engineering physics text book

Table of Contents

1 Elements Of Wave Mechanics 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Black body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 Experimental observation of black body radiation . . .. . . . 2

1.2.2 Laws of black body radiation . . . . . . . . . . . . . . . . . . 3

1.2.3 Stefan Boltzmann radiation law . . . . . . . . . . . . . . . . 3

1.2.4 Wien’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.5 Rayleigh – Jean’s law . . . . . . . . . . . . . . . . . . . . . . 4

1.2.6 Planck’s radiation law . . . . . . . . . . . . . . . . . . . . . 4

1.2.7 Derivation of Wien’s law from Planck’s law . . . . . . . . . . 5

1.2.8 Derivation of Rayleigh – Jean’s law from Planck’s law .. . . 6

1.3 Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Compton effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Matter waves and de Broglie’s hypothesis . . . . . . . . . . . . .. . 10

1.5.1 Davisson-Germer experiment . . . . . . . . . . . . . . . . . 11

1.5.2 G.P. Thomson experiment . . . . . . . . . . . . . . . . . . . 13

1.5.3 Wave packet and de Broglie waves . . . . . . . . . . . . . . . 14

1.5.4 Characteristics of matter waves . . . . . . . . . . . . . . . . 14

1.6 Phase and group velocities . . . . . . . . . . . . . . . . . . . . . . . 15

1.6.1 Relation between phase velocity and group velocity . .. . . . 16

1.6.2 Relation between group velocity and particle velocity . . . . . 17

1.6.3 Derivation of de Broglie relation . . . . . . . . . . . . . . . . 18

1.7 Heisenberg’s Uncertainty principle . . . . . . . . . . . . . . . .. . . 19

1.7.1 Origin and nature of the Principle . . . . . . . . . . . . . . . 19

1.7.2 An illustration of uncertainty principle . . . . . . . . . .. . . 21

1.7.3 Physical significance of uncertainty principle . . . . .. . . . 22

1.7.4 Applications of uncertainty principle . . . . . . . . . . . .. . 22

1.8 Wave mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.8.1 Characteristics of wave function . . . . . . . . . . . . . . . . 25

1.8.2 Physical significance of wave function . . . . . . . . . . . . .25

1.8.3 Schrodinger’s wave equation . . . . . . . . . . . . . . . . . . 26

1.8.4 Eigen values and eigen functions . . . . . . . . . . . . . . . . 27

1.9 Applications Of Schrodinger’s Equation . . . . . . . . . . . . .. . . 28

1.9.1 Case of a free particle . . . . . . . . . . . . . . . . . . . . . 28

1.9.2 Particle in a box . . . . . . . . . . . . . . . . . . . . . . . . 28

1.9.3 Finite Potential well . . . . . . . . . . . . . . . . . . . . . . 32

1.9.4 Tunnel effect . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.9.5 Examples of tunneling across a finite barrier . . . . . . . .. . 37

1.9.6 Theoretical interpretation of tunneling . . . . . . . . . .. . . 39

1.9.7 Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . 40

1.9.8 Practical applications of Schrodinger’s wave equation . . . . . 42

Numerical Examples . . . . . . . . . . . . . . . . . . . . . . 43

Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2 Crystallography and X-rays 55

2.1 Crystal Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2.1.1 Unit cell and space lattice . . . . . . . . . . . . . . . . . . . 55

2.1.2 Crystal systems . . . . . . . . . . . . . . . . . . . . . . . . . 57

2.1.3 Bravais lattices . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.1.4 Miller indices and their uses . . . . . . . . . . . . . . . . . . 60

2.1.5 Interplanar spacing in cubic crystals . . . . . . . . . . . . .. 63

2.1.6 Atomic packing factors . . . . . . . . . . . . . . . . . . . . . 65

2.1.7 Some crystal structures . . . . . . . . . . . . . . . . . . . . . 68

2.2 X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.2.1 Origin of x-rays . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.2.2 Continuous x-ray spectrum . . . . . . . . . . . . . . . . . . . 73

2.2.3 Characteristic x-ray spectrum . . . . . . . . . . . . . . . . . 75

2.2.4 Moseley’s law . . . . . . . . . . . . . . . . . . . . . . . . . 77

2.3 X-ray diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.3.1 Bragg’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.3.2 Bragg’s spectrometer . . . . . . . . . . . . . . . . . . . . . . 79

2.3.3 Structure determination . . . . . . . . . . . . . . . . . . . . . 80

2.4 Electron diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

2.5 Neutron diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3 Electrical Conductivity In Metals 89

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

3.2 Classical Free Electron Theory Of Metals(Drude - Lorentz Theory) . . . . . . . . . . . . . . . . . . . . . . . . 90

3.2.1 Expression for electrical conductivity . . . . . . . . . . .. . 90

3.2.2 Electron - lattice interaction and consequences . . . .. . . . 95

3.2.3 Failure of classical free electron theory . . . . . . . . . .. . 96

3.3 Quantum free electron theory of metals . . . . . . . . . . . . . . .. 97

3.3.1 Density of energy states in a metal . . . . . . . . . . . . . . . 100

3.3.2 Metal as a Fermi gas . . . . . . . . . . . . . . . . . . . . . . 103

3.3.3 Band theory of metals . . . . . . . . . . . . . . . . . . . . . 104

3.3.4 Merits of quantum free electron theory . . . . . . . . . . . . 105

3.4 Electron Scattering Mechanisms . . . . . . . . . . . . . . . . . . . .106

3.4.1 Effect of temperature . . . . . . . . . . . . . . . . . . . . . . 107

3.4.2 Effect of impurities . . . . . . . . . . . . . . . . . . . . . . . 108

3.5 Thermionic emission . . . . . . . . . . . . . . . . . . . . . . . . . . 110


Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4 Superconductivity 117

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

4.2 Characteristic features of superconductors . . . . . . . . .. . . . . . 118

4.2.1 Isotope effect . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.2.2 Meissner effect . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.3 Classification of superconductors . . . . . . . . . . . . . . . . . .. . 121

4.4 Applications of superconductors . . . . . . . . . . . . . . . . . . .. 123

4.5 Theoretical interpretation of superconductivity . . . .. . . . . . . . . 128

4.6 High Temperature Superconductors . . . . . . . . . . . . . . . . . .129


Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

5 Semiconductors 132

5.1 Band Structure of Solids . . . . . . . . . . . . . . . . . . . . . . . . 132

5.2 Intrinsic semiconductors . . . . . . . . . . . . . . . . . . . . . . . . 134

5.2.1 Carrier generation in intrinsic semiconductors . . . .. . . . . 135

5.2.2 Fermi factor and Fermi energy . . . . . . . . . . . . . . . . . 136

5.2.3 Conductivity of an intrinsic semiconductor . . . . . . . .. . 137

5.2.4 Effect of temperature on conductivity . . . . . . . . . . . . . 138

5.3 Extrinsic semiconductors . . . . . . . . . . . . . . . . . . . . . . . . 138

5.3.1 Conductivity of an extrinsic semiconductor . . . . . . . .. . 141

5.3.2 Effect of temperature on the conductivity of extrinsic semicon-ductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

5.3.3 Concentration and mobility of current carriers . . . . .. . . . 143

5.4 Generation and recombination of carriers . . . . . . . . . . . .. . . 146

5.5 Direct and indirect band gap semiconductors . . . . . . . . . .. . . . 146

5.5.1 Semiconductor materials . . . . . . . . . . . . . . . . . . . . 149

5.6 Hall effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

5.6.1 Experimental determination of carrier concentration . . . . . 153

5.6.2 Hall effect in intrinsic semiconductors . . . . . . . . . . . . . 154

5.6.3 Applications of Hall effect . . . . . . . . . . . . . . . . . . . 155

5.7 p-n Junction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

5.7.1 Unbiased p-n junction . . . . . . . . . . . . . . . . . . . . . 156

5.7.2 Semiconductor junction with applied bias . . . . . . . . . .. 160

5.7.3 Incremental junction capacitance . . . . . . . . . . . . . . . .163

5.7.4 Breakdown in a p-n junction . . . . . . . . . . . . . . . . . . 164

5.8 Zener diode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

5.8.1 Zener breakdown mechanism . . . . . . . . . . . . . . . . . 165

5.8.2 Identification of breakdown mechanism in a p-n junction . . . 166

5.9 Applications of p-n junctions . . . . . . . . . . . . . . . . . . . . . .167

5.9.1 Junction diode as rectifier . . . . . . . . . . . . . . . . . . . 167

5.9.2 Zener diode . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

5.9.3 Photo diode . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

5.9.4 Photovoltaic effect and solar cell . . . . . . . . . . . . . . . . 171

5.9.5 Light emitting diode . . . . . . . . . . . . . . . . . . . . . . 174


Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

6 Dielectric Properties Of Materials 182

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

6.2 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

6.2.1 Mechanisms of polarization . . . . . . . . . . . . . . . . . . 186

6.2.2 Temperature dependence of polarization . . . . . . . . . . .. 189

6.2.3 Effect of frequency on polarization . . . . . . . . . . . . . . . 190

6.3 Dielectric Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

6.3.1 Dielectric constant of monoatomic gases . . . . . . . . . . .191

6.3.2 Dielectric constant of polyatomic gases . . . . . . . . . . .. 193

6.3.3 Internal field in solids and liquids . . . . . . . . . . . . . . . 193

6.3.4 Dielectric constant of elemental solids . . . . . . . . . . .. . 196

6.3.5 Dielectric constant of ionic solids without permanent dipoles . 197

6.3.6 Dielectric constant of polar materials . . . . . . . . . . . .. 198

6.4 Ferroelectric materials . . . . . . . . . . . . . . . . . . . . . . . . . 199

6.5 Piezoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

6.6 Dielectric losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203


Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

7 Magnetic Properties 209

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

7.2 Classification of magnetic materials . . . . . . . . . . . . . . . .. . 210

7.3 Origin of permanent dipoles . . . . . . . . . . . . . . . . . . . . . . 212

7.4 Magnetic hysteresis . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

7.5 Hard and soft magnetic materials . . . . . . . . . . . . . . . . . . . .218

7.6 Metallic and ceramic magnetic materials . . . . . . . . . . . . .. . . 219

7.7 Ferrites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

7.8 Applications of magnetic materials . . . . . . . . . . . . . . . . .. . 221


Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

8 Applied Optics 225

8.1 Absorption and emission of radiation . . . . . . . . . . . . . . . .. . 225

8.1.1 Luminescence . . . . . . . . . . . . . . . . . . . . . . . . . 225

8.1.2 Induced absorption . . . . . . . . . . . . . . . . . . . . . . . 226

8.1.3 Spontaneous emission . . . . . . . . . . . . . . . . . . . . . 227

8.1.4 Stimulated emission . . . . . . . . . . . . . . . . . . . . . . 227

8.2 Lasers - basic principles . . . . . . . . . . . . . . . . . . . . . . . . . 228

8.2.1 Einstein’s theory of stimulated emission . . . . . . . . . .. 228

8.2.2 Conditions for laser action . . . . . . . . . . . . . . . . . . . 230

8.2.3 Methods of achieving population inversion . . . . . . . . .. 231

8.2.4 Requirements of a laser system . . . . . . . . . . . . . . . . . 231

8.3 Types Of Lasers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

8.3.1 Ruby Laser . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

8.3.2 Helium-Neon laser . . . . . . . . . . . . . . . . . . . . . . . 236

8.3.3 Semiconductor diode laser . . . . . . . . . . . . . . . . . . . 237

8.4 Applications of lasers . . . . . . . . . . . . . . . . . . . . . . . . . . 239

8.4.1 Industrial applications . . . . . . . . . . . . . . . . . . . . . 239

8.4.2 Medical applications . . . . . . . . . . . . . . . . . . . . . . 240

8.4.3 Estimation of atmospheric pollution . . . . . . . . . . . . . .240

8.4.4 Applications in basic sciences . . . . . . . . . . . . . . . . . 240

8.5 Holography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

8.5.1 Recording of holograms . . . . . . . . . . . . . . . . . . . . 241

8.5.2 Reconstruction of images . . . . . . . . . . . . . . . . . . . . 242

8.5.3 Applications of holography . . . . . . . . . . . . . . . . . . . 243

8.6 Optical fibers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

8.6.1 Materials for optical fibers . . . . . . . . . . . . . . . . . . . 243

8.6.2 Propagation of light through an optical fiber . . . . . . . .. . 244

8.6.3 Modes of propagation in a fiber . . . . . . . . . . . . . . . . 247

8.6.4 Signal distortion in optical fibers . . . . . . . . . . . . . . . .249

8.6.5 Signal attenuation in optical fibers . . . . . . . . . . . . . . .249

8.7 Applications of optical fibers . . . . . . . . . . . . . . . . . . . . . .250

8.7.1 Fiber optic communication . . . . . . . . . . . . . . . . . . . 250

8.7.2 Applications in medicine and industry . . . . . . . . . . . . .252


Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

9 Modern Materials And Methods 256

9.1 Ceramics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

9.1.1 Glasses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

9.1.2 Clay products . . . . . . . . . . . . . . . . . . . . . . . . . . 257

9.1.3 Refractories and abrasives . . . . . . . . . . . . . . . . . . . 257

9.1.4 Cements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

9.1.5 Cermets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

9.2 Composite Materials . . . . . . . . . . . . . . . . . . . . . . . . . . 259

9.3 Smart Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

9.4 Shape Memory Alloys . . . . . . . . . . . . . . . . . . . . . . . . . 261

9.5 Microelectromechanical Systems . . . . . . . . . . . . . . . . . . .. 263

9.5.1 Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

9.5.2 Actuators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

9.6 Nano Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

9.6.1 Synthesis of nano materials . . . . . . . . . . . . . . . . . . 267

9.6.2 Applications of nano materials . . . . . . . . . . . . . . . . . 269

9.6.3 Scaling laws . . . . . . . . . . . . . . . . . . . . . . . . . . 271

9.6.4 Carbon nano clusters . . . . . . . . . . . . . . . . . . . . . . 273

9.6.5 Carbon nano tubes . . . . . . . . . . . . . . . . . . . . . . . 274

9.7 Liquid Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

9.7.1 Classification of liquid crystals . . . . . . . . . . . . . . . . .275

9.7.2 Applications of Liquid Crystals . . . . . . . . . . . . . . . . 278

9.8 Non Destructive Testing Of Materials . . . . . . . . . . . . . . . .. 279

9.8.1 Radiographic methods . . . . . . . . . . . . . . . . . . . . . 279

9.8.2 Ultrasonic methods . . . . . . . . . . . . . . . . . . . . . . . 280

9.8.3 Magnetic methods . . . . . . . . . . . . . . . . . . . . . . . 283

9.8.4 Electrical methods . . . . . . . . . . . . . . . . . . . . . . . 284

9.8.5 Optical methods . . . . . . . . . . . . . . . . . . . . . . . . 284

9.8.6 Thermal methods . . . . . . . . . . . . . . . . . . . . . . . . 284

9.9 Quantum Computation . . . . . . . . . . . . . . . . . . . . . . . . . 285

9.9.1 Properties of quantum bits . . . . . . . . . . . . . . . . . . . 285

9.9.2 Quantum gates . . . . . . . . . . . . . . . . . . . . . . . . . 287

9.9.3 Multiple qubits . . . . . . . . . . . . . . . . . . . . . . . . . 288

Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

10 Special Theory of Relativity 29110.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

10.1.1 Frames of Reference . . . . . . . . . . . . . . . . . . . . . . 291

10.1.2 Galilean transformation . . . . . . . . . . . . . . . . . . . . 292

10.1.3 Michelson-Morley experiment . . . . . . . . . . . . . . . . . 293

10.2 Postulates of Special Theory of Relativity . . . . . . . . . .. . . . . 294

10.3 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

10.4 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

10.5 Twin Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

10.6 Relativity of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

10.7 Massless Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

Chapter 1

Elements Of Wave Mechanics

1.1 Introduction

The nineteenth century was a very eventful period as far as Physics is concerned. Thepioneering work on dynamics by Newton, on electromagnetic theory by Maxwell, lawsof thermodynamics and kinetic theory were successful in explaining a wide variety ofphenomena. Even though a majority of experimental evidenceagreed with the classi-cal physics, a few experiments gave results that could not beexplained satisfactorily.These few experiments led to the development of modern physics. Modern physicsrefers to the development of the theory of relativity and thequantum theory. Inabil-ity of the classical concepts to explain certain experimental observations, especiallythose involving subatomic particles, led to the formulation and development of mod-ern physics. Early twentieth century saw the development ofmodern physics. Thepioneering work of Einstein, Planck, Compton, Roentgen, Born and others formedthe basis of modern physics. The dual nature of matter proposed by de Broglie wasconfirmed by experiments. The wave mechanics and quantum mechanics were latershown to be identical in their mathematical formulation. The validity of classical con-cepts was explained to be the result of an extrapolation of modern theories to classicalsituations. In the present chapter, experimental observations of three important phe-nomena – black body radiation, photoelectric effect and Compton effect – consideredas the beginning of modern physics, are briefly described.

1.2 Black body radiation

When radiation is incident on material objects, it is eitherabsorbed, reflected or trans-mitted. These processes are dependent on the radiation and the object involved. Anobject that is capable of absorbing all radiation incident on it is called a black body.Practically, we cannot have a perfect black body but can haveobjects that are onlyclose to ablack body.

For example, a black body can be approximated by a hollow object with a verysmall hole leading to the inside of the object. Any radiationthat enters the objectthrough the hole gets trapped inside and will be reflected by the walls of the cavity tillit is absorbed.Objects that absorb a particular wavelength of radiation are also

1

2 Elements Of Wave Mechanics

found to be a good emitter of radiation of that particular wavelength. Hence, ablack body is also a good emitter of all radiations it has absorbed.

Emissions from objects depend on the temperature of the object. It has been ob-served that the energy emitted from objects increases as thetemperature of the objectis increased. Laws of radiation have been formulated to explain the emission of energyby objects maintained at specific temperatures.

1.2.1 Experimental observation of black body radiation

Experiments have been carried out to study the distributionof energy emitted by apractical blackbody as a function of wavelength and temperature.

Eλ

λ

T

T

T

T T T

1

2

3

1 2 3> >

Figure1.1Distribution of emitted energy as a function of wavelength and temperaturefor a black body.

Figure 1.1 shows the distribution curves in which the energydensityEλ is plottedas a function of wavelength at different temperatures of the black body. Energy densityis defined as the energy emitted by the black body per unit areaof the surface. Theimportant features of these distribution curves may be summarized as follows:

(i) The energy vs wavelength curve at a given temperature shows a peak indicatingthat the emitted intensity is maximum at a particular wavelength and decreasesas we move away from the peak.

(ii) An increase in temperature results in an increase in thetotal energy emitted andalso the energy emitted at all wavelengths.

Elements Of Wave Mechanics 3

(iii) As the temperature increases, the peak shifts to lowerwavelengths. In otherwords, at higher temperatures, maximum energy is emitted atlower wavelengths.

1.2.2 Laws of black body radiation

The initial attempts to explain black body radiation were based on classical theoriesand were found to be limited in application. They could not explain the entire spectrumof the radiation satisfactorily.

1.2.3 Stefan Boltzmann radiation law

It states that the total energy densityE of radiation emitted from a black body isdirectly proportional to the fourth power of its absolute temperatureT. Energy densityE0 is defined as the total of all the energy emitted at all wavelengths per unit areaof the emitter surface.

E α T4

or E = σT4 (1.1)

whereσ is a constant calledStefan’s constant. It has a numerical value equalto 5.67× 10−8 watt m−2K−4. This law was suggested empirically by Stefan and laterderived by Boltzmann on thermodynamic considerations. Thelaw agrees well with theexperimental results.

1.2.4 Wien’s Laws

Wien’s displacement lawstates that the wavelengthλm corresponding to the maxi-mum emissive energy decreases with increasing temperature.

i.e., λm ∝ 1/T or λmT = b (1.2)

whereb is called theWien’s constant and is equal to 2.9 × 10−3 mK. The energydensity emitted by a black body in the wavelength rangeλ andλ + d λ is given by

Eλ dλ = c1λ−5exp (−c2/λT)dλ (1.3)

wherec1 andc2 are constants. This is known asWien’s distribution law . This lawholds good for smaller values ofλ but does not fit the experimental curves for highervalues ofλ (fig 1.2).


λ

Eλ

Wien′s law

Figure1.2Comparison of experimental distribution curve with Wien’slaw.

1.2.5 Rayleigh – Jean’s law

According to this law, the energy density emitted by a black body in the wavelengthrangeλ And λ + dλ is given by

Eλdλ =8πkTλ4

dλ (1.4)

This equation does not show any peak in the energy value but the energy goes onincreasing with decrease in wavelength. The total energy emitted is infinite for alltemperatures above 0K.

This is not at all in agreement with the experimental observation. The law holdsgood only for large values of wavelength (fig 1.3). At lower wavelengths, the energydensity increases and becomes very large for wavelengths inthe ultra violet region.Such a large increase in the energy emitted at low wavelengthdoes not occur experi-mentally. This discrepancy is known as “Ultraviolet catastrophe” of classical physics.

All the above laws are based on classical thermodynamics andstatistics. They areinsufficient to explain the black body radiation satisfactorily.

1.2.6 Planck’s radiation law

This law is based on quantum theory. Max Planck proposed thatatoms or moleculesabsorb or emit radiation in quanta or small energy packets called photons. Energy of


λ

Eλ

T

R J law

Figure1.3Comparison of experimental distribution curve with Rayleigh-Jean’s law.

each photon can be expressed asE = hν

whereν is the frequency of the radiation corresponding to the energy E, h is a constantcalledPlanck’s constantand is equal to 6.63× 10−34 Js. Light quanta are indistin-guishable from each other and there is no restriction on the number of quanta havingthe same energy. In other words, Pauli’s exclusion principle is not applicable to them.The quantum statistics applicable to photons is Bose-Einstein statistics. Consideringall the energy emitted by the black body in the form of photonsof different energy,Planck applied Bose - Einstein statistics to obtain the energy distribution of photons.Accordingly, the energy density emitted in the wavelength rangeλ and (λ + dλ) isgiven by

Eλdλ =8πhcλ5

1(ehc/λkT − 1)

dλ (1.5)

This distribution agrees well with the experimental observation of black body ra-diation and is valid for all wavelengths. Further, it reduces to Wien’s law for lowerwavelength region and to Rayleigh – Jean’s law for higher wavelength region.

1.2.7 Derivation of Wien’s law from Planck’s law

Whenλ is small, we can consider

ehc/λkT > 1


∴

[

ehc/λkT − 1]

≈ ehc/λkT

Substituting in equation (1.5), we get

Eλ dλ =8πhcλ5·

1(

ehc/λkT) · dλ

i.e., Eλ dλ = c1λ−5. exp (−c2/λT)dλ (1.6)

wherec1 = 8πhcandc2 = hc/k

Equation (1.6) is theWien’s law.

1.2.8 Derivation of Rayleigh – Jean’s law from Planck’s law

Whenλ is large, hcλkT < 1.

[

ehc/λkT − 1]

≈ hc/λkT


Eλ dλ =8πhcλ5·λkThc· dλ

i.e., Eλ dλ =8πkTλ4

dλ (1.7)

Equation (1.7) is the Rayleigh – Jean’s law.

1.3 Photoelectric effect

Emission of electrons from a metal surface when light of suitable energy falls on it iscalledPhotoelectric effect. The experimental setup for observing photoelectric effectconsists of a pair of metal plate electrodes in an evacuated tube connected to a sourceof variable voltage as shown in fig.1.4.

When light of suitable energy is incident on the cathode, electrons are emitted anda current flows across the tube. The characteristic curves for the photoelectric emissionas shown in fig. 1.5.

The important properties of the emission are as follows:

(i) There is no time interval between the incidence of light and the emission of pho-toelectrons.


A

V

Figure1.4Experimental set up to study photoelectric effect.

L3

L2

L1

I

V

Figure1.5Current - voltage characteristics of photocell. The Intensity of illuminationincreases from L1 to L3.

(ii) There is a minimum frequency for the incident light below which no photoelec-tron emission occurs. This minimum frequency, calledthreshold frequency,depends on the material of the emitter surface. The energy corresponding to thisthreshold frequency is the minimum energy required to release an electron fromthe emitter surface. This energy is characteristic of the material of the emitterand is called thework function of the material of the emitter.

(iii) For a given constant frequency of incident light, the number of photoelectronsemitted or the photo current is directly proportional to theintensity of incidentlight.


(iv) The photoelectron emission can be stopped by applying areverse voltage to thephototube, i.e. by making the emitter electrode positive and the collector nega-tive. This reverse voltage is independent of the intensity of incident radiation butincreases with increase in the frequency of incident light.The negative collec-tor potential required to stop the photo electron emission is called thestoppingpotential.

These characteristics of photoelectron emission can not beexplained on the basisof classical theory of light but can be explained using the quantum theory of light. Ac-cording to this theory, emission of electrons from the metalsurface occurs when theenergy of the incident photon is used to liberate the electrons from their bound state.The threshold frequency corresponds to the minimum energy required for the emission.This minimum energy is called the work function of the metal.When the incident pho-ton carries an energy in excess of the work function, the extra energy appears as thekinetic energy of the emitted electron. When the intensity of light increases, the num-ber of photoelectrons emitted also increases but their kinetic energy remains unaltered.The reverse potential required to stop the photoelectron emission, i.e. the stopping po-tential, depends on the energy of the incident photon and is numerically equivalent tothe maximum kinetic energy of the photoelectrons.

When a photon of frequencyν is incident on a metal surface of work functionΦ,then,

hν = Φ +

(

12

mv2

)

max

(1.8)

where (12mv2)max is the maximum kinetic energy of the emitted photoelectrons. This is

known asEinstein’s photoelectric equation. SinceΦ = hv, it can also be written as(

12

mv2

)

max

= hv− Φ = h(v− v) (1.9)

If V is the stopping potential corresponding to the incident photon frequencyv, then,(

12

mv2

)

max

= hv−Φ = eV (1.10)

Then, by experimental determination ofV, it is possible to find out the work func-tion of the metal.

The experimental observation of photoelectric effect leads to the conclusion that theenergy in light is not spread out over wavefronts but is concentrated in small packetscalled photons. All photons of a particular frequency have the same energy. A change


in the intensity of the incident light will change the numberof photoelectrons emittedbut not their energies. Higher the frequency of the incidentlight, higher will be thekinetic energy of the photoelectrons. These observations confirm the particle propertiesof light waves.

1.4 Compton effect

When x-rays are scattered by a solid medium, the scattered x-rays will normally havethe same frequency or energy. This is a case of elastic scattering or coherent scatter-ing. However, Compton observed that in addition to the scattered x-rays of same fre-quency, there existed some scattered x-rays of a slightly higher wavelength (i.e., lowerfrequency or lower energy). This phenomenon in which the wavelength of x-rays showan increase after scattering is calledCompton effect.

Compton explained the effect on the basis of the quantum theory of radiation. Con-sidering radiation to be made up of photons, he applied the laws of conservation ofenergy and momentum for the interaction of photon with electron. Consider an x-rayphoton of energyhν incident on an electron at rest (fig. 1.6.) After the interaction, thex-ray photon gets scattered at an angleθ with its energy changed to a valuehν′ andthe electron which was initially at rest recoils at an angleΦ. It can be shown that theincrease in wavelength is given by

λ = hmc

(1− cosθ) (1.11)

wherem is the rest mass of the electron.

E = hνp = hν/c

E′ = hν′

p′ = hν′/c

θ

Φ

Figure1.6Schematic diagram of the scattering of a photon by a stationary electron.

Whenθ = 90, λ = hmc= 0.0242A.

This constant value is called Compton wavelength. Whenθ = 180, λ = 2hmc


Experimental observation indicate that the change in the wavelength of the scat-tered x-rays is indeed in agreement with equation (1.11), thus providing further confir-mation to the photon model.

Thus, Planck’s theory of radiation, photoelectric effect and Compton effect areexperimental evidences in favour of the quantum theory of radiation.

1.5 Matter waves and de Broglie’s hypothesis

Quantum theory and the theory of relativity are the two important concepts that led tothe development of modern physics. The quantum theory was first proposed by Planckto explain and overcome the inadequacies of classical theories of black body radiation.The consequences were very spectacular. Louis de Broglie made the suggestion thatparticles of matter, like electrons, might possess wave properties and hence exhibitdual nature. His hypothesis was based on the following arguments:

The Planck’s theory of radiation suggests that energy is quantized and is given by

E = hv (1.12)

whereν is the frequency associated with the radiation. Einstein’smass-energy relationstates that

E = mc2 (1.13)

Combining the two equations, it can be written as

E = hv= mc2

Hence, the momentum associated with the photon is given by

p = mc= hv/c = h/λ

Extending this to particles, he suggested that any particlehaving a momentum p isassociated with a wave of wavelengthλ given by

λ = h/p (1.14)

This is calledde Broglie’s hypothesisof matter waves andλ is called the de Brogliewavelength.

The de Broglie wavelength can be calculated for any particleusing the above re-lation. In case of charged particles like electrons, a beam of high energy particles canbe obtained by accelerating them in an electric field. For example, an electron starting


from rest when accelerated with a potential differenceV, the kinetic energy acquiredby the electron is given by

(1/2)mv2 = eV

wherev is the velocity of the electron. The momentum may be calculated as

p = mv = (2meV)1/2

Using the de Broglie equation, the wavelength associated with the accelerated elec-tron can be calculated as

λ = h/p = h/(2meV)1/2 (1.15)

This equation suggests that, at a given speed, the de Brogliewavelength associatedwith the particle varies inversely as the mass of the particle. This concept of mat-ter waves aroused great interest and several physicists launched experiments designedto test the hypothesis. Heisenberg and Schrodinger proceeded on to develop mathe-matical theories whereas Davisson and Germer, G.P. Thomsonand Kikuchi attemptedexperimental verification.

1.5.1 Davisson-Germer experiment

The hypothesis of de Broglie was verified by the electron diffraction experiment con-ducted by Davisson and Germer in the United States. The experimental set up used bythem is shown in the figure 1.7.

Figure1.7Experimental arrangement for Davisson-Germer experiment.

The apparatus consists of a filament heated with a small a.c power supply to pro-duce thermionic emission of electrons. These electrons areattracted towards an anode


in the form of a cylinder with a small aperture maintained at afinite positive potentialwith respect to the filament. They pass through the narrow aperture forming a finebeam of accelerated electrons. This electron beam was made to incident on a singlecrystalline sample of nickel. The electrons scattered at different angles were countedusing an ionization counter as a detector. The experiment was repeated by record-ing the scattered electron intensities at various positions of the detector for differentaccelerating potentials (Fig.1.8).

44ev

φ

0 48ev 54ev

50

64ev 68ev

Figure 1.8 Scattered electron intensity maps at different accelerating potentials.Thevertical axis represents the direction of the incident electron beam andΦ is the scat-tering angle.The radial distance from the origin at any angle represents the intensityof scattered electrons.

When a beam of electrons accelerated with a potential of 54 V was directed per-pendicular to the nickel target, a sharp maximum occurred inthe electron density at anangle of 50 with the incident beam. When the angleΦ between the direction of theincident beam and the direction of the scattered beam is 500, the angle of incidencewill be 250 and the corresponding angle of diffractionθ will be 650. The spacing ofthe planes responsible for diffraction was found to be 0.091nm from x-ray diffractionexperiment. Assuming first order diffraction, the wavelength of the electron beam canbe calculated as

λ = 2dsinθ = 2× 0.091× sin 650 = 0.165nm.

The wavelength of the electrons can also be calculated usingthe de Broglie’s relation as

λ = h/ (2meV)1/2

= 6.63× 10−34/(2× 9.1× 10−31 × 1.6× 10−19× 54)1/2

= 0.166nm.


Thus, the Davisson-Germer experiment directly verifies thede Broglie’s hypothe-sis.

1.5.2 G.P. Thomson experiment

At almost the same time as the Davisson-Germer experiment, G.P.Thomson of Englandcarried out electron diffraction experiments independently using a thin polycrystallinefoil of aluminium metal. The experimental set up is shown in fig. 1.9.

Electron Gun

Aluminium Foil

Screen

θ

Figure1.9Experimental arrangement for G.P.Thomson experiment.

He allowed a beam of accelerated electrons to fall on the aluminium foil and ob-served a diffraction pattern consisting of a series of concentric rings around the di-rection of the incident beam. This pattern was similar to theDebye-Scherrer patternobtained for aluminium using x-ray diffraction. Using the data available on aluminium,he calculated the wavelength of the electrons using the Bragg’s equation,

nλ = 2dsinθ

He also calculated the de Broglie wavelength of the electrons with the knowledge ofaccelerating potential using the relation,

λ = h/(2meV)1/2

The value of wavelength calculated from the two equations matched well thereby ex-perimentally proving the de Broglie’s relation.


A similar experiment was conducted by Kikuchi in Japan in which he obtainedelectron diffraction pattern by passing an electron beam through a thin foil of mica toconfirm the validity of de Broglie’s relation.

The wave nature of particles is not restricted to electrons.Any particle with a mo-mentump has a de Broglie wavelength equal to (h/p). Neutrons produced in nuclearreactors possess energies corresponding to wavelength of the order of 0.1nm. Theseparticles also should be suitable for diffraction by crystals. Neutrons from a nuclearreactor are slowed down to thermal energy of the order ofkT and used for diffrac-tion and interference experiments. The results agree well with the de Broglie relation.Since neutrons are uncharged particles, they are particularly useful in certain situationsfor diffraction studies. Neutron beams have also been used as probesto investigate themagnetic properties of nuclei.

1.5.3 Wave packet and de Broglie waves

We have seen that moving particles may be represented by de Broglie waves. The am-plitude of these de Broglie waves does not represent any parameter directly describingthe particle but is related to the probability of finding the particle at a particular placeat a particular time. Hence, we cannot describe de Broglie waves with a simple waveequation of the type,

y = Acos(ωt − kx) (1.16)

Instead, we have to use an equation representing a group of waves. In other words, awave packetconsisting of waves of slightly differing wavelengths may represent themoving particle. Superposition of these waves constituting the wave packet results inthe net amplitude being modified, thereby defining the shape of the wave group. Thephase velocity of individual waves depends on the wavelength. Since the wave groupconsists of waves with different wavelengths, all the waves do not proceed togetherand the wave group has a velocity different from the phase velocities of the individualwaves. Hence, de Broglie waves may be associated with group velocity rather than thephase velocity.

1.5.4 Characteristics of matter waves

1. Matter waves are associated with a moving body.

2. The wavelength of matter waves is inversely proportionalto the velocity withwhich the body is moving. Hence, a body at rest has an infinite wavelength andthe one traveling with a high velocity has a lower wavelength.


3. Wavelength of matter waves also depends on the mass of the body and decreaseswith increase in mass. Due to this reason, the wavelike behaviour of heavierbodies is not very evident whereas wave nature of subatomic bodies could beobserved experimentally.

4. A wave is normally associated with some quantity that varies periodically withthe frequency of the wave. For example, in a water wave, it is the height of thewater surface; in a sound wave it is the pressure and in an electromagnetic wave,it is the electric and magnetic fields that vary periodically. But in matter waves,there is no physical quantity that varies periodically. We use a wave function todefine matter waves and this wave function is related to the probability of findingthe particle at any place at any instant, which varies periodically.

5. Matter waves are represented by a wave packet made up of a group of waves ofslightly differing wavelengths. Hence, we talk of group velocity of matter wavesrather than the phase velocity. The group velocity can be shown to be equal tothe particle velocity.

6. Matter waves show properties similar to other waves. For example, a beam ofaccelerated electrons produces interference and diffraction effects similar to anelectromagnetic wave of same wavelength.

1.6 Phase and group velocities

A wave is represented by the formula

y = Acos(ωt − kx) (1.16)

wherey is the displacement at any instantt, A is the amplitude of vibration,ω is theangular frequency equal to 2πν andk is the wave vector, equal to (2π/λ). The phasevelocity of such a wave is the velocity with which a particular phase point of the wavetravels. This corresponds to the phase being constant.

i.e., (ωt − kx) = constant

or x = constant+ ωt/k

Phase velocityvp = dx/dt = ω/k

= 2πν/(2π/λ) = λν (1.17)

vp is called the ‘wave velocity’ or ‘phase velocity’.


The de Broglie waves are represented by a wave packet and hence we have ‘groupvelocity’ associated with them.Group velocity is the velocity with which the wavepacket travels. In order to understand the concept of group velocity, consider the com-bination of two waves represented by the formula

y1 = Acos(ωt − kx)

y2 = Acos(ω + ω)t − (k+ k)x

The resultant displacement is given by

y = y1 + y2

= 2Acos(ω + ω + ω)t − (k+ k+ k)x

2cos

(ωt − kx)2

≈ 2Acos(ωt − kx). cos

(

ωt2− kx

2

)

(1.18)

The velocity of the resultant wave is given by the speed with which a referencepoint, say the maximum amplitude point, moves. Taking the amplitude of the resultantwave as constant, we have

2Acos

(

ωt2− kx

2

)

= constant

or

(

ωt2− kx

2

)

= constant

or x = constant+ (ωt/k)

Group velocityvg = dx/dt = (ω/k) (1.19)

Instead of two discrete values for and k, if the group of waveshas a continuousspread fromω to (ω + ω) andk to (k + k), then, the group velocity is given by

vg =dωdk

(1.20)

It can be shown that the group velocity of the wave packet is equal to the velocity ofthe particle with which the wave packet is associated.

1.6.1 Relation between phase velocity and group velocity

We have the mathematical relation for phase velocity given by

vp = ω/k or ω = k.vp


The group velocityvg is given by

vg =dωdk=

d(k · vp)

dk

= vp + k ·dvp

dk

= vp + (2π/λ) ·dvp

d(2π/λ)

= vp + (2π/λ) · (−λ2/2π).dvp

dλ

= vp − λ ·dvp

dλ(1.21)

In the above expression, if (dvp/dλ) = 0, i.e., if the phase velocity does not dependon wavelength, then the group velocity and phase velocity are equal. Such a mediumis called a non-dispersive medium. In a dispersive medium, (dvp/dλ) is positive andhence the group velocity is less than the phase velocity.

1.6.2 Relation between group velocity and particle velocity

(Velocity of de Broglie waves)The phase velocity of waves depend on the wavelength. This isresponsible for the

well known phenomenon of dispersion. In the case of light waves in vacuum, the phasevelocity is same for all wavelengths. In the case of de Broglie waves, we have,

ω = 2πν = 2πmc2/h =2πm0c2

h(1− v2/c2)1/2

(1.22)

and k = 2π/λ = 2πmv/h =2πm0v

h(1− v2/c2)1/2

(1.23)

The group velocity of de Broglie waves is given by

vg = dω/dk=dω/dv

dk/dv

dω/dv = (2πm0c2/h) · d

dv(1− v

2/c2)1/2 =2πm0v

h(1− v2/c2)3/2

(1.24)

dk/dv =2πm0

h(1− v2/c2)3/2

(1.25)


From equations (1.24) and (1.25) we get,

vg = v

Thus, the group velocity associated with de Broglie waves isjust equal to the velocitywith which the particle is moving. If we try to calculate the phase velocity,

vp = ω/k = c2/v = c2/vg (1.26)

Since the group velocity or the particle velocity is always less thanc, the phase velocityof de Broglie waves turn out to be greater thanc. This only indicates that we cannottalk of phase velocity of de Broglie waves since they are madeup of a group of waves.Phase velocity has no physical significance for de Broglie waves.

1.6.3 Derivation of de Broglie relation

The de Broglie relation may be derived as follows. If we assume a particle having akinetic energy equal tomv

2/2 to have a de Broglie wavelengthλ, we can writehν = mv

2/2 (assuming the energy of the particle to be purely kinetic)

or ν =m2h· v2 (1.27)

Differentiating with respect toλ,

dνdλ=

m2h· 2v ·

dv

dλ(1.28)

But we have

vg = v =dωdk=

2πdν2πd(1/λ)

= −λ2 dνdλ

∴dνdλ= − v

λ2(1.29)

Substituting in equation(1.28), we get

mv

h·

dv

dλ= −

v

dλ2

Rewriting this, we havedv

dλ= − h

mλ2(1.30)


Integrating with respect toλ,

v =h

mλ+ c

By applying the boundary condition that the wavelength tends to infinity as thevelocity tends to zero, we find that the constant of integration has to be zero. Hence,we get

λ =h

mv

(1.31)

which is thede Broglie relation.

1.7 Heisenberg’s Uncertainty principle

1.7.1 Origin and nature of the Principle

When we assign wave properties to particles there is a limitation to the accuracy withwhich we can measure the properties like position and momentum.

∆x

Figure1.10A wave packet with an extensionx along x-axis.

Consider a wave packet as shown in fig.1.10. The particle to which this wave packetcorresponds to may be located anywhere within the wave packet at any instant. Theprobability density suggests that it is most likely to be found in the middle of the wavepacket. However, there is a finite probability of finding the particle anywhere withinthe wave packet. If the wave packet is smaller in extension, the position of the particlecan be specified more precisely. But the wavelength of the waves will not be welldefined in a narrow wave packet. Since wavelength is related to momentum through


de Broglie’s relation, the momentum is not precisely known.On the otherhand, awave packet with large extension can have a more clearly defined wavelength andhence momentum at the cost of the knowledge about the position. This leads to theconclusion that it is impossible to know both the position and momentum of an objectprecisely at the same time. This is known asUncertainty principle.

For a wave packet of extensionx with an uncertainty in the wave numberkassuming the uncertainties to be the standard deviation in the respective quantities, itmay be shown that a minimum value of the product of such deviations is given by

x · k =12

(1.32)

This minimum value of the product of uncertainties is for thecase of a gaussian distri-bution of the wave functions. Since the wave packets in general do not have gaussianforms, the uncertainty relation becomes

x · k ≥ 12

(1.33)

But we have

k = 2π/λ (1.34)

Also λ = h/p (1.35)

Hence, k = 2πp/h

k =2πh· p (1.36)


x · p ≥ h4π

or x · p ≥ ~

2(1.37)

This equation states that the product of uncertaintyx in the position of an objectat some instant and the uncertainty in the momentum in the x-direction at the sameinstant is equal to or greater than~/2.

Another form of uncertainty principle relates energy and time. In the atomic pro-cess, if energyE is emitted as an electromagnetic wave during an interval of timet,then, the uncertaintyE in the measured value ofE depends on the duration of thetime intervalt according to the equation,

E · t ≥ ~/2 (1.38)


It may be mentioned that these uncertainties are not due to the limitations of theprecision of the measuring methods or measuring instruments but due to the nature ofthe quantities involved.

1.7.2 An illustration of uncertainty principle

We have the following ‘Thought experiment’ to illustrate the uncertainty principle.Imagine an electron being observed using a microscope (fig.1.11).

Microscope

LightSource

Electron

Figure1.11Schematic diagram of experimental set up to study uncertainty principle.

The process of observation involves a photon of wavelengthλ incident on the elec-tron and getting scattered into the microscope. The event may be considered as a two-body problem in which a photon interacts with an electron. The change in the velocityof the photon during the interaction may be anything betweenzero( for grazing angleof incidence) and 2c (for head-on collision and reflection). The average change in themomentum of the photon may be written as equal to (hν/c) or (h/λ). This difference inmomentum is carried by the recoiling electron which was initially at rest. The changeor uncertainty in the momentum of the electron may thus be written as (h/λ). At thesame time, the position of the electron can be determined to an accuracy limited by theresolving power of the microscope, which is of the order ofλ. Hence, the product ofthe uncertainties in position and momentum is of the order ofh. This argument impliesthat the uncertainties are associated with the measuring process. This illustration onlyestimates the accuracy of measurement, the uncertainty being inherent in the nature ofthe moving particles involved.


1.7.3 Physical significance of uncertainty principle

Uncertainty principle is a consequence of the wave particleduality. It states that it isimpossible to know both the position and momentum of an object exactly and at thesame time. Mathematically, it can be shown that the product of uncertainties in theposition and momentum measured simultaneously will have a value greater that~/2,i.e., (h/4π). If x is the uncertainty in the measurement of the positionx of an objectandpx is the uncertainty in the measurement of momentumpx , then, at any instant,

x · px > ~/2

We can try to estimate the product of the uncertainties with the help of illustrations asthe one mentioned above. The principle is based on the assumption that a moving par-ticle is associated with a wave packet, the extension of which in space accounts for theuncertainty in the position of the particle. The uncertainty in the momentum arises dueto the indeterminacy of the wavelength because of the finite size of the wave packet.Thus, the uncertainty principle is not due to the limited accuracy of measurement butdue to the inherent uncertainties in determining the quantities involved. But we canstill define the position where the probability of finding theparticle is maximum andalso the most probable momentum of the particle.

1.7.4 Applications of uncertainty principle

The uncertainty principle has far reaching implications. In fact, it has been very usefulin explaining many observations which cannot be explained otherwise. A few of theapplications of the uncertainty principle are worth mentioning.

(a) Diffraction of a beam of electronsDiffraction of a beam of electrons at a slit isthe effect of uncertainty principle. As the slit is made narrower, thereby reducingthe uncertainty in the position of the electrons in the beam,the beam spreads evenmore indicating a larger uncertainty in its velocity or momentum.

