molecule will be dissociated.The discrete absorption lines will be observed if the points in the upper state to which transitions

(a) For vertical transitior AB (i) e er r′ ′′= (ii) Velocities of nuclei in the initial and

final states are equal. The (0–0) band is most intense.

(b) For vertical transistion AB (which corresponds to 3–0 band) (i) re

internuclear distance has the same value and (ii) velocities of nucleihave the same value in the initial and final states.

(c) The vertical transition AB meets the requirements of Frank-Condonprinciple.


takes place lies some what below B and continuum is observed when the end point of transition liessome what above B.

Thus, in '-progression with '' = 0, the most intense absorption band will always correspondto the vertically upward transition which starts from the minimum of the lower potential energycurve, if we disregard the zero point energy of vibration of nuclei. The transitions, which do notoriginate from '' = 0, are most probable when they start from the extreme positions (turning points)of nuclei.

Franck-Condon Principle: Emission

For the explanation of intensity distribution in emission bands within a band system of electronicspectrum, refer to the potential energy curves for upper and lower electronic states of Fig. (4.2.4).In the upper state, during vibrational motion represented by horizontal line AB, the molecule spendsmaximum time at the turning points A and B and minimum time at the intermediate positions. Sothe electronic transition starts either from A or from B. When the molecule starts from B, it findsitself at point C, vertically below B, after transition. The point C becomes the new turning point ofthe vibrational motion CD. When the molecule starts from A, it will be at point F, vertically belowA, after transition. The point F will be the new turning point, in this case, of the vibrational motionFE.

Fig. 4.2.4

It is evident that there are two values of '' for which probability of the transition from a givenvalue of ', according to Franck-Condon principle, is maximum and hence there will be two intensitymaxima in ''-progression with ' = constant.

In Fig. 4.2.4, the intensities of PN band are arranged in array. The horizontal rows represent''-progression with ' = constant. Each row contains two intensity maxima (except ' = 0). As 'increases, both the turning points C and F move upward but this upward shift of C is more rapidthan that of F. Meaning thereby, the two intensity maxima get more and more separated with increasing'. Of course, the value of '' corresponding to intensity maxima also increases. The curve joining


the intensity maxima is a parabola with principal diagonal as its axis. This parabola is known asCondon parabola.

If the minima of the potential energy curves have the same value of r, the intensity maxima ineach progression (' = constant) merge together and the Condon parabola degenerates into a straightline, which is coincident with the principal diagonal. In this case, the most intense bands are thosefor which ' = ''.

Quantum Mechanical Treatment

The probability of transition between two energy levels characterized by total wave function ' and'' is proportional to the square of matrix element R of electric dipole moment (also called transitionmoment)

*R d′ ′′= ψ µψ τ∫ ...(4.2.1)

The Cartesian components of vector are , andi i i i i ie x e y e zΣ Σ Σ .

To a first approximation, the total eigen function may be assumed to be the product of theelectronic e, vibrational v and rotational r eigen functions respectively and the reciprocal ofinternuclear distance r.

1

e rr νψ = ψ ψ ψ ...(4.2.2)

To a good approximation, the rotational motion may be neglected and under this approximationthe wave function becomes

e vψ = ψ ψ ...(4.2.3)

The electric moment may be resolved into two parts: electronic part e and nuclear part n.

= e + n ...(4.2.4)Thus

R e e v e v n e v e vd d∗ ∗′ ′ ′′ ′′ ′ ′ ′′ ′′= µ ψ ψ ψ ψ τ + µ ψ ψ ψ ψ τ∫ ∫ ...(4.2.5)

where the volume element d is product of de the volume element of electronic coordinates and dn

the volume element of nuclear coordinates. Keeping in mind that vibration wave functions are real,Eqn. (4.2.5) may be written as

R v v n e e e e n v v n e e ed d d d∗ ∗′ ′′ ′ ′′ ′ ′′ ′ ′′= ψ ψ τ µ ψ ψ τ + µ ψ ψ τ ψ ψ τ∫ ∫ ∫ ∫ ...(4.2.6)

The electronic wave functions ande e∗′ ′′ψ ψ belong to different electronic states and therefore

they are orthogonal

0e e ed∗′ ′′ψ ψ τ =∫ ...(4.2.7)

Equation (4.2.6) then simplifies to

R v v n e e e ed d∗′ ′′ ′ ′′= ψ ψ τ µ ψ ψ τ∫ ∫ ...(4.2.8)


Fig. 4.2.5 Electronic transitions: (a) r'''e = r'e. Maximum value of overlap integral is for 0 0 transition,(b) r''e < r'e. Maximum value of overlap integral is for 0 2 transition

Since the vibrational wave function depends only on internuclear distance r, we can replace dn

by dr,

Therefore R = v v e e e edr d∗′ ′′ ′ ′′ψ ψ µ ψ ψ τ∫ ∫ ...(4.2.9)

The electronic transition probability is proportional to the square of matrix element

Re e e e ed∗′ ′′= µ ψ ψ τ∫ ...(4.2.10)

The electronic wave function e depends to some extent on the internuclear distance r but itsvaries very slowly. If we disregard the slow variation of e on r, we can replace Re by an average

value R .e Thus, the matrix element for the electronic transition between the vibrational levels '

and " can be expressed as

R Re v vdr′ ′′ν ν ′ ′′= ψ ψ∫ ...(4.2.11)

The integral appearing in Eqn. (4.2.11) is called overlap integral.The transition probability and hence the emission and absorption intensities of bands is

proportional to the square of overlap integral.


Refer to the potential energy curves of lower and upper electronic states of Fig. (4.2.5a). Theminima of the potential energy curves lie one above the other. Also shown are the wave functionsover the vibrational levels. The value of the overlap integral is maximum for the transition 0 0and hence, according to Franck-Condon principle, (0, 0) band will be most intense. The potentialenergy curves of Fig. (4.2.5b) are such that their minima are displaced relative to one another. Theoverlap integral is maximum for transition 0 2 and therefore this band (2, 0) will be most intense.

4.3 MOLECULAR STATES

Like atomic states, molecular states are defined by certain quantum numbers, which are defined asdescribed below:

Orbital Angular Momentum Quantum Number

When atoms combine to form a molecule, the inner electrons in each atom can be regarded as remainingassociated with their parent nucleus, but the outer electrons come to belong to the molecule as awhole rather than to any individual nucleus. In a diatomic molecule, there exists a strong electricfield in the direction of inter-nuclear axis. As a consequence of this field the resultant orbital angularmomentum L of all electrons undergo a precession about the direction of the electric field. The spacequantization of vector L permits only discrete values for the component of L along the field direction.The component of L along field direction is represented by vector Λ whose magnitude is given by

, where is quantum number specifying the magnitude of vector .Λ The allowed values of are

= 0, 1, 2, 3, ………..L ...(4.3.1)

where L is quantum number of the resultant orbital angular momentum L for all electrons in themolecule. The vector L is not defined, so L cannot be specified at all. For each value of L, there areL + 1 values of . Each value of corresponds to a distinct energy state. The negative values of are not considered because = + L and = –L represent the same state with identical energies.Therefore, for each value of L, there are L + 1 molecular energy states. Thus, all the states(except = 0 ) are doubly degenerate. Molecular states are represented by symbols , ...according to following scheme:

0 1 2 3 4……

states …..

Spin Quantum Number

The appearance of fine structure of electronic bands necessitates the introduction of spin quantumnumber. The spin of electrons in a molecule outside the closed shells, add up to form a resultant spinangular momentum S. The quantum number associated with S is represented by S. The quantumnumber S is obtained by adding spins of various electrons outside the closed shells. If the total numberof electrons is even, S is zero or integer. If the total number of electrons is odd, S is odd number ofhalf integer. The orbital motion of electron in states other than = 0, produces magnetic field in the


direction of inter-nuclear axis. This field causes precession of vector S about the direction of thefield. The permitted components of S along the inter-nuclear axis is given by quantum number whose allowed values are

= S, S – 1, S – 2, ……..0, ………..– (S – 1), – S. ...(4.3.2)

Thus, there are 2S + 1 different values of for every value of S. The quantum number isnot defined for the state = 0 (i.e., -state). The quantity 2S + 1 is referred to as the multiplicity ofthe state characterized by quantum number S.

Total Angular Momentum Quantum Number

In atom, the orbital angular momenta of electrons strongly couple to form a resultant L and so dothe spin momenta to form a resultant S. The two resultant vectors then combine to give total electronicangular momentum J. Similar phenomena takes place in molecule too. The components of orbitalangular momentum and spin angular momentum in the direction of inter-nuclear axis combineto form a resultant angular momentum . The quantum number associated with is represented by . The allowed values of are obtained by taking magnitude of algebraic sum of quantum numbers and .