Figure 1.12 shows the diffraction of an electron beam by a narrow slit of widthx.The beam traveling alongOX is diffracted alongOY through an angleθ. Due tothe wave nature of the electron, we observe Fraunhoffer diffraction on the screenplaced alongXY. The accuracy with which the position of the electron is known isx since it is uncertain from which place in the slit the electron passes. Accordingto the theory of diffraction, we have

λ = x · sinθ or x = λ/ sinθ


X

Y

θ

O

Figure1.12Diffraction at a single slit.

Further, the initial momentum of the electron alongXY was zero and after diffrac-tion, the momentum of the electron isp · sinθ wherep is the momentum of theelectron along the incidence direction. Hence, the change in momentum of theelectron alongXY is p · sinθ or (h/λ) · sinθ. Assuming the change in the momen-tum as representative of the uncertainty in momentum, we get

x · px =λ

sinθ·

h · sinθλ

= h

(b) Nuclear beta decay: In beta decay, electrons are emitted from the nucleus ofthe radioactive element. Assuming the diameter of the nucleus to represent theuncertainty in the position of electron inside the nucleus,the uncertainty in themomentum can be calculated as follows:

Radius of the nucleus= r = 5× 10−15m

x = 2r = 10−14m.

p = h/2πx = 6.62× 10−34/(2× 3.14× 10−14)

= 1.055× 10−20kgms−1

Assuming that the electron was at rest before its emission, the change in momen-tum can be taken as equal to its momentum. This magnitude of change in mo-mentum indicates large velocity for the electron. Hence, the energy of the emittedelectron will be

E = pc= 1.055× 10−20× 3× 108 = 3.165× 10−12J

= 19.8MeV.


This indicates that the electrons inside the nucleus must have kinetic energy of19.8 MeV. But the electrons emitted during beta decay have kinetic energy of theorder of 1 MeV. This indicates that electrons do not exist in the nucleus of the atombut are ‘manufactured’ by the nucleus at the time of decay.

(c) Binding energy of an electron in an atom: In a hydrogen atom, the electronrevolves round the nucleus in an orbit of radius 5× 10−11m. Assuming this as themaximum uncertainty in position, we can calculate the minimum uncertainty inthe momentum as

(p)min = h/2π(x)max= 2.1× 10−24kgms−1.

Assuming this as the momentum of electron, the kinetic energy of the electron willbe equal to

K.E. = p2/2m= 2.45× 10−18J = 15.3eV.

Thus, the binding energy of an electron in hydrogen atom is nearly 15eV which isfound to be correct experimentally.

(d) Nitrogen doping of silicon: The laws of conservation of energy and momentumrestrict the generation and recombination processes in semiconductors. Silicon,which is an indirect band gap semiconductor, has low efficiency as a material forphoto diode or light emitting diode. Nitrogen doping of silicon will bind the freeelectrons to the lattice thereby restricting the value of uncertainty in position. Thisresults in a larger uncertainty in momentum thereby increasing the probability forgeneration or recombination process.

1.8 Wave mechanics

Quantum theory is based on the quantization of energies. It deals with the particlenature of radiation. It implies that addition or liberationof energy will be betweendiscrete energy levels. It assigns particle status to a packet of energy by calling it‘quantum of energy’ or ‘photon’ and treats the interaction of radiation with matter asa two-body problem. On the other hand, de Broglie’s hypothesis and the concept ofmatter waves led to the development of a different formulation called ‘Wave mechan-ics’. This deals with the wave properties of material particles. It was shown later thatthe quantum mechanics and the wave mechanics are mathematically identical and leadto the same conclusion.


1.8.1 Characteristics of wave function

Waves in general are associated with quantities that vary periodically. For example,water waves involve the periodic variation of the height of the water surface at a point.Similarly, sound waves are associated with periodic variations of the pressure. In thecase of matter waves, the quantity that varies periodicallyis called‘wave function’ .The wave function, represented byΨ, associated with matter waves has no direct phys-ical significance. It is not an observable quantity. But the value of the wave functionis related to the probability of finding the body at a given place at a given time. Thesquare of the absolute magnitude of the wave function of a body evaluated at a partic-ular time at a particular place is proportional to the probability of finding the body atthat place at that instant.

The wave functions are usually complex. The probability in such a case is takenasψ ∗ ψ, i.e. the product of the wave function with its complex conjugate. Since theprobability of finding the body somewhere is finite, we have the total probability overall space equal to certainty.

i.e.,∫

ψ ∗ ψ dV = 1 (1.39)

Equation (1.39) is called the normalization condition and awave function thatobeys the equation is said to benormalized. Further,Ψmust be single valued since theprobability can have only one value at a particular place andtime. Since the probabilitycan have any value between zero and one, the wave function must be continuous. Mo-mentum being related to the space derivatives of the wave function, the partial deriva-tives∂Ψ/∂x, ∂Ψ/∂y and∂Ψ/∂zmust also be continuous and single valued everywhere.Thus, the important characteristics of wave function are asfollows:

1. Ψmust be finite, continuous and single valued everywhere.

2. ∂Ψ/∂x, ∂Ψ/∂y and∂Ψ/∂z must be finite, continuous and single valued every-where.

3. Ψmust be normalizable.

1.8.2 Physical significance of wave function

We have already seen that the wave function has no direct physical significance. How-ever, it contains information about the system it represents and this can be extracted


by appropriate methods. Even though the wave function itself is not directly an ob-servable quantity, the square of the absolute value of the wave function is intimatelyrelated to the moving body and is known as the probability density. This probabilitydensity is the quantum mechanical method of finding the body at a particular positionat a particular time. The wave function carries informationabout the particle’s wave-like behaviour. It also provides information about the momentum and energy of theparticle at any instant of time.

1.8.3 Schrodinger’s wave equation

The motion of a free particle can be described by the wave equation.

Ψ = Aexp−i(ωt − kx) (1.40)

Butω = 2πν = 2π(E/h) = (E/~) andk = 2π/λ = 2π(p/h) = (p/~)whereE is the total energy andp is the momentum of the particle. Substituting in theequation (1.40), we get,

Ψ = A exp−i

~(Et− px)

(1.41)

Differentiating equation (1.41) with respect tox twice, we get,

∂2Ψ

∂x2=−p2

~2Ψ or p2Ψ = −~

2 · ∂2Ψ

∂x2(1.42)

Differentiating equation (1.41) with respect tot, we get,

∂Ψ

∂t= −

iE~· Ψ or EΨ = −

~

i·∂Ψ

∂t(1.43)

The total energy of the particle can be written as

E =p2

2m+ U (1.44)

whereU is the potential energy of the particle. Multiplying both sides of the equationbyΨ

EΨ =p2Ψ

2m+ U Ψ (1.45)

Substituting forEΨ andp2Ψ from equation (1.42) and (1.43)

−~

i∂Ψ

∂t= − ~

2

2m∂2ψ

∂x2+ UΨ (1.46)


This is known asSchrodinger’s time dependent equationin one dimension.The wave functionΨ in equation (1.41) may also be written as

Ψ = A exp i~

(Et− px)

= A exp(−iEt)

~· exp

(ipx)~

Ψ = Φ exp(−iEt)

~(1.47)

whereΦ is a position dependent function. Substituting this form ofΨ in equation(1.45),

EΦ exp(−iEt)

~=

p2

2mΦ exp

(−iEt)~+ UΦexp

(−iEt)~

or EΦexp(−iEt)

~= −

~2

2m·∂2Φ

∂x2· exp

(−iEt)~+ UΦ exp

(−iEt)~

or∂2Φ

∂x2exp

(−iEt)~+

2m~2

(E − U)Φexp(−iEt)

~= 0

or∂2Ψ

∂x2+

2m~2

(E − U)Ψ = 0 (1.48)

This is the Schrodinger’s wave equation in one dimension. Inthree dimensions, theabove equation may be written as

∂2Ψ

∂x2+∂2Ψ

∂y2+∂2Ψ

∂z2+

2m(E − U)Ψ~2

= 0

or ∇2Ψ +2m(E − U)Ψ

~2= 0

This equation is known asthe steady state or time independent Schrodinger waveequation in three dimensions.

1.8.4 Eigen values and eigen functions

These terms come from the German words and mean proper or characteristic values orfunctions respectively. The values of energy for which the Schrodinger’s equation canbe solved are called‘Eigen values’ and the corresponding wave functions are called‘Eigen functions’. The eigen functions possess all the characteristics properties ofwave functions in general (see section 1.8.1).


1.9 Applications Of Schrodinger’s Equation

1.9.1 Case of a free particle

A free particle is defined as one which is not acted upon by any external force thatmodifies its motion. Hence, the potential energyU in the Schrodinger’s equation isa constant and does not depend on position or time. For convenience, the potentialenergy may be assumed to be zero. Then, the Schrodinger’s equation for the particlebecomes

∂2Ψ

∂x2+

2m~2

EΨ = 0 (1.49)

whereE is the total energy of the particle which is purely kinetic. This is of the form,

∂2Ψ

∂x2+ k2Ψ = 0

wherek2 = 2mE/~2. The solution of this equation may be written as

Ψ = Asinkx+ Bcoskx

Solving for the constantsA andB pose some difficulties because we cannot apply anyboundary conditions on the wave function as it represents a single wave which is notlocalized and not normalizable. Since the solution has not imposed any restriction onthe value ofk, the free particle is permitted to have any value of energy given by theequation,

E = ~2k2/2m

Since the total energy is purely kinetic, the momentum of theparticle would bep = ~kor h/λ. This is just what we would expect, since we have constructedthe Schrodingerequation to yield the solution for the free particle corresponding to a de Broglie wave.

1.9.2 Particle in a box

The simplest problem for which Schrodinger’s time independent equation can be ap-plied and solved is the case of a particle trapped in a box withimpenetrable walls.

Consider a particle of massm and energyE travelling along x-axis inside a box ofwidth L. The particle is thus restricted to move inside the box by reflections atx = 0andx = L (Fig. 1.13).

The particle does not lose any energy when it collides with the walls and hence thetotal energy of the particle remains constant. The potential energy of the particle is


considered to be zero inside the box and infinite outside. Since the total energy of theparticle cannot be infinite, it is restricted to move within the box. The example is anoversimplified case of an electron acted upon by the electrostatic potential of the ioncores in a crystal lattice.

X = O X = L

U

Figure1.13Schematic for a particle in a box. The height of the wall extends to infinity.

Since the particle cannot exist outside the box,

Ψ = 0 for x ≤ 0 andx ≥ L (1.50)

We have to evaluate the wave function inside the box. The Schrodinger’s equation(1.48) becomes

∂2Ψ

∂x2+

2m~2

EΨ = 0 for 0≤ x ≤ L (1.51)

Ψ = Asin

(

2mE~2

)1/2

x+ Bcos

(

2mE~2

)1/2

x (1.52)

whereA andB are constants.Applying the boundary condition thatΨ = 0 atx = 0, equation (1.52) becomes

Asin 0+ Bcos 0= 0 or B = 0.

Again, we haveΨ = 0 atx = L. Then,

A. sin

(

2mE~2

)1/2

· L = 0


If A = 0, the wavefunction will become zero irrespective of the value of x. Hence, Acannot be zero.

Therefore, sin

(

2mE~2

)1/2

· L = 0

or

(

2mE~2

)1/2

L = nπ where n = 1, 2, 3, . . . (1.53)

From (1.53), the energy eigen values may be written as

En =n2π2

~2

2mL2where n = 1, 2, 3, . . . (1.54)

From this equation, we infer that the energy of the particle is discrete asn can haveinteger values. In other words, the energy is quantized. We also note thatn cannot bezero because in that case, the wave function as well as the probability of finding theparticle becomes zero for all values ofx. Hence,n = 0 is forbidden. The lowest energythe particle can possess is corresponding ton = 1 and is equal to

E1 =π2

~2

2mL2

This is called ‘ground state energy’ or ‘zero point energy’.The higher excited stateswill have energies like 4E1, 9E1, 16E1, etc. This indicates that the energy levels arenot equally spaced.

The wave functions or the eigen functions are given by

Ψn = A · sin

(

2mEn

~2

)1/2

x

or Ψn = A · sinnπL

x (1.55)

Applying the normalization condition,

i.e.∫

A2 sin2 nπxL· dx= 1 (1.56)

Since the wave function is non-vanishing only for 0≤ x ≤ L, it can be shown that∫

sin2 nπxL

dx=(L2

)

(1.57)


Substituting in equation (1.56), we have

A2(L2

)

= 1 or A =

(

2L

)1/2

(1.58)

The eigen function or wave functions in equation (1.55) becomes

Ψn =

(

2L

)1/2

sin

(

2mEn

~2

)1/2

x

Ψn =

(

2L

)1/2

sinnπxL

(1.59)

x = 0 x = L

n = 1

ψ

n = 2

n = 3

Figure 1.14Variation of wave function associated with an electron confined to a boxin its ground state and excited states.

Figure 1.14 shows the variation of the wave function inside the box for differentvalues of n and Fig.1.15 shows the probability densities of finding the particle at dif-ferent places inside the box for different values ofn. Thus, wave mechanics suggeststhat the probability of finding any particle at the lowest energy level is maximum atthe centre of the box which is in agreement with the classicalpicture. However, theprobability of finding the particle in higher energy states is predicted differently by thetwo formulations.


n = 1

n = 2

n = 3|Ψ|2

x = 0 x = L

Figure1.15Probability density as a function of position.

1.9.3 Finite Potential well

In real life situations, the potential energy is never infinite. The box with impenetrablewalls has no physical significance. However, we come across situations where thepotential energy is finite. Let us try to solve the case of an electron in a finite potentialwell. We can consider two different cases corresponding to the following situations:(i) the total energyE being greater than the potential energyU, and(ii) the total energyE being less than the potential energyU.

The first case may be represented by the figure 1.16. Consider the particle withtotal energyE inside a potential well of heightU. In the region II, where the particleis not influenced by the potential (U = 0), the solution of the Schrodinger’s equationis of the form,

Ψ = Asinkx+ Bcoskx

wherek = (2mE/~2)1/2. This particle may be represented by a wave of wavelengthλ = 2π/k. When the particle is in region I and III, its wavelength changes toλ′ = 2π/k′

wherek′ = [2m(E − U)/~2]1/2. In other words, the effect of the potential energy stepis to reduce the kinetic energy of the particle as evident from an increase in the valueof the wavelength.

In the second case, the total energy of the particle is less than the potential energy.Under this condition, classically, the particle cannot propagate beyond the step sincethis amounts to the kinetic energy being negative. But, wavemechanically, a different


E

U

λ λ

x = 0

Figure1.16Schematic for a particle in a potential well of finite depth (Egreater than U).

solution results. LetU be greater than the total energyE of the electron but finite. Toanalyze this case, we have to consider the three regions separately.

In region II, sinceU = 0, the electron is free and the Schrodinger’s equation is

d2Ψ

dx2+

2m~2

EΨ = 0 (1.60)

In regions I and III, we have

d2Ψ

dx2+

2m~2

(E − U)Ψ = 0 (1.61)

The solutions for these equations can be assumed to be

ΨI = Aeiβx + Be−iβx in region I (1.62)

ΨII = Ceiαx + De−iαx in region II (1.63)

and ΨIII = Feiβx +Ge−iβx in region III (1.64)

where α = [(2mE)/~2]1/2 (1.65)

β = [2m(E − U)/~2]1/2 (1.66)

sinceE is less thanU, (E − U) is (−)ve andβ is imaginary.Let us define a new constant

γ = −iβ (1.67)


Then the equations (1.62) and (1.64) become

ΨI = Ae−γx + Beγx (1.68)

ΨIII = Fe−γx +Geγx (1.69)

To evaluate the constants, we consider the boundary condition in the region I wherethe wave function should reduce to zero asx→ −∞. Then equation (1.68) becomes

0 = A · ∞ + B · 0 orA = 0.

∴ ΨI = Beγx (1.70)

Similarly, in region III, since the wave function should reduce to zero asx → ∞ ,equation (1.69) becomes

0 = F · 0+G · ∞ orG = 0.

∴ ΨIII = Fe−γx (1.71)

This indicates that the wave function decreases exponentially as we move away fromthe potential well on either sides. Inside the potential well the wave function repre-sented by the equation (1.63) varies sinusoidally. Further, since the wave function andits derivative are continuous at the boundaries corresponding to x = 0 andx = L, thewave functions are non-zero at these boundaries. The plots of the wave functions andthe probability densities are shown in Fig. 1.17 and 1.18 respectively.

ψ

x = 0 x = L

n = 3

n = 2

n = 1

Figure1.17Wave function as a function of position.


n = 1

n = 2

n = 3|Ψ|2

x = 0 x = L

Figure1.18Probability density as a function of position.

Thus, we observe that in case of a particle in a potential wellof finite height, theparticle has a finite probability of penetrating into the wall. However, if the wallsof the well are infinitely thick, the particle will be confinedto the well and performsoscillatory motion inside the well.

1.9.4 Tunnel effect

In the previous case of a finite potential well, even though the height of the wall wasfinite, the thickness of the wall was assumed to be infinite. Asa result, the particle wastrapped in the well in spite of penetrating into the wall. Under the same condition of thetotal energy being less than the potential energy, if the thickness of the wall is reducedand made finite, the solution of the Schrodinger’s equation predicts a finite probabilityof the particle passing through the barrier and finding itself on the other side. Thus, aparticle without the necessary energy to pass over the barrier can still penetrate throughthe barrier. This phenomenon is called “Quantum mechanicaltunneling”.

Consider a particle with energy E incident on a potential barrier of heightU andwidth L as shown in Fig. 1.19. The potential energy is zero in the regions I and III, butis finite and equal toU in region II. The Schrodinger’s equation for the three regionswill be

d2Ψ

dx2+

2m~2

EΨ = 0 in region I (1.72)


x = 0 x = L

λλ

IIIIII

U

Figure1.19Electron tunneling across a finite potential barrier.

d2Ψ

dx2+

2m~2

(E − U)Ψ = 0 in region II (1.73)

d2Ψ

dx2+

2m~2

EΨ = 0 in region III (1.74)

The solutions of these equations can be written as

ΨI = Aeiαx + Be−iαx in region I (1.75)

ΨII = Ce−γx + Deγx in region II (1.76)

ΨIII = Feiαx +Ge−iαx in region III (1.77)

where α = [(2mE)/~2]1/2 (1.65)

β = [2m(E − U)/~2]1/2 (1.66)

and γ = −iβ (1.67)

The wavefunction in the region I is made up of two terms as evident from equation(1.75). The first term with a positive exponent represents anincoming or incident wavemoving in the positive x-direction and the second term represents a wave reflected bythe barrier moving in the negative x-direction. Similarly,the first term in equation(1.77) represents the transmitted wave moving in region IIIin the positive x-direction.The wavefunction in the region II is given by equation (1.76). Here, the exponents arereal quantities and hence the wavefunction does not oscillate. The probability density|ΨII |2 is finite and represent the probability of finding the particle within the barrier.Such a particle may emerge into region III. This is called tunneling.


The transmission probabilityT for a particle to pass through the barrier is given by

T =|ΨIII |2

|ΨI |2=

FF∗

AA∗ e−2γL (1.78)

The above equation represents the dependence of tunneling probability on the width ofthe barrier and the energy of the particle.

1.9.5 Examples of tunneling across a finite barrier

There are a few examples of tunneling across a thin finite potential barrier in nature.These observations are in fact proof in favour of the theory of quantum mechanicaltunneling. Let us consider a few of them.

(a) Alpha decay: Alpha particles are made up of two protons and two neutrons. Inradioactive decay, the alpha particle must free itself fromthe attractive nuclearforce and penetrate through a barrier of repulsive coulombic potential to be emittedout of the nucleus(Fig.1.20). A calculation of the energy ofthe particle inside thenucleus and the measurement of the energy of the emitted alpha particle indicatethat it is not possible that the particle has surmounted the barrier of coulombicpotential but must have penetrated through it.

Nuclear surface

Energy of α − particle

Repulsive coulomb potential

Attractive nuclear potential

d

E

Figure1.20Emission of an particle in nuclear decay.


N

HH

H

Figure 1.21Schematic diagram of ammonia molecule. Nitrogen atom oscillates be-tween two symmetric positions across the repulsive plane ofhydrogen atoms.

(b) Ammonia inversion: In a molecule of ammonia, the three hydrogen atoms forma plane with the nitrogen atom placed symmetrically at a finite distance from theplane. It has been observed experimentally that the nitrogen atom oscillates be-tween two positions on either sides of the plane (Fig. 1.21).

Classical calculations show that the nitrogen atom cannot perform such oscillationsince the hydrogen atoms form a barrier against nitrogen atom to prevent it frommoving through the plane formed by the hydrogen atoms. However, nitrogen atomoscillates across the plane with a frequency higher than 1010 per second. This canbe explained only on the basis of tunneling process.

(c) Zener and tunnel diodes:These are diodes made out of heavily doped semicon-ductors with special characteristics. The current–voltage characteristics of thesediodes can be explained only on the basis of quantum mechanical tunneling pro-cess. The high speed of operation of these devices can be explained only as due totunneling since the movement of charge carriers is otherwise by diffusion which isa very slow process. The scanning tunneling microscope is another device operat-ing on the principle of tunneling.

(d) Frustrated total internal reflection: Figure 1.22 shows a beam of light reflectedtotally from the surface of glass. If a second prism of glass is brought close to thefirst, the beam appears through the second glass prism indicating tunneling of lightthrough the surfaces of glass which were otherwise acting asbarriers.


Figure1.22Demonstration of frustrated total internal reflection.

1.9.6 Theoretical interpretation of tunneling

Penetration of a particle into the forbidden region of a stepor a barrier can be explainedwith the help of Heisenbergs uncertainty principle. To enter this region, the particlemust gain an energy of atleast (U − E) and to move in this region, it must have anadditional kinetic energy,K. This is a violation of the principle of conservation ofenergy for the particle. However, according to the uncertainty relation, we may write

∆E · ∆t ≈ ~ (1.79)

According to this, the conservation of energy does not applyfor a time duration∆t ifthe change in energy is not greater than∆E. If we presume that the particle borrows anenergy∆E and returns the borrowed energy within a time interval of∆t, the observerwill still believe that the energy is conserved. The time interval within which the extraenergy must be returned is given by

∆t = ~/∆E = ~/(U − E + K) (1.80)

The particle moves with a velocityv given by

v = (2K/m)1/2 (1.81)

If the particle travels a distance∆x into the forbidden region and returns, then, the totaldistance travelled is 2∆x and hence we can write

∆x = v · ∆t/2

=12

(

2Km

)1/2

·~

(U − E + K)(1.82)

As the kinetic energyK tends to zero, the value of∆x also tends to zero since the ve-locity tends to zero. Also, asK tends to infinity,∆x tends to zero since it is the distance


travelled in a time interval∆t tending to zero. In between these limits, there must bea maximum value of∆x corresponding to a particular value ofK. Differentiating∆xwith respect toK in equation 1.82, we can find the maximum value of∆x as

∆xmax =12

[

~2

2m(U − E)

]1/2

(1.83)

Or ∆xmax = (1/2γ) (1.84)

From equation 1.78, the probability of finding the particle at a distance∆xmax from thestep is

T = e−2γ∆xmax = e−1 (1.85)

Hence, we may define the maximum penetration distance as the distance at which thetransmission probability is (1/e).

It may be mentioned that the particle is never observed in theforbidden region.The particles incident on the potential energy step will be reflected back. Some arereflected at the step itself where as others penetrate a finitedistance before returning.If the barrier width is small, the particle will re-emerge onthe other side of the barrier.This phenomenon is known as quantum mechanical tunneling.

1.9.7 Harmonic oscillator

When a body vibrates about an equilibrium position, the bodyis said to be executingharmonic motion. We have many examples of such motion which we come across, likethe vibration in a spring that is stretched and released, vibrations of atoms in a crystallattice, etc. Whenever a system is disturbed from its equilibrium position, it can comeback to its original position only under the influence of a restoring force. Hence, thepresence of a restoring force is a necessary condition for harmonic motion. The systemoscillates indefinitely if there is no loss of energy.

A special case of harmonic motion is simple harmonic motion.In simple harmonicmotion, the restoring force F acting on a particle of mass m islinear. In other words, therestoring force is proportional to the displacement x of theparticle from its equilibriumposition and is in the opposite direction.

i.e., F = −kx (1.86)

wherek is called theforce constant. The relation is called Hooke’s law. From thesecond law of motion, we have,

F = ma (1.87)


∴ −kx= md2xdt2

ord2xdt2+

k · xm= 0 (1.88)

This is the equation for the simple harmonic oscillator. Thesolution of this equationmay be written as

x = Acos(ωt + φ) (1.89)

where ω = 2πν = (k/m)1/2 (1.90)

ν is called the frequency of the oscillator.φ is the phase angle and depends on the valueof x at t = 0. The potential energyU corresponding to the restoring force F may becalculated as equal to the work done in bringing a particle from x = 0 to x = x againstthe force.

i.e., U(x) = −F(x)dx= k× dx= kx2/2 (1.91)

A plot of the potential energyU as a function of displacementx is a parabola as shownin fig. 1.23. This indicates that an oscillator with energyE vibrates back and forthwith an amplitude from−a to +a. Classically, it appears that the oscillator can haveany value of energy forming a continuous spectrum. Let us examine the quantummechanical modification to this classical picture.

a−a−x xO

E

n = 0

n = 1

n = 2

Figure1.23Energy states in a one dimensional harmonic oscillator.

The Schrodinger’s equation for the harmonic oscillator with a potential energyUequal tokx2/2 may be written as

δ2Ψ

δx2+

2m(E − kx2/2)~2

Ψ = 0 (1.92)


This equation may be rewritten in terms of dimensionless quantitiesa andy as

δ2Ψ

δy2+ (a− y2)Ψ = 0 (1.93)

wherea = 2E/hν andy2 = kmx2/~2.The solution to the equation (1.93) has to satisfy the boundary condition,

Ψ = 0 as y→ ∞

and the normalization condition,

∞∫

−∞

|Ψ|2dy= 1

These conditions will be satisfied when

a = (2n+ 1) wheren = 0, 1, 2, 3, . . .

i.e., a = 2E/hν = (2n+ 1)

or E = (n+ 1/2)hν wheren = 0, 1, 2, 3, · · · (1.94)

This solution leads to the following conclusions:

(i) The allowed energies will form a discrete spectrum and not a continuous spec-trum.

(ii) The least allowed energy is not zero but a finite minimum value.

At n = 0, E0 = hν/2This minimum energyE0 is called thezero point energy. It is also observed that

the higher energy levels are all equally spaced with a spacing of hν. This is in contrastto the result obtained for the case of a particle in a potential well of infinite depth.

1.9.8 Practical applications of Schrodinger’s wave equation

The real life situations are much different from the one considered while deriving theSchrodinger’s wave equation. This is especially true when one is analyzing the motionof a particle like electron traveling at velocities comparable to that of light. Relativisticmodification to the Schrodinger’s equation and its solutionare complex. Further, the


boundary condition of an infinitely high potential barrier is never encountered. In caseof metals, conduction electrons move in crystal lattice under the influence of finitepotentials of the ion cores. The potential energy due to the influence of external forcesacting on it may also be functions of position of the particleand time. Incorporationof these factors while formulating and solving the Schrodinger’s wave equation hasled to accurate prediction of the behaviour of subatomic, atomic, molecular and othermicroscopic systems.

Numerical Examples

1.1 Calculate the velocity of photoelectrons emitted from a metal surface of workfunction 1.5eV when the metal surface is irradiated with a light beam of wave-length 4× 10−7m.

Solution:

Incident energy= hν =hcλ=

6.62× 10−34 × 3× 108

4× 10−7

= 4.97× 10−19J

Threshold energy= hνo = Φ = 1.5eV= 1.5× 1.6× 10−19J

= 2.4× 1019J

Kinetic energy of electrons= (hν − hν0)

= (4.97− 2.40)× 10−19

= 2.57× 10−19J.

Velocity of Photoelectrons= [2(hν − hν0)/m]1/2

= 7.52× 105ms−1 (Ans.).

1.2 In a photoelectric effect experiment, a stopping potential of 4.6V was required tostop photoelectron emission with an incident light of frequency 2×1015Hzand astopping potential of 12.9V when the incident light had a frequency 4× 1015Hz.Evaluate the Planck’s constant.

Solution: If V1 andV2 represent the stopping potentials corresponding to in-cident frequenciesν1 andν2, then

eV1 = hν1 −ΦeV2 = hν2 −Φ


h =e(V2 − V1)(ν2 − ν1)

=1.6× 10−19(12.9− 4.6)(4× 1015− 2× 1015)

h = 6.64× 10−34Js.(Ans.)

1.3 Calculate the maximum change in wavelength that can take place during Comp-ton scattering of a photon.

Solution: Change in wavelength= λ = hmoc

(1− cosθ)

This will be maximum when cosθ = −1, i.e., whenθ = 180.

∴ (λ)max=2hmoc=

2× 6.62× 10−34

9.1× 10−31 × 3× 108

(λ)max= 4.85× 10−12m(Ans.)

1.4 The material of the emitter of a photocell has a work functionof 2eV. Calculatethe threshold frequency.

Solution:

Work functionΦ = hν0 = 2× 1.6× 10−19J

Threshold frequencyν0 = 2× 1.6× 10−19/6.62× 10−34

= 4.83× 1014Hz.

1.5 The threshold frequency for the material of the emitter of a Photocell is 4×1014Hz. What is the stopping potential required to supress photo electrons emis-sion when light of frequency 6× 1014Hz is incident on the emitter?

Solution:

Stopping potentialV0 = h(ν − ν0)/e

= 6.62× 10−34(

6× 1014− 4× 1014)

/1.6× 10−19

= 0.829V (Ans).

1.6 In a photocell, a stopping potential of 2.5V is required to stop the photo electronemission completely. Calculate the kinetic energy of the emitted photo electrons.

Solution:

Kinetic energy= potential energy= e.V

= 1.6× 10−19 × 2.5J = 2.5eV or 4× 10−19J (Ans).


1.7 In a photocell illuminated by light of frequency 5x1014Hz,a reverse potential of2V is required to stop the photo electron emission. Find the work function of thematerial of the emitter.

Solution:Work functionΦ = h(ν − ν0) = (hν − eV)

= 6.62× 10−34 × 5× 1014− 1.6× 10−19× 2

= 0.072eV (Ans.) .

1.8 X-rays of wavelength 1.54A are Compton – scattered at an angle of 60. Calcu-late the change in the wavelength.

Solution:Change in wavelength= λ =

hmc

(1− cosθ)

=h

mc(1− cos 60) = 1.2× 10−12m (Ans.).

1.9 In a Compton scattering experiment, incident photons of energy 10 KeV are scat-tered at 45 to the incident beam. Calculate the energy of the scattered photon.

Solution:Change in wavelength= λ = h

mc(1− cosθ)

= 7.1× 10−13m.

Wavelength of incident photon= λ = hc/eE

= 1.243× 10−10m.

Wavelength of scattered photon= λ′ = λ + λ

= 1.25× 10−10m.

Energy of scattered photon= hc/λ′

= 1.59× 10−15J

= 9.93keV(Ans.).

1.10 Gamma Rays of energy 0.5 MeV are scattered by electrons. Whatis the energyof scattered gamma rays at a scattering angle of 30? What is the kinetic energyof scattered gamma ray?

Solution:

Wavelength of incident gamma rays= λ = hc/E

= 6.62× 10−34 × 3× 108/(1.6× 10−19 × 0.5× 106)


= 2.486× 10−12m.

Change in wavelength= λ = hmc

(1− cosθ)

= 3.24× 10−13m.

Wavelength of scattered photon= λ′ = λ + λ2.81× 10−12m.

Kinetic energy of the scattered gamma ray= hc/λ′

= 0.442MeV (Ans.).

1.11 X-rays of wavelength 1.5A are Compton scattered. At what angle will be scat-tered x-rays have a wavelength of 1.506A?

Solution:

Change in wavelength= λ = hmc

(1− cosθ)

cosθ = (1−mc.λ/h) = (1− 0.247)= 0.753

Angle of scattering,θ = 41.20 (Ans.).

1.12 Calculate the de Broglie wavelength associated with an electron travelling with avelocity of 105ms−1. Assume the mass of the electron to be 9.1 × 10−31kg. andh = 6.62× 10−34Js.

Solution:

De Broglie wavelengthλ =hP=

6.62× 10−34

9.1× 1031× 105

λ = 7.27× 10−9m. (Ans.).

1.13 In an electron diffraction apparatus, the electron beam is accelerated to a poten-tial of 25 kV. Calculate the de Broglie wavelength associated with the electrons.Given:m= 9.1× 10−31kg, h = 6.62× 10−34 Js.e= 1.6× 10−19C.

Solution:De Broglie wavelengthλ =

h(2mE)1/2

=h

(2meE)1/2

λ =6.62× 10−34

(2× 9.1× 10−31 × 1.6× 10−19 × 25× 103)λ = 7.6× 10−12m. (Ans.).


1.14 Calculate the phase velocity and group velocity associatedwith an electron as-suming (i) non – relativistic case and (ii) relativistic case.

Solution:Non – relativistic case.

Phase velocityvp =ω

K=

(E/~)p/~

=Ep=

(p2/2m)p

=p

2m=

v

2

Phase velocity is half the particle velocity.

Group velocityvg =dωdk=

dEdp=

d(p2/2m)dp

=pm= v

Group velocity is equal to the particle velocity.

Relativistic case:

Phase velocityvp =Ep

whereE =(

p2c2 +m2oc

4)1/2

∴ vp = c[

1+m2oc

2/p2]1/2

Phase velocity is greater thanc. This indicates that we cannot talk of phasevelocity of a particle since it is represented by a group of waves or a wavepacket.

Group velocityvg =dEdp=

ddp

[

p2c2 +moc4]1/2

= pc2[

p2c2 +m2oc

4]−1/2

=pc2

E

=mvc2

mc2= v assumingE = mc2

Group velocity is equal to the particle velocity.1.15 An electron is trapped in a one dimensional potential well ofinfinite depth and a

width of 1×10−10 m. What is the probability of finding the electron in the regionfrom x = 0.09× 10−10m to x = 0.11× 10−10m in the ground state.

Solution:Method I: The probability of finding an electron in the region betweenx1 andx2 is given by


Pn =

x2∫

x1

Ψ∗Ψdx= (2/L)

x2∫

x1

sin2(nπx/L)dx

= [x/L − (1/2nπ). sin(2nπx/L)x2

]x1

For ground state,n = 1.

P1 = [x/L − (1/2π). sin(2πx/L)x2

]x1

= [0.11− (1/2× 3.14) sin(2× 180× 0.11/1.0)]

− [0.09− (1/2× 3.14) sin(2× 180× 0.09/1.0)]

= (0.11− 0.10150)− (0.09− 0.08532)

= 0.00382 (Ans.).

Method II: The probability of finding an electron in the region of widthxaroundx is given by

Pn = |Ψn|2 · x = (2/L) sin2(nπx/L) · x

Here, we taken = 1, x = 1× 10−10m and x = 0.02× 10−10m.

P1 = (2/1× 10−10) sin2(1× 180× 0.1) · 0.02× 10−10

= 0.00382(Ans).

Note: The second method is only approximate and gives result close to the one ob-tained by method I only whenx is very small.1.16 An electron is trapped in a one dimensional potential well ofwidth 1× 10−10m

and infinite height. Find the amount of energy required to excite the electron toits first excited state. What is the probability of finding theelectron in its firstexcited state betweenx = 0.4× 10−10mandx = 0.6× 10−10m?

Solution: Energy of the electron in thenth excited state is given by

En =n2h2

8mL2

Energy required to take the electron from ground state (n = 1) to the firstexcited state (n = 2) is given by


E = E2 − E1 =h2

8mL2(22 − 12) =

3h2

8mL2

=3× (6.62× 10−34)

8× 9.1× 10−31× (10−10)2

= 1.81× 10−17J = 112.87eV. (Ans.)

The probability of finding the electron in the first excited state betweenx =0.4A andx = 0.6A is given by

x2∫

x1

Ψ∗Ψdx= (2/L)

x2∫

x1

sin2(2πx/L)dx

= [x/L − (1/4π). sin(4πx/L)x2

]x1

= (0.6− 0.076)− (0.4+ 0.076)

= 0.048 (Ans)

1.17 The velocity of an electron is measured to be 3x106ms−1 in a particular direction.If the velocity is measured with a precision of 1%, what is theaccuracy withwhich its position can be measured simultaneously?

Solution: Momentum of the electron (in non - relativistic calculation) is

Px = mvx = 9.1× 10−31 × 3× 106 = 2.73× 10−24kg.m.s−1

Uncertainty in momentum= px = 1% o f px

= 2.73× 10−26kg.ms−1

∴ Uncertainty in position,x ≈ ~

px

= 3.86× 10−9m (Ans.).

1.18 In an experimental study of nuclear decay, the emitted energy spectrum showsa peak with a spread of energy equal to 120eV. Compute the life time of thedecaying nuclei.

Solution: Life time of the decaying nuclei= t ≈ ~

E


t =6.62× 10−34

120× 1.6× 10−19 × 2× 3.14

= 5.5× 10−18s. (Ans.)

1.19 Calculate the de Broglie wavelength associated with the following:

(a) A car of mass 1000 kg moving with a velocity of 50ms−1.

(b) A tennis ball of mass 50 gms moving at a speed of 75ms−1.

(c) An air molecule of average mass 2× 10−26kg moving with a velociity of300ms−1.

(d) An electron with an energy of 1 MeV.

Solution: The deBroglie wavelength can be calculated asλ = h/p.

(a) λ = 6.62× 10−34/1000× 50= 1.324× 10−38m. This value is too low to beobservable.

(b) λ = 6.62× 10−34/50× 10−3 × 75= 1.765× 10−34m.

(c) λ = 6.62× 10−34/2× 10−26 × 300= 1.103× 10−10m.

(d) λ = h/(2meV)1/2

= 6.62× 10−34/(

2× 9.1× 10−31× 1.6× 10−19× 106)1/2

= 1.23× 10−12m (Ans.).

1.20 Compute the accelerating potential required to produce an electron beam of deBroglie wavelength 0.1A.

Solution:

V = h2/2meλ2

= (6.62× 10−34)2/2× 1.6× 10−19× (0.1× 10−10)2

= 15.1kV (Ans).

1.21 Calculate the deBroglie wavelength associated with a thermal neutron at 270C.


Solution: Kinetic energy of thermal neutrons= (3/2)kT

∴ λ = h/(3mkT)1/2

= 6.62× 10−34/(

3× 1.67× 10−27× 1.38× 10−23 × 300)1/2

= 1.45× 10−10m (Ans).

1.22 A cricket ball of mass 250 gms moves with a velocity of 100ms−1. If its velocityis measured with an accuracy of 1%, what is the accuracy of a simultaneousmeasurement of its position?

Solution:

x · px > ~/2 or x · px ~

px= 1% of 250× 10−3 × 100= 250× 10−3kgms−1

x = ~/px = 6.62× 10−34/250× 10−3 = 2.65× 10−33m (Ans).

1.23 In a gamma decay process, the life time of decaying nuclei is found to be 2 ns.Compute the uncertainty in the energy of gamma rays emitted.

Solution: E · t ≥ ~/2 orE · t ~

E = ~/t = 6.62× 10−34/2× 3.14× 2× 10−9

= 5.27× 10−26J (Ans).

Exercise

1.1 Explain photoelectric effect. Give the assumptions needed, and the laws of pho-toelectric emission. Calculate the velocity of the photoelectrons emitted from ametal surface whose work function is 1.5 eV and the wavelength of the incidentlight being 4× 10−7m. (March 1999)

1.2 Describe the ultraviolet catastrophe. Explain how the Planck’s law of radiationovercomes it. (March 1999)

1.3 Explain Wien’s law and Rayleigh-Jean’s law. Mention their drawbacks.

(August 1999)

1.4 Explain Einstein’s theory of photo-electric effect. (August 1999)


1.5 What is Planck’s radiation law? Show how Wien’s law and Rayleigh-Jean’s lawcan be derived from it. (August 1999)

1.6 The work function of tungsten is 5.4 eV. When light of wavelength 170 nm isincident on the surface, electrons with energy 1.7 eV are ejected. Estimate thePlanck’s constant. (August 1999)

1.7 What is the nature of black body radiation? Explain its significance.

(March 2000)

1.8 In Compton scattering, calculate the maximum kinetic energy of the scatteredelectron for a given photon energy. A metal surface has a photoelectric cut offwavelength of 325.6 nm. What is the stopping potential for anincident light ofwavelength of 259.8 nm? (March 2000)

1.9 State Compton effect and explain the experimental observations of Comptonshift. (August 2000)

1.10 The stopping potential of 5.2 V was observed for light of frequency 3× 1015 Hz.What would be the frequency of light when the stopping potential is doubled?

(August 2000)

1.11 State and explain Planck’s law of radiation. Show that it reduces to Wien’s lawand Rayleigh-Jean’s law under certain conditions. (August2000)

1.12 The photoelectric threshold for a certain metal is 5000 AU. Determine the maxi-mum energy of the photoelectrons emitted when a radiation ofwavelength 3000AU is incident on its surface. (March 2001)

1.13 Explain Planck’s radiation law. Discuss Einstein’s theoryof photoelectric effect.

(March 2001)

1.14 Explain Wien’s Displacement law. Mention its drawbacks. (August 2001)

1.15 Explain Compton effect and give its physical significance. (August 2001)

1.16 What is ultraviolet catastrophe? Show how it can be overcomewith the help ofPlanck’s radiation law. (August 2001)

1.17 What is photoelectric effect? Mention the features of photoelectric emission.

(March 2002)


1.18 What is Compton scattering? With necessary equations, explain Compton scat-tering effect. (March 2002)

1.19 Calculate the change in wavelength in Compton scattering atan angle of 60 tothe incident direction. (March 2002)

1.20 Calculate the wavelength associated with an electron carrying an energy 2000 eV.

(March 1999)

1.21 Set up one dimensional Schrodinger’s wave equation for a free particle.

(March 1999)

1.22 Explain group velocity and phase velocity. Derive a relation between the two.

(August 1999)

1.23 Explain Heisenberg’s uncertainty principle. Give its physical significance.

(August 1999)

1.24 What is de Broglie’s concept of matter waves? An electron hasa wavelength of1.66× 10−10m. Find the kinetic energy, phase velocity and group velocityof thede Broglie wave. (August 1999)

1.25 Explain Heisenberg’s uncertainty principle. Give its physical significance. Theposition of an electron can be measured with an accuracy of 1.5× 10−10m. Findthe uncertainty in its position after 1 sec. (August 2001)

1.26 What are Eigen functions and Eigen values? Find them for a particle in onedimensional potential well of infinite height. (August 2001)

1.27 Obtain the time independent Schrodinger’s wave equation for a particle in onedimensional potential well of infinite height and discuss the solution.

(August 2000)

1.28 Write a note on group velocity and phase velocity. (March 2001)

1.29 Explain group velocity and phase velocity. Derive relationbetween the two.

(August 2001)

1.30 What is a wave function? Give its physical significance. Whatis normalizationof a wave function? (August 2001)


1.31 What is de Broglie concept of matter waves? Derive an expression for de Brogliewavelength. (August 2001)

1.32 Set up time independent Schrodinger’s wave equation and explain Eigen func-tions and Eigen values. (August 1999)

1.33 What is Heisenberg’s uncertainty principle? Discuss its consequences.

(March 2000)

1.34 Explain the behaviour of a particle in a one-dimensional infinite potential wellin terms of de Broglie waves. (March 2000)

1.35 What is the physical interpretation of wave function, nature of Eigen values andEigen functions. (March 2000)

1.36 Discuss phase velocity and group velocity. (August 2000)

1.37 Explain only the conclusions drawn in photoelectric effect and Davisson–Germerexperiment leading to deBroglie hypothesis. (Feb 2005)

1.38 Show that a free electron cannot exist in a nucleus of an atom. (Feb 2005)

1.39 What are matter waves? Show that the electron accelerated bya potential differ-enceV volt is= 1.226/

√Vnmfor non–relativistic case. (July 2005)

1.40 Explain phase velocity and group velocity. Derive the expression for deBrogliewavelength using the concept of group velocity. (July 2005)

1.41 Explain only the conclusions drawn in photoelectric effect and Davisson–Germerexperiment leading to deBroglie hypothesis. (Feb 2005)

1.42 Show that a free electron cannot exist in a nucleus of an atom. (Feb 2005)

Chapter 2

Crystallography and X-rays

2.1 Crystal Structure

The behaviour of solids is much different from that of liquids and the properties ofmaterials differ widely in their liquid and solid states. One of the reasonsfor suchdifferences is that in a solid, depending on the nature of bonding, there will be a ten-dency for the atoms or molecules to arrange themselves in particular order. The extentand range of such an ordering may of course depend on the conditions under whichthe solid state is formed. The valency, the electronic configuration, the coordinationnumber, etc. are factors that prompt the atoms or the molecules to arrange themselvesin a way to result in a solid. If such an ordering extends only to a few atomic spac-ing (and some ordering is necessary due to the conditions of bonding), the material isreferred to be amorphous or non-crystalline. The crystalline solids may be classifiedas polycrystalline and single crystalline depending on therange of ordering. In caseof polycrystalline materials, the arrangement of atoms or molecules along the differentdirections will be disrupted by some defects in the arrangement. Thus, the orderingwill extend only upto certain distance. The material may be considered to be madeup of a large number of small single crystals, oriented randomly in three dimensions.Each single crystalline part is called a grain and the size ofthe grain is used as a mea-sure of crystallinity. On the other hand, the ordering in a single crystal will continuethrough out the sample.