= | | ...(4.3.3)

Fig. 4.3.1 Addition of and S ( = 2, S = 1). Energy level diagram for 3 state


Only the positive values of have significance. For 0, there are 2S + 1 possible values of, hence for a given value of , other than zero, there are 2S + 1 different values of . Thesecorrespond to somewhat different energies of the state. Thus, a molecular term with a given value of ( 0) splits into a multiplet of 2S + 1 components. If = 0, there is no magnetic field in thedirection of inter-nuclear axis, quantum number is not defined, and consequently, the -state doesnot split. So long as the molecule performs no rotation about the inter-nuclear axis, the state ' remainssingle. Nevertheless, 2S + 1 is called the multiplicity of a state.

To determine the molecular state, we consider an example in which = 2 and S = 1. Therelative orientation of vectors ,Λ S and Ω are shown in the Fig. 4.3.1.

The multiplicity 2S + 1 is added to the symbol as a left superscript and the value of assubscript. The components of 3 are designated as 33, 32, 31.

EXAMPLES

(a) Components of 1:

S = 0, = 0, = 1, = | | = 1. State is 11.

(b) Components of 2:

= 1, S = 1/2, = + 1/2, – 1/2, = | | = 1/2, 3/2. States are 21/2, 23/2.

(c) Components of 3:

= 1, S = 1, = –1, 0, 1, = | | = 2, 1, 0. States are 32, 31, 3 0.

(d) Components of 4:

= 1, S = 3/2, = –3/2, – 1/2, 1/2, 3/2. = | | = – 1/2, 1/2, 3/2, 5/2.

In this case four values of correspond to four different equidistant energy levels even thoughtwo of them have the same values ( | = 1/2). Hence , rather than is used todistinguish the multiplet components.

The states are 45/2, 43/2, 41/2, 4– 1/2.

(e) Components of 4:

= 2, S = 3/2, = –3/2, –1/2, 1/2, 3/2. | = 1/2, 3/2, 5/2, 7/2.

The states are 47/2, 45/2,

43/2, 41/2.

(f) Components of 4:

= 3, S = 3/2, = – 3/2, – 1/2, 1/2, 3/2. = | | = 3/2, 5/2, 7/2, 9/2.

The states are 49/2, 47/2, 45/2, 43/2.

CHAPTER

RAMAN SPECTRA

5.1 INTRODUCTION

When a monochromatic light of frequency 0 is made to pass through a cell containing a transparentsubstance, most of the light passes through the substance without suffering any changes. Only a verysmall fraction (0.1%) of light is scattered by the molecules of the sample in all directions. A largepart of the scattered light has frequency 0, which is the same as that of the incident light. This typeof scattering is called Rayleigh (or elastic) scattering. In addition to Rayleigh scattering, the scatteredlight is observed to have both lower and higher frequencies than the incident one. Although thischange in frequency due to scattering of light by molecules, was predicted theoretically by Smekalin 1923 but was experimentally discovered by Sir C.V. Raman and his collaborators in 1928. Forthis discovery Raman was awarded Nobel Prize in 1930. The Raman scattered light constitutes avery small fraction of incident light and hence is very weak. The lines of lower frequencies are calledStokes lines and those of higher frequencies are called Anti-stokes lines.

Fig. 5.1.1 Raman effect

Raman Spectra 583

Fig. 5.1.2 Vibrational spectrum of CCl4. Raman lines are displacedfrom the frequency of any exciting line by 218, 314 and 459 cm–1

To observe Raman spectrum, an intense highly monochromatic radiation in the visible regionis employed as an exciting (incident) radiation. Before 1960, the radiation of wavelength 4358Åemitted from mercury Toronto arc was used but now a days laser sources are extensively used becausethey are capable of producing highly coherent, monochromatic, and very intense narrow beam oflight. With laser sources, multistage photomultipliers are used as detectors for recording of Ramanspectrum.

Fig. 5.1.3 (a) Experimental set up for observing Raman spectrum


Fig. 5.1.3 (b) Raman cells

Laser sources : He-Ne 6328 Å Red

Ar 5145 Å Green

Kr 6471 Å Red

5.2 CLASSICAL THEORY OF RAMAN EFFECT

When a molecule is placed in an electric field E, a dipole moment P is induced in it. The ease withwhich the molecule gets polarized is measured by a quantity , called polarizability and the relationbetween P and E is written as

|P| = |E|

For an isotropic molecule, the induced dipole moment is in the direction of the electric field andthe polarizability is a scalar. In non-isotropic molecule, P is not in the direction of E, the electricfield in one direction produces dipole moment in different directions and is a tensor. In such moleculesthe relation between P (Px, Py, Pz) and E (Ex, Ey, Ez) is expressed as

P E E Ex xx x xy y xz z= α + α + α

P E E Ey yx x yy y yz z= α + α + α

P E E Ez zx x zy y zz z= α + α + α

The polarizability tensor α is defined by nine coefficients , ,..... .xx xy zzα α α Since

, ,xy yx yz zy zx xzα = α α = α α = α , the quantity is defined by six coefficients. From above equations it

is evident that x-component of electric field Ex produces dipole moment in molecule not only in x-direction but also in y-and z- directions. This act of electric field also holds for y and z components.The relation between the six polarizability coefficients and coordinates x, y and z is

2 2 2 2 2 2 1xx yy zz xy yz zxx y z xy yz zxα + α + α + α + α + α =

Raman Spectra 585

This equation represents an ellipsoid.When a non-isotropic molecule is subjected to an electromagnetic radiation, the oscillating electric

field induces time varying dipole moment and the polarizability also becomes a time varying function.For a molecule to exhibit Raman scattering, any component of its polarizability must change in thecourse of rotation or vibration. Therefore, it is necessary for the molecule to be polarizable to differentextents in different directions. For a diatomic molecule whether homonuclear or not, the polarizabilityellipsoid is not spherical and it will also change its dimensions in the course of vibration, hence allsuch molecules exhibit both rotational and vibrational Raman spectra.

If x represents the displacement during the oscillation of molecule, the variation of polarizabilitymay be expressed as

0 A

xα = α + β

where 0 = equilibrium polarizability, = rate of variation of polarizability with displacement, A =amplitude of oscillation of molecule. Assuming the molecule as a harmonic oscillator, the displacementx can be represented as

A cos2x tν= πν

where

is the frequency of oscillation.The polarizability may now be written as

0 cos2 tνα = α + β πν

If 0 0E E cos2 t= πν represents the oscillating electric field of the incident radiation, the dipole

moment induced in the molecule can be expressed as

0 0 0 0 0P E E cos2 E cos2 cos2t t tυ= α = α πν + β πν πν

= 0 0 0 0 0 01

E cos2 E cos2 ( ) cos2 ( )2

t t tν να πν + β π ν + ν + π ν − ν

It is evident from this equation that the dipole moment of molecule oscillates not only with

frequency 0ν of the incident radiation but also with frequencies 0 0and .ν νν − ν ν + ν The first

frequency 0 of the oscillating dipole is interpreted in terms of Rayleigh scattering. The frequencies

0 0andν νν − ν ν + ν are interpreted as the vibrational Raman (Stokes and anti-Stokes) frequencies.

Now, we shall show that rotation of molecule will also give Raman scattering. During the rotationof molecule, its orientation with respect to electric field of incident radiation undergoes continuouschange. If the molecule has different polarizabilities in different directions, its polarization will varywith time. The time variation of polarizability can be expressed as

0 cos2 (2 )r t′α = α + β π ν

where r represents the frequency of rotation of molecule. The presence of multiplication factor 2before r accounts for the fact that the rotation of molecule through angle brings it into an orientationin which its polarizability has the same value as that in the initial state. Thus, the polarizability variesat a rate that is twice as great as the rotation. The dipole moment of the molecule is given by


0 0 0 0 0 01

P E cos2 E cos2 ( 2 ) cos2 ( 2 )2 r rt t t′= α πν + β π ν − ν + π ν + ν

This equation states that the scattered radiation should consist of three frequencies

0 0 0, 2 and 2 .r rν ν − ν ν + ν The frequency 0 is the frequency of incident radiation,

0 02 and 2r rν − ν ν + ν are the Stokes and Anti-stokes frequencies.

5.3 QUANTUM THEORY OF RAMAN EFFECT

The quantum model of radiation treats a monochromatic radiation of frequency ν as a stream of

particles (photons) of energy hcν . When a radiation of frequency iν passes through matter, the

possible events that may occur with incident photon are as follows:(i) The incident photon collides elastically with the molecule of the scattering substance and

is scattered without any change in its frequency. This is called Rayleigh scattering ( ).s iν = ν

(ii) The incident photon collides inelastically with a molecule lying initially in a lower energystate E'' causing the latter to go over to higher energy state E'. In this process the incidentphoton transfers some of its energy to the molecule and is scattered with diminishedfrequency .sν Applying the law of conservation of energy to the collision process, wehave

E Ei shc hc′′ ′ν + = ν +

E E

i s hc

′ ′′−ν − ν = ...(5.3.1)

Since E' > E" , the Raman displacement i s∆ν = ν − ν is positive, s iν < ν . The frequency

sν of scattered light, in this case, is called Stokes frequency.