Many important properties of materials are found to depend on the structure ofcrystals. A study of the crystal structure has led to the understanding of the behaviourof materials and hence a correlation between physical properties and the crystal struc-ture has been possible. The study of the crystal structure and their relation with theproperties has led to the development of new materials for electronic and structuralengineering applications.

2.1.1 Unit cell and space lattice

A crystal is a solid in which the atoms or molecules are arranged in an orderly manner.The periodicity of ordering may be different in different directions. It is convenientto imagine arrangement of points in space to represent atomsor molecules.These

55

56 Crystallography and X-rays

imaginary points are called lattice points and an arrangement of these imaginarypoints in three dimensions constitutes a space lattice. Each point in the space latticerepresents the location of an atom or a molecule or a collection of atoms which repeatthrough out the space lattice in the same shape and orientation. Such a collection ofatoms or molecules or group of atoms which is represented by a point in the spacelattice is called a basis. The space lattice in combination with the basis gives thecrystal lattice(Fig. 2.1).

Lattice + basis = crystal structure

Figure2.1Crystal structure in two dimensions.

A

CB

O

Figure2.2Unit cell in two dimensions.

In order to define a unit cell, let us consider a lattice in two dimensions (Fig. 2.2).In this lattice, we can identify two directions along which the packing of lattice pointsis closest. Choosing any lattice point O as origin, the nearest point A is at a distance OAin one direction and point B is at a distance OB in the other direction. The geometricalfigure OACB is called a unit cell. By repeating such unit cells, it is possible to fill the

Crystallography and X-rays 57

entire space.OA andOB represent the fundamental translation vectors characteristicof the unit cell. In three dimensions, we will require three translation vectors andthree angles between these translation vectors to define thelattice completely. Thesesix parameters, three translation vectors and the three angles, are called thelatticeparameters. The cell formed by the three translation vectors is called aunit cell.Hence,unit cell is the smallest cell which contains the entire information aboutthe arrangement of atoms or molecules in the crystal in threedimensions. It willbe defined by three distances along translation directions and the three anglesbetween them(Fig. 2.3). It is customary to designate the three repeat distances asa,b andc with α, β andγ representing the inter axial angles betweenb andc, c anda anda andb respectively.

x

z

y

c

a

bα

β

γ

Figure2.3Unit cell with cell parameters

2.1.2 Crystal systems

Arrangement of atoms or molecules in any solid can be described with the help of unitcell parameters. The variations in the interaxial angles and the repeat distances leadto seven types of unit cells. These different unit cells are referred to be belonging toseven crystal systems. These seven crystal systems and the characteristic values of thelattice parameters are summarised in Table 2.1. The unit cells for these crystal systemsare shown in Fig. 2.4.


a = b = cα = β = γ = 90

(a)

a = b 6= cα = β = γ = 90

(b)

a 6= b 6= cα = β = γ = 90

(c)

a 6= b 6= cα = γ = 90 6= β

(d)

a 6= b 6= cα 6= β 6= γ 6= 90

(e)

a = b 6= cα = β = 90 γ = 120

(f)

a = b = cα = β = γ = 90 < 120

(g)

Figure 2.4 Unit cells for the seven crystal systems. The crystal systems are (a) cu-bic,(b) tetragonal,(c) orthorhombic,(d) monoclinic,(e)triclinic,(f) hexagonal and (g)rhombohedral.


Table2.1The seven crystal systems and their characterstics.

System Axial lengths Interaxial angles

Cubic a= b = c α = β = γ = 90

Tetragonal a= b , c α = β = γ = 90

Orthorhombic a, b , c α = β = γ = 90

Monoclinic a, b , c α = γ = 90 , βTriclinic a , b , c α , β , γ , 90

Hexagonal a= b , c α = β = 90, γ = 120

Rhombohedral a= b = c α = β = γ , 90 < 120

2.1.3 Bravais lattices

The unit cells of crystals belonging to the seven crystal systems will have differentshapes and sizes decided by the six lattice parameters. The distribution of atoms ormolecules, in the simplest of the cases, will be restricted to the end points of the unitcells, namely the corners. However, it can be seen by simple translation operationsthat addition of atoms or molecules at certain other points within the unit cell willnot disturb the periodic repetition of the lattice. For example, an atom or molecule,similar to the one at the corners of a cubic cell, may be added at the centre of theunit cell. Addition of a body-centre atoms does not alter thelattice parameters and thecrystal system remains the same. Such an unit cell, which is different from the simplecubic unit cell, is called a body centered cubic cell. Similarly, atoms or molecules,similar to the ones at the corners of a simple cubic cell, may be placed at the facecentres in which case the unit cell is referred to as a face centred cubic cell. Thesemodified cells, along with the primitive cell, are said to constituteBravais lattices,named after the French crystallographer, Auguste Bravais.If we consider all the sevencrystal systems, there are only 14 Bravais lattices which can be classified as primitive(P), body centred (I), face centred (F) or base centred (C). Table 2.2 lists all possibleBravais lattices belonging to the seven crystal systems. Itcan be shown by simplegeometric considerations that other Bravais lattices do not exist. For example, we donot have a base centred cubic lattice since it will be the sameas a primitive tetragonal.A face centred tetragonal will be equivalent to a body centred tetragonal cell.


Table2.2Possible Bravais lattices.

Crystal system Bravais lattices

Cubic P, I, FTetragonal P, IOrthorhombic P, I, F, CMonoclinic P, CTriclinic PHexagonal PRhombohedral P

2.1.4 Miller indices and their uses

It is observed that many physical properties of crystallinesolids are dependent on thedirection of measurement or the planes across which the properties are studied. Hence,it becomes necessary to specify directions and planes in a crystal lattice. In order tospecify directions in a lattice, we make use of lattice vectors. For example, in a cubicunit cell shown in Fig. 2.5 the lattice vectorOA can be expressed in terms of the threetranslation vectorsa, b andc of the unit cell as

OA= a + b + c

Or, in general, any directional vectorr may be expressed as

r = n1a + n2b + n3c

wheren1, n2 andn3 are integers.

O

A

x

y

z

Figure2.5Representing directions in a unit cell


The direction of the vectorr is determined by these integers. If these numbers havecommon factors, they are removed and the direction ofr is denoted as [n1 n2 n3]. Thisincludes all other lines parallel to this direction. Generally, square brackets are usedto indicate a direction. A similar set of three integers enclosed in a round bracket isused to designate planes in a crystal. The crystal lattice may be regarded as made up ofan aggregate of a set of parallel, equidistant planes passing through the lattice points.These planes are known aslattice planesand may be represented by a set of threesmallest possible integers. These numbers are called“Miller Indices” , named afterthe crystallographer, W.H.Miller. It can be shown that planes and normal to the planeswill have the same set of Miller indices. Thus, Miller indices can be defined as a setof three integers used to specify planes, directions and positions of atoms or moleculeswith reference to an arbitrary origin.

The steps involved in determining the Miller indices of a setof parallel planes canbe explained using the Fig.2.6.

a 2a3a

4a 5a

b

2b

3b

4b

5b

A

B

C

c

2c

3c

X

Y

Z

Figure2.6Determination of Miller indices for plane ABC.

Let x, y andzrepresent the three crystallographic axes along which the given crystalhas primary translation vectors of magnitudea, b andc respectively. LetABCrepresentthe plane whose Miller indices are to be determined. LetOA, OB and OC be theintercepts made by the plane along the three crystallographic axes.

(i) Find the intercepts along the x y xthree crystallographic axes OA OB OC


(ii) Express the intercepts asmultiples of lattice parameters pa qb rc

(iii) Devide by the lattice parametersalong each direction pa/a qb/b rc/c

(iv) Take the reciprocals 1/p 1/q 1/r

(v) Reduce them to the smallest setof integers and enclose them inround brackets. (h k l)

The Miller indices of the planeABCare (hkl). If d is the distance of the planeABCfrom O, then planes parallel toABCand at perpendicular distances equal to 2d, 3d, 4d,. . . from 0 will have the same Miller indices. Hence, (hkl) represents the Miller indicesfor a set of planes separated by an interplanar distanced. Some important directionsand planes are shown in Fig.2.7 along with their Miller indices.

O

(100)

x

y

z

[100]

Ox

y

z

(110)

[110]

O

[111]

x

y

z (111)

Figure2.7Representation of planes and directions.

The procedure for representing different planes with Miller indices (hkl) in a unitcell is just the reverse of the procedure used for finding the Miller indices of the planes.The steps followed are as follows:

(i) Let the set of planes to be represented have the Miller indices (hkl). Draw a unitcell with the cell dimensionsa, b andc along the directionsx, y andzrespectively.

(ii) Take the reciprocals of the Miller indices and multiplythem with the cell dimen-sion along each direction. This will give us the intercepts the plane makes alongthe three directions. For the planes (hkl), the intercepts will bea/h, b/k andc/lalongx, y andz directions respectively.


(iii) Mark the pointsA, B andC at distancesa/h, b/k andc/l from a reference originO (a corner in the unit cell) alongx, y andz directions respectively.

(iv) Join the pointsA, B andC. ABC is the required plane.

Planes with negative Miller indices may be plotted by shifting the origin accord-ingly. For example, ifh is negative, the origin is shifted to (1, 0, 0) position. Ifk isnegative, the origin is shifted to (0, 1, 0) position and so on. Ifh andk are both nega-tive, the origin is shifted to (1, 1, 0) position. Examples of some planes are representedin Fig. 2.8.

X

Y

Z

(222)

X

Y

Z

(212)

X

Y

Z

(423)

X

Y

Z

(100)

OX

Y

Z

(101)

O

X

Y

Z

(111)

O

Figure2.8Examples of some planes and their Miller indices.

2.1.5 Interplanar spacing in cubic crystals

Consider a planeABC making intercepts (Fig.2.9)pa, qa, ra along x, y, andz axesrespectively (a = b = c since we have considered cubic system).


X

Y

Z

O A

B

C

θ1

θ2

θ3

N

Figure2.9Plane ABC with the normal ON drawn to it from O

Draw a perpendicularON from origin O to the planeABC. By definition,ON =d, the interplanar spacing corresponding to the set of planeshaving the same Millerindices as the planeABC. Let the angles made byON with OA, OB, andOC beθ1, θ2

andθ3 respectively. Then, we have

cosθ1 = d/pa, cosθ2 = d/qa, cosθ3 = d/ra

Squaring and adding, we get

cos2 θ1 + cos2 θ2 + cos2 θ3 =d2

a2

[

1p2+

1q2+

1r2

]

It can be shown by simple geometry that for the case when

1p= h,

1q= k and

1r= l,

cos2 θ1 + cos2 θ2 + cos2 θ3 = 1

Hence,d2

a2(h2 + k2 + l2) = 1

or d =a

(h2 + k2 + l2)1/2

Here,d represents the interplanar spacing for the set of planes having Miller indices(hkl).


2.1.6 Atomic packing factors

In crystalline solids, the atoms are arranged in a particular order and each atom issurrounded by a definite number of nearest equidistant neighbours. Number of suchnearest neighbours that an atom has in the given structure iscalledcoordination num-ber. For example, in a simple cubic lattice where atoms are located at the corners ofthe unit cell, each atom is surrounded by six atoms at a distance equal to the lattice pa-rameter. Hence, the coordination number for simple cubic lattice is six. Similarly, fora body centred cubic lattice, the nearest atom for a corner atom is the body centre atomand hence the coordination number is eight. For a face centred cubic lattice each cor-ner atom has the face centre atom as the nearest neighbour andhence the coordinationnumber is twelve.

The density of atoms in a given crystalline solid depends on the lattice type. Thenumber of atoms per unit cell in each of these cases can be calculated as follows:

In simple cubic lattice, there are eight atoms at the eight corners of the unit cell,each atom being shared by eight unit cells.

∴ No. of atoms/unit cell= 8/8 = 1In body centered cubic lattice, there are eight atoms at the corners of the unit cell,

each atom being shared by eight unit cells. In addition, there is an atom at the bodycentre.

∴ No. of atoms/unit cell= (8/8) + 1= 2In face centred cubic lattice, there are eight atoms at the corners of the unit cell,

each atom being shared by eight unit cells. In addition, there is an atom at the centreof each face which is shared by two neighbouring unit cells. There are six faces thatform the unit cell.

∴ No. of atoms/unit cell= (8/8) +(6/2) = 4If we assume cubic crystal lattices to be made up of identicalatoms that are closely

packed we can work out relations between the lattice parameter and the atomic radius.For example in a close packed simple cubic lattice, the atomstouch each other alongthe cubic axes (Fig.2.10).

So, we havea = 2r

wherea is the lattice constant andr is the atomic radius. In a face centred cubic lattice,there is an additional atom at the centre of the face so that the atoms are in contact witheach other along the face diagonal. Hence,

(4r)2 = a2 + a2 or a = 2√

2r

In a body centred cubic lattice, there is an additional atom at the centre of the unit


cell so that the atoms are in contact with each other along thebody diagonal. Hence,

(4r)2 = 3a2 or a = (4/√

3).r

(a) (b) (c)

Figure 2.10 Atomic arrangements in unit cells showing atoms in contact (a)alongcube axis in simple cubic cell, (b) along face diagonal in face centred cubic cell and(c) along body diagonal in body centred cubic cell.

Theatomic packing factor is defined as the ratio of the volume of the atoms in theunit cell to the volume of the unit cell. It represents the fractional volume of the unitcell filled by atoms. Hence, we can write,

Packing factor=Volume of each atom x No.of atoms in unit Cell

Volume of the unit cell

We can calculate the packing factor for the cubic crystals asfollows :In simple cubic crystals

Number of atoms per unit cell= 1

Volume of each atom= (4/3)πr3

Volume of unit cell= a3 = (2r)3

Packing factor=(4/3)πr3

(2r)3= π/6 = 52%

In body centred cubic crystals,




Volume of unit cell= a3 = (4/√

3)r3

Packing factor=2(4/3)πr3

4/√

3r3= (√

3π/8) = 68%

In face centred cubic crystals



Volume of unit cell= a3 = (2√

2r)3

Packing factor=4.(4/

√3)πr3

(2√

2r)3= (√

2π)/6 = 74%

The various parameters associated with cubic unit cells aresummarized in Ta-ble 2.3.

Table2.3Properties of cubic unit cells.

Type of CellNumber of atoms

per unit cellCoordination

numberAtomicradius

Packingdensity

Simple cubic 1 6 a/2 π/6

Body centred cubic 2 8√

3.a/4√

3.π/8

Face centred cubic 4 12√

2.a/4√

2.π/6

A relation between lattice constanta and the densityρ of the crystal may be derivedas follows. Consider the case of a cubic lattice.

Volume of the unit cell= a3.Mass of all the atoms(molecules)in the unit cell= ρa3

Number of atoms(molecules) per unit cell= zAtomic(Molecular) weight of the material=MAvogadro number= NMass of each atom(molecule)= (M/N)Mass of atoms(molecules) in unit cell= z(M/N)Hence,ρ a3 = z(M/N) or a = (zM/ρN)1/3.


Thus, the lattice parameter can be calculated for any cubic crystal with the help of theavailable data.

2.1.7 Some crystal structures

Materials crystallize in a variety of structures. These structures are said to be simple ifthe arrangement of atoms or molecules can be identified with the Bravais lattices. Butin some cases, the structure becomes complicated with one ormore atoms or moleculesoccupying specific positions within the unit cell, but retaining the periodicity of theunit cell as a whole. Many of these structures can be considered to be a combinationof two or more unit cells having the same cell parameters but translated with respectto each other in three dimensional space through finite space. We consider a few suchexamples.

(a) Sodium chloride: Sodium chloride is an ionic solid. Sodium loses its outer elec-tron to acquire a positive charge while chlorine accepts theelectron to become anegative ion. The crystal structure of sodium chloride has sodium ion and chlorineions arranged side - by - side in a three dimensional orthogonal array. The resultis a face centred cubic lattice (Fig. 2.11). In fact it can be

Na

Cl

Figure2.11Unit cell of sodium chloride.

considered to be a superposition of two sub - lattices, one ofsodium ions and theother of chlorine ions, both the sub - lattices having the same lattice parameters.


The superposition is such that one sub-lattice is translated through a distance equalto half the lattice parameter along the cube edge of the face centred cubic lattice. Ifthe sodium and chlorine ions are replaced by identical atoms, the structure wouldthen be simple cubic with a lattice parameter equal to half that of the NaCl struc-ture. The lattice parameter, being the repeat distance along crystallographic axes,is equal to the distance between two sodium ions or two chlorine ions. Conse-quently, the unit cell of sodium chloride has four sodium ions and four chlorineions in its unit cell. In other words, there are four molecules of NaCl in the crystalstructure of sodium chloride. Assuming one corner ion (sayNa) as origin, thepositional coordinates for the other ions in the unit cell can be written as follows:

Na ions: 8 corner ions (each shared by 8 unit cells) with positional coordinates(0 0 0) (1 0 0) (0 1 0) (0 0 1)(1 1 0) (1 0 1) (0 1 1) (1 1 1)6 face centre ions (each shared by 2 unit cells) with positional

coordinates(1/2 1/2 0) (1/2 0 1/2) (0 1/2 1/2)(1/2 1/2 1) (1/2 1 1/2) (1 1/2 1/2)

Cl ions: 12 edge centre ions (each shared by 4 unit Cells) with positionalcoordinates

(1/2 0 0) (0 1/2 0) (0 0 1/2)(1/2 0 1) (1/2 1 0) (1 1/2 0)(0 1/2 1) (1 0 1/2) (0 1 1/2)(1/2 1 1) (1 1/2 1) (1 1 1/2)one body centre ion at (1/2 1/2 1/2).

Thus, the structure ofNaCl contains four molecules per unit cell.

(b) Caesium chloride:

Ceasium chloride is also an ionic solid and crystallizes with a simple cubic struc-ture (Fig. 2.12).

The lattice ofCsCl is made up of two sub-lattices of caesium ions and chlorineions each being simple cubic in arrangement. The corner of one sub-lattice is thebody centre of the other. In other words, the two sublatticesare related through atranslation along the body diagonal by a distance equal to half the body diagonal. Ifthe caesium and chlorine ions are replaced by identical atoms, the resultant lattice


Cs

Cl

Figure2.12Unit cell of caesium chloride.

will be a body centred cubic lattice with the same lattice parameter as that ofCsCl.The effective number of ions in the unit cell ofCsClare one each of caesium andchlorine. Thus, it contains one molecule ofCsCl per unit cell. Assuming onecorner ion (sayCs) as origin, the positional coordinates can be written as follows.

Cs ions: 8 corner ions each shared by 8 unit cellsWith positional coordinates(0 0 0) (1 0 0) (0 1 0) (0 0 1)(1 1 0) (1 0 1) (0 1 1) (1 1 1)

Cl ion: one body centre ion at (1/2 1/2 1/2)

Thus, the structure ofCsClcontains one molecule per unit cell.

(c) Zinc Sulphide: The nature of bonding in Zinc sulphide is partly covalent andpartly ionic. The structure of cubic zinc sulphide is calledzinc blende structure(zinc blende is the mineral name for cubic zinc sulphide). Itconsists of two inter-penetrating face centred cubic sub lattices. The two sub-lattices, made up of zincand sulphur ions respectively, are displaced from each other along the body di-agonal through a distance equal to one quarter of the body diagonal (Fig. 2.13).Hence, unit cell ofZnSconsists of four ions of zinc and four ions of sulphur on anaverage or four molecules ofZnS.

Assuming one corner ion (sayZn) as origin, the positional coordinates can bewritten as follows.

Zn ions: 8 corner ions (each shared by 8 unit cells) with positional coordinates(000) (l00) (010) (001)(110) (101) (011) (111)6 face centre ions(each shared by 2 unit cells) with positional coordinates


Zn

S

Figure2.13Unit cell of zinc sulphide.

(1/2 1/2 0 ) (1/2 0 1/2) (0 1/2 1/2)(1/2 1/2 1 ) (1/2 1 1/2) (1 1/2 1/2)

S ions: Four interior ions with positional Coordinates(1/4 1/4 1/4) (3/4 3/4 1/4) (3/4 1/4 3/4) (1/4 3/4 3/4)

Thus the structure ofZnScontains four molecules per unit cell.

(d) Diamond: Structure of diamond is similar to the structure of Zinc sulphide. Whenthe Zinc and sulphur ions are replaced by identical carbon atoms, the diamondstructure results. Silicon and germanium also have have diamond structure.

The nature of bonding in diamond is partly covalent and partly ionic. It consists oftwo inter-penetrating face centred cubic sub lattices. Thetwo sub-lattices, madeup of carbon atoms are displaced from each other along the body diagonal througha distance equal to one quarter of the body diagonal (Fig. 2.14). Hence, unit cellof diamond consist of eight atoms of carbon.

Figure2.14Unit cell of diamond.


Assuming one corner atom as origin, the positional coordinates can be written asfollows.

(i) 8 corner atoms (each shared by 8 unit cells) with positional coordinates(000) (l00) (010) (001)(110) (101) (011) (111)

(ii) 6 face centre atoms (each shared by 2 unit cells) with positional coordinates(1/2 1/2 0) (1/2 0 1/2 ) (0 1/2 1/2)(1/2 1/2 1) (1/2 1 1/2 ) (1 1/2 1/2)

(iii) 4 interior atoms with positional coordinates

(1/4 1/4 1/4) (3/4 3/4 1/4) (3/4 1/4 3/4) (1/4 3/4 3/4)

Thus the structure of diamond contains eight atoms per unit cell.

2.2 X-Rays

X rays are electromagnetic radiations, like light, but witha much smaller wavelength.They exhibit all the characteristic properties similar to light. By virtue of their smallwavelength (corresponding to higher frequency and photon energy) they have a higherpenetration power than light and hence are immensely usefulin medical applications.Since the wavelength is of the order of atomic spacings in solids, they provide impor-tant information about the arrangement of atoms in solids through diffraction patterns.

2.2.1 Origin of x-rays

X-rays are produced when electrons with high velocity interact with atoms. When anelectron beam is incident on a solid target material, most ofthe energy of the electronswill be dissipated as heat and a very small amount of energy isemitted in the form ofx-rays. A conventional x-ray tube consists of an evacuated glass tube with a filamentas cathode and the target metal as anode (Fig. 2.15).

The filament is heated with a low voltage source to emit electrons. These electronsare accelerated with a high voltage applied between the filament and the anode. Whenthe high energy electrons strike target, x-rays are emitted. The wavelength of x-raysso emitted will depend on various factors including the initial and final energy of theincident electron, nature of the target material and the nature of interaction. Those x-rays possessing higher penetration power (i.e., x-rays of lower wavelengths) are calledhard x-rays and those with lesser penetration power are called soft x-rays.


Filament with targetWater cooled anode

H. T

X − Rays

Figure2.15A conventional x-ray tube.

2.2.2 Continuous x-ray spectrum

When a target material is bombarded with high energy electrons, some of the electronspenetrate into the cores of the atoms of the target material.Due to the attractive inter-action with the positive nuclei of the atoms, electrons get decelerated. The energy lostduring this deceleration is given off in the form of x-rays (Fig.2.16).

+

v′

v

Figure2.16Schematic representation of emission of continuous x-rays.

Since the energy involved can have all possible values upto the total energy pos-sessed by the electrons, the emitted x-rays produce a continuous spectrum. This spec-trum will have a sharply defined short wavelength limitλmin, corresponding to the totalkinetic energy of the incident electron. If an electron has its velocity reduced from


v to v′ during such an interaction, the change in the energy of the electron will be

1/2m(v2 − v′2) wherem is the mass of the electron. This energy will be emitted as

radiation of wavelengthλ.

∴12

m(v2 − v′2) = hν =

hcλ

(2.1)

The short wavelength limit corresponds to the case when the electron loses all its ki-netic energy.

i.e. v′ = 0

∴12

m v2 = hνmax=

hcλmin

(2.2)

Since the kinetic energy of the electron is acquired due to the application of an electricpotentialV,

∴12

mv2 = eV (2.3)

From equation (2.2) and (2.3), we have

λmin =hceV

(2.4)

This equation is known asDuane – Hunt lawand this equation was used for an ex-perimental determination of Planck’s constant. Substituting the values of the constants,equation (2.4) may be rewritten as

λmin =12400

V(2.5)

whereλmin will be obtained inA whenV is in volts. Fig. 2.17 shows the variationof the intensity of x-rays emitted as a function of wavelength at different acceleratingpotentials.

The following features are characteristic of the continuous x-ray spectrum.

(i) The intensity of continuous spectrum increases at all wavelengths as the potentialis increased.

(ii) The short wavelength limit shifts towards lower wavelengths as the potential isincreased.

(iii) The wavelength corresponding to the maximum intensity also shifts towardslower wavelengths as the potential is increased.


I

10 kV

20 kV

30 kV

λ

Figure 2.17 Intensity of continuous x-rays emitted as a function of wavelength at dif-ferent accelerating potentials (characteristic x-ray peaks are also shown)

2.2.3 Characteristic x-ray spectrum

It is also observed that some of the high energy electrons, during its interaction withthe target atoms, knock off the tightly bound electrons from the innermost shells (likeK-shell, L-shell, etc) of the atoms. The vacancies so produced are filled up by elec-trons from the outer orbits and the energy difference is given out in the form of x-rays(Fig.2.18).

hν

Figure2.18Schematic representation of emission of characteristic x-rays.


Since the transitions are between levels of well defined energies, the emitted x-rayswill form a line spectrum which is characteristic of the material of the target. Fig.2.18 shows an electron from the K-shell of an atom of the target being knocked offby the incident electron. An electron from any of the higher energy levels may fillthe vacancy so created in the K-shell, i.e., from L-shell, M-shell, N-shell, etc. Theemission lines due to these transitions are namedKα,Kβ . . ., etc., and form the K-seriesof x-rays. Similarly, if the transitions are from a higher shell to L-shell, the x-raysbelong to L-series (Fig.2.19).

Kα Kβ

Lα LβLγ

Mα Mβ

K

L

M

N

O∞

Figure2.19Energy level diagram showing the possible emission lines.

Thus, the line spectrum depends on the target material and hence called the char-acteristic x-ray spectrum. The important features of line spectrum are as follows:

(i) The characteristic x-ray spectrum consists of discretespectral lines dependingon the transitions occurring during emission. K-series consists of spectral linesemitted due to transitions from higher shells to K-shell.

(ii) The wavelength of different lines in the spectrum are characteristic of the targetmaterial. This fact is used in the chemical analysis of the target material.

(iii) The wavelength of the K-lines shifts towards lower values as the atomic numberof the target element increases.

(iv) There is a minimum accelerating potential required to excite the emission of dif-ferent spectral lines.

(v) The intensity of a spectral line depends on the probability of that particular transi-tion taking place and hence on the accelerating potential and the incident electronbeam current.


2.2.4 Moseley’s law

A detailed study of the characteristic x-ray spectra for different target materials wascarried out by Moseley. He observed that the wavelength of a given spectral line de-creased as the atomic number of the target metal increased. The square root of thefrequency ofKα line was found to be proportional to the atomic number of the targetmaterial (Fig.2.20). He expressed the relation in the form of the equation.

√ν

Kα

Atomic No. z

Figure2.20Variation of the square root of frequency of emitted x-rays as a function ofthe atomic number of the target material.

√ν = a(z− b) (2.6)

whereν is the frequency of a particular spectral line (sayKα) from a target of atomicnumberz, a andb being constants. It can be shown that this is in accordance with theBohr’s theory with modification for the screening effect of electrons. The law may alsobe expressed in the form,

1λ= R(z− b)2

(

1

n21

−1

n22

)

(2.7)

whereR is called Rydberg constant,b is the screening constantn1 andn2 are the princi-pal quantum numbers corresponding to the final and initial energy levels correspondingto the transition. Hence, for a K-line,n1 = 1 andb = 1

∴1λK= R(z− 1)2

(

1−1

n22

)

Where n2 = 2, 3, .. (2.8)

Similarly for L-line, n1 = 2 andb = 7.4

∴1λL= R(z− 7.4)2

(

14− 1

n22

)

Where n2 = 3, 4, .. (2.9)


One of the important application of Moseley’s law was to resolve the discrepancies inthe arrangement of elements in the periodic table. The law suggests that the physicaland chemical characteristics of elements are more closely related to the atomic numberrather than the atomic weight. Accordingly, Moseley’s law helped in assigning properplaces to the elements in the periodic table. For example, Argon (18A40) should beplaced before Potassium (19K39) and Cobalt (27Co58.9) should be placed before Nickel(28Ni58.7). It has helped in assigning atomic numbers to rare earth elements and also inthe discovery of new elements.

2.3 X-ray diffraction

When a monoenergetic x-ray beam is made to incident on a sample of a single crys-tal, diffraction occurs resulting in a pattern consisting of an arrayof symmetricallyarranged diffraction spots. The single crystal acts like a grating with a grating constantcomparable with the wavelength of x-rays, making the diffraction pattern distinctlyvisible. The arrangement of atoms in a crystal results in theformation of planes witha high density of atoms which diffract x-rays preferentially along specific directions.Hence, a study of the diffraction pattern helps in the analysis of the crystal parameters.

2.3.1 Bragg’s law

Consider the diffraction of x-rays from a single crystal sample as shown in Fig.2.21.Let XX’ and YY’ be two parallel planes with an interplanar spacing d.

A

DF

C

B

EM N

θ d

x x′

y y′

θ

Figure2.21Diffraction of x-rays by crystal planes.

Let AB represent an x-ray incident at an angleθ to the plane which is scatteredalongBC. DE is a parallel ray scattered alongEF. The two raysBC andEF shouldinterfere constructively to produce a resultant diffracted beam. The path difference


between the two rays can be found by drawing perpendicularsBM andBN to the raysDE andEF respectively.

The total path difference= ME + EN

= dsinθ + dsinθ

= 2dsinθ (2.10)

The two reflected rays will be in phase if this path difference is equal to an integralmultiple ofλ, the wavelength of x-rays.

Hence, the condition for producing a maximum becomes

nλ = 2dsinθ (2.11)

wheren is an integer called order of diffraction. This is known as Bragg’s law.

2.3.2 Bragg’s spectrometer

Bragg’s spectrometer (more commonly referred to as x-ray diffractometer) is an in-strument used to study the angles and intensities of diffracted beams for a given crystalsample. A conventional Bragg spectrometer is shown in Fig.2.22.

S

F

ST

C

D

R

S1 2

Figure2.22Bragg spectrometer.

It consists of an x-ray source S producing characteristic x-rays of required wave-length. Unwanted continuous x-rays as well as characteristic x-rays of other wave-lengths are removed from the beam using a suitable absorption filter. The monochro-matic x-ray beam is then passed through a pair of slitsS1 andS2 which will definethe divergence of the beam. The x-rays will then be incident on the crystal sample C


mounted at the centre of a turn table T. The turn table also carries a side arm on whicha detector D for x-rays (usually an ionization counter like G.M. counter) is mounted.The sample holder and the detector can be rotated about an axis perpendicular to thedirection of the incident x-rays. The rates of rotation of the crystal holder and thedetector arm are such that the detector always receives the x-rays reflected from thecrystal. When x-rays are incident on any particular plane atan angleθ, the diffractedbeam which will be at an angle 2θ will be received by the detector. The output of thedetector is measured or recorded on a x-y recorder R to obtaina chart showing thevariation of diffracted intensity of x-rays as a function of diffraction angle.

The verification of Bragg’s law is carried out by mounting a standard sample on theturn table of the spectrometer. The x-rays are allowed to fall on the sample at an angleclose to the glancing angle (θ = 0) and the angle is gradually increased. The detectoroutput observed at angle 2θ in each case is measured and the angles corresponding topeak intensities are noted. For the standard sample, knowing the crystal structure andthe lattice parameters, the interplanar spacings may be calculated. For example, for acubic crystal, the permitted d-spacing can be calculated as

d =a

(h2 + k2 + l2)1/2(2.12)

Hence, Bragg’s law can be rewritten as

sin2 θ =n2λ2

4a2(h2 + k2 + l2) (2.13)

Knowing,λ anda, by assigning different integer values forh, k, l and ordern, allowedvalues ofθ can be computed. These values ofθ are compared with those obtained fromthe Bragg spectrometer recordings. A perfect match betweenthe computed values andthe observed values of Bragg angles will be obtained therebyestablishing the validityof Bragg’s law.

2.3.3 Structure determination

The Bragg spectrometer analysis is a widely used experimental technique for the rou-tine determination of crystal structure. It is very much suitable for identification andstructure determination of crystals possessing high symmetry (eg. cubic crystals). Itis also possible to distinguish between the lattices of the cubic system using extinctionrules.

Theθ values for different planes can be obtained from the spectrometer recording.For a given cubic lattice, it is possible to list of all combinations ofh, k andl. It is seen


that sin2 θ values will increase with (h2 + k2 + l2). Sinceλ is a constant and known, thelattice parameter ‘a′ can be found out for which sin2 values will be in the same ratioas (h2 + k2 + l2).

The distinction between lattices of cubic system (i.e s.c.,b.c.c. and f.c.c.) is possi-ble by using the fact that not all combinations of (h2 + k2 + l2) lead to diffraction for agiven lattice. Some planes may not give diffraction peaks if the net diffracted intensityadd up to be zero.

It is possible to derive the extinction rules for different cubic lattices. These rulesare listed below:

Lattice type Allowed reflections

1. Simple cubic All values of (h2 + k2 + l2)2. Body centred cubic Even values of (h2 + k2 + l2)3. Face centred cubic h, k, l, all odd or all even.

From the extinction rules, we can derive the ratio of (h2+k2+l2) values and comparewith the observed sin2 θ value to identify the lattice type. The ratio of (h2 + k2 + l2)values (orsin2θ values) will be as follows:

Lattice type Ratio

1. Simple cubic 1:2:3:4:5:6:8:9:....2. Body centred cubic (2:4:6:8:10:12:14:16: ...)

1:2:3:4:5:6:7:8:...3. Face centred cubic 3:4:8:11:12:16:19:20:...

It may be mentioned that (h2 + k2 + l2) = 7, 15, . . . etc., correspond to forbiddenreflections as the sum of the squares of three integers can notbe 7, or 15, etc. Thus, if7 appears in the ratio of (h2 + k2 + l2) values, it must correspond to 14 and the latticetype will be b.c.c. If 7 is absent, the lattice is simple cubic. Face centred lattice can beeasily identified by the distinct ratio.

2.4 Electron diffraction

De Broglie’s hypothesis suggests that an electron of massm travelling with a velocityv is associated with a de Broglie wavelengthλ given by

λ =hp=

hmv


It is interesting to know that Davisson and Germer’s experiment on the scatteringof electrons shows that electrons can be diffracted and the diffraction obeys Bragg’slaw.

An electron diffraction apparatus consists of a long evacuated column with afila-ment suitably heated to emit electrons (Fig.2.23).

filament

Anode

sample

Screen

Figure2.23Experimental study of electron diffraction.

These electrons are accelerated by a high positive potential applied to an anode.The accelerated electrons passing through the small aperture in the anode are made tofall on a crystalline sample. The diffraction pattern may be observed on a fluorescentscreen or may be photographed. The measurements made on the diffraction patternindicate that the diffraction of electrons is in accordance with the Bragg’s law.

The advantage of electron diffraction is that the wavelength of the electron beamcan be conveniently modified to suit our requirement by altering the accelerating po-tential. But the electron beam has a much lower penetration power compared to that ofx-rays and hence restricts its use to very thin samples.

2.5 Neutron diffraction

Neutrons produced in a nuclear reactor, after undergoing a large number of collisionswill assume an average energy equal to the thermal energykT which is of the orderof 0.025eV. Such neutrons are called thermal neutrons and the de Broglie wavelengthassociated with them is about 1.8A. Hence, it can be used as an useful probe to studycrystalline materials. The advantage of using neutrons fordiffraction studies is due tothe fact that they are neutral particles and are not influenced by the electron cloud orthe nuclear charge of the atoms. Further, because of the magnetic moment carried bythe neutrons, they have been used to probe the magnetic properties of crystals.


The main difficulty with neutron diffraction is to obtain mono energetic neutronbeams. Neutron monochromators, which are modified versionsof Bragg spectrometer,can be used to obtain neutron beams of thermal as well as higher energies. The detec-tion and recording of diffracted beams is also complicated since ionization countersdonot detect neutrons directly.

Numerical Examples

2.1 A cubic crystal has a lattice parameter of 5.6A. Calculate the interplanar spacingfor (110) and (111) set of planes.

Solution:d =

a(h2 + k2 + l2)1/2

whereh,k,l are the Miller indices of the set of planes under consideration.For (110) planes,

d110 =5.6

(12 + 12 + 02)1/2=

5.6√

2= 3.96A(Ans.)

For (111) planes,

d111 =5.6

(12 + 12 + 12)1/2=

5.6√

3= 3.23A(Ans.)

2.2 A cubic crystal of lattice parameter 4.31A has a set of planes with interplanarspacing of 1.76A. Find the Miller indices of the set of planes.

Solution:

(h2 + k2 + l2)1/2 =ad=

4.311.76

= 2.45

(h2 + k2 + l2) = (2.45)2 = 6

Sum of the squares of three integers is 6. Hence, the Miller indices of theplanes are (211). Since it is cubic system, (211), (121), and(112) are identi-cal.

2.3 Find the Miller indices of a set of parallel planes which makes intercepts in theratio 3a : 4b on the x and y axis , and are parallel to the z axis wherea, b andcare the primitive translation vectors of the lattice.


Solution:

Intercepts as multiples of 3a 4b ∞lattice parameter

Devide by lattice parameter 3 4 ∞Take reciprocals 1/3 1/4 1/∞Reduce to smallest set 4 3 0of integers

Miller indices of the given set of planes are(430) (Ans.).2.4 Show that in a cubic crystal, the interplanar spacing of the (300) set of planes is

equal to the inter planar spacing of the (221) set of planes.Solution:Interplanar spacing is given by

d = a/(h2 + k2 + l2)1/2

d300 = a/(32 + 0+ 0)1/2 = a/3

d221 = a/(22 + 22 + 12)1/2 = a/3

Hence the interplanar spacing of the set of planes (300) and (221) are equal.2.5 A cubic crystal of lattice parameter 5.60A has a set of planes with interplanar

spacing of 1.77A. Find the Miller indices of the set of planes.Solution: We have

d = a/(h2 + k2 + l2)1/2

Or (h2 + k2 + l2)1/2 = a/d

= 5.60/1.77

= 3.16≈ 3

Hence, (h2 + k2 + l2) is equal to 9. The possible values of miller indices areeither (300) or (221).

2.6 In an x-ray tube, an accelerating voltage of 35 kV is applied between the filamentand the target. Calculate the minimum wavelength of continuous x-ray emitted.

Solution: The minimum wavelength of x-rays emitted is given by

λmin =hceV

6.62× 10−34 × 3× 108

1.6× 10−19× 35× 103

= 3.55× 10−11m (Ans.).


2.7 First order Bragg diffraction of x-rays of wavelength 1.54A was observed at anangle of 12. Calculate the interplanar spacing for the set of planes responsiblefor the diffraction. If the crystal has a lattice parameter of 6.41A, find the planesgiving rise to diffraction.

Solution:d =

nλ2 sinθ

=1× 1.54

2× sin 12

d = a/(h2 + k2 + l2)1/2

∴ (h2 + k2 + l2)1/2 = a/d = 6.41/3.70= 1.73

∴ (h2 + k2 + l2) = (1.73)2 = 3.

Hence, the set of planes are (111) (Ans.).2.8 An x-ray diffraction pattern for cubic lead crystal is obtained with x-rays of wave-

length 1.54A. If the first order reflection from (110) planes is observed at12.7,determine the interplanar spacing for (110) planes and alsothe lattice parameterof lead.

Solution:Interplanar spacingd =

nλ2 sinθ

=1× 1.54

2× sin 12.7= 3.50A (Ans.)

Lattice parametera = d(h2 + k2 + l2)1/2 = 3.50×√

2

= 4.95A. (Ans.).

2.9 The Kα line of x-ray spectrum from a molybdenum target (z = 42) has a wave-length of 0.71A. Calculate the wavelength ofKα line when a copper target(z= 29) is used.

Solution:ν =

3cR4(z− 1)2

whereR is the Rydberg constant.

λ(cu)λ(Mo)

=(42− 1)2

(29− 1)2

∴ λ(Cu) = 0.71(41)2

(28)2

= 1.52A (Ans.).

2.10Calculate the minimum accelerating potential to be appliedto an x-ray tube inorder to produce continuous x-ray emission down to a wavelength of 1Acirc.


Solution: We haveλmin = hc/eV

V = hc/eλmin

=6.62× 10−34× 3× 108

1.6× 10−19 × 1× 10−10= 12412.5V (Ans.).

2.11In an x-ray diffraction experiment, the first order diffraction from a particular setof planes was observed at 12. Find the angle at which the second order diffractionoccurs for the same set of planes.

Solution: nλ = 2dsinθ

For n = 1, sinθ1 = λ/2d = sin 120

For n = 2, sinθ2 = 2λ/2d = 2 sin 120

= 2× 0.2079

= 0.4158

θ2 = sin−1(0.4158)= 24.570 (Ans.).

2.12The Bragg diffracted beam of x-rays from a set of planes is observed at an angleof 15 when a molybdenum target (λ = 0.71A) is used. Find the position of thediffracted beam for the same set of planes when a copper target (λ = 1.54A) isused.

Solution: We havenλ = 2dsinθ

Hence,nλ1 = 2dsinθ1 andnλ2 = 2dsinθ2

sinθ2 = λ2 sinθ1/λ1

= 1.54× sin 150/0.71

= 0.5614

θ2 = sin−1(0.5614)= 34.150 (Ans.).

2.13The wavelength ofKα lines from a copper target is found to be 1.54A. Find thewavelength ofLα line from the same target.

Solution: We know that

1α= R(z− b)2

(

1

n21

− 1

n22

)

For Kα line, b = 1, n1 = 1 andn2 = 2


For Lα line, b = 7.4, n1 = 2 andn2 = 3Substituting and taking the ratio ofλKα andλLα

λLα/λKα = 9.074

Hence,λLα = 9.074× 1.54= 13.97× 10−10m (Ans.).Exercise

2.1 State and explain Moseley’s law, and explain on of its important applications.

(March 1999).

2.2 What is a space lattice ? Describe briefly the seven systems ofcrystals.

(March 1999).

2.3 Describe Bragg’s law for x-ray diffraction in crystals. Describe how Bragg spec-trometer is used for determination of wavelength of x-rays. (August 1999).

2.4 Explain Miller indices and indicate the following planes ina cubic unit cell:(110), (101) and (111). (August 2000).

2.5 Explain Moseley’s law and mention its applications. (August 2000).

2.6 Derive Bragg’s law for x-ray diffraction. Describe the determination of wave-length of x-rays using Bragg spectrometer. (August 2000).

2.7 Explain Moseley’s law and its importance. (March 2001).

2.8 If the lattice parameter of a cubic crystal is 3 A.U.,find the interplanar spacingbetween (111) planes. (March 2001)

2.9 Derive Bragg?s law of x-ray diffraction. Explain how Bragg spectrometer is usedin the verification of the law. (March 2001).

2.10 Define unit cell and Bravais lattice. Derive the expression for the interplanarspacing in terms of Miller indices. The interplanar spacingof (110) planes is 2A.U. for a cubic crystal. Find out the atomic radius. (August2001).

2.11 State Moseley’s law. Discuss its applications. Calculate the energy of electronsthat produce Bragg diffraction of first order at an angle of 220 when incident oncrystal planes with interplanar spacing of 1.8 A.U. (August2001).


2.12 What are Miller indices? How do you find the Miller indices of agiven plane?

(March 2002).

2.13 Derive Bragg’s law. (July 2005)

2.14 Explain the structure of NaCl and calculate the packing factor for a bcc structure.

(July 2005)

2.15 Obtain the expression for interplanar distance in terms of Miller indices.

(July 2005)

2.16 Sketch the following planes in a cubic unit cell : (101), (121), (010). (July2005)

Chapter 3

Electrical Conductivity In Metals

3.1 Introduction

Metals have been known for their excellent ability to conduct heat and electricity. Sucha high conductivity is attributed to the presence of free electrons in metals and attemptshave been made to explain the conductivity of metals quantitatively. Earlier theorieswere classical in nature. In the year 1900, Paul Drude proposed a theory of conductiv-ity of metals based on the free electron model which was laterdeveloped and refinedby Hendrik Lorentz. According to the classical free electron theory, a metal can beconsidered to be made up of ion cores comprising of nucleus and inner electrons ex-cluding the valence electrons. These ion cores are immobileand do not take part inthe conduction process. The valence electrons are considered to be free and almostindependent of the ions to which they belong. These valence electrons can move underthe influence of an applied electric field and hence contribute to the conductivity.

Before discussing the classical free electron theory, let us examine the Ohm’s lawwhich is the most fundamental of the laws governing the charge transport in materials.