(iii) The incident photon collides inelastically with a molecule, which is already in excitedstate E', and acquires energy from the molecule causing it to go over to lower state E". In

this case the scattered photon has frequency sν greater than that of the incident photon iν .

The law of conservation of energy is

E Ei shc hc′ ′′ν + = ν +

E E

i s hc

′ ′′−ν − ν = − ...(5.3.2)

The Raman displacement i s∆ν = ν − ν is negative, i.e., s iν > ν . The scattered radiation of higher

frequency is called anti-Stokes frequency. It is evident from equations (5.3.1) and (5.3.2) that theRaman displacement is characteristic of scattering substance and is independent of exciting frequency( iν ). A schematic diagram showing stokes and anti-stokes transitions is given below. It is worth tonotice that the broken horizontal lines do not correspond to any energy level. Actual transitions areshown by heavy line.

Raman Spectra 587

Fig. 5.3.1 Stokes and Anti-stokes transitions

Raman displacement ∆ν lies within the range of 100 cm–1 to 3000 cm–1, which falls in theinfrared region. This indicates that the origin of Raman lines can be traced in the transitions betweenrotational and vibrational levels of molecules of the scattering substance.

Vibrational Raman Scattering

Vibrational Raman effect is observed when the energy levels E' and E" involved in Raman transitioncorrespond to the vibrational energy levels. The vibrational energy levels of a diatomic molecule aregiven by

21 12 2

E ( ) ( )e e ehc hcx= ν + ω − ν + ω

The corresponding term value is

21 12 2

G ( ) ( )e e ex= ν + ω − ν + ω

The allowed transitions are subject to the selection rules 1∆ν = ± . Vibrational Raman shift forStokes line, which corresponds to transition '' = 0 ' = 1, is given by

G ( ) G ( )

(1 2 )e ex

′ ′ ′′ ′′∆ν = ν − ν= − ω

The frequency (1 – 2 xe ) eω is the center of the fundamental band in the infrared spectrum of

the molecule. The intensity of a line is proportional to the population of molecules in the initialstate. Since a majority of molecules are in the ground state ('' = 0) at room temperature, the Stokeslines are obviously more intense. Since a very small number of molecules are in the state with' = 1, the anti-Stokes lines, which correspond to the transition ' = 1 '' = 0, are so weak thatthey are to observe.

It is worth to note that the presence of permanent dipole moment is not a condition for theoccurrence of Raman spectrum. The necessary condition for the Raman line to appear is that thepolarizability of dipole moment of molecule should change during the vibration or rotation. Symmetricmolecules such as H2, N2, O2, F2, Cl2, which do not give infrared spectrum, give Raman spectra.Thus, valuable information about a symmetric molecule can be obtained from the analysis of Ramanspectra which is not possible from infrared spectra.


Fig. 5.3.2 Raman spectrum of HCl molecule

When vibration Raman spectrum of a diatomic molecule is observed with a spectrograph ofsmall dispersive power, only one Stokes line and one Anti-stokes line for each exciting line withrelatively large Raman displacement is found. This is because even if molecules with different

values were present, the vibrational levels of harmonic oscillator are equidistant. The Raman shiftfor some molecules is given below:

HCl HBr HI NO H2 N2 O2

2886 2558 2233 1877 4160 2330 1554.7 cm–1

Rotational Raman Spectrum

The rotational energy levels of a diatomic molecule are given by

E B J (J 1)ch= +

The corresponding term values are

F BJ(J 1)= +

The selection rules for rotational Raman transitions for diatomic molecules in state(i.e., = 0) are

J = 0, ± 2.

+ sign for Stokes lines and – for anti-Stokes lines. [For diatomic molecule in other than state

(i.e., 0Λ ≠ ), the selection rules are J = 0, ± 1, ± 2]The Raman shift of Stokes lines (J = 2, i.e., J' = J'' + 2) is given by

J (J 1) J (J 1)B B′ ′ ′′ ′′∆ν = + − +

= B(J 2)(J 3) J (J 1)′′ ′′ ′′ ′′+ + − +

= 2B(2J 3)′′ +

= 2B(2J 3)+ J = 0, 1, 2, 3, ....,

4B(J 3/2) 6B, 10B, 14B,= + =

Raman Spectra 589

The Raman shift of Anti-stokes lines (J = –2, i.e., J' = J'' – 2) is

2B(2J 3)∆ν = − + , J = 0, 1, 2, 3, …..

4B(J 3/2) 6B, 10B, 14B, = − + = − − −

Notice that the first line ( J = 0 ) is displaced by a frequency of 6B on both sides of the excitingline. The subsequent lines are separated by a frequency of 4B. Thus, the rotational Raman spectrumconsists of two sets of lines, one on the lower frequency side (Stokes lines) and the other on thehigher frequency side of the exciting line. [In ordinary rotational spectrum, the successive lines areseparated by 2B. This difference is due to change in selection rule i.e., change in J by two units inRaman effect.]

Fig. 5.3.3 Rotational Raman spectrum

Since the rotational energy is of diatomic molecule is small, a considerable number of moleculesoccupy the higher rotational energy levels at room temperature because of the thermal motion. Forthis reason Stokes and anti-Stokes lines appear with nearly equal intensity.

Homo-nuclear diatomic molecules do not give infrared spectrum but they do give Ramanspectrum. Therefore, rotational and vibrational constants can be calculated from the analysis of Ramanspectra of such molecules.

The rotational Raman displacements [ 4B(J 3/ 2)∆ν = ± + ] are much smaller than the vibrational

Raman displacements [ (1 2 ) ]e ex∆ν = − ω . Therefore, the Raman spectrum will consist of a very intenseline representing the exciting line. On either side of this line, there will be Stokes and anti-Stokeslines of almost equal intensity corresponding to rotational transitions. At a greater distance from theexciting line, on the low frequency side there will be a strong Stokes line for the Q-branch (J = 0).Close to it there may be weaker lines of O and S branches for J = – 2 and + 2.


Fig. 5.3.4 Pure rotational Raman spectrum of CO2

Vibration-rotation Raman Spectrum

The vibration-rotation energy of a diatomic molecule is given by

21 12 2

E ( ) ( ) B J(J 1)e e ehc hcx hc= ν + ω − ν + ω + +

or the term value is given by

T = 21 12 2

( ) ( ) BJ(J 1)e e exν + ω − ν + ω + +

The Raman shift corresponding to the transition '' ', J'' J' is

21 12 2

21 12 2

[( ) ( ) BJ (J 1)]

[( ) ( ) BJ (J 1)]

e e e

e e e

x

x

′ ′ ′ ′∆ν = ν + ω − ν + ω + + −

′′ ′′ ′′ ′′ν + ω − ν + ω + +

The transitions are subject to the selection rules

1 and J 0, 2∆ν = ± ∆ = ±Designation of branches are made according to the following scheme:

Branch O P Q R S

J∆ –2 –1 0 1 2

Raman Spectra 591

The vibrational transition 1 0, 1 and J J′ ′′ ′ ′′ν = ν + = , which corresponds to Q-branch, gives

the Raman shift

Q (1 2 )e ex∆ν = − ω = 0ν for all J values.

Notice that Q-branch makes its appearance in Raman spectrum whereas it is absent in infraredspectrum.

The selection rules 1 and J 2 with 1 0′ ′′∆ν = ∆ = ν = ← ν = give the wave number of the lines

of S branch with Raman shift

0 [BJ (J 1) BJ (J 1)]S ′ ′ ′′ ′′∆ν = ν + + − +Substituting J' = J'' + 2, we have

S∆ν = 0 B[(J 2)(J 3) J (J 1)′′ ′′ ′′ ′′ν + + + − + ]

= 0 B[4J 6]′′ν + +

= 0 B[4J 6] J 0,1,2,3,...........ν + + =

0 4B(J 3/2)= ν + +

The selection rules 1 and J 2 . ., J J 2, with J 2,3,4,......i e ′ ′′ ′′∆ν = ∆ = − = − = give the lines of

O branch with Raman shift

0 0 B[(J 2)(J 1) J (J 1)′′ ′′ ′′ ′′∆ν = ν + − − − +

= 0 B[4J 2 ], J 2,3,4,......′′ ′′ν − − =

= 0 B[4J 6 ], J 0,1,2........′′ ′′ν − + =

= 0 B[4J 6] J 0,1,2,3.......ν − + =

= 0 4B[J 3/ 2]ν − +From the analysis of O and S branches of Raman spectrum it is possible to determine B from

which moment of inertia and inter-nuclear distance can be determined.The Anti-stokes lines are observed at high frequency side at the same distance from the exciting

line.