Consider a metal sample to which a voltage source is connected. This results in theflow of an electric current through the sample. The magnitudeof the currentI is foundto be proportional to the applied voltageV and the relation is expressed as

I ∝ V or I = GV (3.1)

whereG is a constant of proportionality called the conductance of the sample. Therelation may also be expressed in terms of resistanceRof the sample as

V = IR whereR= 1/G (3.2)

This relation is known asOhm’s law. The quantities involved in this relation aresample specific and depend on the dimensions of the sample. A more general form ofOhm’s law is

J = σE (3.3)

whereJ is the current density representing the charge moving past areference point perunit area of cross section of the sample per unit time,E is the Electric field equivalentto the voltage per unit length of the sample andσ is the electrical conductivity of the

89

90 Electrical Conductivity In Metals

material. The electrical resistivity is the reciprocal of electrical conductivity and isdefined as the electrical resistance per unit area per unit length of the sample and isgiven by the relation

ρ =1σ=

RAl

(3.4)

whereA is the area of cross section andl is the length of the sample. The advantageof using Ohm’s law in its general form is that the quantities involved are not specificto the sample and are independent of sample dimensions. The electrical resistivity isa constant for the material and may be used to specify the relative ease with which anelectric current passes in metals.

3.2 Classical Free Electron Theory Of Metals(Drude - Lorentz Theory)

The classical free electron theory of electrical conductivity of metals proposed byDrude and Lorentz is based on the following assumptions:

1. Atoms in a metal are considered to be made up of ion cores which are nucleisurrounded by inner electrons excluding the valance electrons and the valanceelectrons which are free to move anywhere inside the sample.

2. The effect of electric potential due to positive ion cores on the electrons is con-sidered to be constant and hence neglected.

3. The electrostatic repulsion between the electrons is neglected.

4. The electrons in the metal are considered equivalent to molecules in a gas and thegas laws are made applicable to the electrons in a metal. Accordingly, the distri-bution of energy and velocity is assumed to follow Maxwell-Boltzman statistics.

3.2.1 Expression for electrical conductivity

Consider a metal sample of electrical conductivityσ subjected to an electric fieldE.The effect of this field is to exert a force equal to−eE on each free electron in themetal. As a result of this force, the electrondeltawill be accelerated. The accelerationproduced is given by

ax = −(e/m)E (3.5)

assuming that the electric field is acting along x-axis andm is the mass of the elec-tron. If vx represents the velocity component along x-axis, we may write the above

Electrical Conductivity In Metals 91

equation as[

δvx

δt

]

f ield= −(e/m)E (3.6)

The partial time derivative of velocity and the subscript ‘field’ indicate that the accel-eration mentioned here is due to the effect of the applied electric field. Contributionto the acceleration due to other possible effects will be considered later. Considering alarge group of electrons, the average velocity at any instant may be written as

< vx >=1N

N∑

i=1

vxi (3.7)

whereN is the number of electrons considered. Hence, we may write equation (3.6) as[

δ < vx >

δt

]

f ield= −(e/m)E (3.8)

Considering unit area of cross section of the sample and counting the total number ofcarriers crossing the unit area along the x-axis as shown in Fig. 3.1 in unit time, wefind that all the carriers present in an elemental volume of unit area of cross-sectionand upto a length equal tovx will cross the reference point at origin.

1

1

vx

X

Y

Z

Figure3.1Schematic representation of electron flow for calculation of current density.

In other words, all the electrons present in the elemental volumevx will contributeto the current density. If there aren conduction electrons per unit volume in the givensample, the number of electrons contributing to current density is equal tonvx. Sinceeach electron is associated with a charge (−e), the current density at any instant maybe written as

J = −ne< vx > (3.9)


According to Ohm’s law, we have

J = σE

This means that as long as the applied field is held constant, the current density willremain constant.

i.e.,dJdt= 0

Substituting this in equation (3.9), we get

dJdt=

ne.d < vx >

dt= 0

ord < vx >

dt= 0 (3.10)

Comparing equation (3.8) with equation (3.10), since the time invariance of currentdensity is an experimentally confirmed fact, we conclude that equation (3.8) does notgive the total time rate change of velocity and that there must be other processes whichmake the total time rate change of velocity equal to zero.

i.e.,d < vx >

dt=

[

δ < vx >

δt

]

f ield+

[

δ < vx >

δt

]

other= 0 (3.11)

The other process responsible for the time rate change of velocity of conduction elec-trons has been identified aselectron- lattice interaction. Since the electrons are mov-ing in the crystal lattice of ion cores, such an interaction is a possibility. An experi-mental observation that a sample tends to warm up when an electric current is passedindicates possible transfer of energy (momentum) from electrons to the lattice.

In order to understand the nature of lattice - electron interaction, let us consider asample subjected to an electric fieldE resulting in a current densityJ and an associatedaverage electron velocity< vx >. If the applied electric field is switched off at anyinstant, sayt = 0, the electron flow will stop indicating that the velocity< vx > goes tozero. This decrease in the velocity is attributed to the interaction of electrons with thelattice. The average velocity is assumed to decay exponentially with time (Fig.3.2) inaccordance with the equation

< vx >t=< vx >o exp (−t/τ) (3.12)


< vx >

Figure 3.2 Exponential decay of the average velocity of electrons whenthe appliedfield is removed.

where< vx >t is the instantaneous value of average velocity at timet, < vx >o isthe average velocity when the electric field is turned off. The quantityτ is called therelaxation time of conduction electrons. Differentiating equation (3.12)

[

δ < vx >

δt

]

coll= −< vx >

τ(3.13)

In other words, when the applied electric field is turned off, the velocity componentwill reduce to zero due to electron-lattice collisions and the second component of ac-celeration mentioned as due to other processes in equation (3.11) is now identified asdue to collisions. Substituting for the two components of acceleration from equations(3.6) and (3.13), equation (3.11) becomes

d < vx >

dt= −eE

m− < vx >

τ= 0

or < vx > = −eτEm

(3.14)

The steady state average velocity of electrons in presence of an applied electric field iscalled thedrift velocity of the carriers. This is proportional to the applied electric fieldand the constant of proportionality is called themobility of the carriers. Hence, themobility of carriers is defined as the drift velocity per unitapplied field and is given by

µ =eτm

(3.15)


The current densityJ can be calculated from equation (3.9) as

J = −ne< vx >=ne2τE

m(3.16)

Comparing equation (3.16) with equation (3.3), we have the electrical conductivity ofa metal given by the expression

σ =ne2τ

m= neµ (3.17)

This expression for the conductivity of a metal predicts thedependence of conductivityon the total number of valence electrons per unit volume and their mobility.

Drude made use of the kinetic theory of gases to evaluateτ. Considering electronsas equivalent to molecules in a gas, he evaluated the kineticenergy of the electrons as

12

mv2 =

32

kT

wherev is the root mean square velocity of electrons, which can be calculated as

v = (3kT/m)1/2 (3.18)

At room temperature, the thermal velocity of electrons calculated using the above equa-tion is 1.15× 105ms−1. The drift velocity acquired by the electron due to an appliedfield is much smaller than the thermal velocity. Hence, the relaxation time may bewritten in terms of the mean free path as

τ = λ/v = λ(m/3kT)1/2

Hence, by substituting this value ofτ in equation (3.17), the conductivity of a metalmay be expressed as

σ =ne2τ

m=

ne2λ

(3mkT)1/2(3.19)

This equation suggests that the conductivity of a metal mustbe inversely proportionalto the square root of temperature. This is in contrast to the experimental observationthat the conductivity is inversely proportional to the temperature. This discrepancyis due to the fact thatλ is assumed to be independent of temperature. In fact,λ is afunction of temperature which is not explained or accountedfor by the classical freeelectron theory.


3.2.2 Electron - lattice interaction and consequences

In order to account for the validity of Ohm’s law in the case ofmetals, we have in-troduced the concept of electron-lattice interaction. Letus now try to understand thenature of such an interaction and its consequences.

vx

vx

x

θ

Figure3.3Schematic representation of scattering of an electron by anobstacle.

Suppose an electron moving with a velocityvx along x− direction collides withan obstacle (like an ion core) and gets scattered at an angleθ (Fig.3.3). We shallassume that the magnitude of velocity is same before and after the collision and onlythe direction is changed. (This is not true and, in fact, someenergy is transferred tothe lattice which results in Joule heating of the lattice. However, in a good conductor,this energy is quite small and we may assume the velocity before and after collisionto be same.) After scattering, the electron will have a velocity equal tovx cosθ alongx- direction. Hence, along x- direction, the electron has lost a velocity componentequal tovx(1 − cosθ). If we consider a group of electrons, all moving with the samevelocity vx but scattered in different directions, the average velocity component lostalong x- direction will bevx(1− < cosθ >) where< cosθ > is the average valueof the cosine of the scattering angle. If we consider the electrons to have differentinitial velocities before collision, then, the average velocity component lost for allthese electrons will be< vx > (1− < cosθ >). Let us introduce a termmean collisiontime, τc, which represents the average time between successive collisions. Then, thenumber of collisions per second on the average will be (1/τc). Thus,

average change in velocity per collision=< vx > (1− < cosθ >)

Number of collisions per second= 1/τc

Total change in velocity per second=< vx > (1− < cosθ >)

τc

This change in velocity is negative since the velocity decreases with each collision and


hence,[

δ < vx >

δt

]

coll= −< vx > (1− < cosθ >)

τc(3.20)

Comparing this equation with equation (3.13), we see that the mean collision time isrelated to the relaxation time as per the equation

τ =τc

(1− < cosθ >)(3.21)

When the scattering is isotropic, i.e., without any preferred directions,< cosθ >= 0and

τ = τc

If the scattering is predominantly along the direction of carrier flow, as is usually thecase,< cosθ > will have a positive value and

τ > τc

If v is the total velocity of an electron, we can write the averagedistance travelledbetween successive collisions as

λ = vτc (3.22)

whereλ is called themean free pathof the electron.

3.2.3 Failure of classical free electron theory

1. The classical free electron theory assumes that the valence electrons in a metal,by virtue of being the outermost electrons in the atoms, are almost free and hencetake part in the conduction process. Hence, divalent and trivalent metals, with alarger concentration of valence electrons should possess much higher electricalconductivity than monovalent metals. This is contrary to the experimental ob-servation that silver and copper are more conducting than zinc and aluminium.

2. Assuming that the kinetic energy associated with the electrons to be equal to thethermal energy, Drude obtained the expression for the conductivity as given bythe equation

σ =ne2τ

m=

ne2λ

(3mkT)1/2(3.19)

This equation suggests a temperature dependence of conductivity as

σ ∝ T−1/2


This is once again in contradiction to the observed dependence of conductivityon temperature. The electrical conductivity of metals at normal temperatures hasbeen found to be inversely proportional to the absolute temperature.

3. The mean velocity of electrons can be computed from equation (3.18) and themean collision time from equation (3.17) assuming isotropic scattering. A sub-stitution of these values in equation (3.22) gives a value ofmean free path forthese electrons. If we assume that the ion cores are responsible for the scatteringof electrons, the mean free path must be of the order of atomicspacing in thesolid. But the calculated value of the mean free path is much higher than theatomic spacing in the metal.

It may be mentioned that there are many physical and chemicalproperties whichdepend strongly on the behavior of electrons in materials. Any theory can be consid-ered to be successful if it can explain all these properties satisfactorily. For example,the molar specific heat of a gas at constant volume is given by

Cv = (3/2)R

If we consider the electrons in a metal to behave similar to gas molecules, theabove equation should hold good for metals as well. However,it is observed that theexperimental value of specific heat of metals is much lower than the value predictedby the above equation. Further, specific heat of metals is found to be a function oftemperature which is not reflected in the above equation.

These observations, along with many others, point to the deficiency of the classicalfree electron theory in explaining the electrical conductivity and other properties ofmetals. The quantum free electron theory is an attempt to modify the classical theoryto account for these discrepancies.

3.3 Quantum free electron theory of metals

According to the kinetic theory of gases, a gas is assumed to consist of a large numberof rigid and perfectly elastic particles, namely molecules. These molecules are all ofsame mass and are moving freely inside the container. Duringtheir random motion,they collide with each other and acquire all possible velocities. At a given temperature,in a state of thermal equilibrium, the distribution of molecular velocities and energyamong the molecules of a gas is given by Maxwell-Boltzman statistics. This statisticscan be applied to a system of identical particles that can be distinguished from oneanother. This means that the wavefunctions associated withthe particles do not overlap


considerably. The distribution function that denotes the fraction of the total number ofmolecules possessing an energyE is given by,

FMB(E) =1

A exp (E/kT)

whereA is a factor dependent on the density of the gas and temperature.Quantum statistics was first formulated by Bose in an attemptto derive the Plank’s

radiation law from statistical methods. He regarded radiation to be composed of lightquanta with energyhν whereν is the frequency of the radiation. The light quanta areindistinguishable from each other and there is no restriction on the number of quantahaving the same energy. In other words, Pauli’s exclusion principle is not applicableto them. Particles that do not obey Pauli’s exclusion principle are those having inte-gral spin like photons and phonons. They are called ‘Bosons’. The wavefunctions ofBosons overlap considerably and hence the particles are indistinguishable. The wave-function of a system of Bosons is not affected by the exchange of any pair of them.A wavefunction of this kind is called asymmetric wavefunction. For example, con-sider a system of two particles 1 and 2 in the states ‘a’ and ‘b’respectively. Whenthe particles are distinguishable, there are two possibilities for the occupancy of states,described by the wavefunctions

ΨI = Ψa(1)Ψb(2)

and ΨII = Ψa(2)Ψb(1)

The wavefunctions of Bosons overlap considerably and hencethe particles are indis-tinguishable. We cannot tell which of them is in which state.The wavefunction of asystem of Bosons is not affected by the exchange of any pair of them and hence, mustbe a combination ofΨI andΨII . A wavefunction of this kind is called asymmetricwavefunction. The system is described by the wavefunction

Ψ = (1/√

2) [Ψa(1)Ψb(2)+ Ψa(2)Ψb(1)]

Because the particles are identical, the probability density |Ψ|2 of the system willremain same even after the particles are interchanged.

The distribution function for Bosons, which represents theprobability of findingthat Boson in an energy stateE at a given temperatureT, is known as Bose-Einsteindistribution function and is given by

FBE(E) =1

A exp (E/kT) − 1


We may recall that this was the function used in Planck’s theory of radiation for ex-plaining the black body radiation.

Later, Fermi and Dirac found it necessary to modify the Bose-Einstein statisticsfor the case of particles that obey Pauli’s exclusion principle. These are particles hav-ing half integral spin like electrons and nucleons. They arecalled ‘Fermions’. Thewavefunctions of Fermions also overlap considerably and hence the particles are in-distinguishable. However, the wavefunction of a system of Fermions is affected by theexchange of any pair of them. A wavefunction of this kind is called aantisymmetricwavefunction. The system is described by the wavefunction

Ψ = (1/√

2) [Ψa(1)Ψb(2)−Ψa(2)Ψb(1)]

However, since the particles are identical, the probability density|Ψ|2 of the system willremain same even after the particles are interchanged. The distribution function forfermions is known as Fermi-Dirac distribution function or ‘Fermi factor’ and representthe probability of finding a Fermion in an energy stateE at a given temperatureT. Thisis given by

FFD(E) = F(E) =1

A exp (E/kT) + 1

The classical free electron theory, based on Maxwell-Boltzman statistics, failed toexplain the electrical conductivity of metals quantitatively. The theory was modifiedby Sommerfeld using some quantum mechanical principles. Inthe classical picture, allthe valence electrons forming the electron gas were permitted to have identical energyvalues and all the electrons were considered to contribute to the conduction process.In the quantum free electron theory, the quantization of electron energy levels and theapplicability of Pauli’s exclusion principle have been incorporated. Consequently, theMaxwell - Boltzman distribution can no longer be used to describe the distribution ofelectron energies. The electron energy levels are quantized and are occupied in accor-dance with the Pauli’s exclusion principle. Drawing similarities between electrons in ametal and electron inside a potential well of infinite depth,the energy values permittedfor the electrons are given by the relation,

En =h2n2

8ma2

wheren is an integer indicating the different energy states permitted and ‘a’ is theatomic spacing in the metal lattice. According to the Pauli’s exclusion principle, onlytwo electrons may occupy any orbital state. Hence, two electrons will occupy theground state, two will go into the next higher energy state and so on till all the elec-trons in the metal are accommodated in their lowest energy state. Since the number


of electrons per unit volume of the sample is very large, we can expect some elec-trons to have relatively large kinetic energy even at ordinary temperatures. In order tomake a theoretical estimation of the electrical conductivity of a metal, it is necessaryto know the distribution of electrons in various energy levels and also a knowledge ofthe number of electrons actually contributing to the conductivity.

3.3.1 Density of energy states in a metal

The electrical conductivity of a metal depends on the total number of electrons con-tributing to conductivity. This can be evaluated with the knowledge of the total numberof energy states available for occupation at different energy levels and their occupation.The number of allowed energy states per unit energy intervalat a particular energyvalue is called thedensity of energy states. We will now proceed to obtain an expres-sion for the same.

∆x

∆y

∆x

X

Z

Y

(a)

∆Px

∆Py

∆Pz

Px

Pz

Py

PM

P

(b)

Figure 3.4 (a) An elemental cell of metal having dimensions∆x = ∆y = ∆z = 1. (b)Momentum cell having dimensions∆px = ∆py = ∆pz = h.

Consider a cube of a metal having unit sides in three coordinate directions as shownin Fig.3.4 (a). The electrons are free to move within the limits of the unit cube. Thus,an electron cannot be located more accurately than the uncertainty in its position given


byx = y = z= 1

If px, py andpz represent the components of momentum in the three directions, usingHeisenberg’s uncertainty principle,

px = py = pz = h

Figure 3.4 (b) shows the momentum diagram where the three axes indicate the compo-nents of momentum along the three directions. An electron with a momentump willlie inside a cell of sideh and volumeh3 at the end of the momentum vectorp. Thiscell is known as unit momentum cell.

Pauli’s exclusion principle states that no two electrons can have the same set off ourquantum numbers in an atom. Because of the close interactionamong the electrons, theexclusion principle applies in the metal crystal as well. Since the spin quantum numbercan take only two values,+1/2 and -1/2, the Pauli’s exclusion principle leads to theconclusion that only two electrons can have the same three quantum numbers. Sincethese three quantum numbers determine the momentum of the particle, the principle isapplicable to the momentum diagram also. Thus, each unit momentum cell accountsfor two electrons.

Let M represent the number of free electrons available in the material. At 0K,since there is no thermal energy available, these electronsoccupy their lowest energystates( and hence the lowest momentum states). These electrons are accommodated in(M/2) momentum cells which are arranged spherically in the momentum space. LetpM represent the radius of the sphere upto whichM electrons will be distributed in(M/2) momentum cells. Since the volume of each cell ish3, the total volume occupiedby (M/2) cells is given by

M2· h3 =

4πp3M

3

or M =8πp3

M

3h3(3.23)

It is possible to relate the maximum momentumpM with the corresponding energyE.

E =12

mv2 =p2

M

2mor pM = (2mE)1/2 (3.24)

Substituting the value ofpM from equation 3.24 in equation 3.23, we get,

M =8π3h3

(2mE)3/2


=29/2πm3/2E3/2

3h3(3.25)

If M is increased by a small valuedM, the radius of the sphere will increase frompM

to (pM + dpM) and the highest energy level fromE to (E + dE). By definition, thedensity of energy statesg(E) is given by

g(E) =dMDE

=27/2πm3/2E1/2

h3(3.26)

E

g(E)

Figure3.5Energy density of states as a function of energy.

Figure 3.5 shows the variation ofg(E) with energyE. The actual number of elec-trons present in an energy rangedE around any energyE can be calculated as

N(E)dE = g(E) · F(E)dE

whereF(E) is the Fermi factor representing the probability of occupation of the energystateE.


3.3.2 Metal as a Fermi gas

The distribution of electron energies in a metal can be described by Fermi-Dirac statis-tics. The fraction of the total number of electrons having anenergyE at any tempera-ture is given by Fermi-Dirac distribution function or ‘Fermi factor’. This is given by

FFD(E) = F(E) =1

A exp (E/kT) + 1

The coefficientA is a function of temperature and may be expressed as

A = exp (−EF/kT)

whereEF is a characteristic energy called ‘Fermi energy’. Hence, wehave

F(E) =1

1+ exp [(E − EF)/kT](3.27)

Fermi factor represents the probability that a particular energy levelE is occupiedat a temperatureT. At absolute zero, we find the probability function has a value equalto 1 for all energies less thanEF and a value equal to zero for all energies higher thanEF. Thus,EF serves as a reference level and is called the Fermi energy. Fermi energycan be defined as that energy at which the probability of occupation is half. This can beshown to be valid at all temperatures. The distribution of electrons at different energylevels is shown in Fig. 3.6.

E

T = OK

T1

T2

EF

N

Figure3.6Distribution of electrons as a function of energy at different temperatures ina metal.


At O K, all the energy levels uptoEF are completely filled and all the energy levelsaboveE f are empty. Thus, Fermi energy can also be defined as the energyof the topmost filled level at 0 K. An increase in temperature results ina few electrons acquiringenergy greater thanEF. However, the change is prominent only at energy values aroundEF. Thus, only a small fraction of valence electrons lying at energy levels close toEF

will contribute to the electrical conductivity. Since the thermal energykT at any finitetemperature is very small compared to the Fermi energy, the effect of temperature onthe number of electrons contributing to conductivity is negligible. The velocity ofconduction electrons can be calculated as

12

mv2F = EF or vF = (2EF/m)1/2 (3.28)

It may be mentioned that the relaxation time of conduction electrons refers to thevalue calculated for electrons with velocityvF. Hence, the electrical conductivity givenby the equation (3.17) may be expressed as

σ = ne f fe2τ

m(3.29)

wherene f f refers to the number of valence electrons at the Fermi energyand with avelocityvF. The above equation may also be obtained by considering the movement ofelectron in a metal to be equivalent to the propagation of a wave in a uniform potentialresulting from the ion cores. Thus quantum free electron theory is capable of explain-ing a few experimental observations which could not be explained on the basis of theclassical theory. The higher electrical conductivity of monovalent noble metals overthat of some bivalent and trivalent metals can be explained on the basis of the effectivenumber of electrons contributing to conductivity. Further, since the relaxation time andthe mean free path of the conduction electrons are inverselyrelated to the temperature,it follows that the conductivity varies inversely with temperature at high temperatures.However, the large values of mean free path, of the order of hundreds of atomic dis-tances, cannot be explained on the basis of quantum free electron theory. The theoryalso fails to explain other experimental observations likepositive Hall coefficient forsome metals and the temperature dependence of conductivityof semiconductors andinsulators.

3.3.3 Band theory of metals

The quantum free electron theory of metals assumes that a conduction electron is sub-jected to a constant potential due to the ion cores comprising of nucleus and inner


electrons. This amounts to the conclusion that the ion coreshave no influence on themovement of electrons that are free to move within the sample. The band theory treatsthe metal as an infinite array of positively charged ions through which the conductionelectrons move. The potential energy due to coulombic attraction between the electronand the positive ion may be represented by periodic square well type potential. Then,we observe that the solution of the Schrodinger’s wave equation leads to the followingconclusions:

The motion of electrons in the metal lattice is restricted bytwo factors. The firstfactor is the restriction on the energy of the electrons. Thedistribution of energy isdetermined by the Fermi-Dirac statistics which predicts discrete energy levels for theelectrons. The second factor is the restriction on the energy possessed by the electronsas a result of the formation of allowed and forbidden energy bands.

Thus, the quantum theory leads us to the Band theory of solids, which is used veryeffectively to explain various properties of metals such as electrical conduction, heatcapacity and heat conduction.

3.3.4 Merits of quantum free electron theory

(i) The discrepancy in the electrical conductivity of metals of different valencescould not be explained by the classical free electron theory. Quantum mechan-ically, it is observed that all valence electrons are not conduction electrons andonly those valence electrons having an energy equal to or greater than the fermienergy will take part in conduction. On this basis, it is observed that aluminium,in spite of having three electrons per atom, have a fewer electrons available forconduction as compared to silver and hence a lower electrical conductivity.

(ii) The temperature dependence of electrical conductivity of metals as explained bythe classical free electron theory is based on the expression,

σ =ne2λ

(3mkT)1/2(3.19)

This expression indicates that the conductivity is inversely proportional to thesquare root of absolute temperature. However, this conclusion assumes that themean free path of the conduction electrons is a constant and independent of tem-perature. Quantum theory is capable of explaining the temperature dependenceby taking into account the temperature dependence of mean free path of the elec-trons. According to the quantum free electron theory, the conductivity can beshown to be inversely proportional to temperature which is in agreement with theexperimental observations.


(iii) The large value of mean free path experimentally observed is not explained byclassical free electron theory. The quantum free electron theory, with the helpof the band model, assumes that the movement of an electron inthe metal lat-tice is equivalent to the propagation of an electron wave in aperiodic potential.The scattering of electron is equivalent to the modificationof the amplitude ofthis electron wave due to the deviation from the periodic potential at the sites ofimpurities and imperfections. This explains the large value of mean free path ofconduction electrons. Further, the relation between the concentration of defectsor impurities and the mean free path has been experimentallyestablished.

In addition, many other physical properties associated with electrons in metals canbe successfully explained on the basis of quantum theory.

3.4 Electron Scattering Mechanisms

The large values of the mean free path of conduction electrons of the order of hundredsof atomic distances have been computed for the case of many metals at room temper-ature. The quantum free electron theory tries to explain thesituation by attributingwave nature to the conduction electrons moving in a periodicpotential of the lattice.The obstacles responsible for the scattering of the electron wave come in the form ofdeviations from the periodicity of the lattice. The deviations from periodicity arisesdue to the impurities and imperfections. Thus, the scattering of conduction electronsmay occur due to the following reasons:

(a) vacancies in the lattice,

(b) atoms occupying “wrong” positions in the lattice,

(c) lattice atoms replaced by impurity atoms

(d) thermal vibrations of atoms about their mean position, etc.

These various kinds of obstacles have to be considered in thescattering of electrons,each of them giving rise to certain relaxation time separately. The probability forscattering by a particular kind of obstacle for unit time is equal to the reciprocal of therelaxation time associated with the obstacle and the total probability of scattering of anelectron will be the sum of the individual probabilities. Hence, we can write

1/τ = 1/τi + 1/τT (3.30)


whereτi andτT are the effective relaxation times due to temperature independent andtemperature dependent scattering events. The resistivityof the metal may be writtenas

ρ =m

ne2(1/τi + 1/τT) (3.31)

or ρ = ρi + ρT (3.32)

whereρi is called theresidual resistivity which is independent of temperature andis caused by defects and impurities whose characteristics are independent of temper-ature. The partρT is the temperature dependent resistivity with contributions fromthermal vibration of the lattice and thermal defects. This part of resistivity is calledthermal resistivity or ideal resistivity. The equation (3.31) and (3.32) are referred toasMatthiessen’s rule.

3.4.1 Effect of temperature

It is well known that the electrical resistivity of metals increases with increase in tem-perature. Fig. 3.7 shows the variation of resistivity with temperature. The curve maybe divided into three regions.

T/θD

ρ

Figure3.7Variation of resistivity of a metal with temperature

Region I: This region correspond to very low temperature, i.e., close to absolute zero.In this region of temperature, the contribution from the thermal vibrations will be verysmall.

i.e., ρT ≪ ρi


Hence, the thermal resistivity may be neglected in comparison with the residual resis-tivity. The total resistivity will be almost equal to the residual resistivity and almostindependent of temperature.Region II: In the intermediate temperature region, resistivity is a complex function oftemperature. The resistivity may be expressed as

ρ ∝ (T/θD)n (3.33)

whereθD is the Debye temperatureand the exponentn can have values between5 and 1. (The Debye temperatureθD corresponds to the temperature at which thefrequency of vibrationνD of the lattice has a maximum value given by the relationhνD = kθD. Above Debye temperature, all the modes of vibration are excited.) Thevalue ofn is equal to 5 at the low temperature end of the region and equalto 1 at thehigh temperature end.Region III : This region corresponds to temperature above the Debye temperature. Inthis region, the value ofn is 1 and hence, resistivity may be expressed as

ρ = ρi + aT (3.34)

whereρi is the residual resistivity anda is a constant.This is because the thermal resistivity will be linearly increasing with temperature

for temperature above the Debye temperature and the contribution from the residualresistivity will be very small.

i.e., ρT ≫ ρi

Most of the common metals have their Debye temperature closeto room tempera-ture. Neglecting the contribution from the residual resistivity, the temperature depen-dence of resistivity may be expressed as

ρ ∝ (T/θD)

This indicates that the resistivity increases linearly with temperature in the regionabove the Debye temperature.

3.4.2 Effect of impurities

The effect of impurities on the electrical resistivity of metals may be demonstrated bythe following experiment. Small amounts of impurity likeNi is added gradually to ametal likeCu and the electrical resistivity is monitored (Fig. 3.8).


%Ni

ρ

0 1 2 3

Figure3.8Effect of nickel impurity concentration on the electrical resistivity of copper.

It is found that the resistivity increases with increase in the concentration of im-purity added. Now, alternatively, small amount ofCu is added to the metalNi. Itis observed that the resistivity increases irrespective ofthe electrical resistivity of theimpurity added. This is due to the fact that the impurity atoms destroy the periodicityof the lattice and thus act as scattering centres. The changein the resistivity due toimpurity is related to the concentration of impurity according to the relation

ρi ∝ x(1− x) (3.35)

whereρi is the resistivity due to impurities andx is fraction of impurity atoms inthe metal. This is known asNordheim’s rule and is valid in most cases. For smallconcentration of impurities, we can write

ρi ∝ x (3.36)

In the case of alloys with higher concentration of the impurity the resistivity versusconcentration curve shows a peak. An example ofCu−Ausystem is shown in Fig. 3.9.TheCu−Aualloy has a maximum resistivity at 50% concentration of eachcomponent.This is called the disordered phase of the alloy. If the atomsin the alloy are ordered (byannealing at elevated temperature) the resistivity drops at finite composition indicatingthe formation of an ordered phase. In an ordered phase, the lattice is again periodic andhence the resistivity reduces. This is in fact a proof for theassumption that impuritiesenhance scattering probability.


0 50 100

ρ

Cu3Au

CuAu

% Au

Figure 3.9 Electrical resistivity of disordered and ordered Cu− Au alloys. Orderedphases like Cu3Au and CuAu appear on annealing.

3.5 Thermionic emission

In any metal, at any temperature above 0 K, the energy states at the top of the bandare not completely full. A few electrons close to the Fermi level will find themselvesat energy levels above the Fermi energy and are free to move about within the metal(fig 3.6). But these electrons are not free enough to move out of the metal. For a givenmetal, we can define an energy levelEs, above the Fermi level, at which the electronwill be completely free from the metal. This energy levelEs is called thevacuumlevel, which represents the energy of an electron at rest outside the metal (Fig.3.10).

EF

ES

Φ

Figure 3.10 Energy level diagram for a metal surface showing Fermi level, vacuumlevel and barrier energy for electron emission.


The difference in the energy levelsEF andEs is known as thework function φ of thematerial. It is defined as the energy required to raise an electron from the Fermi levelto the vacuum level of the metal and is characteristic of the metal.

As the temperature of the metal is raised, the distribution of electrons in variouselectron energy levels near the Fermi energy is modified (Fig.3.11).

E

T = OK

T > OK

EF

ES

N

Figure3.11Distribution of electrons in energy at any temperature T. Electrons havingan energy higher than Es should be thermionically emitted.

A few electrons will be found above the Fermi level in accordance with the Fermi- Dirac distribution. At sufficiently high temperatures, an appreciable number of elec-trons will be having energies greater thatEs. These electrons will have sufficient en-ergy to escape from the metal surface and are thermionicallyemitted into vacuum.This is called‘Thermionic emission’ as the emission of electrons has been achievedby thermal excitation process.

It can be shown that for a given metal, there is a maximum thermionic current atany temperature. This saturation current density can be calculated on the basis of freeelectron theory and is given by

J = AT2 exp (−Φ/kT) (3.37)

whereA is a constant dependent on the metal andΦ is the work function of the metal.Equation (3.37) is calledRichardson - Dushman equation. An experimental study ofthe variation of saturation current density with temperature helps in the determinationof the work function of the metal surface.


An estimation of the electron emission by calculation of number of electrons abovethe vacuum level does not match well with the experimentallyobserved values mea-sured at any temperature. The lack of agreement may be attributed to the followingreasons:

(i) All the electrons having an energy greater thanEs will not be necessarily emittedfrom the surface. Only those electrons travelling with a velocity normal to thesurface sufficient to overcome the energy barrierΦ will be emitted.

(ii) Measurement of current density is by the application ofan electric field whichmodifies the work function of the metal.

Thermionic emission is widely used in electron tubes and other electronic deviceshaving a heated cathode.

Numerical Examples

3.1 What is the mobiity of electrons in copper which has resistivity 1.6× 10−8ohm mand electron density of 8.5× 1028m−3?

Solution:σ = neµ

∴ µ = σ/ne= 1/neρ whereρ is resistivity.

=1

8.5× 1028× 1.6× 10−19 × 1.6× 108

µ = 4.6× 10−3m2v−1s−1. (Ans.).

3.2 In the above example, calculate the relaxation time of conduction electrons.Solution: We have,

σ = ne2τ/m

∴ τ = σm/ne2 = m/ne2ρ

=9.1× 10−31

8.5× 1028× (1.6× 10−19)2 × 1.6× 10−8

= 2.6× 10−14s. (Ans.).

3.3 The relaxation time of conduction electrons in copper is 2.6 × 10−14s. If copperhas a resistivity of 1.6 × 10−8Ohm m, calculate the number of free electrons perunit volume and Fermi velocity. Assume Fermi energy as 7 eV.


Solution: We have,σ = ne2τ/m

n = σm/e2τ = m/e2ρτ

=9.1× 10−31

(1.6× 10−19)2 × 2.6× 10−14 × 1.6× 10−8

n = 8.5× 1028m−3 (Ans.).

Assuming Fermi energy as the kinetic energy of the electronsat the Fermilevel,

E f =12

mv2f

∴ Fermi velocity, v f = [2E f /m]1/2

=[2 × 7× 1.6× 10−19]1/2

[9.1× 1031]= 1.57× 106ms−1. (Ans.).

3.4 Estimate the relaxation time of conduction electrons in silver from the followingdata:

Resistivity= 1.6× 10−8ohm m

Density= 10.5× 103kg/m3

Atomic weight= 107.88

Avogadro number= 6.023× 1023 per gm mole.(6.023× 1026 per kg mole)

Solution:

Concentration of electron, n =(10.5× 103) × (6.023× 1026)

107.88= 5.86× 1028m−3.

Relaxation time, τ = m/ne2ρ

=9.1× 10−31

(5.86)× 1028× (1.6× 10−19) × x1.6× 10−8

= 3.79× 10−14s. (Ans.).

3.5 The resistivity of copper at 0oC is 1.6×10−8ohm mand changes to 2.2×10−8ohm mat 100oC. By what factor does the relaxation time of conduction electron change?

Solution: Relaxation time,τ = m/ne2ρ


τ100/τ0 = ρ0/ρ100 = 1.6× 10−8/2.2× 10−8 = 0.727

The relaxation time reduces by a factor 0.727 (Ans.).3.6 Copper contains 8.5×1028 conduction electronsm−3 and has an electrical conduc-

tivity 5.5× 107ohm−1m−1. What is the electric field required to maintain a currentdensity of 2× 106Am−2?

Solution:Electric fieldE = J/σ

= 2× 106/5.5× 107

= 0.036Vm−1 (Ans.).

3.7 What is the drift velocity of conduction electrons in a copper wire of cross sectionalarea 1× 10−6m2 when a current of 2A flows through it? Assume the electrondensity as equal to 8.5× 1028m−3.

Solution:Current densityJ = nev or v = J/ne

v = 2/(1× 10−6 × 8.5× 1028 × 1.6× 10−19)

= 1.47× 10−4ms−1 (Ans.).

3.8 A metal has an electronic conductivity of 5× 106ohm−1m−1 at 0oc. Assuming acarrier concentration of 6× 1028m−3, evaluate the relaxation time of conductionelectrons in the metal.

Solution:Conductivityσ = ne2τ/m or τ = σm/ne2

τ =5× 106 × 9.1× 10−31

6× 1028 × (1.6× 10−19)2

= 2.96× 10−15s (Ans.).

3.9 A uniform wire has a resistivity of 2× 10−8ohm mat room temperature. For anelectric field of 200 V/m along the wire compute the average drift velocity ofthe electrons assuming a carrier concentration of 6× 1028m−3 Also calculate themobility and relaxation time of electrons.

Solution:Drift velocity v = µE = σE/ne= E/neρ

=200

6× 1028× 1.6× 10−19× 2× 10−8

= 1.0417ms−1 (Ans.).

Mobility = v/E = 1.0417/200


= 5.2× 10−3m2v−1s−1 (Ans.).

Relaxation timeτ = mµ/e

=9.1× 10−31 × 5.2× 10−3

1.6× 10−19

= 2.96× 10−14s (Ans.).

Exercise

3.1 Elucidate the difference between classical free electron theory and quantum freeelectron theory. (March 1999).

3.2 Explain the classical free electron theory of metals. Mention its shortcomings.

(August 1999).

3.3 Explain the motion of a free electron in a periodic lattice. (March 2000).

3.4 What are the effects of temperature and impurity on the electrical resistivity ofmetals? (March 2000).

3.5 What are relaxation time and collision time? On the basis of free electron theoryof metals, obtain an expression for the electrical conductivity of a metal in termsof collision time. (March 2000).

3.6 For a metal having 6.5 × 1028 conduction electrons perm3, find the relaxationtime of conduction electrons if the metal resistivity is 1.43× 10−8ohm.m.

(August 2000).

3.7 Based on free electron theory, derive an expression for electrical conductivity inmetals and discuss the failure of classical free electron theory. (August 2000).

3.8 Calculate the mobility and the relaxation time of electronsin copper with thefollowing data:

Resistivity of copper= 1.73× 10−8ohm m.

Atomic weight= 63.5.

Density= 8.92gm cc−1

Avogadro number= 6.02× 1023per gm mole.

(March 2001).


3.9 Assuming the electron-lattice interaction to be responsible for scattering of con-duction electrons in a metal, obtain an expression for the conductivity in termsof relaxation time. (March 2001).

3.10 Obtain an expression for electrical conductivity with the help of classical freeelectron theory. (August 2001).

3.11 Give a qualitative account of thermionic emission. (March 2002).

3.12 Find the mobility of conduction electrons in copper using the data given:

Resistivity of copper= 1.7× 10−8ohm m.

Atomic weight of copper= 63.54

Density of copper= 8.96× 103kgm−3

Avogadro number= 6.025× 1023 per gm mole

Charge of the electron= 1.6× 10−19C.

(March 2002).

3.13 Based on free electron theory, derive an expression for the electrical conductivityin a metal. Explain the deficiencies of classical free electron theory.

(March 2002).

3.14 Explain density of states. Derive the expression for numberof allowed energystates for a unit volume of a solid. (Feb 2003).

3.15 Explain the terms drift velocity, mobility and relaxation time. (Feb 2004).

3.16 Find the temperature at which there is 1% probability that a state at an energylevel 0.5 eV above Fermi level will be occupied. (Feb 2004).

Chapter 4

Superconductivity

4.1 Introduction

The electrical resistivity of all metals decreases as the temperature is reduced. Thisis because the thermal vibration of atoms decreases with decrease in temperature. Atvery low temperature, the thermal vibrations contribute very little to the scattering ofconduction electrons and the electrical resistivity will be dominated by the effect ofimpurities and defects. The electrical resistivity may tend to zero as the temperatureapproaches O K only in the case of pure and perfect single crystals. However, somemetals show a remarkable behaviour. They lose their electrical resisistivity completelybelow a certain temperature. For example, the electrical resistance of pure mercurysuddenly drops to zero (Fig.4.1) when cooled below 4.2 K. Thematerial is said tobe in the superconducting state below the critical temperature and the phenomenon isknown assuperconductivity.

R

T, K

Figure4.1Resistance of a metal as a function of temperature showing superconductingtransition.

The critical temperature or transition temperature,Tc, is characteristic of the ma-terial. Above the transition temperature the material shows normal resistivity witha finite temperature dependence. Many elements, compounds and alloys have beenfound to exhibit superconducting behaviour at low temperature. Table 4.1 gives thetransition temperature of some superconducting materials.

117

118 Superconductivity

Table4.1Superconducting transition temperature of some selected materials

Element Tc,K Compound Tc, K

Al 1.1 Pb2Au 7In 3.4 MoN 12Ta 4.4 NbN 15V 5.1 V3S i 17Pb 7.2 Nb3S n 18Nb 9.2 Nb3Ge 21

4.2 Characteristic features of superconductors

When a material undergoes a transition from normal to the superconducting phase,significant changes take place in the electrical and magnetic properties. Even the ther-mal properties and the mechanical properties show abrupt changes. In other words, allthe physical and chemical properties dependent on the electrons in the material showmodifications. These changes indicate that the conduction electrons undergo certainchange in the superconducting phase.

(i) The most important characteristic feature of the superconducting phase is theabsence of resistance to the flow of electric current throughthe material. Thetransition from the normal phase to the superconducting phase occurs at a welldefined critical temperature. For pure materials, the transition occurs over a tem-perature range as low as 0.01 K.

(ii) The materials in their superconducting phase repel magnetic lines of force. Whena transition takes place from the normal phase to the superconducting phase inpresence of a magnetic field, the magnetic flux lines will be expelled from withinthe body of the superconductor.

(iii) The presence of magnetic materials like iron, cobalt or nickel, even in minutequantities as impurity, destroys superconductivity in metals.

(iv) The property of superconductivity is not restricted tometals or good conductors.It has been observed that certain semiconducting and insulating mixed oxides aswell as some polymers exhibit superconductivity.

(v) Interestingly, good conductors like copper, silver, gold and alkali metals like

Superconductivity 119

lithium, sodium and potassium do not exhibit superconductivity even at temper-atures as low as 0.1 K.

4.2.1 Isotope effect

When an element exhibiting superconductivity exists in various isotopic forms, it isobserved that the transition temperature decreases as the isotopic mass increases. Thevariation of transition temperature with isotopic mass is described by the relation,

M1/2Tc = constant. (4.1)

This dependence of transition temperature on the isotopic mass is known as “Isotopeeffect”. This indicates an involvement of ion cores in the phenomenon of supercon-ductivity since the natural frequency of vibration of the lattice atoms is proportional toM−1/2. However, experimental observations have indicated that the relation is not validfor many materials. A more general form of the isotope effect may be written as

Tc∞M−β (4.2)

whereβ varies from 0 to 0.5

4.2.2 Meissner effect

An important property of the superconducting phase is the expulsion of magnetic fluxlines from within the bulk of the superconductor. This is known asMeissner effect.Consider a material in the normal state(Fig.4.2a).

T > Tc

T < Tc

B > Bc

(a) (b) (c)

Figure 4.2 Demonstration of Meissner effect. (a) The magnetic flux lines passingthrough the sample for T> Tc, (b)flux lines being expelled for T< Tc and (c) the fluxlines penetrating when the field is increased above the critical value.


When a magnetic field is applied to the material, the magneticflux lines passthrough the material. Now, if the temperature is reduced below the critical tempera-ture the magnetic flux lines will be expelled from inside the superconductor (Fig 4.2b).Hence, we have,

B = µo(H + M) = 0 (4.3)

whereM is the magnetization in the material due to an applied magnetic field H. Themagnetic susceptibility is given by

χ =MH= −1 (4.4)

which indicates that the material in its superconducting state is a perfect diamagneticmaterial.

If a superconductor is subjected to a strong magnetic field, the material loses itssuperconducting property and becomes normal. As the magnetic field is increased instrength, at a particular value of the magnetic field, the magnetic flux starts penetratinginto the material and makes it a normal material (Fig.4.2c).Hence, we can define acritical magnetic field, corresponding to a temperature, upto which the material re-mains in the superconducting phase and above which the material becomes normal. Atany temperatureT belowTc, the critical magnetic fieldHc(T) upto which the materialremains in the superconduting phase, is given by the relation

Hc(T) = Hc(0)

[

1− T2

T2c

]

(4.5)

whereTc is the critical temperature in the absence of a magnetic fieldandHc(0) is aconstant representing the critical magnetic field atT = 0K (Fig.4.3). The magneticfield that destroys superconductivity in any material need not be an externally appliedfield. It may also be an induced magnetic field due to an electric current flowingthrough the material. The magnetic fieldH induced in a wire of radiusr when acurrent I flows through it is given by

I = 2πrH (4.6)

If this induced magnetic field becomes equal to the critical magnetic field, the materialbecomes a normal conductor. Hence, we can define acritical current Ic correspondingto the critical magnetic fieldHc that is sufficient to destroy superconductivity in thewire.


T, K

HC

Figure4.3The variation of critical magnetic field with temperature.

Ic = 2πrHc (4.7)

Equation (4.7) is calledSilsbee’s rule. This puts a limit on the electric current that canbe passed through a superconductor.

4.3 Classification of superconductors

Superconductors are classified as Type I (or soft) superconductors and Type II (or hard)superconductors. The classification is based on the magnetic behaviour of supercon-ductors. Type I superconductors have a small critical magnetic field and the transitionfrom the superconducting to the normal phase at the criticalfield is abrupt. Assum-ing that there is a magnetization in the superconductor in a direction opposite to thedirection of the applied magnetic field, the variation of magnetization with the appliedfield will be as shown in Fig.4.4. At the critical magnetic field, there will be an abrupt

H

−M

HC

Figure4.4Effect of magnetic field on Type I superconductor.


decrease in the magnetization and the material becomes normal. For all values of mag-netic field above the critical field, the material shows finiteresistivity and the magneticflux penetration is complete. In other words, Type I superconductors display Meissnereffect completely.