Raman Effect and Fluorescence

(1) Raman effect is a light scattering phenomenon. It can occur for any frequency of incidentradiation. The frequency of Raman lines (Stokes and Anti-stokes) depends on the frequencyof the exciting line. The Raman displacement depends on the nature of scattering substance.

Fluorescence is a phenomenon in which incident light quantum is completely absorbedby the molecule and it is raised to an excited state from which it makes transition to thelower energy levels after a certain time. The downward transition may take place in severalsteps with emission of radiation at each step. The frequency of emitted radiation is alwaysless than that of the incident radiation i.e., only Stokes frequencies are observed. It occurs


only for the absorbed frequency. The frequency of the Stokes line does not depend on thefrequency of the incident radiation but depends on the nature of the fluorescent material.

(2) Raman lines are strongly polarized whereas fluorescent lines are not.

(3) Raman lines are weak in intensity whereas fluorescent lines have considerable intensity.

SOLVED EXAMPLES

Ex. 1. For oxygen molecule, the internuclear distance is 1.21 × 10 –10 m, mass of oxygen atom is2.7 × 10 –26 kg. Calculate moment of inertia I, rotational constant B, separation of energy levels inm–1 and the wave number of the line corresponding to the transition J = 0 J = 1 in absorptionspectrum.

Sol. Moment of inertia of molecule

26

2 261 2

1 2

2.7 10I where kg 1.35 10 kg

2 2

m m mr

m m

−−×= µ µ = = = = ×

+

26 10 2 46 2I (1.35 10 kg)(1.21 10 m) 1.976 10 kg m− − −= × × = ×

Rotational constant 34

2 46 8 1

6.626 10 JB

8 I 8 9.86 (1.976 10 kg) (3 10 ms )

h s

c

−

− −×= =

π × × × × ×

= 141.5 m–1

Energy in m–1 or rotational term value is given by

F (J) = B J (J + 1), J = 0, 1, 2, …….

Separation of two consecutive rotational terms is

F F(J 1) F(J) B (J 1)(J 2) J(J 1) 2B(J 1)∆ = + − = + + − + = + Frequency of line corresponding to the transition J = 0 J = 1 is

1F(1) F(0) 2B 2 141.5 283.0 m−ν = − = = × =

Ex. 2. For hydrogen molecule the internuclear distance is 0.74Å, mass of hydrogen atom is 1.6738× 10 –27kg, calculate the rotational energy levels in electron volts.

Sol. Moment of inertia of molecule

2 2 27 10 21 12 2

I (1.6738 10 kg)(0.74 10 )r mr − −= µ = = × × kg m2

= 4.60 × 10–48 kg m2

Rotational energy of molecule

2 34 2

48 2

(1.054 10 Js)E(J) J (J 1) J (J 1)

2I 2(4.60 10 kgm )

−

−×= + = +

×

Raman Spectra 593

3E(J) (7.54 10 eV)J(J 1)−= × +This gives E0 = 0, E1 = 1.51× 10–2 eV, E2 = 4.54 × 10–2 eV, E3 = 9.05 × 10–2 eV.

Ex. 3. The OH-radical has a moment of inertia of 1.48 × 10 – 47 kg m2. Calculate its internucleardistance. What are its angular momentum and angular velocity in the state J = 5. Determine the energyabsorbed in the transition J = 6 J = 5 and corresponding wave number of the absorption line.

Sol. Reduced mass of the molecule 271 2

1 2

16 1 16 161.68 10 kg

16 1 17 17

m mu u

m m−×µ = = = = × ×

+ +

= 1.563 × 10–27 kg

Moment of inertia of molecule 47

227

I 1.48 10I

1.563 10r r

−

−×= µ ∴ = =

µ ×

119.73 10 m 0.973r −= × = Å

Magnitude of angular momentum | J | J(J 1)= +

In the state with J = 5, we have 34 34| J | 5(5 1) (1.054 10 ) 5.77 10 Js− −= + × = ×

Angular velocity 34

1347

| J | 5.77 103.90 10 rad/s

I 1.48 10

−

−×ω = = = ××

Wave number of line corresponding to transition J = 5 J = 6 is

5 6

34

2 47 2 8 1

2B(J 1), J 5

12 6.63 10 Js12B 12

8 I 8 9.86 (1.48 10 kgm ) (3 10 ms )

h

c

→−

− −

ν = + =

× ×= = = π × × × × ×

= 2.27 × 104 m –1

Energy of corresponding photon 21E 4.5 10 J.ch −= ν = ×

Ex. 4. Rotational spectrum of CO shows a strong absorption line at frequency 1.153 × 10 11 Hz.Calculate the internuclear distance.

Sol. Rotational term F (J) = B J ( J + 1 )Wave number of absorption line

0 1 F(1) F(0) 2B→ν = − =

Sincec

νν = , therefore, 22B or B or

2 28 I

h

c c cc

ν ν ν= = =π

This gives34

46 22 11 1

6.63 10 JsI 1.46 10 kgm

4 4 9.86 1.153 10 s

h −−

−×= = = ×

π ν × × ×


Internuclear distance46

1027

I 1.46 101.13 10 m 1.13

11.38 10r

−−

−×= = = × =

µ ×Å.

Ex. 5. The experimental values of wave number of line of P and R branches in absorption spectrumof HCl are given below. Find equilibrium internuclear distance and force constant of the molecule.

ν (P) cm–1 ν (R) cm–1

2906.3 3012.2

2927.5 3033.4

2948.7 3054.6

2969.9 3078.8

Sol. The separation of successive line of P-branch or R-branch is 2B. From the given data,

12B 21.2 cm .−∆ν = = B = 10.6 cm–1

= 1.62 × 10–27 kg

Moment of inertia 34

2 1 8

6.62 10 JsI

8 B 8 9.86 (1060m )(3 10 m/s)

h

c

−

−×= =

π × × ×

= 2.64 × 10–47 kg m2

47 210

27

I 2.64 10 kg m1.276 10 m 1.27

1.62 10 kger

−−

−×= = = × =

µ ×Å

Equilibrium frequency in wave number unit 01

2

k

cν =

π µ

Therefore, 2 2 204k c= π µ ν where 1

02969.9 3010.2

2991cm2

−+ν = =

From these values, we find k = 530 N/m.

Ex. 6. The ground state vibrational energy of hydrogen molecule is 0.0273 eV. Find the forceconstant of the molecule.

Sol. Ground state energy of molecule corresponds to = 0, therefore,

01

E ,where2

k= ω ω =µ

Raman Spectra 595

Therefore,

22280

16

2E 2 0.273eV(8.37 10 kg)

6.58 10 eVsk −

−

× = µ = × ×

= 576 N/m.

Ex. 7. In the near infrared spectrum of HCl molecule, there is a single intense band at 2885.9cm–1. Assuming that this band is due to the transition between vibrational levels, find the force constantof the molecule.

Sol. Reduced mass of the molecule

271 2

1 2

1 35 351.68 10 kg

1 35 36

m mu

m m−×µ = = = × ×

+ +

= 1.63 × 10–27 kg

Wave number of line 1288590 m−ν =Force constant of molecule

2 2 24k c= π ν µ

= 8 2 1 2 274 9.86 (3 10 m/s) (288590m ) (1.63 10 kg)− −× × × × = 481.86 N/m.

Ex. 8. Determine the vibrational energy levels of CO molecule in eV and cm –1. kCO =1870 N/m.

Sol. Reduced mass of CO molecule 1 2

1 2

12 16 48

12 16 7

m mu u

m m

×µ = = =+ +

= 1.14 × 10 –26 kg

Frequency of vibration of molecule 26

1 1 1870 N/m

2 2 3.14 1.145 10 kg

k−ν = =

π µ × ×

= 6.45 × 10 13 Hz

Energy of molecule 12

E ( ) , 0,1,2,.....h= ν + ν ν =

34 13 2012

E ( )(6.6 10 J 6.45 10 H ) (4.25 10 J)s z− −= ν + × × × = × 12

( )ν +

= 12

(0.265 eV)( )ν +

= 0.132 eV, 0.396 eV, 0.660 eV…….. = 1064.5, 3193.7, 5322.5 cm–1 (1 eV = 8065 cm–1).

Ex. 9. HCl molecule absorbs strongly infrared radiation of wavelength 3.465 µ. Calculate the forceconstant of the molecule.

Sol. Reduced mass of HCl µ = 1.61 × 10 –27 kg


Frequency of radiation 8

13 16

3 10 m/s8.658 10 s

3.465 10 m

c −−

×ν = = = ×λ ×

This is the frequency of molecule. Therefore,

2 2 13 1 2 274 4 9.86 (8.658 10 s ) (1.61 10 kg)k − −= π ν µ = × × × × = 475.9 N/m.