In the case of Type II superconductors, the material is a perfect superconductorupto a magnetic fieldHc1 (Fig.4.5). The magnetic flux penetration begins atHc1 andis complete at a magnetic fieldHc2. The material becomes a normal conductor formagnetic field greater thanHc2. The material is said to be in a‘mixed state’ or ‘vortexstate’ between the two critical magnetic fieldsHc1 andHc2. Inspite of the fact that themagnetic flux lines penetrate the material in the vortex state,

Hc1 Hc2 H

−M

Figure4.5Effect of magnetic field on Type II superconductor.

the electrical resistivity continues to be zero uptoHc2. Since the values ofHc2 are rel-atively larger, Type II superconductors are more useful forpossible application basedon the resistanceless state of materials. A comparison of Type I and Type II supercon-ductors is given in Table 4.2.

Table4.2Comparison of the characteristics of Type I and Type II superconductors.

Type I Type II

1. These are usually elements in theirpure form.

These are impure elements, alloys orcompounds.

2. The transition to normal state occursabruptly at a critical magnetic field,Hc

The transition to normal state beginsat Hc1 and is complete only atHc2.


3. The value of critical magnetic fieldHc

is usually very small.Though the value ofHc1 is small,Hc2

is quite large.

4. They exhibit complete Meissner effectuptoHc

Meissner effect is complete uptoHc1

when the magnetic flux starts pene-trating into the superconductor. Theflux penetration is complete only atHc2 when the material becomes nor-mal conductor.

5. The superconductivity or the zero re-sistance state is observed upto criticalmagnetic fieldHc.

The material remains in the resis-tanceless state even in the intermedi-ate state betweenHc1 andHc2 and be-comes normal only atHc2.

6. The small value ofHc, whether ap-plied or induced, restricts their use

The large value ofHc2 makes it suit-able for applications.

4.4 Applications of superconductors

The major applications that are based on the properties of the superconducting phaseare

(i) Lossless power transmission

(ii) Superconducting magnets

(iii) Magnetic levitation

(iv) Superconducting memories and switches

(v) Superconducting quantum interference devices.

(i) Lossless power transmission: The transmission of electrical power on a largescale is limited due to the restrictions on the current carrying capacity of trans-mission lines and their maintenance. Loss of power due to power dissipation intransmission lines due to the finite resistance offered by the cables is quite consid-erable. Further, due to the passage of the current, the metallic cable gets heatedup thereby increasing the electrical resistance of the cables. This increases thejoule heating losses further. If the transmission lines aremade of materials intheir superconducting state, their current carrying capacity increases. Since theselines do not offer any resistance to the flow of electric current, there is no power


dissipation due to joule heating. Thus, power losses are minimised and currentcarrying capacity is improved. This is the principle used inlossless power trans-mission.

(ii) Superconducting magnets:The problem of joule heating again puts a limit onthe magnetic field that can be obtained in an electromagnet. The magnetic fieldgenerated in a coil depends on the strength of the direct current flowing through it.Any attempt to increase the magnetic field by increasing the current through thecoil results in increasing the joule heating . This puts an upper limit on the currentcarrying capacity of the wire used and hence on the magnetic field produced. Useof superconducting coils in the electromagnets can enhancethe magnetic fieldgenerated as they can carry large currents with practicallyno dissipation. This isthe principle used in the construction ofsuperconducting magnets. However,the limitation now comes in the form of the effect of magnetic field on the su-perconducting state of the coils. When the magnetic field generated exceeds thecritical value at the operating temperature, the metal wireused in the coils willreturn to their normal state and start offering large resistance to current flow anddrastically reducing the magnetic field generated.

(iii) Magnetic levitation: The principle of the repulsion of magnetic flux from asuperconductor can be used inmagnetic levitationapplications. When a magnetis brought near a superconductor, there will be a repulsion and the superconductortries to move away from the magnet. One of the reasons for a limitation to thespeed of a train, for example, is the friction between the wheels of the train andthe tracks on which the train moves. Superconducting wheelsexperience reducedfriction on magnetic tracks. Alternatively, magnetic levitation may be used toreduce friction between the wheels of a train and the track onwhich it moves.It is well known that like poles of magnets repel and the friction between thewheels and the tracks get reduced when they are made like poles of magnets.The large magnitude of repulsive force required may be realized with the help ofsuperconducting magnets. The principle can also be used in making friction-lessbearings.

Practically, Maglev is a vehicle that runs levitated from the guideway, which issimilar to the tracks in railways. This is achieved by using electromagnetic forcesbetween superconducting magnets on the vehicle and coils onthe ground. Levita-tion coils shaped like figure 8 are installed on the sidewallsof guideway (Fig.4.6a). When the superconducting magnet attached to the vehiclepasses below thecentre of these coils, electromagnetic force pushes the magnets upwards thereby


N

S

S

N

S N S N S

NSNSN

N S

GUIDWAY(a) (b)

Figure4.6Effect of magnetic field on Type II superconductor.

levitating the vehicle. A repulsive force and an attractiveforce induced betweenthe magnets are used to propel the vehicle ( Fig.4.6 b). The propulsion coils arelocated on the sidewalls of the guideway energized by alternating current. Thesuperconducting magnets are attracted and pushed by the changing field therebypropelling the vehicle.

(iv) Superconducting memories and switches:The disappearance of superconduc-tivity for magnetic fields higher than the critical field is the principle used in theconstruction of acryotron .

Control, Nb(9.2K)

Core.Ta(4.4K)

Liq.He(4.2K)

Figure4.7Construction of cryotron.

It consists of a core wireA around which a control coilB is wound (Fig.4.7). ThecoreA is made of tantalum (Tc = 4.4K) and control coilB is made of niobium(Tc = 9.2K) or lead (Tc = 7.2K). The whole assembly is maintained at liquidhelium temperature (4.2 K). At this temperature, both the control coil and the core


wire are in the superconducting phase. Hence, the resistance of the core wire willbe zero. A current can be passed through the control coil to produce a magneticfield sufficient to make the core wire ‘normal’. Thus, the core wire can be madeto possess zero or finite resistance depending on the controlcurrent being ‘off’or ‘on’ respectively. These two states of the core wire may beconsidered as theON and OFF states. Thus, cryotrons can be used as switches.

Based on the same principle is the construction of superconducting memory. If acurrent is induced in a superconducting ring, the current persists and continues toflow until a magnetic field is applied to make the ring normal, thereby allowingthe current to decay. The direction of current flow may be usedto represent thestate of memory, namely ‘0’ or ‘1’.

(v) Superconducting quantum interference devices:The phenomenon of quantummechanical tunneling has been used in the construction of some superconductingdevices. Consider two metal electrodes separated by an insulator as shown inFig. 4.8. The insulator normally acts like a barrier and doesnot allow the flow ofelectrons from one metal to another.

METAL 1 METAL 2

INSULATOR

Figure4.8Metal-insulator-metal structure to study tunneling.

However, if the barrier is made sufficiently thin, there will be a finite probabilitythat an electron will pass through the insulator. This is called tunneling. Whenboth the metals are normal conductors, the current-voltagerelation will be ohmicat low voltage levels. If the metals are in superconducting state, then, we ob-serve some interesting effects. A dc current may flow across the junction evenin the absence of any electric or magnetic field. This is knownasdc Joseph-son effect. When a fixed dc voltageV is applied across the junction,the phasewill vary linearly with time and an alternating current witha frequency (2e/~)Vis generated. This is known asac Josephson effect. Hence, a Josephson junc-tion can act as a voltage-to-frequency converter. Josephson effect may be usedto detect magnetic fields. It may also be used in the generation and detection of


electromagnetic radiations. A dc magnetic field applied through a superconduct-ing circuit containing two junctions causes the maximum supercurrent to showinterference effects as a function of the applied magnetic field intensity. This isused in the construction ofSuperconducting Quantum Interference Devices(SQUIDs).

B

JOUTJIN

AJ1

J2

Figure4.9Construction of a SQUID.

Figure 4.9 shows the construction of a SQUID.A andB are two Josephson junc-tions connected in parallel to form a ring. Any current that enters atJIN will bedivided into two componentsJ1 andJ2 which will pass through the junctionsAandB respectively and recombine to produce the output currentJOUT. If δa andδb represent the phase difference between the input current and the output currentwhile passing through the insulator junctions A and B respectively, then,

δa = δb = δ0

Application of a magnetic field results in the modification ofthe phases of thetwo current components. When a magnetic field of flux densityφ is applied atthe centre of the ring as shown in the figure, the phase term forthe current out ofthe two junctions will be modified as

δa = δ0 − (eφ/hc)

δb = δ0 + (eφ/hc)

Hence the total current out of the ring will be

JOUT = J0[sin(δ0 + eφ/hc) + sin(δ0 − eφ/hc)]

= 2J0 sinδ0 · cos(eφ/hc)


This expression indicates that the output current varies with the applied magneticflux and shows oscillations(Fig.4.10).

Φ

J

Figure4.10Output current oscillations as a function of Magnetic flux.

The maximum value of current is given by

JOUT,MAX = 2J0 sinδ0 when (eφ/hc) = nπ

wheren is an integer. The sensitivity of the output current to variations in themagnetic fluxφ is the basis for the applications of SQUIDs. In addition to beinga very sensitive tool to detect and measure minute magnetic fields, they also findapplication as storage devices for magnetic flux, in magnetometry for geologicalexploration for identifying magnetic ore deposits, oil exploration, under-waterexploration, in medical field for magnetic resonance imaging(MRI),etc.

4.5 Theoretical interpretation of superconductivity

The most satisfactory interpretation of the occurrence of superconductivity in materialsat low temperatures is due to the BCS Theory put forward by thescientists Bardeen,Cooper and Schrieffer. According to this theory, the electrons in a superconductor ex-ist in pairs called‘Cooper pairs’ . In the superconducting phase, when an electron ap-proaches an ion core of the lattice, the electron experiences a force of attraction due tothe electrostatic interaction between the electron and thepositive ion core. As a resultof this interaction, the ion core will be displaced from its position thereby distorting thelattice. A second electron passing that way interacts with the distorted lattice and thereis an exchange of energy in the process. This may, in effect, be considered as equiv-alent to an interaction between two electrons with the assistance of the lattice. This


is equivalent to an attractive force between the two electrons which becomes maxi-mum when the electrons have opposite spins and associated momentum. The electronsovercome the electrostatic force of repulsion and are boundin pairs with a weak forceof attraction assisted by the lattice. Due to this interaction, the scattering felt by oneelectron is exactly nullified by the response of the other electron of the pair, so thatthe electrons move without experiencing any effect of scattering. The resulting Cooperpairs are ‘bosons’ with zero spin and are not bound by the Pauli’s exclusion principle.Hence, they condense into a common ground state. Such a condensation results in theformation of an energy gap with the Cooper pairs existing below the gap and the nor-mal electrons above it. The energy gap represents the binding energy of these Cooperpairs and is found to be related to the transition temperature. When the temperature isincreased above the transition temperature, the energy gapdisappears and the Cooperpairs break up to form normal electrons. The BCS theory also predicts a close relationbetween the electron lattice interaction and superconductivity. The isotope effect is insupport of this view. Further, the absence of superconductivity in highly conductingnoble metals like gold, silver and copper , even at very low temperature, is attributedto the absence of strong lattice-electron interaction. In other words, the BCS theoryhas been able to explain the observed characteristic features of superconductors.

4.6 High Temperature Superconductors

Superconductivity has been observed in a few elements, but in a large number of alloysand compounds. Until 1986, the highest transition temperature was about 23 K. In1986, several materials were discovered withTc in the range 30-40 K. In a couple ofyears, spectacular results were reported withTc upto 125 K. This led to the hope that itmight be possible to develop materials that are superconductors at room temperature.Discovery of such materials would mean large scale application of the phenomenon.Further, the BCS theory predicts a superconducting transition temperatureTc given by

Tc = 1.14θD exp −1/N(EF)U (4.8)

whereθD is the Debye temperature,N(EF) is the density of states at the Fermi levelandU is a measure of the electron-phonon interaction. This equation suggests thatmaterials, which are characterized by high resistivities at normal temperatures, havea better probability of becoming superconductors at highertransition temperatures.Accordingly, some mixed oxide systems were studied and werefound to have transi-tion temperatures above liquid nitrogen temperature (77 K). These are known as hightemperature superconductors.


Table4.3Some oxide superconductors.

Compound Tc, K

Yba2Cu3O7 90

Bi2S r2Ca2Cu3O10 110

TlBa2Ca3Cu4O11 120

Tl2Ba2Ca2Cu3O10 125

The discovery of high temperature superconductivity is an important achievementas it eliminates the difficulty of maintaining low temperatures using liquid helium.Some of these superconducting materials are listed in Table4.3.

Numerical Examples

4.1 The critical temperature and critical magnetic field for superconducting lead are7.2 K and 800 gauss respectively. What will be the temperature up to which leadwill be in superconducting state in a magnetic field of 400 gauss?

Solution:

Hc(T) = Hc(0)[

1− T2/T2c

]

400= 800[

1− T2/7.22]

∴ T = 5.09K. (Ans.).

4.2 Superconducting tin has a critical magnetic field of 217 gauss at 2 K. If the crit-ical temperature for superconducting transition for tin is3.7 K, find the criticalmagnetic field at 3 K.

Solution:Hc(T1) = Hc(0)

[

1− T21/T

2c

]

Hc(T2) = Hc(0)[

1− T22/T

2c

]

Hc(T1)Hc(T2)

=

[

1− T21/T

2c

]

[

1− T22/T

2c

] =

[

T2c − T2

1

]

[

T2c − T2

2

]

Hc(T2) =

[

T2c − T2

2

]

[

T2c − T2

1

]Hc(T1)


=

[

(3.7)2 − (3)2]

[

(3.7)2 − (2)2] · 217

= 105.03 Gauss (Ans.).

4.3 The critical temperature forHg199.5 is 4.185 K. Calculate the critical temperaturefor Hg203.4.

Assumeβ = 0.5.Solution:

TC(199.5)1/2 = TC′(203.4)1/2

TC′ =TC · (199.5)1/2

(203.4)1/2= 4.185·

(199.5)1/2

(203.4)1/2= 4.145K (Ans.).

Exercise

4.1 Compare the dependence of resistance on temperature of a superconductor withthat of a normal conductor. Describe briefly the formation ofCooper pairs.

(March 99)

4.2 Describe different types of superconductors. (March 99)

4.3 Discuss the Meissner’s effect in superconductors. Mention a few applications ofsuperconductors. (March 2000)

4.4 Explain the phenomenon of superconductivity and Meissner effect. Discussbriefly BCS theory of superconductivity. (March 2001)

4.5 What are superconducting magnets? Give their uses. (August01)

4.6 What is Meissner effect? Explain the validity of Meissner effect in Type I andType II superconductors. (March 02)

4.7 What are SQUIDS ? Give a brief account of their applications. (Feb 2003)

4.8 Explain Meissner effect. Illustrate with an example. (Aug 2003)

4.9 Explain any two applications of superconductivity. (Aug 2003)

4.10 Distinguish between Type I and Type II superconductors. Discuss briefly theBCS theory of superconductivity. (Aug 2004)

4.11 Explain Type II superconductor. Discuss Maglev vehicle. (Feb 2004)

Chapter 5

Semiconductors

5.1 Band Structure of Solids

A solid contains a large number of atoms packed closely together. Each atom, whenisolated, has a discrete set of electron energy levels 1s, 2s, 2p, 3s, 3p, 3d. . . . . . and soon. Each of these energy levels can hold a finite number of electrons with differentmagnetic and spin quantum numbers. A given atom has some of these levels filled andsome empty depending on the total number of electrons associated with the atom. Forexample, an atom of carbon which has six electrons will have two electrons each in 1s,2s, and 2p levels. However, 2p level can hold a total of six electrons out of which fourlevels are empty. When atoms are brought nearer, as in the case of solids, the orbitals ofthe outer electrons overlap indicating that these electrons have a sort of an interaction.In a solid, because of the close packing of atoms the outer electrons from many atomsinteract. The result of these interactions is that these electrons cannot be considered tobelong to any particular atom. In other words, the electronsare no longer localized at aparticular atom and its wave function is altered due to the influence of the neighbouringatoms. This results in shifting and splitting of energy levels into energy bands. Theeffect of reducing the inter atomic distance in carbon is illustrated in Fig.5.1.

2p

2s

1s

E

rr0

Figure5.1Formation of energy bands in carbon (diamond).

132

Semiconductors 133

The discrete energy levels 2s and 2p will broaden into bands which overlap as theinter atomic distance is reduced. With further decrease in the distance a split occursproducing two bands with four electron states each per atom.The lower band contain-ing four electron states occupied by four electrons is called thevalence bandand theupper band containing four empty electron states is called the conduction band. At theinter atomic distance equal to that observed in diamond crystal, the gap between thevalence and conduction bands is quite large and of the order of 6 eV. This gap is calledthe forbidden energy gapas there are no permitted energy levels for occupation byelectrons. In diamond, all the energy levels in the valence band are full and the en-ergy levels in the conduction band are empty at O K. Hence, diamond is an electricalinsulator (Fig.5.2a).

6eV

CB

VB

(a)

1.1eV

CB

VB

(b)

Figure5.2Energy band diagrams for (a) insulator and (b)semiconductor.

In silicon, the 3s and 3p bands overlap and result in valence and conduction bandsseparated by a forbidden energy gap of 1.1.eV (Fig.5.2b). Germanium has a forbiddenenergy gap of 0.7 eV. Silicon and germanium are classified as semiconductors becauseat room temperature quite a few electrons can cross the forbidden energy gap andcontribute to electrical conductivity.

In the case of metals (or conductors), the outermost band containing valence elec-trons is either partially filled(as in Na, K, Al, etc) or overlaps with the next higherband(as in Be, Mg, Ca, etc). For example, in sodium there is only one electron in the3s state. Since the 3s state can accommodate two electrons, the band formed with theseelectrons is said to be half-filled. In magnesium, the 3s state is full with two electrons.

However, 3p states overlap with 3s states making it a partially filled band. In eithercase, the electrons can easily take part in electrical conduction as they need not cross aforbidden energy gap (Fig.5.3).

Thus, materials can be classified as conductors, semi-conductors and insulators onthe basis of band theory of solids. The order of magnitude of electrical conductivity

134 Semiconductors

3s

2p

2s

1s 1s

2s

2p

3s

(a) (b)

3p

Figure 5.3 Energy band diagrams for (a) sodium showing half-filled 3s band and (b)magnesium showing 3s and 3p bands overlapping.

serves as an indicator for the classification. There are a fewexceptions like conductorswith incomplete inner orbitals(like Sc, Mn, Zr, Re, Rh, etc)showing higher resistivitiesand narrow band gap semiconductors(likeInS b, Bi2Te3, etc) with exceptionally highconductivity. Hence, conductivity alone should not be the basis for classification ofmaterials under these categories. The temperature coefficient of resistivity helps us toclassify and distinguish between highly conducting semiconductors and less conduct-ing metals. The positive value of temperature coefficient of resistivity for conductorsand negative value for semiconductors can be explained on the basis of band theory.

5.2 Intrinsic semiconductors

These are semiconducting materials which are chemically pure, structurally perfectand stoichiometric in composition if compound. They have properties characteristicof the material alone. The carriers contributing to electrical conduction are generateddue to the breaking of covalent bonds by thermal activation process. Pure germaniumor silicon are examples of intrinsic semiconductors. Thesematerials have four valenceelectrons and require four more to complete the subshell. Hence, they have a tendencyto form four covalent bonds with neighbouring atoms. This isshown schematically intwo dimensions in Fig.5.4.

It may be mentioned that the four covalent bonds are equally inclined to each otherin three dimensions and are referred to as tetrahedral bonding. The two dimensionalrepresentation is used only for ease of explaining the generation and motion of carriersin the semiconductor lattice.

Semiconductors 135

5.2.1 Carrier generation in intrinsic semiconductors

Carrier generation in an intrinsic semiconductor is due to the breaking of covalent

Figure5.4Formation of covalent bonds in silicon and germanium.

bonds. Number of covalent bonds breaking is a strong function of temperature. Break-ing of a covalent bond result in the release of an electron into the conduction band.Consequently, an electron energy state becomes vacant in the valence band. This va-cant site provides an opportunity for the valence electronsto move under the influenceof an applied electric field. The movement of electrons in thevalence band will thusequivalent to the movement of vacant site in the opposite direction (Fig.5.5). The va-cant sites thus created by the transfer of an electron to the conduction band are called‘holes’.

E

Figure5.5Breaking of covalent bond and generation of electron - hole pairs.

The movement of electrons in the valence band is conveniently explained as equiv-alent to the movement of holes. Since the direction of movement of these holes isopposite to that of electrons, holes behave as if they are positively charged. Thus,breaking of each covalent bond results in the formation of anelectron-hole pair, andthey contribute to the electrical conductivity of the material. Since the electron and thehole are created together in pair, the number of electrons inthe conduction band willbe equal to the number of holes in the valence band at any giventemperature.

136 Semiconductors

5.2.2 Fermi factor and Fermi energy

In order to understand the electrical conductivity of semiconductors, it is necessary toknow the distribution of electrons over the available energy states. Electrons in solidsobey Fermi-Dirac statistics. Fermi-Dirac distribution function orFermi factor whichgives the probability that an available energy state atE will be occupied by an electronat a temperatureT is given by

F(E) =1

1+ exp (E − EF)/kT(5.1)

The quantityEF is calledFermi energy and it represents a reference level in asemiconductor. Fermi energy may be defined as the energy corresponding to a levelwhose probability of occupation by an electron is half. The variation of Fermi factorwith energy is shown in Fig. 5.6. It is easy to show that

At T = 0, for E < EF, F(E) = 1

At T = 0, for E > EF, F(E) = 0

At T , 0, for E = EF, F(E) = 0.5

EF

E

T=0 K

T>0 K

10.50

F(E)

Figure5.6Fermi distribution factor as a function of energy.

At absolute zero temperature, a semiconductor behaves likean insulator since allthe covalent bonds are intact. The valence band is full and the conduction band is

Semiconductors 137

empty. It can be shown that for an intrinsic semiconductor, the Fermi level lies at thecentre of the band gap and remains invariant with temperature. We notice that theFermi factor has a non-vanishing value even for the energy values corresponding tothe forbidden energy gap. However, no carriers are present in the forbidden gap sincethere are no sites available for occupation by the charge carriers.

5.2.3 Conductivity of an intrinsic semiconductor

When an electric field is applied to an intrinsic semiconductor, there will be movementof electrons in the conduction band and of holes in the valence band. Hence, there willbe two currents, an electron current in the conduction band and a hole current in thevalence band. As these currents are constituted by oppositely charged carriers and arein opposite directions, the net current will be sum of the twocurrents.

If ve represents the drift velocity of electrons due to the applied field E, the currentdensity due to electrons is given by

Je = neve (5.2)

wheren is the number of conduction electrons available per unit volume ande is thecharge associated with these electrons (compare this with equation (3.9) for the caseof a metal). Similarly, the current density due to holes is given by

Jh = pevh (5.3)

The total current density is given by

J = (Je + Jh) = neve + pevh = ne(ve + vh) (5.4)

since the number of conduction electrons will be equal to thenumber of holes in anintrinsic semiconductor. Further,it is observed that the drift velocity is directly propor-tional to the applied fieldE.

i.e. v α E or v = µE (5.5)

whereµ is called themobility of carriers . Hence the above equation becomes

J = neE(µe + µh) (5.6)

whereµe andµh are the the electron and hole mobilities respectively. The conductivity,which is defined as the current density per applied field, is given by

σ = ne(µe + µh) (5.7)

138 Semiconductors

This equation gives the conductivity of an intrinsic semiconductor. The mobility ofholes will be usually less than that of the electrons. This isdue to the fact that themovement of holes in the valence band is nothing but the movement of electrons in theopposite direction. Being tightly bound, they move with a smaller drift velocity.

5.2.4 Effect of temperature on conductivity

The conductivity of an intrinsic semiconductor depends strongly on temperature. Thehigher the temperature, more will be the number of electronsexcited to the conduc-tion band. Thermal energy much lower than the forbidden gap energy is sufficient toexcite a large number of electrons from valence band to the conduction band. Hence,the electrical conductivity of a semiconductor increases with increase in temperature.This accounts for the negative value of the temperature coefficient of resistivity ofsemiconductors. The carrier concentration may be expressed as

n = k1T3/2 exp (−Eg/2kT) (5.8)

whereEg is the energy gap of the semiconductor andk1 is a constant. The mobility ofelectrons and holes generally depend on temperature as

µ = k2T−3/2 (5.9)

wherek2 is a constant. It may be mentioned that equation (5.9) is valid for a case wherethe scattering of charge carriers is predominantly by phonons rather than ionised im-purities. Substituting from equations (5.8) and (5.9) in equation (5.7), the conductivitycan be expressed as

σ = k1k2e exp (−Eg/2kT)

σ = K exp (−Eg/2kT) (5.10)

whereK = k1k2e

A plot of lnσ versus (1/T) will be a straight line (Fig.5.7) with a slope equal to(Eg/2k). Hence, the band gap energy of the semiconductor can be calculated.

5.3 Extrinsic semiconductors

Pure or intrinsic semiconductors are of theoretical importance because of their low con-ductivity. The electrical conductivity of a semiconductoris very sensitive to the purityof the material. Addition of small amounts of impurities results in drastic changes

Semiconductors 139

ln σ

(1 / T)

Figure5.7Variation of lnσ with (1/T) for an intrinsic semiconductor.

in the properties of semiconductor. For example, the conductivity of silicon is en-hanced by three orders of magnitude by addition of a few partsper million of boron.Intentional addition of small amounts of impurities to semiconductor with the inten-tion of modifying its properties is called‘doping’ . Such semiconductors in which theelectrical conductivity is dominated by the presence of impurities is calledextrinsicsemiconductors. The added impurities are called dopants. The impurities normallyemployed for germanium and silicon are elements of group IIIor group V of the pe-riodic table. Boron, gallium and indium from group III and phosphorus, arsenic andantimony from group V are the common dopants.

When impurity atoms are added into the lattice of semiconductors, new energy lev-els are introduced in the band structure. Consider the result of adding a small amountof phosphorus to silicon (Fig.5.8).

CB

VB

DL

Figure 5.8 Effect of adding impurity atoms of Group V showing free electronand thedonor energy level.

140 Semiconductors

The impurity atom occupies one of the lattice sites and sincethe sizes of the atomsare not much different, the crystal is not distorted. However, each phosphorus atombrings five valence electrons with it whereas only four are required to form covalentbonds with the neighbouring silicon atoms. The fifth electron will be loosely boundto the parent atom due to the high permittivity of silicon. The energy state of thiselectron is represented by an energy level just below the bottom of the conduction band(Fig.5.8) since the electron has a very small binding energyof the order of 0.05 eV.The electron can be easily released into the conduction bandby means of thermalexcitation. Hence, phosphorus atom or any group V element isreferred to as a donorsince it contributes additional electron for conduction. The semiconductor is said to ben-type semiconductor.

When a semiconductor is doped with a group III element, the impurity atom willhave three valence electrons but occupies the lattice position of a tetravalent semicon-ductor atom (Fig.5.9). It will have a tendency to capture an electron to complete thefourth covalent bond. Hence the impurity is called anacceptor.

CB

VB

AL

Figure5.9Effect of adding impurity atom of Group III showing a hole and theacceptorenergy level.

Acceptor levels lie close to the top of the valence band and capture electrons fromthe covalent bonds thereby generating holes in the valence band. The captured elec-trons are not available for conduction and the conductivityis due to the movement ofholes in the valence band. The semiconductor is calledp-type semiconductor.

In case of compound semiconductors, a perfectly stoichiometric compound willbe intrinsic in nature. Any deviation from stoichiometry will make the compoundextrinsic. For example, semiconducting compoundZnS will be intrinsic in nature ifthe compound is stoichiometric, i.e., the atomic composition ofZnand S is 50:50. Anydeviation from this composition, will result in an excess concentration of one of thecomponents over the stoichiometry and gives the compound extrinsic nature. Hence,by controlling the composition of the compound, it is possible to obtain intrinsic aswell as extrinsic (n-type and p-type) semiconductors.

Semiconductors 141

5.3.1 Conductivity of an extrinsic semiconductor

We have seen that the conductivity of an intrinsic semiconductor is due to the move-ment of electrons in the conduction band and holes in the valence band. These carriersare generated together and hence their numbers are equal. Further, the process respon-sible for the generation of these carriers is breaking of covalent bonds which has afinite probability at any given temperature. In case of extrinsic semiconductors, thecarriers are generated due to two processes, namely

(i) ionization of impurity levels

(ii) breaking of covalent bonds.

In an n-type semiconductor, for example, the donor levels are close to the bottom ofconduction band and require only a small thermal energy to release an electron into theconduction band. This results in a mobile electron in the conduction band and an im-mobile positive ion of the donor impurity atom. Since the donor levels are very close tothe conduction band, a large number of these levels will be ionised even at low temper-ature. At the same time, there are a few covalent bonds also being broken resulting inthe generation of electron - hole pairs. Thus, there will be alarge number of electrons(called majority carriers)and a few holes (called minoritycarriers) contributing to theconductivity of n-type semiconductor. The conductivity can be expressed as

σ = neµe + peµh (5.11)

wheren represents the total number of electrons contributing to conductivity generatedby both the mechanisms of carrier generation.

At normal temperature, most of the donor levels will be ionised and each donoratom contributes an electron to the conduction band. Further, the number of electronsgenerated due to breaking of covalent bonds is very small andnegligible comparedto the number of electrons generated due to ionization of donor levels. Hence, thenumber of electrons contributing to conductivity may be taken as equal to the donorconcentration and the conductivity will be given by

σ = Ndeµe + peµh (5.12)

whereNd represents the donor concentration in the semiconductor. Since the donorconcentration is quite high compared to the number of covalent bonds broken at roomtemperature, the second term in the conductivity equation may be neglected

∴ σn ≈ Ndeµe (5.13)

142 Semiconductors

Similarly, the conductivity of a p-type semiconductor willbe dominated by majoritycarrier holes and may be represented as

σp ≈ Naeµh (5.14)

whereNa is the acceptor concentration in p-type semiconductor.

5.3.2 Effect of temperature on the conductivity of extrinsic semiconductors

Of the two mechanisms responsible for carrier generation inextrinsic semiconductors,the ionisation of impurity levels predominates at low temperatures as the probabilityof the covalent bonds being broken will be small.

CB

VB

CB

VB

T = 400 KT = 200 KT = 0 K

T = 0 K T = 200 K T = 400 K

(a)

(b)

DL

FL

AL

FL

Figure 5.10Effect of temperature on the Fermi level in (a) n-type semiconductor and(b) p-type semiconductor.

However, as the temperature is increased, more and more covalent bonds will be bro-ken thereby increasing the minority carrier concentration. The effect of temperature isillustrated by the energy band diagrams shown in Fig.5.10. In an n-type semiconduc-tor, at 0 K, all the donor levels will be neutral; i.e., no electrons are contributed to theconduction band by the impurity atoms. The Fermi level, which represents the energylevel for which the probability of occupation is half, lies mid-way between the impuritylevel and the bottom of the conduction band. As the temperature is increased, the donorlevels will be ionised to a greater extent as compared to breaking of covalent bonds.At elevated temperatures, when most of the donor levels are ionised, the carrier gen-eration due to breaking of covalent bonds will dominate the variation of conductivity

Semiconductors 143

(1 / T)

ln σ

Figure5.11Variation of lnσ with (1/T) for an extrinsic semiconductor.

with temperature. The Fermi level gradually shifts closer to the centre of the forbid-den energy gap and the semiconductor behaves more like an intrinsic semiconductor.This effect is called‘intrinsic effect’. The temperature at which the intrinsic conduc-tivity becomes dominant is called thetransition temperature at which the extrinsicsemiconductor starts behaving like an intrinsic semiconductor.

This transition temperature decides the maximum temperature at which the extrin-sic semiconductor or the devices made out of the extrinsic semiconductor can be used.For germanium, this temperature is around 70C and for silicon it is about 150C. Thetemperature dependence of conductivity of an extrinsic semiconductor is shown inFig.5.11. The graph shows a low temperature region, called the extrinsic region, wherethe conductivity is dominated by the ionisation of impuritylevels and a high temper-ature region, called the intrinsic region, where the carrier generation is dominated bythe breaking of covalent bonds.

The temperature dependence of conductivity of a p-type semiconductor will be ex-actly identical to that of an n-type semiconductor except ofthe fact that in the extrinsicregion, the conductivity is decided by the concentration and mobility of majority carri-ers (holes) generated due to the ionisation of acceptor atoms and the minority carriers(electrons) generated due to the breaking of covalent bonds.

5.3.3 Concentration and mobility of current carriers

In an intrinsic semiconductor, the carriers are generated due to the breaking of covalentbonds. In an extrinsic semiconductor, an additional mechanism namely ionisation of

144 Semiconductors

impurity levels will contribute current carriers. In an intrinsic semiconductor, electronsand holes are produced as pairs and hence are equal in number.At a given temperature,the concentration of electrons, and holes is constant and isknown as the‘intrinsiccarrier concentration’ of the given semiconductor.

i.e. at constant temperature,ni = pi = constant.In an extrinsic semiconductor, the carriers are generated due to ionisation of impu-

rity atoms producing mobile majority carriers (electrons in an n-type semiconductorsand holes in a p-type semiconductor) along with immobile impurity ions (positive ionsof donor atoms in the case of n-type semiconductors and negative ions of acceptoratoms in the case of p-type semiconductors). In addition, there will be a finite num-ber of covalent bonds also being broken at a given temperature. It is observed that theconcentration of majority carriers and the concentration of minority carriers are relatedto each other and the intrinsic carrier concentration of thesemiconductor material at agiven temperature as

np= n2i (5.15)

This relation is called‘Law of mass action’. An attempt to increase one type ofcarrier prompts a recombination resulting in a decrease of both carriers to maintainthe product constant. Further, since the semiconductor is electrically neutral, the totalcharge, summing of both mobile and immobile charges, must bezero. In a hypotheticalcase of a semiconductor having both acceptor and donor atomswith concentrationNa

andNd respectively which are ionised at a given temperature, we can write

n+ Na− = p+ Nd+ (5.16)

This is called the‘equation of charge neutrality’ .The conductivity of a semiconductor is decided not only by the number of mobile

charge carriers available but also by their mobility. Underthe influence of an appliedelectric field E the charge carriers will move with an averagevelocity vd called thedrift velocity . According to Ohm’s law, conductivity is given by

σ = J/E = nevd/E = neµ (5.17)

whereJ is the current density andvd is the drift velocity at an applied electric fieldE.The quantity (vd/E) is a constant specific to a particular type of carrier and is calledthemobility of the carrier. Since we have both electrons and holes contributing to theconductivity,

σ = neµe + peµh (5.7)

It may be mentioned here that the mobility of holes is usuallyless than the mobilityof electrons since the mobility of holes is nothing but the mobility of electrons in the

Semiconductors 145

valence band. The mobility is dependent on the scattering processes in the semicon-ductor. The effect of scattering by the atomic vibrations is different from the effect ofionised impurity atoms on the mobility of the carriers. The temperature dependence ofmobility is a combination of the effect of temperature on the two scattering processes.

A comparison of the important characteristics of intrinsicand extrinsic semicon-ductors is given in Table 5.1.

Table5.1Comparison between intrinsic and extrinsic semiconductors

Intrinsic semiconductors Extrinsic semiconductor1. It is a chemically pure, structurally

perfect and if compound, stoichio-metric material.

It is an impure material, obtained byadding a group III or group V ele-ment as dopant into an intrinsic semi-conductor or structurally imperfect ornonstoichiometric, if compound.

2. It behaves as an insulator at 0 K It behaves as an insulator at 0 K.

3. At normal temperature, conductivityis low.

At normal temperature, the conductiv-ity is high.

4. Carrier generation is due to breakingof covalent bonds.

Carrier generation is due ionisation ofimpurity atoms (forming majority car-riers) and breaking of covalent bonds(forming minority carriers).

5. Current carriers, viz., electrons andholes are equal in number

Current carriers are essentially major-ity carriers at normal temperature

6. Fermi level lies at the centre of the for-bidden energy gap at all temperatures

At 0 K, the Fermilevel lies close to theconduction band in n-type semicon-ductor and close to the valence bandin p-type semiconductor. Fermi levelmoves towards centre of the band gapas temperature increases.

7. Hall coefficient is negative due tohigher mobility of electrons

Hall coefficient is negative for n-typesemiconductor and positive for p-typesemi-conductor.

146 Semiconductors

5.4 Generation and recombination of carriers

At a given temperature, electron-hole pairs will be continually generated due to break-ing of covalent bonds. This is equivalent to the excitation of electrons from the valenceband to the conduction band thereby creating two mobile charge carriers, electron inthe conduction band and hole in the valence band. At the same time, the reverse pro-cess of recombination is also taking place in which electronrecombines with a holeannihilating each other (Fig.5.12).

CB

VB

Figure 5.12Energy level diagram showing the processes of generation and recombi-nation of electron – hole pairs.

The direct transition of an electron from the conduction band to the valence bandwill have to satisfy the conservation laws of energy and momentum. Since the prob-ability that an electron and hole have an equal and opposite momentum is very less,a direct recombination is a rare event. Most of the recombinations take place withthe help of an impurity atom or a lattice defect. These sites where a recombination isfavoured are called‘traps’ or ‘Recombination Centres’. In a semiconductor, theselevels are located in the forbidden energy gap.

5.5 Direct and indirect band gap semiconductorsThe energy band diagrams which we have used to describe semiconductors are toosimplified and are not sufficient to explain the behaviour of semiconductors. For ex-ample, the recombination process is dependent both on the presence of electron-holepairs but also requires the conservation of momentum. The energy and momentum dis-tribution of carriers may be explained using band diagrams in k-space. These are calledE-k diagrams. In the reduced zone scheme,k represents the propagation vector and isrelated to the momentum of the carrier. It is observed that the E − k diagrams will bedifferent for different directions ofk, i.e the direction of travel of the carrier. Fig.5.13shows typicalE − k diagrams. The forbidden energy gap for this semiconductor will

Semiconductors 147

E

CB

VB

k

E

CB

k

VB

(b)(a)

Figure5.13E-k diagrams for (a) direct band gap semiconductor and (b) indirect bandgap semiconductor.

be the energy difference between the valence band maximum and the conduction bandminimum.

If these two extreme points lie at the samek-value, sayk = 0 as in Fig.5.13(a),the semiconductor is said to be adirect band gap semiconductor. If the valence andconduction band edges do not lie at the same value ofk (Fig.5.13b), the semiconductoris calledindirect band gap semiconductor. Silicon and germanium are indirect bandgap semiconductors whereas gallium arsenide, cadmium selenide, cadmium tellurideetc are examples of direct band gap semiconductors. In case of a direct band gap semi-conductor, when an energy equal to the band gap energy is supplied, a direct transitionof an electron from the valence band maximum to the conduction band minimum canoccur. However, in case of an indirect band gap semiconductor, in addition to the en-ergy requirement, a transition can occur only with the assistance of a phonon (latticevibration). Thus, we see that in case of a direct band gap semiconductor, the energy andmomentum conservation requirement for the transition are met in the interaction of anelectron and a photon whereas in case of an indirect band gap semiconductor, an inter-action of an electron with a photon is assisted by a phonon to satisfy the conservationconditions.

The band gap of a semiconductor can be deduced from optical absorption studies.Such studies also give us information whether the band gap isa direct one or an indi-rect one. When radiation of suitable energy is incident on a semiconductor sample ofthicknessx, the intensity of radiation transmitted is given by

I = I exp (−αx) (5.18)

148 Semiconductors

whereI is the incident intensity andα is the absorption coefficient. In a typical plotof the variation ofα with the energy of the incident radiation, we observe an increasein the absorption coefficientα near the band gap energy. In case of a direct band gapsemiconductor, the increase atEg is very sharp and in case of an indirect band gapsemiconductor the increase will be less steep (Fig.5.14).

This is due to the fact that in case of indirect band gap semiconductors, transitionsof electrons from valence band to the conduction band requires the assistance of aphonon in order to achieve the momentum conservation.

The major differences between direct and indirect band gap semiconductors aresummarized in Table 5.2.

gΕ gΕ E E

αα

(a) (b)

Figure 5.14Variation of absorption coefficient as a function of energy for (a) indirectband gap semiconductor and (b) direct band gap semiconductor.

Table5.2Distinction between direct band gap and indirect band gap semiconductor.

Direct band gap semiconductor Indirect band gap semiconductor

1. The valence band maximum and theconduction band minimum are at thesame value of propagation vector

The valence band maximum and theconduction band minimum have dif-ferent values of propagation vector.

Semiconductors 149

2. Transition between valence band andconduction band involves an electronand a photon of energy equal to theband gap energy.

Transition between valence band andconduction band involving an elec-tron and a photon of energy equal tothe band gap energy is assisted by aphonon to conserve momentum.

3. Presence of traps or recombinationcentres is not necessary for the tran-sitions to occur.

Presence of traps or recombinationcentres is necessary for the transitionsto occur.

4. Absorption increases drastically at anenergy equal to the band gap energy.

Absorption does not increase drasti-cally at an energy equal to the bandgap energy.

5.5.1 Semiconductor materials

There are many semiconductor materials but silicon and germanium are the most im-portant ones. They are elemental semiconductors belongingto group IV. The otherelement of group IV namely carbon which crystallizes in f.c.c. structure is called di-amond & has a band gap of 6 eV. It is an insulator. Silicon (1.1 eV) and germanium(0.72 eV) are semiconductors. Gray tin, which is crystallized in f.c.c. structure hasa band gap of 0.01 eV. Lead is a metal with no forbidden energy gap. Apart fromthese elemental semiconductors, there are a large number ofcompound semiconduc-tors. Of them, binary compounds belonging to III-V and II-VIare quite important.Binary semiconducting compounds like GaAs, InAs, InSb, GaPetc. are examples ofIII-V compounds. Zns, CdS, ZnTe, CdTe, etc. are examples of II-VI semiconductingcompounds. The nature of bonding in these binary compounds is partly covalent andpartly ionic. The properties of these compounds are seriously modified not only bythe presence of impurities but also by the changes in the proportion of the two con-stituent elements in the compound. Hence, preparation of single crystals of compoundsemiconductors is more complicated than preparing single crystals of elemental semi-conductors. Availability of new material, ease of purification and preparation of singlecrystals, monitoring of the various physical properties are some considerations thatdecide the suitability of semiconductor materials for applications. From these pointsof view, silicon and germanium have dominated the semiconductor scene and havewell developed technologies. However, silicon and germanium are both indirect bandgap semiconductors and hence are not very efficient for certain application. Galliumarsenide (GaAs) with a direct band gap energy of 1.4 eV has been found to be an im-

150 Semiconductors

portant compound semiconductor. Some important properties of these semicondutorsare summarized in table 5.3.

Table5.3Some important properties of Ge, S i and GaAs

Property Germanium Silicon Gallium Arsenide

1. Atomic number 32 14 –

2. Atomic weight 72.6 28.1 144.6

3. Crystal structure Diamond Diamond Zinc blende

4. Lattice parameter, nm 0.565 0.543 0.565

5. Band gap, eV 0.67 1.1 1.43(indirect) (indirect) (direct)

6. Intrinsic resistivity,Ω m (300 K) 45 2300 3.7× 106

7. Intrinsic carrierconcentration,m-3 (300K) 2.5× 1019 1.5× 1016 1.4× 109

8. Electron mobility,m2v−1s−1 0.39 0.14 0.85

9. Hole mobilitym2v−1s−1 0.19 0.05 0.045

10 dielectricconstant. 16 11.8 11.1

11. melting pointoC 958 1412 1240

12. Max. devicetemperatureoC 70 150 350

Both germanium and silicon have diamond structure. It is a combination of twointerpenetrating face centered cubic sublattices, displaced one quarter of the lengthalong the body diagonal. In other words, if one sublattice has its origin at (0,0,0), theother has it at (a/4, b/4, c/4) or (a/4, a/4, a/4) since a= b= c. Gallium arsenide haszincblende structure (Fig.5.15), named after the crystal structure of zinc blende (ZnS).

This structure is identical to diamond structure except that the two face centeredcubic sublattices are made up of different atoms namely Ga and As. Consequently Si

Semiconductors 151

As

Ga

(b)(a)

Figure5.15Unit cell of (a) germanium and silicon, and (b) gallium arsenide.

and Ge have 8 atoms per unit cell where as GaAs has four Ga and four As atom perunit cell. Because of the identical crystal structure and the lattice parameter all thethree semiconductors have an identical number of atoms per unit volume. However,the intrinsic carrier concentration and hence the intrinsic resistivity which are functionsof the band gap energy vary considerably. Germanium has the lowest value of bandgap and hence the highest value of intrinsic carrier concentration with an associatedlowest intrinsic resistivity. This low value of intrinsic resistivity leads to large leakagecurrents in reverse biased p-n junctions which may not be desirable in some deviceapplications. Further, germanium has the lowest transition temperature at which thematerial looses its extrinsic behaviour and becomes practically an intrinsic material.Hence, p-n junction devices fabricated using germanium cannot be used above thistransition temperature of 70C. On the other hand, silicon proves to be a very usefulmaterial for many device applications. Gallium arsenide ispreferred over silicon insome applications like light emitting diodes, Gunn diodes,etc.