Ex. 10. HCl shows a strong absorption line in infrared spectrum at 2.886 × 10 5 m–1. Find theamplitude of vibration in the ground state.

Sol. Force constant of molecule 2 2 24k c= π µ ν

8 2 27 5 1 24 9.86 (3 10 m/s) (1.62 10 kg)(2.886 10 m ) 480.7 N/mk − −= × × × × × =

The amplitude of vibration in the ground state 20

1 1E

2 2hc k= ν = ξ

34 8 5 1(6.6 10 Js)(3 10 m/s)(2.886 10 m )

480.7 N/m

hc

k

− −ν × × ×ξ = =

= 0.11×10–10 m = 0.11 Å.

Ex. 11. The fundamental band for CO molecule is centered at 2143.3 cm–1, and the first overtone

at 4259.7 cm–1. Calculate .e e eand xω ωSol. The wave numbers of fundamental and first overtone are given by

1 (1 2 )e exν = − ω = 214330 m–1

2 (1 3 )2e exν = − ω = 425970 m–1

From these equations, we find 1 10.006, 217020 m , 1302me e e ex x− −= ω = ω = .

Ex. 12. The spectrum HCl shows a very intense absorption at 2886 cm–1 and a weaker at 5668 cm –1

and a very weak one at 8347 cm–1. Find the equilibrium frequency eω , anharmonicity constant and

force constant.

Sol. 1 (1 2 ) 2886e exν = ω − = ...(1)

2 2 (1 3 ) 5668e exν = ω − = ...(2)

3 3 (1 4 ) 8347e exν = ω − = ...(3)

From the first two equations, we get

( )1

2

1 2 2886

2 1 3 5668e

e

x

x

−ν= =

ν −

0.0174ex =

Raman Spectra 597

From (1) 1 12990 cm , 52 cme e ex− −ω = ω =

Force constant 2 2 24 ek c= π µω

8 2 27 1 24 9.86 (3 10 m/s) (1.61 10 kg)(299000 m )k − −= × × × = 510.9 N/m.

Ex. 13. The vibration-rotation absorption spectrum of CH molecule shows two peaks at8.657×1013 Hz and 8.483×1013 Hz on either side of the central frequency. Calculate the equilibriumseparation and force constant of the molecule.

Sol. Wave numbers of the absorption lines in vibration-rotation spectrum is given by

0 2B , 1,2,.......m m±ν = ν ± =

0 02B and 2B , for 1c c m+ −ν = ν + ν = ν − =

Therefore, B4c

+ −ν − ν=

2 48 I

h

cc

∆ν=π

, = 0.174 × 1013 s–1

34

2 13 1

6.6 10 JI

2 2 9.86 (0.174 10 s )

h s−

−×= =

π ∆ν × × ×

= 1.929 × 10– 47 kg m2

Equilibrium internuclear distance

I

r =µ

=47 2

1027

1.929 10 kg m1.12 10 m

1.544 10 kg−

−× = ×

×

Central frequency 0 is given by 13 10 8.570 10 s

2−+ −ν + ν

ν = = ×

Since 01

2

kν =π µ

2 2 27 13 1 204 4 9.86 (1.544 10 kg)(8.570 10 s )k − −= π µ ν = × × × ×

= 448 N/m.

Ex. 14. Find the frequencies corresponding to the two peaks on either side of the central frequencyin the absorption spectrum of a gas of CH molecules is the carbon is 12C and hydrogen is 3H. Mass of Catom is 12u and of H atom is 3.016u. For CH molecule r = 0.112 nm, µ = 4.002 × 10–27kg andk = 448 N/m. 1 u = 1.66 × 10 –27 kg.


Sol. Central frequency 13 1

01

5.325 10 s2

k −ν = = ×π µ

Moment of inertia of molecule I = r 2 = 5.02 × 10– 47 kg m2

Frequencies of the lines on either side of central line are given by

13 1

0 0 22 B 2 (5.325 0.033) 10 s

8 I

hc c

c−

ν = ν ± = ν ± = ± × π

= 5.358 × 10 13 Hz, 5.292 × 1013 Hz.

Ex. 15. The vibration-rotation spectrum of CO molecule is shown in the figure. Calculate theequilibrium internuclear distance, central frequency and force constant of the molecule.

Sol. From the spectrum 4B = 7.72 cm–1 B = 1.93 cm –1

Central frequency 10

2174.07 2166.352170.21 cm

2 2−+ −ν + ν +ν = = =

From the relation 46 2

2 2B , I 1.4567 10 kg m .

8 I 8 B

h h

c c−= = = ×

π π

Internuclear distance 46 2

26

I 1.4567 10 kg m1.13

1.145 10 kgr

−

−×= = =

µ × Å

Force constant 2 2 204 1910N/mk c= π µ ν = .

Ex. 16. The fundamental band for DCl 35 is centered at 2011.00 cm –1. Assuming the internucleardistance to constant at 1.288 Å, calculate the wave number of the first two lines of each of the P and Rbranches.

Sol. Reduced mass of molecule = 3.142 × 10–27 kg.

Rotational constant 1

2B 536 m

8 I

h

c−= =

π 1 1

R(0) 0 2B (201100 1072)m 202172m− −ν = ν + = + =

1R(1) 0 4B 203244m−ν = ν + =

1

P(1) 0 2B 200028m−ν = ν − =

1

P(2) 0 4B 199056 m.−ν = ν − =

Raman Spectra 599

Ex. 17. Assuming that there is no interaction between vibration and rotation, calculate the separationof the lines R(0) and P(1) of the fundamental band of HCl 35. The mean internuclear distance forHCl 35 in the = 0 and = 1 level is 1.293 Å.

Sol. Since r is the same in the two states, B also has the same value in the two states. = 1.62 × 10–27 kg.

1

2B 1033m

8 I

h

c−= =

π

R(0) 0 (1) 02B and 2Bpν = ν + ν = ν −

Separation between the two lines

14B 4132m .−∆ν = =

Ex. 18. For I2 molecule 1 1,e e e214.6 cm and x 0.6 cm− −ω = ω = calculate the temperature at whichthe number of molecules in the state = 1 is 1/e of that in the state ground state. At what temperature thenumber of molecules in the state = 1 state will be 10% of that in the ground state. Assume that themolecule behaves like anharmonic oscillator.

Sol.

(i) 1 0

2 21 1 1 12 2 2 2

E E E

1 ) (1 ) (0 ) (0 )e e e e e ehc x x

ν= ν=∆ = −

= + ω − + ω − + ω − + ω

= 2e e ehc xω − ω = 1.986 × 10–25 Jm [21460 – 120] m–1 = 4.238 × 10–21 J

The fraction of molecules in the state = 1 is given by

1

0

N Eexp

N Tk

∆ = −

21

23 1

4.238 10 Jexp( 1) exp

1.38 10 JK T

−

− −

×− = − × ×

T 300K=

(ii) 21

23 1

1 E E 4.238 10 Jexp T 133.5 K.

10 T 2.30 2.30 1.38 10 JKk k

−

− −∆ ∆ × = − ⇒ = = = × ×

Ex. 19. With exciting line 4358Å the pure rotation Raman spectrum of a sample gives Stokes lineat 4458 Å. Deduce the wavelength of Anti-stokes line.

Sol. 6 110

12.2946 10 m

4358 10 mexciting

−−ν = = ×

×


6 1

10

12.2431 10 m

4458 10 mstokes

−−ν = = ×

×

Raman shift 6 10.0515 10 m−∆ν = ×

6 1- 2.3461 10 manti stokes exc

−ν = ν + ∆ν = ×

Å.10- 4262.6 10 m 4262.6anti stokes

−λ = × =Ex. 20. In the rotational Raman spectrum of HCl, the displacements from the exciting line are

given by 1(62.4 41.6 ) .J cm−∆ν = ± + Calculate the moment of inertia of molecule.

Sol. Rotational Raman shift is given by

2B(2J 3)∆ν = ± +

Given that 1(62.4 41.6J) cm−∆ν = ± +Therefore 4B = 41.6 B = 10.4 cm–1

34

2 1 8 1

6.62 10 JsI

8 B 8 9.86 1040 m 3 10 m s

h

c

−

− −×= =

π × × × ×

= 2.7 × 10–47 kg m2.

Ex. 21. For the ground state of O2 molecule the values of e e eand xω ω are 1580.36 and

12.07 cm–1 respectively. Calculate the zero point energy and the expected vibrational Raman shift.Sol. The vibrational energy of molecule is

21 12 2

E ( ) ( ) , 0,1,2,e e ehc hc xν = ν + ω − ν + ω ν =

Zero point energy ( )1 10 2 4

E e e ex hcν= = ω − ω

34 8 1

0158036 1207

E 6.62 10 J 3 10 ms2 4

s− − = − × × × × = 1.563 × 10–20 J

Majority of the molecules will be in the ground state ( = 0). Vibrational Raman shift is

1 01 0

E E

hc←−

∆ν = ( = ± 1)

( ) ( )3 9 1 12 4 2 4

2e e e e e e e e ex x x∆ν = ω − ω − ω − ω = ω − ω

= (1580.36 – 24.14) cm–1

= 1556.22 cm–1.

Raman Spectra 601


1. (a) Explain the origin of far infrared (rotational) absorption spectrum of diatomic molecules.

(b) Why far infrared spectrum is not obtained for homonuclear diatomic molecules?

(c) What information can be obtained from the analysis of pure rotation spectra.

(d) Discuss the effect of centrifugal stretching of bond on the spectrum.

2. Describe the characteristics of far infrared absorption spectrum of diatomic molecules. Whatchanges occur in the spectrum when

(a) the molecule is assumed to be non-rigid rotator.

(b) one of the atom is replaced by its heavier isotope.

Discuss the variation of intensity of rotational lines with temperature.

3. Describe the salient features of infrared (vibration) spectrum of diatomic molecules. Explain thenature of spectrum assuming the molecule to be an anharmonic oscillator.

(a) What information about the molecule can one obtain from the analysis of infrared absorptionspectrum?

(b) Explain the variation of intensity of infrared bands.

(c) Why is infrared spectrum not obtained for homonuclear diatomic molecule?

4. How does a diatomic molecule as rotating-vibrator explain the main features of near infraredabsorption spectrum? Why is infrared spectrum not obtained for homonuclear diatomic molecule?

5. Discuss the effect of interaction (coupling) of vibrational and rotational motion of molecule onvibration-rotation spectrum.

6. Describe the coarse structure and fine structure of electronic spectra of diatomic molecules. Explainthe terms progressions and sequence.

7. (a) In the electronic spectrum of a diatomic molecule the bands are degraded either towards redor towards violet. On the basis of the shading of the bands what information will be obtainedabout the internuclear distance of the molecule in both of the electronic states?

(b) Write down the components of the 3 and 3 molecular states.

8. On the basis of Franck-Condon principle discuss the intensity distribution within a band systemobtained in the process of emission and absorption.

9. What is Raman effect? Give quantum theory of Raman scattering.

10. Describe and explain the main features of

(a) Vibrational Raman scattering.

(b) Rotational Raman scattering.

(c) Vibration-rotation Raman scattering.

11. Give the quantum mechanical theory of Franck-Condon principle.


12. Write notes on the following:

(a) Isotopic shift in vibrational spectrum of diatomic molecule.

(b) Formation of band head in electronic spectra of diatomic molecule.

(c) Shading or degradation of bands in electronic spectra of diatomic molecule.

(d) Raman effect and fluorescence.

13. Calculate the rotational energy levels of oxygen molecules in electron volt. Mass of oxygenatom is 2.7 × 10–26 kg, internuclear distance is 1.2 Å. Compare these energy levels with thermalat room temperature 27°C. What conclusions do you draw regarding the occupation of higherrotational energy levels? [Ans. EJ = (6.0 × 10–5 eV) J ( J + 1 ) and kT = 0.0258 eV]

14. Calculate the rotational energy levels of HCl molecule, given that: reduced mass of the molecule = 1.62 × 10–27 kg, internuclear distance r = 1.30Å. [Ans. EJ = (1.25 × 10–3 eV) J (J + 1) ]

CHAPTER

LASERS AND MASERS

6.1 INTRODUCTION

The words MASER and LASER are acronyms for Microwave Amplification by Stimulated Emissionof Radiation and Light Amplification by Stimulated Emission of Radiation respectively. The workingof these devices is based on the phenomenon of stimulated emission, which was first suggested byEinstein in 1917. In this process an assembly of atoms or molecules, initially in the excited state, arestimulated (induced) by radiation of appropriate frequency to drop to lower energy state therebyemitting radiation of the same frequency as that of the stimulating radiation. In 1953, Soviet scientistsN. Basov and A. Prokhorov and American scientists C. Townes and J. Weber independently developedMASERS. In 1964 Basov, Prokhorov and Townes were awarded Nobel Prize for this work. In 1955Gordon, Zeiger and Townes fabricated ammonia maser. Soon after this discovery the principle ofmaser was extended to optical range and in 1960 T. Meiman (USA) developed the first maser in theoptical range called LASER. In 1961 A. Javan operated the first gas laser, helium-neon gas laser.Since then a large number of masers and lasers have been developed. This chapter is devoted to anelementary account of basic principles of these devices.

6.2 STIMULATED EMISSION

The Einstein’s Coefficients

Consider an assembly of identical atoms in equilibrium with radiation at temperature T. Let E1 andE2 be two energy levels of the atoms. The frequency of radiation is such that

2 1E E−ω =

...(6.2.1)

Prior to 1917, it was supposed that the thermal equilibrium of atomic system and radiation wasdetermined by only two processes.

(i) stimulated (induced) absorption in which atoms are raised under the action of radiationfrom the lower energy level to the higher one.


(ii) spontaneous emission in which the atoms return of their own from higher energy level tothe lower one.

In 1917, Einstein while investigating this problem realized that two processes mentioned earlierviz stimulated absorption and spontaneous emission alone are not sufficient to maintain the equilibriumof the atomic system and the radiation. He introduced the concept of stimulated (induced) emissionin which the external radiation persuades the excited atoms to jump from the higher energy level tothe lower one by giving up energy. Thus, according to Einstein the equilibrium of matter and radiationgoverned by three processes:

(i) stimulated (induced) absorption.

(ii) spontaneous emission.

(iii) stimulated (induced) emission.

Let N1 and N2 be the number of atoms in the energy levels E1 and E2 respectively and u() bethe density of external radiation. The number of stimulated absorption per unit time per unit volumeis proportional to the number of atoms in the initial state and the energy density of the radiation i.e.,

R12 = B12 N1 u () ...(6.2.2)

where B12 is proportionality constant.The number of spontaneous emission per unit time/volume depends only on the number of

atoms in the excited state i.e.

R21 = A21 N2 ...(6.2.3)

where A21 is a constant.

Fig. 6.1.1 Induced, spontaneous and stimulated transitions

The number of stimulated emission per unit time/volume is proportional to the number of atomsin the excited state and the density of stimulating radiation i.e.

R*21 = B21 N2 u (w) ...(6.2.4)

The coefficients A and B’s are called the Einstein’s coefficients.In thermal equilibrium the number of upward transitions must be equal to the number of

downward transitions i.e.R12 = R21 + R*

21

B12 N1 u() = A21 N2 + B21 N2 u()

Lasers and Masers 605

Whence

21

21 12 1

21 2

A 1( )

B B N1

B N

u ω =

−

...(6.2.5)

According to Boltzmann equation

1

2

E / T1

E / T2

N .

N .

k

k

c e

c e

−

−

=

=

where c is a constant. From these equations, we find

1

2

N

N2 1(E E )/ T / Tk ke e− ω= =

...(6.2.6)

Making use of this result we can write eqn. (6.2.5) as

21

/ T1221

21

A 1( )

BB 1B

ku

e ωω =

− ...(6.2.7)

According to Planck’s radiation law the energy density of a radiation, which is in equilibriumwith matter, is given by

3

2 3 / T

1( )

1ku

c e ωωω =

π −

...(6.2.8)

Comparison of (6.2.7) and (6.2.8) gives

3

212 3

21

A

B c

ω=π

...(6.2.9)

and B12 = B21 ...(6.2.10)

The Einstein’s coefficients represent the transition probabilities per unit time. Eqn.(6.2.10) statesthat the stimulated absorption and stimulated emission are equally probable. The ratio of number ofspontaneous transitions to that of stimulated emissions is given by

21 21

*2121

R A

B ( )R u=

ω

3

2 3

1

( )uc

ω=ωπ

/ T 1ke ω= − ...(6.2.11)


In microwave region ( = 0.1m) at room temperature T = 300 K

( )( )Å

Å

4 4

9

12400eV.4.96 10 5 10

T T 10 0.025eV

ch

k k− −ω = = = × ≈ ×

λ

/ T 1ke ω ≈

whence

21

*21

R0

R≈

This means that in the microwave region the rate of stimulated emission is much higher thanthe spontaneous emission. Also

/ T2

1

N1

Nke− ω= ≈

i.e., the energy levels E1 and E2 are nearly equally populated.In the optical region ( = 5000 Å), we have

Å

Å

12400 eV100

T T (5000 )(0.025 eV)

ch

k k

ω = = ≈λ

10021*

21

R1

Re= − ≈ a very large number

That is, the spontaneous emission is more predominant in the optical region.The photons or the wave trains emitted in spontaneous emission move in random directions

and have no definite phase relationship with each other. In other words, the radiation is incoherent.On the other hand, the photons or the wave trains emitted in stimulated emission have the samefrequency, the same direction of propagation, the same phase and the same state of polarization i.e.,the stimulating and stimulated radiation are strictly coherent. This feature of stimulated emissionunderlies the action of a laser – a device in which the number of stimulated emissions predominatethe spontaneous emission. Since, the number of stimulated emissions is proportional to the numberof atoms in the upper level, it is essential to increase the number of atoms in the upper level.