5.6 Hall effect

If a conductor carrying a current is placed in a transverse magnetic field, an electricfield is produced in the conductor in a direction perpendicular to both the current andthe magnetic field. This phenomenon is calledHall effect. The electric field generatedis called Hall field and the corresponding voltage, Hall voltage. Hall effect was firstused to study metals but because of the large number of carriers present in metals, theeffect is not as spectacular as it is for semiconductors. The phenomenon has turned outto be a valuable tool for the evaluation of many semiconductor parameters.

Consider a rectangular slab of an n-type semiconductor sample. Let I be the con-ventional, current flowing through the sample along positive x-direction (Fig.5.16). A

152 Semiconductors

b

t

Face 1 (−)

Face 2 (+)

l

I

B

Figure5.16Schematic representation of Hall effect.

magnetic field B is applied perpendicular to the direction ofcurrent flow, say alongpositive z-direction. The effect of the magnetic field is to exert a force on the currentcarriers, namely electrons, flowing in the negative x-direction. As a result of this force,the electrons will be forced downwards in the negative y-direction. This results inthe lower face (Face 1 in the figure) becoming negatively charged with respect to thetop face (Face 2). This constitutes the Hall fieldEH. This Hall field will oppose thedownward movement of electrons and an equilibrium will be set up when

eEH = Bev (5.19)

wherev is the drift velocity of the electrons.

EH = Bv (5.20)

For a given semiconductor, theHall coefficient RH is defined as the Hall field generatedin the material per unit magnetic field applied when the current density is unity.

i.e., RH = EH/BJ (5.21)

The current densityJ can be expressed in terms of concentration of carriers n and theirdrift velocity as

J = nev (5.22)

From equation (5.20) and (5.22) we get

RH = Bv/Bnev = 1/ne (5.23)

Semiconductors 153

In the present case of n-type semiconductor,

RH = −1/ne (5.24)

since the Hall field developed is in negative Y- direction. For a p-type semiconductor,RH will be positive and is given by

RH = 1/pe (5.25)

Since all the three quantitiesEH, B and J can be measured, the Hall coefficient andthe carrier concentration can be found out.

Equations (5.23) and (5.24) are valid for the case of free electrons moving withsteady drift velocity. However, in practice, assuming average drift velocity for carriers,the Hall coefficient is given by

RH =3π8·

1pe

for p-type semiconductor (5.26)

= −3π8· 1

nefor n-type semiconductor (5.27)

For an n-type material, the conductivity is given by

σn = neµe

∴ µe =σn

ne= −

83π· σnRH (5.28)

Similarly, for a p-type material

µh =83π· σpRH (5.29)

5.6.1 Experimental determination of carrier concentration

Consider a semiconductor sample in the form of a rectangularslab of lengthl, breadthb and thicknesst. Let the current flowing result in a current densityJ across the crosssectionbt. Let B be the magnetic field acting along the thicknesst. We have, for ann-type semiconductor, from equation (5.21) and equation (5.27),

EH

BJ= −

38·

1ne

(5.30)

But EH = VH/b andJ = I/bt

154 Semiconductors

whereVH is the Hall voltage generated across the breadth andI is the current flowingalong the length. Substituting these values in equation (5.30), we get

n = −3π8·

BIetVH

(5.31)

Here,n represents the number of charge carriers per unit volume of the semicon-ductor and the negative sign signifies that the carriers are electrons.

By maintaining currentI constant, the variation of Hall voltageVH with magneticfield is studied. The graph ofVH versusB will be a straight line (Fig. 5.17). The carrierconcentration can be determined as

n =3π8·

Iet·

1slope

(5.32)

The experiment may also be repeated by studying the variation of Hall voltage withcurrent I at a constant magnetic field.

V

B

H

Figure5.17Variation of Hall voltage with the applied magnetic field.

5.6.2 Hall effect in intrinsic semiconductors

In the above discussion of Hall effect in extrinsic semiconductors, the motion of ma-jority carriers is analysed neglecting that of minority carriers. In an intrinsic semicon-ductor, the number of electrons and holes will be same and hence the motion of both

Semiconductors 155

type of carriers is to be considered. The expression for the current densityJ used inequation (5.22) has to be modified for the intrinsic semiconductor as

J = neve + pevh

or J = (neµe + peµh)Ex (5.33)

whereve andvh are the drift velocities of electrons and holes respectively due to theapplied fieldEx. This yields an expression forRH as

RH =1e·

pµ2h − nµ2

e

(pµh + nµe)2(5.34)

Even though the number of electrons is equal to the number of holes in an intrinsicsemiconductor, since the mobility of electrons is greater than that of holes, the Hallcoefficient will be negative.

5.6.3 Applications of Hall effect

Hall effect can be used to identify the semiconductor type, namely n-type or p-type.It is useful in evaluating the carrier concentration. Knowing the conductivity, it ispossible to find the mobility of the carriers. Using a standard sample of known Hallcoefficient, it is possible to measure magnetic fields. This is the principle of construc-tion of a gauss meter.

5.7 p-n Junction

The p-n junctions are among the most simple semiconductor devices. They are usefuldevices in the electronic circuits to perform functions like rectification, amplification,switching, etc. A p-n junction may be obtained by different methods and the junctionsare known by the methods of fabrication. They are:

(a) Grown-in-junction(b) Alloy junction(c) Diffused junction(d) Ion implanted junction(e) Deposited junction.

A grown – in – junction is obtained by abruptly changing the dopant during crys-tal growth process from melt. Initially the melt may containn-type impurities in thesemiconductor melt. At some stage of growth, p-type impurities are added in excess to

156 Semiconductors

compensate for the n-type impurities already present so that a net p-type character isachieved. The process is called “counterdoping” and the junction is known as “grown-in junction”. Another technique for obtaining p-n junctionis alloying a semiconductorwith a metal acting as an opposite type of dopant. For example, an n-type germaniumsample may be alloyed with indium by placing a pellet of indium metal on it and melt-ing it. This results in a region of germanium rich in indium giving it a p-type character.The most common technique for forming p-n junctions is byimpurity di ffusion. Itis achieved by selective masking and exposing semiconductor regions to vapours ofimpurities at elevated temperatures. If a wafer of n-type silicon is placed in a furnaceat 10000C in a gaseous atmosphere containing a high concentration of boron atoms,the boron atoms diffuse into the wafer. If the concentration of the diffused boron atomsis higher than the concentration of the n-type impurities inthe silicon, the top layerbecomes p-type and hence the p-n junction is obtained. The junction obtained will bea graded one with the depth and concentration of doping depending on the temperatureand duration of diffusion process.Ion implantation is a technique in which the im-purity ions are accelerated and made to incident on semiconductor surfaces. With thistechnique, it is possible to control the depth of the junction and also obtain abrupt junc-tions. Another technique widely used in integrated circuitfabrication is the depositionof thin films of semiconductor materials with the required impurity and concentrationto obtain sharp junctions.

Junctions may be abrupt or graded. Anabrupt junction , also known as astepjunction is one in which the acceptor concentration and the donor concentration areconstant and extend upto the junction. In agraded junction, the impurity concentra-tion varies with distance from the junction in some manner. Both types of junctionsare used for different applications.

5.7.1 Unbiased p-n junction

Consider a junction between an n-type and p-type semiconductor which has just beenformed. The n-type semiconductor contains some of the semiconductor atoms replacedby donor atoms of groupV. Similarly, the p-type semiconductor contains some of thesemiconductor atoms replaced by acceptor atoms belonging to group III. The concen-tration of the donor atoms may be different from the concentration of acceptor atomsin a given junction.

Fig. 5.18 shows a p-n junction so formed where only the impurity atoms are shown.A n-type impurity atom is shown to have an additional electron weakly attached toit. Similarly, p-type impurity atoms are associated with holes. When the junction isformed, the electrons from the n-type region diffuse into the p-type region because of

Semiconductors 157

Group V atomswith electrons

Group III Atomswith holes

p n

np

Space charge region

(a)

(b)

Figure5.18Formation of a p-n junction (a) before charge transfer and (b) after chargetransfer.

the large concentration of electrons on the n-side and the lack of it on the p-side. Hence,the electrons from the donor atoms close to the junction cross the junction to combinewith the holes on the p-side. When the electrons leave the donor atoms, they leavebehind positively charged donor ions. Similarly, the acceptor atoms become negativelycharged acceptor ions when the electrons recombine with theholes. Consequentlythere is positively charged region of immobile donor ions close to the junction on then-side and a negatively charged region of immobile acceptorions on the p-side. Theseare referred to asspace charge regions. They are also calledcharge depletion regionssince there are no mobile charges in these regions. However,such diffusion of carriersstops when the space charge produced is sufficient to prevent the carriers from crossingover. The potential existing due to the space charge regionsat equilibrium is called thebarrier potential .

Fig.5.19 shows the variation of impurity concentration, charge density and poten-tial across the junction. Since the total charge density in the space charge region oneither sides of the junction is equal, we have

NdXn = NaXp (5.35)

158 Semiconductors

Xp

Xp

nX

nX

Nd

Nd

Na

Na

V2

V1

(c)

(b)

(a)

p n

e

e

d

d

d

V

Figure5.19Variation of (a) impurity concentration (b) space charge and (c) potentialas a function of distance for an abrupt p-n junction.

whereNd andNa represent the concentration of impurities in the n-type andp-typematerials,Xn andXp are the widths of the charge depletion regions on the two sides.This is called thecharge neutrality condition. Thus, we see that if the concentrationof impurities is higher on any side, the width of the depletion region on that side will beproportionately smaller. The effective width of the junction is the sum of the depletionwidths on the two sides. In this region of the junction, the bulk properties of the p-typeand n-type materials are not valid.

Figure 5.20 shows the energy level diagrams for the p-n junction before and afterthe diffusion of carriers. When the equilibrium is established after the transfer of car-riers, the Fermi levels on the n-side and p-side will be levelled since the Fermi level

Semiconductors 159

p

FE

EF

n

eVB

p

p

n

n

(a)

(b)

Figure 5.20Energy level diagrams for a p-n junction (a) before the formation of thejunction and (b) after the formation of the junction.

represents the energy level where the probability of occupation is half and is same oneither sides of the junction close to it. As a result, the energy bands on the p-type sidewill be at a higher energy level as compared to those on the n-side. An electron at thebottom of the conduction band on the n-side has to acquire an energy equal toeVB tocross over to the p-side, whereVB represents the barrier potential. It can be shown thatthe barrier potential is related to the concentration of impurities on the two sides of thejunction and is given by

VB =kTe

loge

NaNd

n2i

(5.36)

The width of the charge depletion regions is also a function of the impurity concentra-tion and are given by

Xp =

[

2 ∈ VB

eNa(1+ Na/Nd)

]1/2

(5.37)

and Xn =

[

2 ∈ VB

eNd(1+ Nd/Na)

]1/2

(5.38)

160 Semiconductors

where∈ is the permittivity of the semiconductor material.Thus, we see that in spite of the potential barrier, because of the large concentra-

tion of majority carrier electrons on the n-type side of the junction, an electron currentcalled thediffusion current will flow from n-type side to the p-type side. At the sametime, the minority carrier electrons in the conduction bandof the p-type side havehigher energy compared to the electrons on the n-type side (Fig.5.20b) and will be as-sisted by the barrier potential to flow down the barrier constituting thedrift current .At equilibrium for an open circuited junction the net flow of current is zero; i.e., thediffusion current of majority carriers will be equal to the driftcurrent of minority carri-ers across the junction. In a similar way, we can also consider the currents constituteddue to the movement of holes in the valence bands. The diffusion current of majoritycarrier holes from p-side to n-side will be equal to the driftcurrent of minority carrierholes from n-side to p-side at equilibrium. We also note thatthe net current flow due tothe movement of carriers in the conduction band and the valence band are separatelyequal to zero at equilibrium.

5.7.2 Semiconductor junction with applied bias

A semiconductor junction may be subjected to an applied voltage in two ways. If thep-type material of the junction is made positive with respect to the n-type, the junctionis said to beforward biased. When the p-type is made negative with respect to the n-type, the junction is reverse biased. When an external voltage is applied to the junction,the equilibrium of current flow is disturbed. Let us considerthe two biasing conditionsseparately and study the characteristics of the p-n junction under these conditions.

Consider a p-n junction with a forward biasV applied to it (Fig.5.21). Since then-type region is made negative, the applied field assists themajority carriers electronsto cross over to the p-side of the junction. Energetically, this can be represented byan energy band diagram of the p-n junction with all the energylevels (including theFermi level) raised by an energyeV whereV is the applied forward bias (Fig.5.21b).In other words, the effective barrier is now reduced to a value (VB−V) which results inan increased diffusion current of majority carriers. However, the minority carrier cur-rent due to the drift remains unaltered. Thus a net current ofmajority carrier electronswill flow from n-type region to the p-type region. Also flowingwill be a net currentof majority carrier holes from p-type to n-type region which, with the electron current,constitutes the total current across the junction. As a result, the p-n junction will offera low resistance to the flow of current when forward biased.

Semiconductors 161

Figure5.21The energy level diagram for forward biased p-n junction.

The junction is said to bereverse biasedwhen the p-type region is made negativewith respect to the n-type region (Fig.5.22a). The applied voltage now adds to thebarrier potential to present an increased barrier (VB + V) to the diffusion of majoritycarriers (Fig.5.22b). If the reverse bias is large enough, the majority carriers will beprevented from moving across the junction. The minority carrier current of electronsfrom p-type to n-type and of holes from n-type to p-type will now be constituting thetotal current across the junction. Since the drift current is independent of the barrierheight, this reverse current is called the reverse saturation current. Thus, the p-n junc-tion offers high resistance to the flow of current when reverse biased. The voltage-current characteristics of a p-n junction under forward andreverse bias conditions isshown in Fig.5.23. The total current across the junction is given by the diode equation,

Figure5.22The energy level diagram for reverse biased p-n junction.I = Io[exp (eV/kT) − 1] (5.39)

whereV is the applied bias.

162 Semiconductors

V

I

Figure5.23Current-voltage characteristics for a p-n junction.

In forward bias, the diffusion current,Io exp (eV/kT) increases exponentially withthe applied forward bias and is very large compared to the constant drift currentIo. It isobserved that the total forward current is very small upto a certain voltageVk and thenincreases exponentially with the applied voltage. This voltageVk is called the cut-involtage or threshold voltage. Hence, we can approximate thetotal current in forwardbias as

IF = Ioexp (eV/kT) (5.40)

In reverse bias, sinceV is negative, the exponential term becomes negligible and thenet current across the junction is given by

Ir = −Io (5.41)

The diode equation (5.39) may also be written in the form

I = Io[exp (V/VT) − 1] (5.42)

whereVT = (kT/e) is called the volt equivalent of temperature.At 300 K, VT = 0.026VThe diode equations given above are called Ideal diode equations since they do not

take into account the processes of carrier generation and recombination taking place in

Semiconductors 163

the space charge region of the diode. In general, we can write

I = Io[exp (V/nVT) − 1] (5.43)

wheren is called the ideality factor. The value of ideality factor is close to one forgermanium diode but is close to 2 for silicon diode under small current operation.

5.7.3 Incremental junction capacitance

The expression for the depletion width on the two sides of thep-n junction under biasedcondition may be written as

Xp =

[

2 ∈ (VB − V)eNa(1+ Na/Nd)

]1/2

and Xn =

[

2 ∈ (VB − V)eNd(1+ Nd/Na)

]1/2

The total charge in the depletion regions is given by

Q = eNdXn = eNaXp

=

[

2e ∈ NaNd(VB − V)(Na + Nd)

]1/2

The variation in the junction chargeQ with the applied voltageV results in an incre-mental junction capacitance given by

C = dQ/dV

By differentiation of the above expression forQ with respect toV, we get

C =

[

e ∈ NaNd

2(Na + Nd)

]1/2

V−1/2

=∈

(Xn + Xp)

Thus, we see that for an abrupt junction, the capacitance varies as the square root ofthe reverse biasVr . In a graded junction, the capacitance can be written in the form

C ∝ V−nr

164 Semiconductors

The value of the exponentn is equal to (1/3) for a linearly graded junction. Thus, thevoltage sensitivity of capacitance is greater for an abruptjunction than for a linearlygraded junction. By choosing various doping profiles, it is possible to control thedependence ofC onVr .

m = − 3/2

m = 0

m = 1

P

Na

N

Ox

n

Figure5.24A few doping profiles of p-n junction.

Figure 5.24 shows a few examples of graded junction profiles.Normally, the dopingprofile on one side of the junction, say p-side, is kept uniform and the concentration isvaried on the n-side. The donor distribution on the n-side may be represented by

Nd(x) = kxm

wherek is a constant and the exponentm is 0(abrupt),1(linear) or -3/2(hyperabrupt).It can be shown that the exponent n is given by 1/(m+ 2) and will be equal to (1/2),(1/3) and 2 for abrupt, linearly graded and hyperabrupt junctions respectively. Thehyperabrupt junction is particularly interesting since the capacitance is proportionalto V−2

r and when such a junction is used with an inductorL in a resonant circuit, theresonant frequency varies linearly with the applied voltage.

5.7.4 Breakdown in a p-n junction

In a reverse biased p-n junction, as the applied voltage is increased the reverse currentremains practically constant. As the voltage is increased further, at a particular voltage,

Semiconductors 165

the current begins to increase very rapidly (Fig.5.23). Thereverse biased p-n junctionwhich was offering a high resistance to the flow of current suddenly startsconductingheavily. This phenomenon in which the current increases very rapidly at a particularreverse voltage is called breakdown. The mechanism of breakdown is due to impactionisation of lattice atoms in the depletion region by the energetic minority carriers.For example, an electron from the p-side may acquire enough kinetic energy from theapplied electric field to cause ionising collision with lattice atoms. This results in thebreaking of covalent bonds leading to the generation of electron-hole pairs. The carri-ers so generated are further accelerated by the applied fieldand may cause ionsiationevents thereby resulting in an avalanche process and hence alarge current. This typeof breakdown is called“Avalanche breakdown” and is normally observed in lightlydoped p-n junctions.

5.8 Zener diode

A zener diode is a heavily doped p-n junction. When the concentration of impurities inthe p and n type semiconductors used for the junction is high,the width of the chargedepletion region reduces. The voltage-current characteristics of a zener diode is iden-tical to that of an avalanche diode with a forward region where the current increasesexponentially with voltage, a reverse region where the current shows a saturation anda breakdown region. However, the mechanism of breakdown in azener diode is dis-tinctly different.

5.8.1 Zener breakdown mechanism

The generation of a large current at breakdown is due to a quantum mechanical phe-nomenon calledtunneling. Tunneling can take place when the following two condi-tions are satisfied.

(i) The width of the depletion region is very small.

(ii) Under the biased condition, there are vacant sites in the conduction band of n-side at the same energy level where valence electrons are present in the valenceband of p-side.

Application of reverse bias will result in the lowering of all energy levels on then-side (Fig.5.25). At the breakdown voltage, the bottom of the conduction band onthe n-side will be just below the top of the valence band on thep-side. Since thebarrier width is very small, electrons will tunnel from the valence band of p-side to

166 Semiconductors

EF

p

EF

n

p n

CB

VB

VB

CB

Figure5.25Energy level diagram for a Zener diode operating in the breakdown region.

the conduction band of n-side. In other words,zener breakdowncan be consideredto be equivalent to field ionsiation where the applied field islarge enough to pull theelectrons out of the covalent bonds and accelerate them to the n-side of the junction.An electric field of the order of 106V/cm is required for zener breakdown. However,such a high field is realised even at low voltages (usually below 6V) since the width ofthe depletion region is very small. If the breakdown occurs at a higher voltage (above6V) the mechanism is avalanche breakdown. The avalanche breakdown voltage is verystable and is useful as a reference voltage in electronic circuits. The p-n junction diodesused for this purpose are called zener diodes even though they are avalanche diodes.

5.8.2 Identification of breakdown mechanism in a p-n junction

The temperature coefficient of breakdown voltage is helpful in identifying the mecha-nism of breakdown in a given diode. If the breakdown voltage is below 6V, the mech-anism is zener breakdown and the temperature coefficient of breakdown voltage isnegative. In other words, the breakdown voltage decreases as the temperature is in-creased. This is because an increase in temperature increases the energy of the valenceelectrons and hence lesser voltage is sufficient to pull them out of the covalent bonds. Ifthe breakdown voltage is above 6V, the breakdown is due to avalanche mechanism andthe temperature coefficient of breakdown voltage is positive. The breakdown voltageincreases with an increase in temperature. This is due to thefact that the vibrationalmotion of the lattice atoms increases with temperature thereby increasing the proba-bility of the charge carriers being scattered due to collisions. Thus, the charge carriershave lesser probability gaining sufficient energy to start the avalanche multiplication.

Semiconductors 167

The distinction between avalanche diodes and zener diodes is summarized in Table 5.4.

Table5.4Distinction between avalanche diodes and zener diodes.

Avalanche Diodes Zener Diodes1. The doping level is low. The concen-

tration of impurity atoms is usually ofthe order of 1018 to 1020m−3 in silicon.

The doping level is high. The concen-tration of impurity atoms is of the or-der of 1022 to 1024m−3 in silicon.

2. The width of the depletion region (orspace charge region) is quite large andis of the order of 10−6m in silicon.

The width of the depletion region (orspace charge region) is small and is ofthe order of 10−8m in silicon.

3. When biased, electric field existing atthe junction is low because of largerdepletion width.

When biased, electric field existing atthe junction is high because of smallerdepletion width.

4. In the reverse bias condition, break-down occurs at relatively large appliedvoltage (> 6V).

The breakdown occurs at relativelylow reverse bias voltage (< 6V).

5. The dynamic resistance of the diodeinthe breakdown region is large.

The dynamic resistance in the break-down region is relatively small.

6. The breakdown occurs as a result ofavalanche multiplication of minoritycarriers due to impact ionization ofsemiconductor carriers.

The breakdown occurs as a result oftunneling of minority carriers due tofield ionization of semiconductor car-riers atoms at the junction.

7 Breakdown voltage increase with in-crease in temperature.

Breakdown voltage decreases with in-crease in temperature.

5.9 Applications of p-n junctions

5.9.1 Junction diode as rectifier

When an alternating voltage is applied to a diode circuit, the diode will conduct easilywhen it is forward biased , but will allow negligible currentwhen reverse biased. Asa result, the output will be a unidirectional current. The process of converting analternating signal into a unidirectional signal is calledrectification. When a singlediode is used in the circuit, only half the cycle of the input will be allowed to passthrough and the output will have unidirectional pulses. A full wave rectifier makes useof two p-n junctions and hence, the output will be available during both the half cycles

168 Semiconductors

(Fig.5.26). However, the output will be pulsating and far from being a substitute forthe d.c power supply. The out put may be made more smooth usingfilters. Capacitorsare normally used as filters in rectifier circuits.

D1

D2

RL

Figure5.26Full wave rectifier circuit.

5.9.2 Zener diode

Zener diodes are most frequently used in regulator circuitsor as a reference voltage.Fig.5.27 shows a simple regulator circuit designed to maintain a fixed voltage acrossthe loadRL.

Vi

R

R+

−L

−

+

Figure5.27Zener diode as an voltage regulator.

For applied voltages greater than the breakdown voltage of the zener diode, thevoltage across the load will be maintained atVz. Since the voltage across the zenerdiode is constant and equal toVz in this case, it can be used as a reference for compar-ison with other voltages.

Semiconductors 169

Vi

V

V

Z1

Z2

(V + V )Z1 Z2

+

−

−

+

Figure5.28Zener diode as an voltage reference.

Two or more reference voltages can be obtained by using zenerdiodes in seriesas shown in Fig.5.28. As long as the input voltage is greater than the sum of thebreakdown voltages of the zener diodes connected in series,the voltages across thediodes will serve as reference voltages. Two zener diodes can be used back-to-back tomake a simple square wave generator (Fig.5.29) using the clipping action.

Vz

VzV

V

Z

Z

+

−

(b) (c)

t

(a)

Figure 5.29Use of zener diode for squarewave generation. (a) input waveform (b)circuit diagram (c) output wave form.

5.9.3 Photo diode

When bulk semiconductors are irradiated by light of energy higher than the energyband gap of the semiconductors, they show change in their electrical conductivity.This is termed as photoconductivity and the excess conductivity is proportional to theoptical absorption rate. Thus, semiconductors can be used to detect and measure inci-dent photon intensity through a measurement of the electrical conductivity. However,junction devices show a better sensitivity and speed of response as detectors. A pho-todiode is basically a reverse biased p-n junction embeddedin clear plastic with allsides painted black except a small aperture through which light is concentrated on to

170 Semiconductors

Figure5.30Biasing circuit for a photodiode.

the junction with the help of a lens. The circuit symbol and the biasing circuit is shownin fig.5.30. Under dark condition, the diode current is very small and constant, almostindependent of the applied reverse voltage. Under reverse bias condition, the currentacross the junction is due to the minority carriers which drift across the junction toconstitute the reverse saturation current. When the junction is illuminated, the elec-tron? hole pairs produced in the depletion region are separated by the junction field,the electrons being collected in the n-region and the holes in the p-region. The numberof electron – hole pairs produced being related to the numberof photons absorbed, aphoto-current proportional to the intensity of incident radiation will be added to the re-verse saturation current (Fig.5.31). Hence, the change in the reverse saturation currentwill be linearly related to the intensity of illumination. Thus, the photodiode is usefulfor detecting as well as measuring the intensity of incidentradiation.

0

L1

L2

L3

0

−1

−V

Figure 5.31 Current - voltage characteristics of a photodiode showing the effect ofintensity of illumination (L2 > L1 > 0).

Semiconductors 171

The advantages of using a reverse biased p-n junction as a photo detector are asfollows:

(i) The reverse saturation current is almost independent ofthe applied reverse biasvoltage and hence variations in this voltage will not affect the measurements.

(ii) The change in the magnitude of the reverse saturation current due to incidentlight is linearly related to the intensity of radiation. Thedark current, which isnormally very small at low temperatures, may even be neglected.

(iii) The electron-hole pairs produced in the depletion region due to illumination areseparated by the field existing at the junction and the recombination probabilityis reduced. The reverse bias condition results in wider depletion regions ensuringcomplete collection of photo generated carriers.

Applications of photodiodes include detection and measurement of radiations in-cluding infrared, visible, laser, x-ray,γ-rays etc. They are used in high speed countingapplications like reading punched cards and tapes for computers. The are used in highspeed switching applications to convert time varying optical signals to electrical sig-nals. They are also used in reading optical tracks of recorded sound, light operatedswitches, burglar alarms, etc.

5.9.4 Photovoltaic effect and solar cell

It is observed that in case of a photodiode, a constant reverse saturation current flowsacross the junction for large reverse voltages. As the reverse voltage is reduced, thepotential barrier existing at the junction is also reduced.This increases the probabilityof majority carriers diffusing across the junction. This flow of majority carriers re-duces the reverse saturation current as the reverse voltageapproaches zero. However,under illuminated condition, the reverse saturation current does not reduce to zero evenwhen the applied voltage goes to zero (Fig.5.31). In fact, the reverse current can bereduced to zero only with an applied forward bias. As the forward bias is increased, apoint is reached when the majority carrier current in the forward direction just equalsthe minority carrier current in the reverse direction. The net current becomes zero.This forward voltage at which the net current becomes zero iscalled ‘the photovoltaicpotential’. Under open circuit condition (i.e., when no load is connected across thep-n junction), the net current is zero and hence, the photovoltaic potential or the opencircuit photovoltage,Voc, is observed across the p-n junction. The reverse current ob-served when no bias is applied is called the short circuit photo currentIsc(Fig.5.32).

172 Semiconductors

−1

V0 VOC

A−I

ISC

Figure5.32Current - voltage characteristics of a photovoltaic cell.

When a load resistanceRL is connected across the illuminated p-n junction, theload line follows the relation,

V = −IRL

Thus, the load line is a straight lineOA with slope (−1/RL). The load line intersectsthe V-I curve at A. The current and voltage values across the load R can be determinedfrom the graph. The power that can be drawn from the p-n junction can be calculatedas

P = VI

when RL = 0,V = 0 ∴ P = 0

when RL = ∞, I = 0 ∴ P = 0

We find that the output power is maximum for a certain optimum load resistance.When the diode is used for converting radiant energy into electrical energy, the diodeshould be operated with this optimum load resistance.

Since power can be delivered to an external circuit by an illuminated junction, itis possible to convert solar energy into electrical energy.The p-n junction used forthe purpose is called a solar cell. In order to improve the efficiency of conversionfrom optical energy to electrical energy, it is necessary toconstruct solar cells with thefollowing characteristics:

1. The junction area should be large

2. The junction should be close to the top surface so that it can be illuminatedeffectively.

Semiconductors 173

3. The thickness of the n and p regions should be small compared to the diffusionlengths of the minority carriers in the respective regions to reduce recombination.

4. The series resistance of the device should be small so thatthe power is not lostdue to ohmic losses in the device itself.

5. It is desirable to have a large contact potential to have large photovoltage. Opencircuit photovoltage is less than the contact potential which in turn is less thanthe energy band gap equivalent of the material. The contact potential can beincreased by suitable doping of n and p regions. But, heavy doping reduces thecarrier lifetime, which in turn reduces diffusion length of carriers.

LightTop contact

Ohmic contact

Anti−reflectioncoating

n−Si−p−Si−

(b)

(a)

Figure5.33Construction of a solar cell. (a) cross-sectional view b) top view.

Fig. 5.33 shows the construction of a solar cell. The p-n junction has appropriatethickness of p and n regions. The p- region is made thin enoughto allow sufficientnumber of photons to reach the junction. An antireflection coating is given on topof the p- region to reduce possible reflection of the incidentradiation. The top metalcontact is made in the form of a grid to ensure effective collection of carriers.

The conversion efficiency of a solar cell is defined as

η =Output electrical power

Input radiant power(5.44)

174 Semiconductors

The efficiency of a solar cell is also expressed in terms ofquantum efficiencywhichis defined as

η∗ =number of electron-hole pairs generated

number of photons absorbed(5.44)

A well made silicon solar cell can have a conversion efficiency of 15% (η = 0.15)and can provide about 150 W of electrical power perm2 of cell area.

In space applications, solar cells are used for satellite power supply. In terrestrialapplication, they are used in earth communication equipments, for lighting, heating,water pumping, etc.

5.9.5 Light emitting diode

These are p-n junctions which emit visible light when forward biased (Fig.5.34).

Figure5.34Biasing circuit for a light emitting diode.

When a p-n junction is forward biased, the majority carriers(electrons) diffuse fromthe conduction band of n-type region into the conduction band of the p-type region.This is referred to as minority carrier injection since electrons are minority carriers inthe p-type region. This disturbs the equilibrium between electrons and holes in the p-type region and in order to regain equilibrium concentration, the electrons recombinewith holes thereby releasing energy. In all p-n junctions, some of this energy is givenoff as heat and some energy as light depending on the energy released. In indirectband gap semiconductors like germanium and silicon, greater percentage of the energyis given off as heat. In gallium arsenide phosphide (GaAsP), a large fraction of theenergy is released as light. The process of emission of lightfrom a forward biased

Semiconductors 175

p-n junction is called‘injection electroluminescence’. A diode made of GaAs with aband gap of 1.43 eV emits infrared radiation of about 9000A. When GaP (band gap= 2.26 eV) is used, the emitted radiation is in the green region. A mixed compound ofthe typeGaAs1−xPx with varying concentration of As and P may be used to select theemitted radiation in the visible region.

A commercial light emitting diode is made out of a suitable material and the diodeis embedded in a clear plastic cover with a parabolic reflector provided at the back todirect the emitted radiation. The emitted intensity is found to be linearly related tothe forward current flowing across the junction. The application of a forward bias is anecessary condition for the operation of a light emitting diode because minority carrierinjection takes place in a forward biased junction. The increase in minority carrierswill disturb the equilibrium and result in recombination associated with emission ofradiation. The recombination probability is high in the transition region, i.e., the regionjust outside the depletion region where the carrier concentration is modified due tocarrier injection. The emission can be concentrated by keeping the depletion region toa minimum using a forward bias.

Light emitting diodes are used in visual display units, calculators, clocks, panelmeters, pilot lamps, etc. They are used in optical communication applications usingfiber optics like data transmission and telephones. They arealso used in card and papertape readers, burglar alarms, etc.

Numerical Examples

5.1 Calculate the conductivity of silicon doped with 1021 atomsm−3 of boron if themobility of holes is 0.048m2v1s1.

Solution: Conductivity of the extrinsic semiconductor with an acceptor con-centration ofNa is given by

σ = Naeµh

= 1021× 1.6× 10−19× 0.048

= 7.68 ohm−1m−1 (Ans.).

5.2 Calculate the resistivity of intrinsic germanium if the intrinsic carrier density is2.5 × 1019m−3 assuming electron and hole mobilities of 0.38 and 0.18m2v−1s−1

respectively.Solution: Resistivity is given by

ρ =1σ=

1ne(µe+ µh)

176 Semiconductors

=1

2.5× 1019× 1.6× 10−19 × (0.38+ 0.18)= 0.45 ohm m(Ans.).

5.3 Calculate the electron and hole densities in n-Si doped with1019 donorsm−3 ifthe intrinsic carrier concentration in silicon is 1.4× 1016m−3.

Solution: Density of electrons,

n = 1019+ 1.4× 1016 ≈ 1019m−3

Density of holes,p = n2

i /n

whereni is the intrinsic concentration.

∴ p =(1.4× 1016)2

1019

= 1.96× 1013m−3 (Ans.).

5.4 Calculate the conductivity of germanium with the given data. What is the effectof doping germanium with donor - type impurity to the extent of one atom per108 germanium atoms?Given data: Intrinsic carrier concentration in germanium at 300 K,

ni = 2.4× 1019 m−3

Electron mobility,µe = 0.39m2v−1s−1

Hole mobility,µh = 0.19m2v−1s−1

Number of atoms/m3 = 4.4× 1028

Electron Charge,e= 1.6× 10−19C

Solution: Conductivity of intrinsic germanium is

σi = nie(µe + µh)

= 2.4× 1019× 1.6× 10−19× (0.39+ 0.19)

= 2.23 ohm−1m−1

Semiconductors 177

Conductivity of n-type germanium is

σn = neµe + peµh

n ≈ Nd ≈ 4.4× 1020m−3;

p = n2i /Nd = 1.3× 1018m−3

Sincep is very small compared ton, the second term in the conductivity equa-tion may be neglected.

∴ σn = Ndeµe= 27.46 ohm−1m−1 (Ans.).

5.5 The electrical conductivity of an intrinsic semiconductorincreases from 19.96ohm−1m−1 to 79.44 ohm−1m−1 when the temperature is increased from 60 to100C. Find the band gap energy of the semiconductor.

Solution: The electrical conductivity of an intrinsic semiconductoris givenby

σ = k.exp(Eg/2kT)

Neglecting its weak temperature dependence,k can be considered to be a con-stant. A graph oflnσ versus (1/T) will have a slope given by

(lnσ)(1/T)

=−Eg

2k

Eg = −(lnσ2 − lnσ1)(1/T2 − 1/T1)

× 2k

=(−4.3750+ 2.9937)

(3.003− 2.681)× 10−3× 2× 1.38× 10−23J

= 1.184× 10−19J.

=1.184× 10−19

1.6× 10−19eV

= 0.74 eV (Ans.).

5.6 A semiconductor sample of thickness 1.2×10−4m is placed in a magnetic field of0.2T acting perpendicular to its thickness. Find the Hall voltage generated whena current of 100 mA passes through it. Assume the carrier concentration to be1023m−3.

Solution: Hall voltage generated is given by the formula

VH =3π8·

BInet

178 Semiconductors

=3× 3.14× 0.2× 100× 10−3

8× 1023× 1.6× 10−19× 1.2× 10−4

VH = 0.0123Volt or 12.3 mV (Ans.).

5.7 Calculate the barrier potential existing at the junction ofp-type silicon with anacceptor concentration of 1023m−3 and n-type silicon with a donor concentrationof 1020m−3 if the intrinsic carrier concentration at 300 K is 1.4× 1016m−3.

Solution: The barrier potential is given by

VB =kTe

lnNaNd

n21

=1.38× 10−23 × 300

1.6× 10−19ln

1023× 1020

(1.4× 1016)2

VB = 0.64 V (Ans.).

5.8 Intrinsic silicon has a carrier concentration of 1.1× 1016m−3. If the mobilities ofelectrons and holes are 0.17 and 0.035m2v−1s−1 respectively at room temperature,compute the resistivity of silicon.

Solution: Conductivity is given by

σi = nie(µe + µh)

= 1.1× 1016 × 1.6× 10−19(0.17+ 0.035)

= 3.608× 10−4ohm−1m−1

Resistivity= 1/σ = 2.77× 103 ohm m(Ans.).

5.9 The compound gallium arsenide has an intrinsic conductivity of 10−6 ohm−1m−1

at 20oC. How many electrons have jumped the forbidden energy gap? [given:µe = 0.88 m2V−1s−1 andµh = 0.04 m2V−1s−1]

Solution: Number of electrons that have crossed the forbidden energy gapis the number of electrons contributing to conductivity, which is the carrierconcentration,ni.

ni = σ/e(µe + µh)

= 1× 10−6/1.6× 10−19(0.88+ 0.04)

6.79× 1012 m−3( Ans.).

Semiconductors 179

5.10 An electric field of 100Vm−1 is applied to a sample of n-type semiconductorhaving a Hall coefficient - 0.0125 m3C−1. Determine the current density in thesample if the electron mobility is 0.36 m2V−1s−1.

Solution:

Current densityJ = nev = neµE = −µE/RH

= 0.36× 100/0.0125

= 2880Am−2 (Ans.).

5.11 The resistivity of germanium at 20C is 0.5 ohmm. What will be its resistivity at40C if the band gap of germanium is 0.7 eV?

Solution: We have the relation,

(lnρ)(l/T)

=Eg

2k(lnρ) = Eg.(1/T)/2k

=0.7× 1.6× 10−19× (1/313− 1/293)

2× 1.38× 10−23

lnρ40− lnρ20 = −0.885

lnρ40 = lnρ20− 0.885

= −0.693− 0.885

= −1.578

ρ40 = 0.206ohmm(Ans.).

5.12 sample of silicon is doped with 1022 phosphorus atomsm−3. What would youexpect to measure for its resistivity? What Hall voltage would you expect ina sample 100µm thick when a current of 1 mA is passed perpendicular to amagnetic field of 0.1 T. [Given:µe = 0.07m2V−1s−1]

Solution:

Conductivityσ = Ndeµe (5.1)

= 1022× 1.6× 10−19× 0.07 (5.2)

= 112ohm−1m−1. (5.3)

Resistivityρ = 1/ρ = 8.93× 10−3 ohmm(Ans.). (5.4)

Hall voltageVH =3π8

BInet

(5.5)

180 Semiconductors

=3× 3.14× 0.1× 1× 10−3

8× 1022× 1.6× 10−19 × 100× 10−6(5.6)

= 0.735mV (Ans.). (5.7)

5.13 An intrinsic semiconductor has an energy gap of 0.7 eV. Calculate the probabilityof occupation of the lowest level in conduction band at 0, 50 and 100C.

Solution: The lowest level in conduction band corresponds to

(E − EF) = 0.35 eV.

The probability of occupation of this level is given by

F(E) =1

1+ exp (E − EF)/kT

Substituting the values, we get

At T = 00C = 273K, F(E) = 3.5× 10−7

T = 500C = 323K, F(E) = 3.5× 10−6

T = 1000C = 373K, F(E) = 1.88× 10−5 (Ans.).

Exercise

5.1 Explain the formation of a potential barrier in a p-n junction. (March 1999).

5.2 What is the difference between p-type and n-type semiconductors?

(March 1999).

5.3 What are zener diodes? Explain breakdown voltage in the caseof such diodes.

(August 1999).

5.4 Distinguish between conductors, insulators and semiconductors on the basis ofband theory. (August 1999, August 2000, March 2001).

5.5 Describe a p-n junction and explain the effect of forward and reverse biasing onthe barrier potential of the junction. (August 1999).

5.6 The band gap in germanium is 0.68 eV. Assuming that the numberof electron-hole pair is proportional to exp(−Eg/2kT), find the percentage increase in thenumber of charge carriers, when the temperature increases from 300K to 320K.

Semiconductors 181

5.7 Distinguish between avalanche diode and zener diode. (March 2000).

5.8 What is Hall effect? How is it useful in the study of conduction in materials?

(March 2000).

5.9 What is Hall effect? Derive an expression for the Hall coefficient. (August 2000).

5.10 Derive an expression for the conductivity of an intrinsic semiconductor. Theelectron and hole mobility of silicon are 0.14 and 0.05 m2v−1s−1 respectivelyat a given temperature. If the electron density is 1.5 × 1016 m−3, calculate theresistivity of silicon. (August 2000).

5.11 Discuss the I-V characteristics of a zener diode. Discuss the different types ofbreakdown at the p-n junction. (March 2001).

5.12 Describe a p-n junction and explain its voltage-current characteristics. What isthe effect of temperature on the conductivity of semiconductors.

(August 2001).

5.13 Explain how a potential barrier is formed in a p-n junction. (March 2002).

5.14 Distinguish between direct band gap and indirect band gap semiconductors.

(March 2002).

5.15 Distinguish between zener breakdown and avalanche breakdown.

(March 2002).

Chapter 6

Dielectric Properties Of Materials

6.1 Introduction

In this chapter we study the properties of insulator materials in static electric fields. In-sulators possess very few free charge carriers and hence arebad conductors of electriccurrent. However, application of an electric field results in some interesting processesinvolving rearrangement of charges in the material. These characteristic features ofinsulators are called dielectric properties. These properties and the behaviour of di-electric materials in a static electric field may be explained reasonably well with thehelp of a simple atomic theory.

Gauss theorem is one of the most fundamental theorems related to static electricfields. The theorem may be explained by considering a closed surface containing alarge number of charges. IfQ1, Q2, Q3, . . . . . .Qn represent the charges enclosed by asurface, then the total electric flux emerging from the surface is given by

Φ =

n∑

i=1

Qi =

∫

D · ds (6.1)

We note that the total flux is expressed in units of charge, namely coulombs whichmay also be written as equal to the surface integral whereD represents the flux densityexpressed in coulombsm−2 and ds is the surface element where the flux density isevaluated. The flux density in vacuum is proportional to the electric field strength Eaccording to the relation

D = ǫoE (6.2)

Since the electric field is measured in units of voltsm−1, ǫo will have the units faradm−1.This constant of proportionality is called the permittivity of vacuum. If the electricfield and the flux density are considered in any other medium, the relation betweenthem will be modified as

D = ǫE (6.3)

whereǫ is the permittivity of the medium. This relation may also be expressed as

D = ǫoǫrE (6.4)

182

Dielectric Properties Of Materials 183

whereǫr is a dimensionless quantity called “relative permittivity” or “dielectric con-stant” of the medium. For a dielectric medium, this is an important parameter and canbe easily evaluated by simple measurements. Consider a parallel plate condenser ofareaA and separation between the plates,d (Fig.6.1).

− − − − − − − − − − − − − −

+ + + + + + + + + + + + + +

E

Figure 6.1 A parallel capacitor with an applied electric field E. The fluxlines will bedirected from positive plate to the negative plate.

Let q be the charge density on the plates which will be numericallyequal to theflux densityD. Hence, we have the electric field existing between the plates given by

E = D/ǫǫr = q/ǫǫr (6.5)

The capacitance of the parallel plate condenser is given by

C =Total charge on the plate

Potential difference=

qAEd

(6.6)

Substituting forE from equation (6.5) we get

C = ǫǫr A/d (6.7)

If the dielectric medium between the two parallel plates is removed and the spaceevacuated, the capacitance will be

Cvac = ǫA/d (6.8)

From equation (6.7) and (6.8), the dielectric constant of the medium is given by

ǫr = C/Cvac (6.9)

Thus, the value ofǫr can be determined experimentally by measuring the capacitanceof a parallel plate condenser with and without the dielectric.

184 Dielectric Properties Of Materials

6.2 Polarization

An electric dipole, in its simplest form, consists of two equal point charges of oppositesign, Q and−Q, separated by a distance d (Fig.6.2). The dipole moment associatedwith this electric dipole is equal toQd and is represented by a vector pointing from thenegative charge towards the positive charge.

i.e. Dipole momentµ = Qd (6.10)

− Q + Q

d

= Qdµ

Figure 6.2 Dipole constituted by two equal and opposite charges separated by a dis-tance. The direction and magnitude of the dipole moment are indicated.

In a dielectric material, such electric dipoles may be present even in the absence ofan applied electric field. For example, in a diatomic molecule AB, if the atomA has atendency to give up an electron toB during the formation of a bond, atomA is said tobe more electropositive than atomB. Hence, the moleculeABcarries a dipole momenteven in the absence of an applied electric field. This is called a permanent dipole. Onthe other hand, the dielectric material may not have a permanent dipole but a dipolemay be induced under the influence of an applied electric field. For example, in anatom, the centre of the electron cloud coincides with the nucleus and hence there is nodipole moment associated with it. On application of an electric field, the electron cloudand the nucleus move in opposite directions and the centre ofthe electron cloud will beshifted through a finite distance from the nucleus, thus resulting in a dipole. The dipolemoment associated with this atom is said to be induced dipolemoment. Depending onwhether a dielectric contains permanent dipoles or not, thematerials are called polarmaterials and non-polar materials respectively. The dipole moment per unit volume ofthe material is called ‘Polarization’.

Thus we see that when a dielectric medium is subjected to an electric field, therewill be polarization occurring in the medium due to the presence of permanent dipolesor induced dipoles or both. If we consider a parallel plate condenser with an appliedelectric fieldE there will be a total charge+Q on the positive plate and−Q on thenegative plate (Fig.6.3).


We haveE = Q/ǫA (6.11)

+ + + + ++ + + + + + ++ + + + + + +

+ + + + ++ + + + + + ++ + + + + + +

− − − − − − − − − − − − − − − − − − −

− − − − − − − − − − − − − − − − − − −

− Q

+ Q

+ Q

− Q

Figure 6.3 A parallel plate capacitor showing the induced charges on the oppositefaces of the dielectric due to polarization.

whereA is the area of the plate. When a dielectric medium is introduced between theparallel plates, due to polarization in this medium, there will be induced charges at thesurface of the dielectric. The charge on the surface near thepositive electrode will benegative (−Q′) and vise versa. These induced charges will tend to weaken the appliedfield and the resultant electric field in presence of the dielectric medium is given by

E = (Q/ǫA) − (Q′/ǫA) (6.12)

The resultant electric fieldE can also be written as equal to

E = Q/(ǫoǫr A) (6.13)

whereǫr is the dielectric constant of the medium. From equation (6.12) and (6.13) wehave

Qǫǫr A

=QǫA− Q′

ǫA(6.14)

Q/ǫr = Q− Q′ (6.15)

This indicates that the induced charge is less than the charge on the electrodes due tothe applied field andQ′ = 0 for ǫr = 1.


From equation (6.13), we have

(Q/A) = ǫǫrE = D

Further, (Q′/A) = (Q′d/Ad) = µ/V

whered is the separation between the plates andµ/V is the dipole moment per unitvolume of the dielectric medium which is defined as the polarization. Substituting inequation (6.12), we have

E = D/ǫ − P/ǫor D = ǫE + P (6.16)

Substituting forD from equation (6.4) we get

P = ǫ(ǫr − 1)E (6.17)

Thus, we see that the polarization produced in the dielectric medium is proportionalto the applied electric field as long as the dielectric constant ǫr is independent ofE.Further, equation (6.17) is valid for gaseous, liquid and solid dielectrics. The quantity(ǫr − 1) is sometimes referred to as theelectric susceptibility and denoted by thesymbolχ. Hence, we can write

χ = (ǫr − 1) = P/ǫE

or P = ǫχE

The term electric susceptibility is rarely used where as theterm dielectric constant ismore popular and widely used.

6.2.1 Mechanisms of polarization

There are four mechanisms by which polarization can be produced in dielectric mate-rials. They are

a) Electronic polarization

b) Ionic polarization

c) Orientational polarization

d) Space charge polarization


The effective polarization produced in a dielectric material is the net result of the con-tributions and can be explained on the basis of atomic theory.

a) Electronic polarization: Consider a single atom consisting of a positive nucleusof chargeZe andz electrons moving around the nucleus. Taking into account thesize of the nucleus and that of the electron orbits we can consider the atom to bemade up of a point positive charge surrounded by a negative electron cloud. In theabsence of any external influence, the centre of the electroncloud coincides withthe central positive charge (Fig.6.4).

Ze−+

Ze−+

(b)(a)

E

Figure6.4Effect of an applied electric field E on an isolated atom.

When the atom is subjected to an electric fieldE, the nucleus and the electron cloudwill try to move in opposite directions. The nucleus moves inthe direction of theapplied field and the electron cloud in the direction opposite to that of the appliedfield. The centre of the negatively charged electron cloud nolonger coincides withthe positive nucleus and hence results in an induced dipole.The polarization pro-duced due to this induced dipole is called “Electronic polarization”.

b) Ionic polarization: In an ionic material, there will be an arrangement of positiveand negative ions. If their arrangement is symmetrical, there will be no permanentdipole present (Fig.6.5).

When such a material is placed in an electric field, the positive ions will move inthe direction of the applied field and the negative ions in theopposite direction.This results in an induced dipole. For example in sodium chloride the effect of anapplied electric field produces ionic polarization.

c) Orientational polarization: This type of polarization occurs in molecules havingpermanent dipole moments. Consider the case of a moleculeABAwhereA is moreelectropositive thanB. (Fig.6.6).


µ µ

Ε

(b)

µ µ

(a)

Figure6.5Effect of an applied electric field E on an ionic solid. Open circles indicatenegative charges and closed circles indicate positive charges. A dipole moment isinduced due to the field.

AB

A

+ − +

µ µ

Α Α

Β

+

−

+

µµ

(b)

(a)

Figure 6.6Dipole moments in a molecule of the type ABA. In (a), the resultant dipolemoment is zero whereas in (b), the resultant dipole moment isthe vector sum of the twoindividual dipole moments.

The molecule consists of two dipole moments directed fromB to A. It is observedthat the resultant dipole moment may be zero if the molecule has a centre of symme-try. Molecules of carbon dioxide (CO2) and carbon disulphide (CS2) are examples.However, in the absence of a centre of symmetry, the moleculecarries a resultantdipole moment given by the vector sum of the dipole moments ofthe individual


dipoles. An example is water molecule (H2O) where the twoOH bonds and thedipoles associated with them are inclined at an angle of about 105. When an exter-nal field is applied to a molecule carrying a permanent dipolemoment, the dipoleswill tend to align in the direction of the field. The contribution to the polarizationfrom the process of orientation of permanent dipoles is called orientational polar-ization or molecular polarization.

d) Space charge polarization:In dielectric materials, there can be deviations fromperiodic arrangement of atoms due to vacancies or presence of impurity atoms.There can be a few inclusions of an additional phase. When an electric field isapplied, these charges may move through the material to its surfaces. Immobileinclusions get polarized. This results in the development of charges near the elec-trodes (Fig.6.7).

−−−

+++

++++

−−−

−

−−− + +

+

+

−−−

+++

+ −

−

−

−

−

+

+

+

+

E

Figure6.7Space charge polarization in a solid.

The surface of the material near the positive electrode acquires a negative chargeand the surface near the negative electrode acquires positive charge. The resultingpolarization is called space charge polarization or interfacial polarization.

6.2.2 Temperature dependence of polarization

The electronic polarization and the ionic polarization areprocesses involving the shiftof electron cloud or the ions respectively under the influence of the applied electricfield. These processes have contributions determined entirely by the electronic struc-ture of the atom or molecule. The electronic structure is essentially independent oftemperature unless the temperature is extremely high. Hence, the contributions fromelectronic and ionic polarizations are independent of temperature. The orientational


polarization is due to the presence of permanent dipoles. Inthe absence of an elec-tric field, the dipoles are randomly oriented with no net polarization. When a field isapplied, the permanent dipoles tend to align parallel to thefield with the result thatthere will be a net polarization. However, its contributiondecreases at higher tem-perature due to greater thermal agitation. The orientational polarization is inverselyproportional to the temperature and is given by

P =Nµ2

pE

3kT(6.18)

whereµp is the permanent dipole moment associated with each molecule, N is thenumber of molecules per unit volume andE is the applied electric field.

6.2.3 Effect of frequency on polarization

When an electric field is applied to a dielectric the dipoles tend to align parallel to thefield. However, if the electric field is alternating the dipoles will have to keep changingtheir directions of alignment. At low frequencies, the dipoles will have sufficient timeto follow the applied field. As the frequency of the applied electric field increases,the dipoles cannot keep up with the changing field. This results in a decrease in thepolarization with increasing frequency of the applied field.

The ability of the dipoles to follow the applied alternatingfield depends on themechanisms by which the dipoles are produced. Space charge polarization involvesthe movement of ions in the dielectric medium and hence is theslowest process. Itscontribution is normally observed in static (or dc) fields and may vanish in alternatingfields of even very low frequency. Orientational polarization involving the rotationof molecules is also a slow process and its contribution is observed upto frequenciesof the order of 106 −108Hz. Ionic polarization involving displacement of ions hasconsiderable contribution upto frequencies of 1013Hz. Electronic polarization, whichis the fastest mechanism of polarization as it involves the motion of electron cloud,is observed even upto 1015Hz. Assuming that all the four types of polarization arepossible in a dielectric medium, the variation of total polarization with the frequencyof the applied electric field may be as shown in Fig.6.8.

Thus, the space charge polarization will contribute at power frequencies, orienta-tional polarization in audio and radio frequencies, ionic polarization in infrared regionand electronic polarization in the optical region of frequency. Accordingly, the ma-terial will have different dielectric constants in different regions of frequency of theapplied electric field.


P

4 8 12

log f

Figure 6.8 Dependence of polarization on the frequency of the applied electric fieldshowing the contributions from different polarization mechanisms.

6.3 Dielectric Constant

6.3.1 Dielectric constant of monoatomic gases

The rare gases like helium and neon are the simplest examplesof monoatomic gasessince the interaction between them is negligible. Let us tryto obtain an expression forthe dielectric constant of such a gas using the atomic model.

Consider an atom consisting of a positive nucleus of charge+zeandz electronsforming an electron cloud of radiusR (Fig.6.9).

+ Ze

E = 0 E

R

x+ Ze

Figure 6.9 Shift of the electron cloud in an inert gas atom due to the applied electricfield.

When this atom is placed in an electric fieldE, the nucleus and the electron cloudwill try to move in opposite directions. However, the coulombic force of attraction


between the electron cloud and the nucleus will prevent themfrom pulling apart fromeach other and an equilibrium will be obtained. Letx be the displacement of thecentre of the electron cloud from the nucleus at equilibrium. The electron cloud can bedivided into two regions by drawing a sphere of radiusx with the centre of the electroncloud as the centre. The nucleus of chargezewill lie on the surface of this inner sphere.The force exerted on the nucleus will be due to the negative charge inside the sphereof radiusx since the charges in the region between the two spherical surfaces of radiusx andR do not exert any force, according to the Gauss theorem. The negative chargepresent inside the sphere of radiusx is equal to−zex3/R3 and the force exerted by thischarge on the nucleus is

Fcoulomb=1

4πǫ· Ze(Zex3/R3)

x2(6.19)

At equilibrium, this must be equal to the force acting on the nucleus due to the appliedfield.

∴ F f ield = ZeE (6.20)

From equation (6.19) and (6.20) we get

x = (4πǫR3/Ze)E (6.21)

This equation indicates that the displacement of the electron cloud is directly propor-tional to the strength of the applied electric field. The dipole moment induced due tothe applied field is given by

µind = Ze.x = 4πǫR3E = αeE (6.22)

The induced dipole moment is also proportional to the applied electric field and theproportionality factorαe is called the electronic polarizability of the atom.

If the gas containsN atoms per unit volume, the polarization produced in the gasis given by

P = NαeE (6.23)

Comparing this with equation (6.17) which relates polarization with the dielectric con-stant, we obtain

ǫ(ǫr − 1) = Nαe = 4πǫNR3 (6.24)

or ǫr = 1+ 4πNR3 (6.25)

This is the expression for the dielectric constant of the gas. An experimental verifi-cation of the dielectric constant of an inert gas matches very well with the calculated


value from equation (6.25), thereby indicating that the simple atomic interpretation isquite valid. Further, the electronic polarizabilityαe, which is equal to 4πǫR3, is foundto be independent of temperature in agreement with the experimental observations.

6.3.2 Dielectric constant of polyatomic gases

The case of mono atomic gases considered above is the simplest of the cases where thepolarization of the gas is only due to the shift of the electron cloud under the influenceof the applied electric field. In other words, only electronic polarizability contributes tothe polarization and hence to the dielectric constant of thegases. In the case of poly-atomic gases, we have to consider the possibility of other polarization mechanismsalso contributing to the dielectric constant. We have seen that the electronic and ioniccontributions to the total polarization are independent oftemperature. Hence, a studyof the temperature dependence of polarization will help us in identifying the presenceof permanent dipoles in the molecules. For example, the dielectric constant of gaseslike CO2, CS2, CH4 etc., is independent of temperature whereas gases likeHCl, H2O,H2S, CH3Cl show temperature dependence of dielectric constant. Hencethe net per-manent dipole moment for molecules likeCO2, CS2, CH4 etc., is zero and moleculeslike HCl, H2O, H2S, CH3Cl etc., have a net permanent dipole moment which will alignitself parallel to the applied electric field thereby contributing to the dielectric constantof the gas. This, measurement of dielectric constant has been very much useful in theinvestigation of molecular structure.

6.3.3 Internal field in solids and liquids

The atoms or molecules in a gas when subjected to an electric field will have dipolemoments, induced or permanent, aligning parallel to the applied field. These dipoleswere considered to be independent of each other in a gas. However, in a solid or aliquid, these dipoles are close enough to influence the localfield acting at the locationof each atom. The field seen by each atom or molecule in a solid or a liquid is calledthe local field or internal field,Ei. This internal field is larger than the applied fieldE.Let us evaluate the internal field for the one dimensional case using Epstein’s model ofa linear chain of similar, equidistant atoms (Fig.6.10).

Let αe be the polarizability of atoms. When they are subjected to anelectric fieldE, along the chain of atoms, each atom will carry an induced dipole moment ind givenby

µind = αeEi (6.26)


Figure 6.10A linear chain of atoms carrying induced dipole moments in presence ofan applied electric field.

whereEi is the internal field which is written as

Ei = E + E′ (6.27)

E′ being the additional field due to effect of neighbouring dipoles. Our task is toevaluateE′ so thatEi can be determined. Consider a dipole of dipole momentµ besituated at the pointO (Fig. 6.11).

θ

Ο

E r

P

θE

Figure6.11Electric field due to a dipole.

Let P represent a point at a distancer and along a directionOP inclined at an angleθ. According to field theory, the potential measured at the point P is given by

V(r, θ) =1

4πǫ· µ cosθ

r2(6.28)


The electric field due to this potential has two components given by

Er =−δVδr=

14πǫ

·2µ cosθ

r3(6.29)

Eθ =−1rδVδθ=

14πǫ

·µ sinθ

r3(6.30)

The electric field intensity at the pointA due to the dipole on the atom atB can becalculated by puttingr = a andθ = 0 in the equations (6.29) and (6.30). This yieldsthe componentEθ to be zero. Hence, electric field intensity at the atomA due to theinduced dipole at atomB is given by

EB =µind

(2πǫa3)(6.31)

wherea is the inter-atomic spacing along the chain of atoms. Similarly, the electricfield intensity at the atomA due to the induced dipole atB′ is also equal to the valuegiven by equation (6.31). The total electric field due to atoms B andB′ will be the sumof the two since the dipoles on atomsB andB′ are similarly directed. Therefore, theelectric field atA due to dipoles on atomsB andB′ is

EBB′ =µind

πǫa3(6.32)

Similarly, considering the pairs of atoms on either side ofA, we can write

ECC′ =µind

πǫ(2a)3(6.33)

EDD′ =µind

πǫ(3a)3(6.34)

and so on. The net additional fieldE′ due to the induced dipoles is given by

E′ = EBB′ + ECC′ + EDD′ + . . .

i.e., E′ =µind

πǫa3·∞∑

n=1

1/n3 =1.2µind

πǫa3(6.35)

Substituting forE′ andµind from equations (6.35) and (6.26) respectively in equa-tion (6.27), we get

Ei = E +1.2αeEi

πǫa3(6.36)


or Ei =E

(1− 1.2αe/πǫa3)(6.37)

Since the term (1.2αe/πǫa3) is positive, we conclude that the internal field is largerthan the applied field. In the three dimensional case, the calculation of the internal fieldwill be complicated. Proceeding similarly, we may show thatfor a three dimensionalcase (1.2/π) may be replaced byγ whereγ is called the internal field constant. Aparticular case which is often encountered is the one in which the atoms are cubicallyarranged. For this particular case, the internal field constant γ has a value equal to(1/3). The internal field for the case of such materials is calledthe Lorentz field.Further, (1/a3) will be equal toN whereN represents the number of atoms/unit volume.Hence,

Ei = E + P/3ǫ (6.38)

It should be noted that only electronic polarization has been considered in this deriva-tion. However, it can be extended to include ionic polarization but will not be valid forthe case of polar liquids.

6.3.4 Dielectric constant of elemental solids

Elemental solids are those containing only one kind of atomsand the polarization isessentially due to the electronic polarizability of the atoms. The polarization in such asolid is given by

P = NαeEi (6.39)

whereEi is the internal field andN is the number of atoms per unit volume of the solid.Substituting forEi from equation (6.38), we have

P = Nαe[E + (1/3ǫ)P] (6.40)

Or P =NαeE

1− (Nαe/3ǫ)(6.41)

Polarization is known to be related to the dielectric constant of the material by theequation

P = ǫ(ǫr − 1)E (6.17)

From equations 6.38 and 6.17, we get

ǫ(ǫr − 1) = Nαe/(1− Nαe/3ǫ) (6.42)

ǫ(ǫr − 1)(1− Nαe/3ǫ) = Nαe (6.43)


Further simplification of the above equation leads to

(ǫr − 1)(ǫr + 2)

=Nαe

3ǫ(6.44)

This is known as the Clausius-Mosotti equation which relates to the polarizability ofthe atoms with the dielectric constant of the material.

6.3.5 Dielectric constant of ionic solids without permanent dipoles

In the case ionic dielectrics without permanent dipoles, like alkali halides, the totalpolarization will include contributions from electronic and ionic polarization, i.e.,

P = Pe + Pi

Such solids do not carry a permanent dipole moment since the sum of the dipole mo-ments of ionic pairs for the entire solid vanishes.

Further, the internal field at the site of a positive ion is different from that at thesite of a negative ion. Hence, a quantitative interpretation of the dielectric behaviourof these solids is more complicated than that of elemental solids. However, the contri-bution of the ionic polarization to the dielectric constantmay be evaluated as follows:Let the static dielectric constant of the material beǫrs. Then, the polarization may bewritten as given by

ǫ0(ǫrs − 1)E = Pe + Pi

We can also determine the dielectric constant of the material at optical frequencies bydetermining the refractive indexn. According to Maxwell’s theory, we have

ǫre = n2

whereǫre is the dielectric constant measured at optical frequencies. Since the ionicpolarization does not contribute to the dielectric constant at optical frequencies, wehave

ǫ0(ǫre − 1)E = Pe

The difference betweenǫrs andǫre is a measure of ionic polarization. For example, thevalues ofǫrs andǫre for sodium chloride are 5.62 and 2.25 respectively. This clearlyindicates that the static dielectric constant has an appreciable contribution from ionicpolarization.


6.3.6 Dielectric constant of polar materials

In the case of polar materials, there may be additional contribution from orientationalpolarization. This contribution is temperature dependentand will increase with de-crease in temperature. Further, the orientational polarization is expected to be presentonly in the liquid state since the permanent dipoles are frozen in the solid state andcannot be aligned by the applied field. However, some solids show finite contributionfrom orientational polarization due to the possible movement of ions permitted by thecrystallographic structure of the material.

Nitrobenzene(C6H5NO2) is an example of a polar material. The variation of di-electric constant with temperature is shown in Fig.6.12(a).

It is observed that the dielectric constant decreases abruptly as the material freezesinto a solid and remains constant with further decrease in temperature. This is due tothe fact that the permanent dipoles are frozen in the solid state and hence do not con-tribute to the polarization. The dielectric constant is decided by the electronic and ioniccontributions. In the liquid state, the dielectric constant is large but decreases with in-crease in temperature. This is because of the additional contribution from orientationalpolarization which is inversely related to temperature.

Τ (Κ) Τ (Κ)

ε r

270 280 290 300 100 150 200

ε r

(a) (b)

Figure6.12Variation of dielectric constant with temperature for (a) nitrobenzene and(b) hydrogen chloride.

In the case of hydrogen chloride(HCl), a transition from liquid state to the solidstate results in a small, abrupt increase in the dielectric constant as shown in Fig. 6.12(b).


This increase is explained as due to a change in the density ofthe material on solidi-fication. In the solid state, the dielectric constant increases with decrease in tempera-ture indicating the continued contribution from orientational polarization. At a lowertemperature, the dielectric constant decreases suddenly indicating that the permanentdipoles have become immobile.

Hence, a study of the variation of dielectric constant with temperature is helpful inevaluating the contribution from orientational polarization and also to gain insight intothe molecular structure of materials.

6.4 Ferroelectric materials

The polarization produced in a dielectric medium is found tobe linearly related to theapplied electric field according to the equation,

P = ǫ(ǫr − 1)E (6.17)

There are some substances for which the polarization is not alinear function of theapplied field. These materials exhibit hysteresis in the variation of polarization with theapplied field. Such materials are said to be ferroelectric. Fig.6.13 shows the hysteresisloop exhibited by a ferroelectric material.

sP

Pr

E

P

E c

Figure6.13Hysteresis shown by a ferroelectric material.


When an electric field is applied to an unpolarized sample, the polarization in-creases with the applied electric field in a non-linear manner. When the field is re-duced, the polarization also reduces but will not retrace its path. It is observed that acertain amount of polarization is still present when the field is reduced to zero. This iscalled remanent polarization,Pr . In order to make the polarization zero, a field in theopposite direction has to be applied. The electric field required to bring the remanentpolarization to zero is called the coercive field,Ec. The existence of hysteresis indi-cates that the material must be undergoing spontaneous polarization. This behaviourof ferroelectric materials is explained on the basis of domain theory. According to thistheory, the specimen may be considered to be made up of a largenumber of smallregions called domains within which the dipoles are oriented parallel resulting in aspontaneous polarization. In a virgin specimen the direction of polarization variesfrom one domain to another and the resultant polarization iszero. When an electricfield is applied, the domains with parallel orientation growin size at the expense ofothers. Ultimately, the entire specimen may become a singledomain resulting in sat-uration polarization,Ps. Beyond this value of polarization, the increase is due to thenormal dielectric behaviour.

Hysteresis in polarization is the most characteristic property of ferroelectric mate-rials. Material show this property only upto a certain temperature called ferroelectriccurie temperature,θ f . Above this temperature, the material no longer exhibits hystere-sis but behaves like a normal dielectric. The dielectric constant is given by

ǫr = C/(T − θ) for T > θ f (6.45)

whereC is a constant called Curie constant andθ is a characteristic temperature usuallya few degrees smaller thanθ f . This relation is called Curie-Weiss Law. The tempera-ture dependence of dielectric constant is illustrated in Fig.6.14.

Ferro-electric materials are classified into three groups.Tartrate group consists ofsalts of tartaric acid. A typical example of this group is Rochelle salt which is sodium-potassium tartrate,NaK(C4H4O6)4H2O. This material exhibits a ferroelectric phase ina narrow range of temperature from−18C to 23C (Fig.6.15a).

Other materials belonging to this group are those in which sodium is replaced inpart by NH4, Rb or Ti ions. Members of the second group of ferroelectric materi-als are dihydrogen phosphates and arsenates of alkali metals. They exhibit a Curietemperature upto which they remain in ferroelectric phase.For example potassiumdihydrogen phosphate,KH2PO4, remains in the ferroelectric phase upto a temperatureof 123 K (Fig.6.15b). The third group is the oxygen octahedron group with bariumtitanate,BaTiO3, as the best known ferroelectric material. The titanium ionpresent at


θ θ

1/ε

fT

r

Figure6.14Temperature dependence of dielectric constant for a ferroelectric material.The material is ferroelectric at temperature belowθ f and a normal dielectric above thistemperature.

Ps Ps Ps

400300200120110100290270250

T, KT, KT, K

(a) (b) (c)

Figure 6.15 Spontaneous polarization as a function of temperature in (a)NaKC4H4O6.4H2O (b) KH2PO4 and (c) BaTiO3.

the octahedral vacancy, with a charge+4 and large space for displacement accountsfor the large value of spontaneous polarization observed. Barium titanate becomesferroelectric at a temperature below 120C. As the temperature is decreased, the spon-taneous polarization decreases in steps at 5C and−80C associated with the changesin the direction of spontaneous polarization and crystal structure (Fig. 6.15c). Similarto barium titanate other compounds likeKNbO3, PbTi03 etc. are also known to beferroelectric materials with attractive properties.

Barium titanate and other ferroelectric materials with their high value of dielectricconstant are very much useful for storing electrical energyas capacitors in electroniccircuits.


6.5 Piezoelectric Effect

Application of an electric field generally displaces ions and consequently, the dimen-sions of the sample may undergo changes. On the other hand, mechanical stress alsochanges the dimensions of the sample but normally does not produce polarization. Inmost of the materials, polarization is observed to produce deformation but the defor-mation does not lead to polarization. This electro-mechanical effect, present in allmaterials, is called ‘Electrostriction’. A simple exampleis shown in Fig. 6.16.

+ +

+ +

+ +

+

+

+

+

+ +

+ +

+

E

(a) (b) (c)

Figure6.16Illustration of electrostriction showing (a) an ionic lattice, (b) effect of anapplied electric field and (c) the effect of an applied compressive force.

Application of an electric field along x-axis results in the movement of positive ionsalong x-direction and the negative ions in the opposite direction. Hence, a distortionof the lattice occurs. However, when a tensile or a compressive force is applied to thesample, there will be no polarization. This property of electrostriction is characteristicof all crystalline materials which have a centre of symmetry.

There are some dielectric materials in which application ofan electric field pro-duces deformation and conversely, a deformation of the sample produces a polariza-tion. This effect is calledpiezoelectric effect. An example is shown in Fig. 6.17.

Application of an electric field along x-axis produces a deformation. A tensile or acompressive force will result in a change in the angle , thereby resulting in polariza-tion. A tensile force increases the angle and a compressive force decreases the angle.The direction of polarization will change depending on the force being tensile or com-pressive. This effect is observed only in materials which lack a centre of symmetry.Piezoelectric materials are used as transducers to convertmechanical energy into elec-trical energy or electrical energy into mechanical energy.Some applications are in themaking of microphones, strain gauges and ultrasonic generators.


+

+

+ +

+θ

(a)

+ +

+ + +

E

θ

+ ++

++

(c)(b)

Figure6.17Illustration of piezoelectric effect showing (a) an ionic lattice, (b) effect ofan applied electric field and (c) the effect of an applied compressive force.

6.6 Dielectric losses

When a dielectric material is subjected to an alternating electric field, a part of theenergy is lost each time the field changes its direction. Thisis due to the fact thateach time the field is reversed, the direction of the dipoles also has to change and theloss can be considered to be the frictional effect involved. Hence, the dielectric lossdepends on the frequency and the mechanism by which polarization is produced in thematerial.

A capacitor with a dielectric loss can be considered to be equivalent to an idealcapacitor with a parallel resistor as shown in (fig. 6.18a).

ΙC

ΙC

ΙR

ΙR

δ

I

V

(a) (b)

Figure6.18Equivalent circuit for a capacitor and the corresponding phasor diagram.

The current through the capacitor leads the voltage by 90 and hence is 90 out ofphase with the current through the resistor arm. Hence, the resultant current will be


leading the voltage by an angle less than 90. The change in the lead angle is called theloss angle,δ (Fig.6.18b). The dielectric loss is expressed as loss factor which is equalto tanδ.

tanδ = IR/Ic (6.46)

whereIR andIc are the components of current that would have flown through anequiv-alent parallel resistance and the capacitor respectively.

i.e., tanδ =(V/R)(V/Xc)

=Xc

R(6.47)

whereXc is the reactance of the capacitor and is equal to (1/ωC)

∴ tanδ = 1/(ωRC) (6.48)

Hence, the power loss in the capacitor will be equal to VI orV2/R which may bewritten in terms of tanδ as

P = ωCV2 tan δ (6.49)

Thus, the dielectric loss depends on the value of capacitance, the equivalent resistanceand the frequency of the applied field. A good dielectric would have a loss factor ofthe order of 10−5.

Numerical Examples

6.1 A parallel plate capacitor of area 650mm2 and a plate separation of 4mmhas acharge of 2× 10−10C on it. What is the voltage required when a material ofdielectric constant 3.5 is introduced between the plates.

Solution:

Capacitance= ǫǫA/d

=8.85× 10−12 × 3.5× 650× 10−6

4× 10−3

= 5.033× 10−12F

Voltage requiredV = Q/C =2.0× 10−10

5.033× 10−12

V = 39.73V. (Ans.).

6.2 Find the polarization produced in a dielectric medium of relative permittivity 15when it is subjected to an electric field of 500 V/m.


Solution:

P = ǫ(ǫr − 1)E

= 8.85× 10−12 × (15− 1)× 500

P = 6.195× 10−8Cm−2 (Ans.).

6.3 The polarizability of a neon atom is 3.5× 10−41Fm2. Find the dielectric constantof neon gas atOC and normal pressure.

Solution: No of atoms in neon gas per unit volume at NTP

=Avogadro number

Molar volume

i.e., N =6.023× 1023

22.4× 10−3atomm−3

= 2.69× 1025 atomsm−3

We have

ǫ(ǫr − 1) = Nαe

Or (ǫr − 1) =Nαe

ǫ=

2.69× 1025× 3/5× 10−41

8.85× 10−12

= 1.06× 10−4

ǫR = 1.000106 (Ans.).

6.4 The electronic polarizability of argon atom is 1.43× 10−40Fm2. Find the induceddipole moment and the relative shift of the electron cloud when it is subjected toan electric field of 105Vm−1.

Solution:

Induced dipole moment= µind = αeE

= 1.43× 10−40× 105

= 1.43× 10−35cm.

Also, µind = ·x

∴ x =µind

Ze=

1.43× 10−35

18× 1.6× 10−19

x = 4.97× 10−18m (Ans.).


6.5 A solid contains 2× 1028 identical atoms perm3 each with a polarizability of5 × 10−40Fm2. Assuming Lorentz force field to be operative, calculate theratioof the internal field to the applied field.

Solution:

Ei = E + (γ/ǫ)P = E + (γ/ǫ)NαeEi

∴Ei

E=

11− (γ/ǫ)Nαe

=

[

1− 2× 1028 × 5× 10−40

3× 8.85× 10−12

]−1

= (1− 0.377)−1

Ei

E= 1.604 (Ans.).

6.6 An elemental solid containing 2× 1028 atomsm−3 shows an electronic polariz-ability of 2×10−40Fm2. Assuming a Lorentz force field to be operative, calculatethe dielectric constant of the material.

Solution: From Clausius - Mosotti equation,

(ǫr − 1)(ǫr + 2)

=Nαe

3ǫ

=2× 1028× 2× 10−40

3× 8.85× 10−12

= 0.1506

(ǫr − 1) = (0.1506)(ǫr + 2)

ǫr(1− 0.1506)= (1+ 2× 0.1506)

ǫr =1.30120.8494

= 1.52 (Ans.).

6.7 Capacitor consists of two conducting plates of area 200cm2 each separated by adielectric material (ǫ = 3.7) of thickness 1 mm. Find the capacitance and theelectric flux density when a potential of 300 V is applied.

Solution:

CapacitanceC = ǫǫrA/d

=8.85× 10−12× 3.7× 200× 10−4

0.1× 10−2

= 6.549× 10−10F or 654.9pF (Ans.).

Flux densityD = ǫǫrE


= 8.85× 10−12× 3.7× 300× 103

= 9.83× 10−6C.m−2 (Ans.).

6.8 What is the polarization produced in sodium chloride by an electric field of 500Vm−1 if it has a relative permittivity of 6?

Solution:

PolarizationP = ǫ(ǫr − 1)E

= 8.85× 10−12× (6− 1)× 500

= 2.21× 10−8C.m−2 (Ans.).

6.9 The electronic polarizability of argon is 1.5 × 10−40Fm−2. If the gas contains2.7× 1025 atomsm−3, calculate the relative permittivity of the gas.

Solution: We have

ǫ(ǫr − 1) = Nαe

Or (ǫr − 1) =Nαe

ǫ=

2.7× 1025 × 1.5× 10−40

8.85× 10−12

= 4.58× 10−4

R= 1.000458 (Ans.).

6.10 A solid containing 8× 1028 atomsm−3 shows an internal field which is 1.5 timesthe applied field. If the solid has an internal field of Lorentztype, calculate thepolarization associated with each atom.

Solution: Polarization associated with each atom is the polarizability of theatom.

Ei

E=

11− (γ/ǫ)Nαe

Or αe =(1− E/Ei)ǫ

γN=

(1− 1/1.5)× 8.85× 10−12

(1/3)× 8× 1028

= 1.106× 10−40Fm2 (Ans.).


Exercise

6.1 Describe different polarization mechanisms. (March 1999)

6.2 A parallel plate capacitor consists of two plates each of area 5× 10−4m2. Theyare separated by a distance of 1.5× 10−3m and filled with a dielectric of relativepermittivity 6. Calculate the charge on the capacitor if it is connected to a 100 VD.C. supply. (March 1999)

6.3 Explain the properties of ferroelectric materials. (March1999)

6.4 What are dielectrics? Arrive at an expression for internal field in the case ofsolids and liquids. (March 1999)

6.5 What are dielectric materials? Explain the dependence of polarizability and di-electric loss on frequency. (March 2000)

6.6 What should be the voltage required to introduce a material of dielectric constant4 between the plates of a parallel plate capacitor of area 1000mm2, having plateseparation of 5 mm and a charge of 3× 10−10C. (August 2000)

6.7 Define the term ‘Internal field’ and derive an expression for the same in the caseof liquids and solids. (August 2000)

6.8 Explain with theory how static dielectric constant of a material is determined.

(March 2001)

6.9 Write a note on dielectric loss in a dielectric material. (March 2001)

6.10 What is electrical polarization of an atom? Obtain an expression for the internalfield developed in solids when an electric field is applied to it. (March 2001)

6.11 Obtain an expression for internal field in a one-dimensionalarray of atoms insolids. Discuss the properties of ferro-electric materials. (August 2001)

6.12 Discuu different polarization mechanisms. (Feb 2003)

6.13 Define dielectric polarization. Derive an expression for internal field in case ofsolid and liquid dielectric by considering one dimensionalarray of atoms.

(Aug 2003)

6.14 What is polarization and dielectric loss in dielectric materials? Explain the de-pendence of polarizability and dielectric loss on frequency. (Aug 2004)

Chapter 7

Magnetic Properties

7.1 Introduction

The magnetic behaviour of materials is due to atomic magnetic dipoles present in thesematerials. The orbital motion or spin of charged particles is equivalent to an electriccurrent and hence is associated with magnetic effects. For example, the orbital motionof an electron around a nucleus, the spin of an electron and the spin of a nuclear par-ticle will all constitute magnetic dipoles and carry a dipole moment. An atom may beconsidered to contain many such magnetic dipoles carrying different dipole moments.These are called permanent dipoles as they exist even in the absence of an applied mag-netic field. However, when a magnetic field is applied, the orbital angular velocity ofelectrons is modified resulting in a change in the magnetic dipole moment. This can beconsidered to be equivalent to an induced dipole moment acting in a direction oppositeto the applied magnetic field. The magnetic behaviour of all materials is the result ofthe contributions of the induced and permanent magnetic dipoles. A reasonably cor-rect interpretation of the magnetic behaviour is possible on the basis of a simple atomictheory.

The magnetic flux density B due to a magnetic field H applied in vacuum isgiven by

B = µH (7.1)

whereµcirc is calledpermeability of vacuum (or free space). The magnetic flux densityis also known asmagnetic induction and is expressed in units of tesla or weber persquare metre. Magnetic field strength has the units ampere per metre. The units ofpermeability are henry per metre.

When the magnetic field is applied to a solid medium, the magnetic flux densitywill have an additional contribution from the magnetic dipole moments present in thesolid. The net flux density is given by

B = µH + µM (7.2)

whereM is the magnetic dipole moment per unit volume of the material. It can beconsidered to be the internally produced magnetic field and is calledintensity of mag-netisation. The magnetic flux density may also be expressed as

B = µH = µµr H (7.3)

209

210 Magnetic Properties

whereµ is the permeability of the solid andµr is called therelative permeability.From equations (7.2) and (7.3) we have

µr =H + M

H= 1+

MH= 1+ χ (7.4)

Or (µr − 1) = χ (7.5)

The termχ which is equal to (M/H) is called magnetic susceptibility of the material.The relative permeability and magnetic susceptibility areimportant quantities used todescribe the magnetic behaviour of materials.

7.2 Classification of magnetic materials

Materials are classified into various types depending on themagnetic properties of thedipoles present and the interaction between them. The first distinction that can be madeis between materials possessing permanent dipoles and those without them. Materi-als which lack permanent dipoles are calleddiamagnetic. These materials, however,show weak magnetisation when placed in a magnetic field due toinduced dipoles. Themagnetisation is very small and is opposite in direction to the applied magnetic field.Hence, the diamagnetic materials have a small negative susceptibility value. Exam-ples of diamagnetic materials are metals like copper, gold,silver, semiconductors likesilicon, germanium, ionic slats likeNaCl, oxides like alumina, organic solids like ben-zene, napthalene, rare gases like helium, neon, argon, etc.All these materials have amagnetic susceptibility of the order of 10−5. An important exception is the case of ma-terials in their superconducting phase when the diamagnticsusceptibility will be equalto−1. Superconductors are called perfect diamagnetic materials. Materials which havepermanent dipoles are further classified on the basis of the interaction between them.If the interaction between the permanent dipoles is absent or negligibly small, then thematerial is calledparamagnetic(Fig. 7.1).

Metals like aluminium, magnesium, tungsten, titanium, platinum, oxides likeCoO,FeO, Cr2O3, chlorides likeCrCl2, FeCl2, sulphates likeCoS O4, MnS O4, NiS O4 andgases like oxygen, nitrogen etc. are examples of paramagnetic materials.

If the dipoles interact strongly leading to their parallel alignment, the material isferromagnetic. Examples offerromagnetic materials are the elements of iron group(iron, cobalt, and nickel) and alloys containing these elements. The strong interac-tion among the permanent dipoles leads to large value of magnetisation and hencehigh value of magnetic susceptibility. If the result of strong interaction is to align theneighbouring dipoles antiparallel, the material is antiferromagnetic or ferrimagneticin nature. In the case ofantiferromagnetic materials, the dipole moments on the

Magnetic Properties 211

FERRI

ANTIFERRO

FERRO

PARA

Figure7.1Schematic representation of the alignment of permanent dipoles in magneticmaterials.

neighbouring dipoles are equal and opposite in orientationso that the net magnetisa-tion vanishes. In these materials, the distance between theneighbouring atoms is sosmall that the exchange forces produce a tendency for antiparallel alignment of electronspins.MnO, MnO2, MnF2, MnS, FeO, CoOare some examples of anti-ferromagneticmaterials. Variation of susceptibility with temperature shows a sharp maximum at aparticular temperature called Neel temperature, TN (Fig.7.2). Below this temperature,the susceptibility drops due to antiparallel alignment. Above this temperature, thematerial behaves like a paramagnetic material showing a gradual decrease in its sus-ceptibility with increase in temperature.

TN

T

χ

Figure7.2Variation of susceptibility with temperature for an antiferromagnetic mate-rial.


Ferrimagnetic materials, on the other hand, show relatively large magnetisation,since the antiparallel alignment is of unequal dipole moments. They are similar to fer-romagnetic materials and are characterized by large valuesof magnetic susceptibility.Ferrites and garnets, which are mixed metal oxides containing iron, are examples. Asummary of classification of magnetic materials is given in table (7.1).

Table7.1Classification of magnetic materials

Class Permanent Interaction of χ

dipole neighbouring dipoles

Diamagnetic Absent Nil − 10−5

Paramagnetic Present Negligible 10−3

Ferromagnetic Present Strong interaction leading to parallelorientation.

Large

Anti-ferromagnetic

Present Strong interaction leading to anti-parallel orientation

Negligible

Ferrimagnetic Present Strong interaction leading to antiparallel orientation of unequal dipolemoments.

Large

7.3 Origin of permanent dipoles

Permanent dipole moments arise due to the angular momentum associated with thecharged particles. Accordingly, permanent dipoles will bepresent in an atom due tothe following three contributions to the angular momentum:

(i) Orbital angular momentum of electrons

(ii) Spin angular momentum of electrons

(iii) Spin angular momentum of nucleons

Consider the simplest case of a hydrogen atom in which an electron is going rounda proton in an orbit of radiusR (Fig. 7.3).

The current associated with this orbital motion is equal to−e f where f is thefrequency of rotation given by

f = ω/2π (7.6)


R

e−

+ e

Figure7.3Schematic representation of an electron moving in a circular orbit of radiusR around a proton in hydrogen atom.

Here,ω represents the angular velocity of the electron and is assumed to be constant.The magnetic dipole moment associated with the orbital motion of the electron is

µm = Current× Area of orbit.

= −(eω/2π) · πR2 = 1/2eωR2 (7.7)

The angular momentumMa of the electron is defined as

Ma = R×mv (7.8)

wherev is the velocity of the electron given by

v = ωR (7.9)

Substituting the value ofv from equation (7.9) in equation (7.8) we have

Ma = mωR2 (7.10)

From equation (7.7) and (7.10), we can write

µm = −(e/2m) · Ma (7.11)

The negative sign included in equation (7.11) is due to the fact thatµm andMa haveopposite direction as a consequence of the negative charge of the electron. It can beshown that equation (7.11) is valid not only for the simplestcase of a hydrogen atom


but also for any charge distribution. The angular momentum of an electron can beexpressed in terms of the magnetic quantum numberml as

Ma = (h/2π) ·ml (7.12)

It may be mentioned here that the orbital motion of an electron may be described bythe quantum numbersn, l andml. The principal quantum number n accounts for theenergy of the electron; the orbital quantum numberl determines the orbital angularmomentum and the magnetic quantum numberml represents the component of theangular momentum along the direction of the applied external magnetic field. Thequantum numbers are interrelated and have discrete values as follows:

n = 1, 2, 3, . . .

l = 0, 1, 2, . . . (n− 1)

ml = l, (l − 1), (l − 2), . . . . . .0,−1,−2, . . . − l.

Hence, it is possible to compute the permanent dipole momentassociated with an atom.For example, the electrons in the K-shell corresponding ton = 1 will have l value zeroandml value also zero indicating that these electrons do not contribute to the permanentdipole moment. For the electrons in the L-shell (n = 2), the possible l values are zeroand one. Forl = 0, ml = 0. Forl = 1, ml can have values 1,0 and−1. Each of theseml

values can be possessed by two electrons with their spin quantum numbersms equal to1/2 and−1/2. Hence, the distribution of electrons in different shells can be as follows:

n = 1 l = 0 ml = 0 ms = ±1/2 2 electrons.

n = 2 l = 0 ml = 0 ms = ±1/2 2 electrons.

l = 1 ml = 1 ms = ±1/2 2 electrons.

ml = 0 ms = ±1/2 2 electrons.

ml = −1 ms = ±1/2 2 electrons.

The sub-shell corresponding tol = 0 is calleds and the one corresponding tol = 1 iscalledp. Thus, s-subshell can hold 2 electrons, p-subshell can hold6 electrons and soon.

The resultant permanent dipole moment of an atom can be calculated using equa-tions (7.11) and (7.12). It is evident that a resultant dipole moment from a partiallyfilled shell can have a non-vanishing value.

A minimum value of dipole moment is for the caseml = 1 and is equal to

µm =eh

4πm(7.13)


This is an atomic unit of magnetic moment and calledBohrmagneton, β. The numeri-cal value of magnetic moment corresponding to one Bohr magneton is equal to

β = 9.27× 10−24ampm2

Thus, contribution to the permanent dipole moment due to orbital angular momentumof electrons can be expected only from partially filled electronic shells. Hence, the irongroup (atomic number 21 to 28 ) and rare earths (atomic numbers 39 to 45 and 58 to71) are expected to show contributions from orbital angularmomentum. However, inthe solid state, these dipole moments will be ‘frozen in’ andwill not be in a position toorient themselves in presence of an external magnetic field.In the case of iron groupof elements, the incompletely filled shells are close to the outside of the atoms andhence interact strongly with the neighbouring atoms. Thus,their contribution to themagnetic dipole moment is negligible. In the case of rare earths, the incompletelyfilled shells are relatively deep inside the atoms and interact to a lesser degree withthose of neighbouring atoms. Hence, they do contribute to the magnetic properties ofthe material.

The second and the most important contribution to the magnetic behaviour of ma-terials comes from the spin angular momentum of electrons. It can be shown that thedipole moment due to the spin of an electron is equal to

µm(spin) = −(e/m) · Ma(spin) (7.14)

where Ma(spin) = ±h(1/2)

2π= ±(h/4π)

∴ µm(spin) = ±(eh/4πm) = ±β (7.15)

Thus, each electron contributes a magnetic moment equal to±β depending on the spinof the electron. In an atom with many electrons, the resultant magnetic moment due tospin depends on the number of electrons with spin up and thosewith spin down. Thecalculated values of resultant magnetic moments for some atoms is given in Table 7.2.These values are valid for individual free atoms. In the metallic state, the observedvalue differs from the calculated value.

The angular momentum associated with the spin of the nucleons is of the same or-der as that due to the spin of electrons. However, since the mass of the nucleon is largerthan that of an electron by a factor of 103, the magnetic dipole moment associated withthe spin of nucleon is smaller by the same factor. Hence, the effect of nuclear spin isusually negligible.


Table7.2Dipole moments of atoms due to spin of electrons

AtomicNo.