6.3 POPULATION INVERSION

When an atomic system is in thermodynamic equilibrium, Boltzmann’s law determines the distributionof atoms in different energy states

E / TN C. i ki e−= ...(6.3.1)

where Ni is the number of atoms in energy state Ei and T is the temperature of the system. It isevident from the above formula that the population in a state diminishes with increase in energy of


that state. If an atomic system has two characteristic energy states E1 and E2 (> E1) with populationsN1 and N2 respectively then

1 2(E E )/ T1

2

N

Nke− −= ...(6.3.2)

Fig. 6.3.1 Population inversion

Fig. 6.3.2 A photon stimulates an excited atom causing it to emit a photon. The incident and the emittedphotons induce other exited atoms to emit photons. This process rapidly multiplies and an intense

laser beam builds up

At equilibrium, the lower energy state will be more populated than the upper state. Consequentlyin such system the absorption of radiation will predominate over the stimulated emission. A lightbeam while passing through such medium will get attenuated. A medium having this property is saidto have positive absorption coefficient. To obtain amplification of incident light, a condition has tobe created in which stimulated emission predominates over the stimulated absorption. Obviously thiscan be achieved if we can bring the system in a state with greater number of atoms in the upper state


than that in the lower state. A system having N2 > N1 is said to have inverse population. From

equation (6.3.2), we can see that the states of population inversion (N2 > N1, E2 > E1) correspond tonegative value of temperature T and therefore such states are called states with negative temperature.The population inversion is obtained by what is called optical pumping, which is a process ofimparting energy to the working substance of a laser to transfer the atoms to excited states. In asubstance with inverse population the stimulated emission may exceed the absorption of light andhence a light beam while passing through the medium will be amplified. Such a medium is calledactive medium. Allowing the light beam to traverse the same active medium many times before itemerges may further enhance its amplification.

6.4 THREE LEVEL LASER

In case of two levels laser, the method of pumping fails to produce the desired population inversionbecause the excited atoms residing in the excited state for a very short time interval lose their energythrough spontaneous emission and through collision with electrons and drop to the lower level. Toovercome this difficulty a three levels scheme was suggested by Basov, Prokhorov and Townes in1955 and the ruby laser was developed in 1960 by T. Meiman. In order to understand the workingprinciple of three-levels laser let us consider the energy level diagram of atom participating in lasingaction. Such an atom has three energy levels shown in the figure. By means flash light of appropriatefrequency the atoms are lifted from the ground state E1 to excited state E2 where their life time isextremely small (10–8sec). Some of the atoms spontaneously revert to the ground state whoseprobability is small. But most of the atoms rapidly pass through non-radiative transition to themetastable state (E3) where their life time is considerable long (=105 times) and stay there for long.

Fig. 6.4.1 Three level laser

In this way the population in the level E3 increases and that in E1 decreases. The state ofpopulation inversion is thus achieved. The photon emitted in the spontaneous transition (E3 E1)although its probability is small but not zero, induces the atoms in the metastable state to drop to theground state. The photon emitted in this way further induces other excited atoms. The stimulatedemission builds up rapidly.


6.5 THE RUBY LASER

A ruby is aluminium oxide (Al2O3) crystal in which some of the aluminium atoms are substitutedby chromium ions (Cr3+). In such a crystal, stimulated transitions occur in the chromium ions. Thechromium ion has two wide energy bands E2 and E3 very close to the ground level E1 and also adouble level E4 and E5. Light tube, which produces light with a broad band of frequencies, is usedto illuminate the chromium ions. Under the action of this light the chromium ions are raised fromthe ground state to E2 and E3, which is a group of closely spaced energy levels. In these energystates the life-time of ions is very small (=10–8 sec.). During this time some of the ions pass to theground state (spontaneous emission). Most of the ions, however, pass to the metastable state E4 andE5. The probabilities of these transitions are much greater than the spontaneous transitions(E2 E1, E3 E1). The energy of non-radiative transitions (E2 E4, E3 E5) is transferred tothe crystal lattice. In the metastable states the life-times of ions is about 10–3 sec. which is about 105

times greater than the life-time in ordinary excited state. In this way the population of metastablestate may exceed that of the ground state E1. In other words, the population of these two states willbe inverted. The population inversion is promoted still more by the low probability of the spontaneoustransition of ions from metastable states to the ground state.

The probability of spontaneous emission from the metastable states to the ground state is smallbut not zero. A photon emitted in spontaneous transition. (E4 E1, E5 E1) may cause stimulatedemission producing additional photons of wavelengths 6927 Å and 6943 Å, which subsequently furtherstimulate other excited ions to jump to the ground state. This process repeats again and again and acascade of photons is formed. The photons whose direction is parallel to the axis of the ruby rodsuffer multiple reflections at its ends. In their way, these photons stimulate the excited ions to returnto the ground state by emitting photons. The process of formation of cascade results in increase inintensity of the beam.

Fig. 6.5.1 Action of Ruby laser


6.6 HELIUM-NEON LASER

The Helium-Neon gas laser consists of a mixture Helium and Neon in the ratio of 7: 1. The gaseousis kept at low pressure (1 mm of Hg) in a discharge tube of about 1m long. At the both ends of thetube parallel mirrors are placed one of which is partly transparent. The spacing of mirrors is equalto integral multiple of half-wavelength of the laser radiation. An electric discharge is produced inthe gas by connecting the electrodes to a high frequency a.c. source. The electrons from the dischargecollide with helium atoms and the latter are excited to metastable states of energies 19.81 eV and20.5 eV. These excited states of helium are very close to the excited states of neon. When excitedhelium atoms collide with neon atoms in the ground state, a resonant energy transfer takes place andthe neon atoms are raised from their ground states to excited states. If the rate of upward transitionsis greater than the radiative decay of the excited atoms, the population in the excited state exceedsthat in the ground state. In this way, population inversion is achieved. Thus, the purpose of heliumatoms is to create population inversion in neon atoms. The important LASER transitions in neonatoms are:

E4 E6 3s 3p = 3.39 m E4 E5 3s 2p = 6328 Å E3 E5 2s 2p = 1.15 m

The wavelengths 3.39 µm and 1.15 µm are not in visible region.

Fig. 6.6.2 Transitions in He-Ne Laser

Fig. 6.6.1 Helium-Neon laser


6.7 AMMONIA MASER

In 1955 Gordon, Zeiger and Townes first developed the ammonia maser. The vibrational energylevels of ammonia molecule consist of pairs of energy levels with small separation compared to theseparation of one pair from the other. The lowest pair of energy levels, which has a separation of10 – 4 eV, is used in the fabrication of ammonia maser. This energy difference corresponds to frequency23870 MHz or wavelength 1.25 cm.

By heating the ammonia molecules in an oven a collimated beam of molecules is obtained.The beam consists of molecules in the upper and the lower excited states. In order to separate thesetwo kinds of molecules, the beam is passed through an inhomogeneous electric field produced byfour metallic rods placed symmetrically around the beam and connected to a d.c. source of 15 kV.This arrangement acts as a focuser and separator both. Due to different electric properties(polarizabilities) the two kinds of molecules behave differently in the inhomogeneous electric field.The polarizabilities of molecules in the lower and the upper states are opposite in sign. Therefore,the molecules in the upper state are repelled away and those in the lower state are attracted towardsthe electrodes. Thus the molecules in the lower state are dispersed and those in the upper state proceedundeviated along the axis and enter a cavity. All the molecules in the cavity are in the upper state. Asignal of frequency 23870 MHz is fed to the cavity that triggers the stimulated emission. The amplifiedradiation comes from another aperture of the cavity.

Fig. 6.7.1 Ammonia maser

6.8 CHARACTERISTICS OF LASER

The most striking features in which laser differs from conventional sources are following:1. Directionality: The light from a conventional source spreads in all direction whereas the

radiation from a laser travels in one direction only. Owing to this property the light froma laser can be transmitted over a very long distance without appreciable spread.

2. Intensity: Due to diverging nature of ordinary light, its intensity falls rapidly with distancewhereas the intensity of laser radiation remains almost unaltered after traversing a verylong distance.


3. Monochromatic nature of radiation: The photons emitted in stimulated emission haveessentially the same frequency and is therefore is strictly monochromatic.

4. The wave trains emitted during stimulated emission possesses definite phase relationshipwith each other and hence the laser light is highly coherent.

Applications

The directional and the coherence properties of laser light allows it to be used where tremendousspatial concentration of power is required. The extremely concentrated power may be used inconstructive and destructive both ways. In constructive way, it may be used in cutting hard material,drilling metal plates, producing high temperature for nuclear fusion reaction, etc. In destructive waysit may be used to destroy enemy installations, planes, war-heads missiles, etc.

In medicine laser light is used as a very useful surgical tool.In communication it is used to transmit information more conveniently than radio and

microwaves. In fact laser has revolutionized the field of communication.In scientific research it offers as an extraordinary light source for investigating molecular structure

(Raman effect).


1. Explain the terms: induced absorption, spontaneous emission and stimulated emission.

Obtain expressions for Einstein’s A and B coefficients and discuss their physical significance.

2. Distinguish between spontaneous and stimulated emission of radiation. Obtain a relationbetween the transition probabilities of two emissions.

3. Explain population inversion and optical pumping with suitable examples.

4. Explain what do you understand by meta-stable states and population inversion? How is thepopulation inversion achieved and why is it necessary for producing laser beam? Describebriefly the characteristics of Laser.

5. What is the principle of three level laser? Describe the principle, construction and working ofthree level Ruby laser.

6. Describe the principle and working of Helium-Neon laser with suitable diagrams.

7. Describe the principle and working of ammonia maser giving appropriate diagrams.

INDEX

Aberration of light 21Absolute activity 322Active medium 608Alpha decay 163Alpha-particle scattering experiment 379Angular momentum 179Anharmonicity constants 551Anti-stokes frequency 586Anti-stokes lines 582Antistokes’ frequencies 72Aufbau’s principle 415

Balmer formula 388Balmer series 384, 389Band head 560, 569Band origin 570Band system 565Basis functions 112Black body radiation 50, 328Bohr magneton 417Bohr orbit 386Bohr’s theory of hydrogenic atoms 385Bose gas 321Bose-Einstein condensation 324Bose-Einstein distribution function 358Bose-Einstein statistics 251, 305Bosons 252, 305Bracket series 384, 389Bragg’s spectrometer 534Braking radiation 70Breit’s scheme 427

Bremsstrahlung 70Bremsstrahlung process 523Burger-Dorgello-Ornstein sum rule 482

C.J. davisson 80Calcium triads 493Canonical distribution 284Canonical ensemble 284Central force 225Characteristic radiation 71Characteristic temperature 365Characteristic X-rays 521Chemical potential 279, 305, 321Classical principle of relativity 6Compound doublet 476Compound triplet 491Compton shift 65Compton wavelength 66Compton’s effect 65Condensation temperature 324Condon parabola 577Continuous 521Correspondence principle 397Critical temperature 323Cut-off frequency 62Cut-off potential 60

Debye model 362Debye T3 law 364Degeneracy 197, 309


Degeneracy temperature 324Degenerate 112, 252, 253, 257, 312, 324Degenerate states 309Degeneration of fermi gas 313Degree of degeneracy 112, 257Density function 271Density of states 198, 309Deslandre table 565Diffuse series 471Dirac formalism 178Doppler’s effect 19Dual nature of radiation 75Duane and Hunt law 523Dulong-Petits law 361

Effect or screening effect 527Einstein frequency 359Einstein temperature 361Einstein’s coefficients 603Electronic spectra 562Ensemble average 273Enthalpy 279Equipartition theorem 288Equivalent electrons 431Ergodic hypothesis 273Ergodic surface 281Exchange interaction 420Expectation value 113, 121

Fermi energy 310Fermi gas 309Fermi level 305, 311Fermi-Dirac distribution 358Fermi-Dirac statistics 251, 302Fermions 252, 302Fine structure 443Fine structure constant 387, 404Fine-structure 477Fine-structure levels 427

First overtone 552Fitzgerald contraction 15Fluorescence 591Fortrat diagram 571Fourier’s transform 85Frame of reference 3Franck and Hertz experiment 396Franck-Condon principle 573Fugacity (absolute activity) 322Fundamental (Bergmann) series 471Fundamental band 552

Galilean transformation 4G-factor spectroscopic splitting factor 437Gibb’s free energy 279Gibbs canonical probability distribution 284Gibbs paradox 283, 338Grand canonical ensemble 351Grand partition function 351Grand potential 354Group velocity 84, 86-space 268, 271Gyromagnetic ratio 417

Hamilton’s equations 268Heat capacity 297Heisenberg’s uncertainty principle or the principle

of indeterminacy 87Heliocentric frame 4Helmholtz free energy 279, 280, 346Hermite polynomials 210Homogeneity 3Hot bands 552Hund’s rule 416, 434

Incoherent scattering 65Inertial frames 3Interaction energy in J-J coupling 455

Index 615

Interaction energy in L-S coupling 451Internal energy 279Interval 18Inverse photoelectric effect 523Inverse population 608Isotopic shift 394, 547, 553Isotropy 3

J-J coupling 425

L.H. Germer 80Ladder operators 120, 180, 184Lamb shift 446Lande Interval rule 459, 491Laser 603Linear absorption coefficient 529Lorentz contraction 15Lorentz number 501Lorentz transformation equations 10Lorentz transformations 10Lowering operator 181L-S coupling 420Lyman series 383, 389

Macroscopic state 256Magnetic quantum number (ml) 413Magnetic sub-levels 500Maser 603Mass-energy equivalence 26Maxwell-Boltzmann or classical statistics 251Meson 305Metastable state 608, 496Michelson-Morley experiment 7Microscopic state 257Modified radiation 65Momentum of photon 28Monoatomic gas 337Monoatomic ideal gas 344

Morse potential 551µ-space 268Multiplet 427Multiplets 477Muonic (mesic) atom 393

Negative temperature 608Non-degenerate 112, 253, 312Non-equivalent electrons 427Non-inertial frame 4Nuclear motion 391Number space 255

Operators 106Optical pumping 608Orbital 414Orbital (azimuthal) quantum number 413Orbital g-factor 417Orthohelium 495Overlap integral 578

Pair production 30Parahelium 494Parity 186Particle velocity 86Partition function 286Paschen series 384, 389Pauli exclusion principle 342Pauli principle 414Pauli’s exclusion principle 309, 416P-branch 568Periodic boundary conditions 199Pfund series 384, 389Phase point 266Photoelectric effect 60, 523Photons 305Planck’s radiation law 328, 367, 54Polarizability 584


Positronium atom 392Postulate of equal a priori probability 272Potential well 189Principal quantum number 412Principal series 471Probability amplitude 102Probability density 102Probability current density 103Progression 565

Q-branch 568Quantization of phase space 269Quantum defect 471

Radiant emittance 50Raising operator 181Raman effect 72, 582Rayleigh (or elastic) scattering 582Rayleigh and Jeans law 52R-branch 560, 568Regular doublet 527Relativistic dynamics 24Representative point 266Resonance scattering 157Retarding potential 60Rotational characteristic temperature 348Rotational constant 544Rotational raman spectrum 588Rotational spectra 543Ruby laser 608, 609Runge’s law 477, 491Russell-Saunders coupling 420Rydberg constant 388Rydberg-Schuster law 477, 491

Sackur-Tetrode 283Sackur-Tetrode equation 347Satellite 476

Second overtone 552Sequence 565Shading off 570Sharp series 471Shell 414Simultaneity 15Singlet 478, 487Sommerfeld’s free electron theory 309Space quantization 413, 441Spectral distribution of energy 51Spectral series of hydrogen atom 383Spectral terms 427Spin g-factor 417Spin orbit interaction energy 443Spin quantum number (m

s) 413

Spin-orbit interaction 420, 425, 443, 478Spin-relativity doublet 527Spontaneous emission 604Stationary state 100Statistical weight 258Stefan’s law 56Step barrier 147Stern and gerlach experiment 441Stimulated (induced) absorption 603Stimulated (induced) emission 604Stokes frequency 586Stokes lines 582Stokes’ frequencies 72Sub-shell 414

Thermionic emission 318Thermodynamic probability 258, 278Thomas precession 444Three dimensional potential well 195Threshold 62Threshold frequency 60Threshold wavelength 60, 62Time averaged value 273Time dilation 16Transformation of acceleration 5

Index 617

Transformation of length 5Transformation of momentum and energy 28Transformation of velocity 5Triplet 487Two-body problem 225

Unmodified radiation 65

Variation of mass with velocity 22Vector model 420Vibrational characteristic temperature 350Vibrational constant 549Vibrational raman scattering 587

Vibrational spectra 549

Wave function 98, 102Wein’s law 52White radiation 71, 521Wien’s displacement law 57Work function 61

X-rays 520

Zeeman levels 500Zero-point energy 210, 550

introduction to modern physics-r.b.singh

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