ElementNo. of IIIelectrons

Spinalignment

Dipolemoment

20 Calcium 0 0 0

21 Scandium 1 ↑ β

22 Titanium 2 ↑↑ 2β

23 Vanadium 3 ↑↑↑ 3β

24 Chromium 4 ↑↑↑↑ 4β

25 Manganese 5 ↑↑↑↑↑ 5β

26 Iron 6 ↑↑↑↑↑↓ 4β

27 Cobalt 7 ↑↑↑↑↑↓↓ 3β

28 Nickel 8 ↑↑↑↑↑↓↓↓ 2β

7.4 Magnetic hysteresis

Materials with a large value of magnetic susceptibility namely ferromagnetic and fer-rimagnetic materials exhibithysteresisin theB versusH curve (orM versusH curve).When a magnetic fieldH is applied to a typical sample, the magnetic flux densityB(or the magnetisationM) will vary in a non-linear manner. The permeability or themagnetic susceptibility will no longer be constant but willbe functions of the appliedmagnetic field. As the magnetic field is increased the flux density (Magnetisation)increases and reaches a saturation value Bsat (Msat). When the field intensityH is re-duced to zero, the flux density will not go to zero but will havea finite value calledremanent flux density, Br (or remanent magnetisation,Mr). This remanent flux den-sity may be reduced to zero by applying a magnetic field in the opposite direction. Thefield required to reduce the flux density to zero is called thecoercive field, Hc. Thevariation of magnetic flux density (or magnetisation) with the applied magnetic field isrepresented by the hysteresis loop shown in Fig. 7.4.

This is the prominent and characteristic feature of ferromagnetic and ferrimagneticmaterials. It is exhibited by materials below a characteristic temperature. For example,a ferromagnetic material will show hysteresis in its magnetisation below a temperatureθ f called the ferromagnetic transition temperature. Above this temperature, the ma-terial behaves like a paramagnetic material. In the paramagnetic phase the magneticsusceptibility may be expressed as


B sat

B r

H c H

Figure7.4Hysteresis observed in ferromagnetic and ferrimagnetic materials.

χ =C

(T − θ)for T > θ f (7.16)

whereC is called theCurie constantandθ is a characteristic temperature calledpara-magnetic Curie temperature. This expression is referred to as Curie - Weiss law andis not valid at temperatures close toθ f . For a truly paramagnetic material,

χ = C/T (7.17)

and the relation is called theCurie law. The paramagnetic Curie temperature is usuallyhigher than the ferromagnetic Curie temperatureθ f (Fig. 7.5).

The spontaneous magnetisation and hysteresis observed in ferromagnetic and fer-rimagnetic materials are explained on the basis of domain theory. A virgin sample ofthese materials consists of a large number of regions ordomains which are sponta-neously magnetised; i.e., the magnetic dipoles within the domains are aligned paralleland have a resultant dipole moment. However, the direction of spontaneous magneti-sation varies from domain to domain so that the resultant magnetisation in the sampleis zero (Fig.7.6). When an external magnetic field is appliedto this sample, the do-main with parallel orientation will grow in size at the expense of others. As the fieldis increased, all the domains may orient in one direction leading to a saturation valueof magnetisation. Removal of the applied field leaves the sample in its magnetisedcondition and a field in the opposite direction will be neededto re-orient the domains.


Antiferro

Para

Ferro

θ θ Tf

1/χ

Figure7.5The reciprocal of susceptibility as a function of temperature for para, ferroand antiferromagnetic materials.

(a) (b)

Figure7.6Domain orientation in a ferromagnetic sample (a) in the absence of appliedmagnetic field and (b) after application of a magnetic field.

The saturation magnetisation and the coercive field are important parameters thatdefine the characteristics of magnetic materials and their possible applications.

7.5 Hard and soft magnetic materials

Ferromagnetic and ferrimagnetic materials are known for their high magnetic suscep-tibility and high value of saturation magnetisation. Thesematerials are classified ashard or soft depending on the coercive field associated with them. Materials with a


low value of coercive field are calledsoft magnetic materialsand those with a highvalue of coercive field are calledhard magnetic materials. A study of the hysteresisloop will help us in identifying these materials (Fig.7.7).

Hard

Soft

B

H

Figure7.7Hysteresis loops for hard and soft magnetic materials.

Hard magnetic materials with their large values of magnetisation and also coercivefield have fat hysteresis loop. They require a large magneticfield in the opposite direc-tion to alter their parallel alignment. For soft magnetic materials, on the other hand, alow coercive field is sufficient to reduce the magnetisation to zero.

7.6 Metallic and ceramic magnetic materials

Depending on the chemical composition, magnetic materialsare further classified asmetallic magnetic materials and ceramic magnetic materials. Metallic magnetic ma-terials are usually metals and their alloys. For example, commercial iron or iron with3 to 4% silicon, alloys like permalloy (55%Fe, 45%Ni), supermalloy (80%Ni, 15%Fe, 5%Mo) are soft magnetic materials. But carbon steel, tungsten steel, chromiumsteel, cobalt steel and Al-Ni-Co alloys are hard magnetic materials. Because of the lowelectrical resistance of the metallic magnetic materials,the eddy current loss is quiteconsiderable.

Ceramic magnetic materialsare usually oxides belonging to the ferrite or garnetfamily. Ferrites have a general chemical formulaMFe2O4 whereM is a bivalent metal


such asFe, Co, Mn, Zn, Cd, Mg, etc. Garnets have a chemical formulaM3Fe5O12

whereM is a rare earth ion such asS m, Eu, Gd, Y etc. Y − Al garnet,Y −Gd garnet,Y−Al−Gd garnet, etc., are soft magnetic materials useful for microwave applications.

Ceramic magnetic materials, because of their high electrical resistivity, have loweddy current losses.

7.7 Ferrites

Ferrites are solid solutions of two oxides with formulaMO andFe2O3. The magneticproperties of ferrites are intimately related to their structure. By a suitable selectionof the composition, it is possible to design materials with the required magnetisation.Ni − Zn ferrite andMn− Zn ferrite are examples of soft magnetic materials, where asbarium ferrite is an example of hard magnetic material.

Let us consider the case ofmagnetite. This has a chemical formulaFe3O4 andmay be written as a solid solution of two oxidesFeOandFe2O3. Each formula unitof the ferrite has oneFe2+ ion and twoFe3+ ions. A unit cell of the compound haseight molecules with oxygen ions forming a close packed facecentred cubic structure.It is observed that there are two types of interstitial sitesavailable for the metal ionsto occupy. They are octahedral holes called B-sites (an interstitial site surrounded bysix oxygen ions) and tetrahedral holes called A-sites(an interstitial site surrounded byfour oxygen ions). Corresponding to one formula unit of magnetite, there are twooctahedral sites and one tetrahedral site available for occupation by the metal ions.There exists an anti-ferromagnetic interaction between the metal ions present in A-sites and B-sites. In other words, the spin on the ion presentin A-site is oppositelyoriented as compared to the spin on the ion present in B-site.In magnetite, Fe2+ ionsoccupy half the octahedral sites.Fe3+ ions occupy the other half octahedral sites andthe tetrahedral sites (Fig.7.8).

Fe2+

Fe3+

Fe3+

Octahedral

B − Sites

A − Sites

Tetrahedral

4β 5β5β

Figure7.8Distribution of Fe2+ and Fe3+ ions in octahedral and tetrahedral holes.


Hence the magnetic moments due toFe3+ ions get cancelled and the magnetizationof Fe3O4 is due to the magnetic moment ofFe2+ ions alone, i.e., 4β per molecule.

The magnetic properties of ferrites are very sensitive to atomic arrangements. Itis observed that replacing some of theFe2+ ions withZn2+ or Cd2+ ions will lead toan increase in the magnetization. This is becauseZn2+ ions go preferentially into thetetrahedral holes there by forcingFe3+ ions to occupy octahedral holes. Figure 7.9shows the situation when 50% of theFe2+ ions are replaced byZn2+ ions (i.e., oneFe2+ ion from two formula units). SinceZn2+ ions have no magnetic dipole moment,

Octahedral

B − Sites

A − Sites

Tetrahedral

Fe3+

Fe3+

Zn2+

Fe2+

Fe3+

Fe3+

5β 5β 5β5β 4β

Figure 7.9 Distribution of Fe2+, Fe3+ and Zn2+ ions in the case of a compound with50% of the Fe2+ ions replaced with Zn2+ ions.

the net magnetization increases. Thus it is possible to control the magnetic behaviourof ferrites by suitably modifying the chemical composition.

7.8 Applications of magnetic materials

Diamagnetic materials are those which do not possess permanent dipoles and the in-duced dipole moment is extremely small. Hence, these materials find applicationwhere the magnetic effects are undesirable.

Paramagnetic materials possess permanent dipoles but there is negligible interac-tion among them. The magnetic susceptibility is small ( 10−3) but positive. Further, thesusceptibility is inversely proportional to the temperature and decreases with increasein temperature. Paramagnetic salts are used in obtaining very low temperatures (lessthan 1 K) by adiabatic demagnetisation. They are also used insolid state masers.

Ferromagnetic and ferrimagnetic materials show large magnetisation and henceare characterised by large values of magnetic susceptibility. They are classified ashard or soft magnetic materials depending on the coercive field required to remove the


spontaneous magnetisation produced in them. They are also classified as metallic orceramic magnetic materials depending on their chemical composition. Hard magneticmaterials are usually associated with large hysteresis losses and hence are not suitablefor a.c applications. Further, metallic magnetic materials due to their low electricalresistivity compared to that of ceramic magnetic materialsshow large eddy currentlosses. So, the magnetic material for a particular application needs to be selectedcarefully.

Hard magnetic materials(and hard ferrites)are used in permanent magnets. Theyfind application in the construction of instruments like ammeters, galvanometers, fluxmeters, speedometers, compasses, etc. They are also used inmotors and generators.They find use in electronic devices like tape recorders, loud-speakers, telephones, hear-ing aids, TV tubes etc.

Soft magnetic materials (and soft ferrites) are widely usedas transformer cores inall sorts of transformers used in different frequency operations. They are used in chokecoils, recording heads and for the pole pieces in electromagnets. Some soft ferrites areused for computer memory cores.

Numerical Examples

7.1 A magnetic field of 2000Am−1 is applied to a material which has a susceptibilityof 1000. Calculate

(i) relative permeability of the material,(ii) intensity of magnetisation and

(iii) flux density.

Solution:(i) Relative permeability,

µr = 1+ χ

= 1+ 1000= 1001 (Ans.).(ii) Intensity of magnetisation,

M = χH

= 1000× 2000= 2× 106Am−1 (Ans.).

(iii) Flux density,

B = µoµr H

= 4× 107 × 1001× 2000

= 2.52Wbm−2 (Ans.).


7.2 Iron shows a saturation magnetisation of 1.75×106Am−1. Calcualte the contribu-tion of each atom to the magnetisation if iron has a b.c.c structure with a latticeparameter of 2.86A.

Solution:

Volume of unit cell = a3 = (2.86× 10−10)3 = 2.34× 10−29m3

Number of atoms

per unit cell = 2

Number of atoms

per unit volume= (2/a3) = 8.54× 1028m−3

Magnetisation due

to each atom=1.75× 106

8.54× 1028

= 2.05× 10−23Am−1

=2.05× 10−23

9.27× 10−24= 2.2β (Ans.).

7.3 Evaluate the magnetic moment corresponding to one Bohr magneton.

Solution: Bohr magneton is given by

β = (eh/4πm)

=1.6× 10−19 × 6.62× 10−34

4× 3.14× 9.1× 10−31

= 9.27× 10−24Am2 (Ans.).

7.4 Atomic number of nickel is 28. Calculate the dipole moment per atom due tospin of electron. If the dipole moment for an atom in the boundstate is just halfthe value for the free atom, find the saturation magnetisation in nickel. Assumenickel to have 5× 1028atom m−3.

Solution: The electron configuration forNi is

1s22s22p63s23p63d84s2

There are eight 3d electrons with their spins aligned as 5 UP and 3 DOWNproducing a resultant dipole moment of 2β. Hence, in the bound state, eachatom contributes a dipole moment ofβ.


Saturation magnetization is the product of the dipole moment on eachatom and the total number of atoms per unit volume.

MS = 9.27× 10−24 × 5× 1028

= 463500Am−1 (Ans.).

Exercise

7.1 Describe the nature of hard and soft magnets. (March 1999).

7.2 What are ceramic magnets? Discuss their properties and the reason for classify-ing ferrites as ceramic magnets. Explain two important applications.

(March 1999)

7.3 Explain magnetic hysteresis on the basis of domain theory. (August 1999).

7.4 What are ceramic and metallic magnets? Give their properties and applications.

(August 1999).

7.5 Explain the properties and applications of ceramic magnets. (March 2000).

7.6 What are the contributions to the magnetic dipole moment of an atom? How arethe magnetic materials classified? Discuss the changes in magnetic properties ofa ferromagnetic material as a function of external magneticfield. (March 2000).

7.7 Explain the classification and properties of magnetic materials. (August 2000).

7.8 Explain orbital angular momentum, spin angular momentum and nuclear spin.

(August 2001).

7.9 What are ceramic and metallic magnets? Give their properties. (August 01).

7.10 How are magnetic materials classified on the basis of susceptibilities? Discussthe contribution of angular momentum to the magnetic properties of materials.

(March 2002).

7.11 Discuss the properties of hard and soft magnetic materials with the help of hys-teresis loop. (Feb 2003, Aug 2003, Feb 2004).

7.12 Explain the properties and applications of ferrites. (Aug 2004).

Chapter 8

Applied Optics

8.1 Absorption and emission of radiation

When radiation interacts with matter under appropriate conditions, it leads to an abrupttransition of the quantum system such as an atom or molecule from one energy stateto another. If the transition is from a higher state to a lowerstate, the system givesout a part of the energy and if the transition is in the reversedirection, then, it absorbsthe incident energy. When a substance absorbs energy, a partof the energy may bere-emitted in the form of electromagnetic radiation in the visible or near-visible regionof the spectrum. This phenomenon is known asluminescence.

8.1.1 Luminescence

Luminescence involves the following two steps:

(i) The excitation of the electrons in the atoms of the solid,and

(ii) The subsequent emission of photons.

These two major steps may be associated with some intermediate steps. Depending onthe time interval between the excitation and the emission processes, luminescence isclassified as fluorescence and phosphorescence.Fluorescenceis a process in whichthe emission occurs during excitation or the time interval between the two processes ofexcitation and emission being less than 10−8s. The emission of light after the excitationhas ceased or with a lapse of time larger than 10−8s is calledphosphorescence. It isobserved that in majority of cases, the decay time is independent of temperature forfluorescence and is temperature dependent for phosphorescence.

Luminescence is found to be associated with the presence of activators in materi-als. These activators are either impurity atoms present in minute quantities or a smallexcess of one of the constituent elements in a compound. It isinteresting to note thatthe presence of certain other type of impurity may inhibit luminescence. Such impuri-ties are called killers. Addition of activator impurity in acrystalline solid will give riseto localized energy levels in the forbidden energy gap (Fig.8.1).

225

226 Applied Optics

h ν

Figure 8.1 Absorption and emission processes and the role of localizedenergy statesin the band gap.

These levels may be due to the activator atoms themselves or due to the host atomsinfluenced by the presence of activator atoms. They may also be associated with latticedefects induced due to the addition of activator atoms. In case of fluorescence, theexcitation and emission processes involve these localizedenergy levels in addition tointer band transition. In case of phosphorescence, the localized energy levels presentin the band gap may be metastable states. A transition to a metastable state results intrapping of the electron and an emission can occur when the electron is released backinto the conduction band. This accounts for the delay in the emission process.

The process of luminescence is further classified on the basis of the mechanismof excitation. If the carriers are excited by photon absorption, the resulting emissionis called photoluminescence. In cathodo-luminescence, the excitation is by bombard-ment with high-energy electrons. If the excitation is due tothe passage of an electriccurrent, the resulting luminescence is called electro-luminescence. Other types of lu-minescence are tribo-luminescence, chemi-luminescence,radio-luminescence,magneto-luminescence, etc. In all these cases, the nomenclature refers to the mech-anism by which the excitation is achieved. On the other hand,thermo-luminescencerefers to the case of luminescence in which, irrespective ofthe method of excitation,the emission is stimulated by thermal process. Luminescence is an important propertyexhibited by certain materials, which makes them useful fordevice applications.

8.1.2 Induced absorption

Induced absorption is the absorption of an incident photon by a system as a resultofwhich the system is excited from a lower energy state to a higher energy state wherein

Applied Optics 227

the difference in energy of the two states is precisely the energy of the photon. Consideran atomic system in which an outer electron has a ground stateenergy ofE1 and excitedenergy stateE2. An absorption process takes the electron fromE1 to E2 when a photonof energyE, precisely equal to (E2 − E1)) is incident on the atom. As a result, itsenergy becomes

E2 = E1 + E

The atom is said to have made a transition to the excited stateand is indicated asatom∗.This phenomenon is calledinduced absorptionwhich is represented as

Atom+ photon→ atom∗

8.1.3 Spontaneous emission

Spontaneous emissionis the emission of photon when a system (an atom) transitsfrom an excited energy state to the ground state without the aid of any external agency.The rate at which a transition from a higher energy level to a lower energy level takesplace will be proportional to the number of electrons present in the higher energy levelat any instant. In the absence of simultaneous excitation, the number of electrons inthe excited state goes on reducing exponentially with increase in energy. Consider anatom in the excited state of energyE2. A spontaneous emission results in a photon ofenergyhν when the electron makes a transition from excited stateE2 to the groundstateE1. The energy of the photon is given by

hν = E2 − E1

8.1.4 Stimulated emission

The above-mentioned process of emission of radiation occurs randomly and on itsown. Therefore it is classified as spontaneous emission. However, the electrons inthe excited state need not wait for spontaneous emission to occur if it is stimulatedto emit a photon and go to a lower energy level. This stimulation may be broughtabout by photons themselves of proper energy. Such an emission of radiation is calledstimulated emission. The emitted photon will be in phase with the photon stimulatingthe emission. If the process continues, a monochromatic, coherent radiation will result.

Consider an atomic system in which an outer electron has a ground state of en-ergyE1 and an excited state of energyE2. An absorption process involving supply ofnecessary energy takes the electron fromE1 to E2. The excited electron returns to theground state by a spontaneous emission of a photon of energyhν given by

hν = E2 − E1 (8.1)

228 Applied Optics

The emission may also be stimulated with the help of photons of similar energy. Ifn1 andn2 represents the number of electrons present in the energy levelsE1 andE2 atany instant at any temperature, then, by Boltzmann’s distribution,

n2/n1 = exp −(E2 − E1)/kT = exp −hν/kT (8.2)

This equation indicates that the population at energy levelE2 is much smaller thanthat atE1 at equilibrium and most of the electrons are in the lower energy state. Ifby some mechanism, the population atE2 is increased, a stimulating radiation can beused to produce emission of radiation in the form of a strong coherent beam. This isthe principle used in the construction of a laser.

8.2 Lasers - basic principles

Laser is an opto - electronic device based on the principle oflight amplification bystimulated emission of radiation. (Laser is an acronym for light amplification by stim-ulated emission of radiation.) The light output from a laserdepends on the type oflaser and may be a continuous beam of low or medium power or a pulsed beam ofhigh power. Laser is a source of monochromatic, coherent, directional and intenselight. The principle of operation, as the name indicates, isby stimulated emission ofradiation. This may be achieved under the following conditions:

(i) An excess concentration of electrons in the higher energy states, and

(ii) A stimulation by a radiation to bring about the de-excitation process.

The first condition is usually referred to aspopulation inversion since under normalconditions, the concentration of electrons in excited states is much less than that inlower states. An examination of equation (8.2) reveals thatthe population inversion isequivalent to a negative temperature condition. The secondcondition is achieved byusing a radiation field of appropriate frequency.

8.2.1 Einstein’s theory of stimulated emission

Let ρ represent the energy density of the radiation field. It is defined as the total energyin the radiation field per unit volume per unit frequency. This may produce absorptionand subsequent emission in an atomic system (Fig. 8.2).

Applied Optics 229

Figure 8.2Steady state processes of (a) absorption, (b) spontaneous emission and (c)stimulated emission. The rates of the three processes are mentioned.

The rates of different processes that can occur may be written as follows:

(i) The rate of absorption depends on the number of electronsn1 available at thelower levelE1 and the energy densityρ of radiation field assisting absorption,

(ii) The rate of spontaneous emission depends on the number of electronsn2 presentin the excited energy stateE2, and

(iii) The rate of stimulated emission depends on the number of electronsn2 in theexcited energy stateE2 and the energy densityρ of the radiation field assistingemission.

At steady state, we can write

Rate of absorption= Rate of spontaneous emission+ Rate of stimulated emission

K1n1ρ = K2n2 + K3n2ρ (8.3)

This relation is calledEinstein’s equationand the constantsK1, K2 andK3 are calledEinstein’s coefficients. These coefficientsK1, K2 andK3 represent the probability ofthe processes of absorption, spontaneous emission and the stimulated emission takingplace respectively.

In the absence of any stimulated emission, the above equation reduces to

K1n1ρ = K2n2

Or ρ =K2n2

K1n1=

K2

K1· exp −hν/kT (8.4)

230 Applied Optics

But, according to the Planck’s theory of radiation, (Pleaserefer equation (1.5) ofChapter 1), we have,

ρ =8πhν5

c4

1(ehν/kT − 1)

(8.5)

It is clear that the two equations, (8.4) and (8.5) are not in agreement. In order to rec-tify the discrepancy, Einstein proposed another kind of emission, known as stimulatedemission. Equation (8.3) may be rewritten as

ρ =K2n2

K1n1 − K3n2

=K2

K1ehν/kT − K3(8.6)

Comparing equation (8.6) with equation (8.5), we get

K1 = K3

andK2

K1=

8πhν5

c4

Hence, the ratio of the rate of stimulated emission to the rate of spontaneous emissionis given by

K3n2ρ

K2n2=

1(ehν/kT − 1)

(8.7)

Thus, the existence of stimulated emission is proved. The energy density of the radia-tion field may be written as

ρ =8πhν5

c4

1(ehν/kT − 1)

8.2.2 Conditions for laser action

At thermal equilibrium, the ratio of the stimulated emission rate to spontaneous emis-sion rate is very small. Efficient laser emission can be achieved under the followingconditions:

(i) Rate of stimulated emission should be more than the rate of spontaneous emis-sion,

(ii) Rate of stimulated emission should be more than the rateof absorption.

Applied Optics 231

The first condition can be achieved by using a radiation field of very large energydensity. This is made possible by using an optical resonant cavity in which the photondensity can be enhanced by multiple reflections. The second condition requires a pop-ulation inversion in which the number of electrons in the excited state is higher thanthat in the lower state. This is a non-equilibrium conditionand is facilitated by thepresence ofmetastable states. A metastable state is one which has a relatively longerlifetime and electrons excited to these levels will come down to lower levels at a muchsmaller rate than the rate at which they are excited. However, a stimulated emissioncan be induced using a suitable radiation field.

8.2.3 Methods of achieving population inversion

The process of exciting the active atoms in a medium to higherenergy states in orderto achieve population inversion is calledpumping. There are several ways of pumpingbut the most commonly used methods are the following:

(i) Optical pumping or excitation using photons: In this case, an external opticalsource like a flash lamp is used to produce radiation field fromwhich the energymay be absorbed to produce excited atoms. This type of pumping is used in rubylaser and Nd:YAG laser.

(ii) Use of electric discharge:In this case, direct electron excitation in a gaseousdischarge is used. This type of pumping is used in gas ion lasers like argon laser.

(iii) Use of atomic collisions:In this type of pumping, a mixture of two gases havingsame excited state is subjected to an electric discharge. Anexcited atom uponcollision with an atom of the second gas transfers the energyand sends it to theexcited state. This type of excitation is used in Helium-Neon laser.

(iv) Use of chemical reactions:In this method, a chemical reaction results in a prod-uct whose atoms or molecules will be left in an excited state.Under such a con-dition, population inversion may occur. This type of pumping is used in chemicallasers like Hydrogen fluoride laser.

8.2.4 Requirements of a laser system

A laser requires three major components for its operation. They are:

(i) Pumping mechanism,

(ii) Active medium and

(iii) Resonant cavity.

232 Applied Optics

The first one is related topumping. A suitable method of pumping is selectedto produce population inversion. The choice of the method depends on the type oflaser under consideration, the nature of the active medium and the laser output. Forexample, in solid state lasers, optical pumping method is used with a flash lamp as thesource of light. Most often, the output from the source of light will be in the form ofshort flashes. The method may be suitable for pulsed lasers inwhich the output fromthe laser is also in pulses.

The second requirement is anactive medium. It is the medium in which the atomsor molecules will be excited to higher energy states in orderto produce populationinversion condition. The condition of population inversion can be achieved with thehelp of metastable states with which the atoms or molecules of the active medium areassociated. Depending on the active medium used, lasers areclassified into varioustypes. For example, we have solid state lasers making use of crystals like ruby orgarnets as active medium, gas lasers using different gases or their mixtures as activemedium, semiconductor laser making use of a p-n junction, dye and chemical lasersusing chemicals, etc.

Transitions to lower states by excited atoms are governed byselection rules andhence all transitions are not allowed. For atoms with one valence electron, the selectionrules are as follows:

n = 0,±1,±2,±3, . . .

l = ±1

wheren andl are principal quantum number and the orbital quantum numberrespec-tively. For atoms with more than one valence electron, like alkaline earths, a differentset of selection rules apply. The selection rules for two electron systems may be writ-ten as

l1 = ±1,l2 = 0,±2

If a single electron jumps, thel value of one change by 1 and the other remains un-changed. If the two electrons jump simultaneously, thel value of one change by 1 andthat of the other by 0 or 2. There are no restrictions on the total quantum numbern ofeither electron.

The third requirement is aresonant cavity. The resonant cavity essentially consistsof two mirrors facing each other (Fig. 8.3).

One of the mirrors is fully reflecting and the other is partially transparent to allowsome radiation to pass through. This cavity helps to increase the stimulated emissionthereby increasing the intensity of the laser beam. There are two commonly usedmirror configurations for optical cavity resonators. They are plane parallel mirrors and

Applied Optics 233

ACTIVE MEDIUM LASER

PUMPING MEDIUM

100 %

REFLECTOR

99.8 %

REFLECTOR

Figure8.3Basic components of a laser system.

confocal mirrors. The radiation will propagate to and fro within the cavity betweenthe two mirrors. Because of diffraction effects, there will be loss of collimation in thelaser beam. Such losses can be minimized by using concave mirrors. Absorption inthe mirrors is another source of power loss. In order to reduce this, high reflectancemultilayer dielectric coatings are provided on the mirrors.

The efficiency of laser emission depends on the nature of the active medium andthe energy levels between which the laser action takes place. For example, if there areonly the two energy states, a ground state and a metastable state, the laser is said to bea two level laser(Fig. 8.4).

E3

E2

E1

E2

E1

E3

E2

E1

E4

LASER

LASER

(a) (b) (c)

Figure8.4Transitions in a (a) two level, (b) three level and (c) four level lasers.

In the case of two level laser, the population inversion willnot be possible. Thisis because, as more and more electrons are pumped to the metastable state, there willbe more stimulated emission and at any instant of time, therewill be no more thanhalf the number of electrons in each state. As a result, no laser amplification occurs.

234 Applied Optics

Population inversion will be possible only when the absorption is to a level higher thanthe metastable state as in the case of athree level laser. Pumping takes the electronsfrom the ground state to the excited state. From this excitedstate, the electron makes atransition to the metastable state by spontaneous emission. The laser emission occurswhen the electron in the metastable state is stimulated to undergo a transition to theground state. However, a reverse transition is also possible from the ground state tothe metastable state by a re absorption of the emitted radiation. This removes somephotons from the laser beam and hence the efficiency of the laser decreases. Thisproblem is overcome in afour level laser. In this case, the laser transition occurs toan unstable intermediate state rather than the ground state. The electron decays to theground state rapidly. Thus, the initial and the final laser levels are separated from thehigher excited state and the ground state respectively. Theprobability of re absorptionof laser radiation is thus avoided which improves the efficiency of the laser.

8.3 Types Of Lasers

Lasers are generally classified as continuous wave lasers and pulsed lasers. Continuouswave lasers give a continuous output of constant intensity and the pulsed lasers giveout pulses of output with a definite frequency. Pulsed laserscan have very high peakpower and also very short pulse durations. These lasers havetheir own advantages anddisadvantages and are selected depending on the application for which they are used.Further, depending on the active material used for laser action, the lasers are classifiedas solid state lasers, gas lasers, chemical lasers, semiconductor lasers, etc.

8.3.1 Ruby Laser

The first working laser was constructed using a ruby crystal.Ruby, chemically alu-minium oxide with about 0.05 weight percent of chromium impurity, in a single crys-talline form is used as an optical cavity. A cylindrical rod of the crystal, 5 to 20 cm inlength and 0.5 to 2 cm in diameter, with its ends cut parallel and polished is surroundedby a helical xenon flash lamp (Fig. 8.5).

The ends of the rod are coated with a highly reflecting material like silver so thatthe rod acts like a resonant cavity in which light intensity can build up by multiplereflections. One end of the rod is made highly reflecting (100%reflectivity) and theother partially transmitting (50 to 80% reflectivity) so that the laser beam emerges outof the ruby rod.

In ruby laser, chromium ions are the active centres responsible for laser transition.The energy level diagram for chromium ions in ruby crystal isshown in Fig. 8.6. The

Applied Optics 235

reflectingFully

surface

Power supply

Ruby rod Flash lamp

transmittingPartially

surface

Figure8.5Schematic diagram of a ruby laser.

absorption of energy from the flash lamp occurs in a band of energies in the greenregion. This results in electrons from the ground stateE1 being raised to the bandE3. These excited levels are highly unstable and the electronsdecay to the levelE2.During this decay process, energy equivalent to (E3 − E2) will be given out as heat.The levelE2 is a metastable state with a lifetime of the order of 5 ms. Thisresults ina population inversion when the rate of absorption is large enough and a stimulatedemission from the metastable stateE2 to ground stateE1 produces the laser outputwith a peak intensity at 6943A. E2 is a doublet and hence the output has an additionalemission line at 6928A.

Figure8.6Energy level diagram for chromium ions in ruby laser.

Once the stimulated emission begins, the metastable state is quickly depopulated.Thus, the output from ruby laser is in pulses, closely related to the rate of build-up anddepletion of population in the metastable state. Further, the flash lamp used for opticalpumping produces light of many wavelengths not useful for pumping and hence is a

236 Applied Optics

waste of energy. Thus, ruby laser is inefficient and also unsuitable for many applica-tions.

8.3.2 Helium-Neon laser

Helium-Neon laser is a gas laser useful for continuous operation. In gas lasers suchas He-Ne laser, the atoms are characterized by sharp energy levels and an electricdischarge is useful for pumping. The laser consists of a discharge tube, 10 to 50 cmin length and 0.5 to 2 cm in diameter, filled with Helium and Neon gases in the ratio10:1 upto a pressure of 1 torr. The electrodes are connected to a high voltage sourceof a few kilovolts dc supply (Fig. 8.7). The reflecting mirrors are placed outside thedischarge tube to make the alignment easier.

Power supply

Partial

Discharge tube

reflectingmirror

Fullyreflectingmirror

windowBrewster

Figure8.7Schematic diagram of He-Ne laser.

In a laser, the beam is built by multiple reflections. Reflection of radiation bya denser medium always results in polarization. In order to obtain a pure state ofpolarization, laser tubes are often provided with Brewsterwindows. The ends of thelaser tube are inclined at an angleθ (called Brewster’s angle) to the axis of the tubesuch that tanθ is equal to the refractive index of the material of the window. Under thiscondition, the ray reflected away will be polarized normal tothe plane of incidence andthe transmitted ray will be polarized in the plane of incidence. Hence, the laser beamwill be plane polarized.

Fig. 8.8 shows the energy level diagram for the Helium and Neon atoms. Due tothe applied electric field, the accelerated electrons travel across the tube colliding withthe gas atoms to excite them to higher energy states.

Since Helium is in excess, the probability of excitation of Helium atoms is morethan that of Neon atoms. Thus, the population at energy levels E2 andE3 of Helium

Applied Optics 237

Laser

Transitions

Spontaneous

emission

Neon

Figure8.8Energy level diagrams for Helium and Neon atoms.

atoms increases in number. As these levels are metastable having relatively longerlifetime, they remain in excited state till they collide with Neon atoms. This results inthe transfer of electrons fromE1 to E4 andE6 levels in Neon atoms. In fact, all theexcited states of Neon consist of many closely spaced sub-levels and the energy ofE4

andE6 levels in Neon is close toE2 andE3 levels in Helium respectively. Thus,E4 andE6 levels in Neon will be preferentially populated over other levels. Laser emissionoccurs due to three allowed transitions, namelyE6 to E5, E6 to E3 andE4 to E3. Ofthese, the transition fromE6 to E3 is the prominent one with an emission wavelength of6328A. The other two laser transitions are in the infrared region.The laser radiationis also associated with a spontaneous emission of wavelength about 6000A due totansition fromE3 to E2 levels. The deexcitation fromE2 to E1 level of Neon occurs byatomic collisions with the walls of the tube.

8.3.3 Semiconductor diode laser

The basic principle of working of a semiconductor diode laser is exactly the same asthe principle of operation of a light emitting diode. In caseof a light emitting diode, thephenomenon of injection electroluminescence results in the emission of radiation froma forward biased p-n junction. However, the emission is incoherent and less intense.As the current across the junction is gradually increased, astage will be reached atwhich stimulated emission also begins along with spontaneous emission.

In a diode laser, the p-n junction is formed using degeneratesemiconductors. Whenthe dopant concentration is increased, the impurity levelswill expand into bands and

238 Applied Optics

then merge with the nearest energy band. In an n-type semiconductor, the impurityband overlaps with the conduction band and for a p-type semiconductor, the impurityband overlaps with the valence band. Such semiconductors behave like metals and arecalleddegenerate semiconductors. When p-n junctions made out of these degeneratesemiconductors are forward biased, electrons and holes areinjected across the junctionin large concentration (Fig. 8.9).

p n

E

EE

EE

n

c

v

Fc

v

pE F

Inversionregion

Laserp − Ga As

n − Ga As

Figure8.9Diode laser (a) basic structure and (b) energy level diagram.

This results in a condition identical to population inversion in the region close tothe junction and transitions from the bottom of the conduction band to the top of thevalence band can take place with the emission of radiation ofenergy equal to the bandgap energy.

In selecting the material for the fabrication of a diode laser, it is essential that therecombination is a direct process without the involvement of traps or recombinationcenters. Hence, direct band gap semiconductors are suitable materials for the diodelaser. Further, it must be possible to dope these semiconductors to very high impurityconcentration and to form a p-n junction. An efficient resonant cavity has to be con-structed at the junction to ensure laser emission. For this purpose, the front and backsurfaces are made flat and parallel.

The device described above contains a single p-n junction made out of a single ma-terial. Hence, it is called a homojunction laser. The efficiency of laser may be increasedby using multiple layers(junctions) of different materials. Such devices are called het-erojunction lasers. For example, GaAs-AlGaAs heterojunctions and InGaAsP/InP sys-tems are particularly well suited for lasers used in fiber optic communication. Bychoice of composition, lasers can be made in the infrared region of wavelength 1.3to 1.55µm required for fiber optics. In many other applications, different wavelength

Applied Optics 239

ranges may be required for laser output. In pollution diagnostics, ternary alloy Pb-SnTe is used to provide laser output wavelength from 7 to 30µm. For intermediatewavelengths, InGaSb system can be used.

8.4 Applications of lasers

In contrast to ordinary light, laser radiation is highly directional, monochromatic, co-herent and very intense. The applications are base on these distinguishing features.One of the major applications is in pure science to investigate the interaction of matterwith intense electromagnetic radiation. The directionality and coherence of the laserbeam are useful in the measurement of distances based on interferometric methods.In the field of communication, laser has been used with propermodulation for infor-mation transmission. Because of the high intensity and energy associated with laserbeams, they can be utilized for applications such as welding, cutting and ablation ofmaterials. Laser has also been used for medical applications like treatment of dentaldecay, destruction of tumors, treatment of skin diseases and eye surgery. Hologra-phy is another important application which helps in recording the amplitude as wellas the phase of light reflected from objects thereby preserving the three dimensionalinformation.

8.4.1 Industrial applications

Two lasers widely used for industrial applications are Nd:YAG laser andCO2 laser.They are used for a variety of applications in industries.Laser drilling is used toproduce holes with very small diameters. In general, pulsedlaser is used for drillingapplications. Each pulse of laser heats the surface very rapidly leading to the evap-oration locally. Since there is no physical contact, there is no wear and breakage ofthe tool. The holes may be located with high precision and even hard materials can bedrilled easily. It is also possible to drill holes in area difficult to reach and at difficultangles.

Another industrial application of lasers is inlaser cutting. It involves heating ofthe metal to its melting point with the laser and applying a jet of a gas to remove themolten metal. Laser cutting involves local heating and hence, the heat affected zone isminimum in size.

A focused high intensity beam can be used forlaser weldingapplications. Whenfocused to a small spot size, the heat generated melts the metal producing a weldedjoint. It is necessary to control the laser power precisely during welding application torestrict the molten zone and also to avoid loss of material through evaporation. On the

240 Applied Optics

other hand, laser welding has the advantage that the heat affected area is small, offersbetter control of the welding process and can be used in otherwise inaccessible areas.

8.4.2 Medical applications

Laser has been used in medical sciences for surgery. The advantage of a laser beamis that it can be focused into a fine beam of very small size thereby concentrating theoptical energy to a narrow spot. This can be used for the vapourization of tissueswhich need to be removed. Typically continuous wave lasers like CO2 laser orArlaser of 50-100 watt power are used for surgical purposes. Lasers are widely used inophthalmology for welding detached retina and also in cataract surgery. It is also usedto correct the defects of the eye. Recently, lasers are finding applications in dentistryand dermatology.

8.4.3 Estimation of atmospheric pollution

Pollution has become a major concern, especially in urban areas where industries andautomobiles have contributed to the presence of various harmful chemicals in atmo-sphere. There are various methods available for the analysis of atmospheric pollutants.A laser beam also may be used to estimate the concentration ofparticulate suspensionlike dust or smoke in atmosphere. For this purpose, a beam from a pulsed laser is madeto incident on the particulate matter. The back-scattered rays are collected using widearea concave mirror and analysed for intensity. The intensity of the back-scatteredradiation is a measure of the density of the suspended particulate matter. It is also pos-sible to locate the pollutants by measurement of the time delay between the emissionof the laser pulse and its receipt after scattering.

A chemical analysis of the pollutants is more involved and complex. It can becarried out by a spectral analysis of the emitted or scattered radiation. Changes inthe wavelength of the received radiation are indicative of the possible interactions andhence characteristic of the pollutants and their composition.

The method is neither convenient nor accurate to deserve a more detailed discus-sion.

8.4.4 Applications in basic sciences

Lasers have been used for various applications in basic sciences. This is mainly be-cause lasers are monochromatic sources of high intensity and hence are useful toolsin the study of interaction of radiation with matter. High temperature environments,

Applied Optics 241

which are otherwise difficult to produce, may be easily achieved in localized regionsto study the physical and chemical changes. Processes like surface modification andsurface ablation can be studied with the help of laser radiation. Laser has been used toaccelerate certain chemical reaction and to synthesize certain new compounds. Morerecently, lasers have been used to simulate situations similar to the ones existing in sunand stars, and hence to study the thermonuclear fusion reactions in laboratories. Theconstruction of various types of lasers has widened the scope of their use for applica-tions. It will not be surprising if we find them useful in our daily life. Laser printersand scanners for reading bar codes are some examples.

8.5 Holography

Holography is a method of producing three dimensional images of objects using theprinciple of interference. The word holo means complete andholography meansrecording a pattern containing all information about the object. Conventional pho-tography can give two dimensional images since it is a recording of the variations inthe intensities of the light scattered by the object. Holography carries additional infor-mation as it records both the amplitude and the phase of the scattered radiation.

The basic principle involved in holography is as follows:The object is illuminated by light from a coherent source like a laser. This generates

secondary waves from each point on the object which will spread in all directions andinterfere with each other. This interference pattern is recorded on a photographic plate.The recorded pattern is again illuminated with coherent light of the same wavelengthto reproduce the original object in three dimensions.

8.5.1 Recording of holograms

Consider an objectXY illuminated by a coherent source of light (Fig. 8.10). In orderto produce high quality holograms, the source of light must be monochromatic andcoherent. Hence, laser beams are used for the purpose. The beam is divided into twoparts using the principles of division of wavefront or division of amplitude. Figureshows the use of a beam splitter for the purpose. One part of the beam falls on thephotographic plate directly and is called the reference beam. The second part falls onthe object to be recorded and the reflected waves fall on the photographic plate. Thesetwo parts of the beam interfere and produce a pattern. This pattern is recorded andprocessed using standard photographic techniques to obtain a hologram. The actualexperimental arrangement is shown in Fig. 8.10.

242 Applied Optics

Figure8.10Recording of a hologram.

8.5.2 Reconstruction of images

For the reconstruction of the image, the recorded hologram is illuminated by a mono-chromatic and coherent beam of light of the same wavelength as the one used forrecording (Fig. 8.11).

VIRTUAL

IMAGE

LASER

REAL

IMAGE

HOLOGRAM

Figure8.11Reconstruction of image using a hologram.

This produces two images. One image, called the primary image, is virtual and thesecond image, called the conjugate image, is real. The real image may be recorded ona screen or photographed.

Applied Optics 243

8.5.3 Applications of holography

Holography has many important applications in science and technology. One of the ap-plications is in interferometers useful in the study of surface quality, stress and strain,etc. Holographic techniques can be used more effectively in all interferometric applica-tions. A technique called double exposure holographic interferometry is used to studythe distribution of strain in objects. We can record an object holographically before andafter application of stress on the same hologram by double exposure. A reconstructionof the image reveals an interference pattern in which there is an overlap of informationfrom the strained and unstrained sample. A study of the shapeand number of fringesin the pattern provides information on the distribution of strain in the object.

Holographic technique is used to produce diffraction gratings. The conventionalmethod of producing gratings using ruling engines is prone to errors and such errorscan be completely avoided in the holographic method.

Holography can be used in microscopy also. The magnificationeffect can be ob-tained by using an illuminating radiation of a higher wavelength in the image recon-struction step.

Holography can also be used in pattern recognition. It is used for data storagedevices in computer technology, for making photographic masks for lithographic tech-nique in microelectronics.

8.6 Optical fibersThe use of optical fibers for transmitting light signals overlarge distances is now awell known phenomenon. With the development of laser and flexible fibers, opticalfibers are being used extensively for various communicationapplications. Fiber op-tic communication has significant advantages over the transmission by conventionalcoaxial cables. The loss of signal strength is considerablyless in optical fibers andhence permits transmission over long distances. Use of light waves in place of radioand microwaves has improved the speed of communication.

8.6.1 Materials for optical fibers

An optical fiber consists of a central core surrounded by a cladding (Fig. 8.12). Therefractive index of the core material is higher than that of the cladding. The coreand cladding are made of either glass or plastic. Additionalcovering is provided toimprove the strength and also to protect it from moisture andmechanical wear. Withall the protective coverings, the optical fibers are still very thin, less than 0.1 mm indiameter typically.

244 Applied Optics

CladdingCoreprotective coveringsReinforcing and

Figure8.12Construction of an optical fiber.

There are mainly two types of materials used for the construction of optical fibers.They are glass and plastic. The glass fibers are made of silicates like sodium borosili-cate, sodium calcium silicate and lead silicate. Pure silica (S iO2) has a refractive indexof 1.45 at a wavelength of 1µm. Additives likeB2O3 can be used to lower the refractiveindex while additives likeGeO2 can be used to raise it. Hence, a typical fiber may haveS iO2 : GeO2 as the core and pureS iO2 as cladding. These glass fibers may be manu-factured with a wide range of refractive indices and are known for their characteristicsof low scattering. They are good for communication applications. Fibers can also bemade from plastic, but these fibers are associated with high attenuation due to scatter-ing. Alternatively, plastic coated silica fibers may be madewith core of pure silica andcladded with a suitable polymer. These optical fibers are economical, but are suitablefor medium distances with moderate bandwidth.

8.6.2 Propagation of light through an optical fiber

Propagation of light through a fiber is by total internal reflection. For a light ray trav-eling from one medium of refractive indexn1 to a second medium of refractive indexn2, we have, by Snell’s law, (Fig. 8.13)

n1 sini = n2 sinr (8.8)

wherei andr are the angle of incidence and the angle of refraction respectively. Whenthe ray of light is traveling from a medium of higher refractive index to a medium oflower refractive index, i.e.,n1 > n2, we can define a critical angle of incidence at whichthe refracted ray just grazes the interface, i.e.,r = 90. The critical angle of incidenceic is given by

sinic = n2/n1 for n1 > n2 (8.9)

The ray will be totally internally reflected into the denser medium when the angle ofincidence is greater than the critical angle. Thus, if we have a fiber consisting of a core

Applied Optics 245

n 2

n1

n 2

n1

n 2

n1

γ

t c

(a) (b) (c)

i

Figure 8.13 Propagation of light from a medium of higher refractive index n1, to amedium of lower refractive index n2. When the angle of incidence is less than thecritical value, it is refracted (a), grazes the interface when incident at the criticalangle (b) and totally internally reflected when the angle of incidence is more than thecritical value (c).

of higher refractive index surrounded by a cladding, the light ray may be confined totravel inside the fiber.

Consider a ray of light incident on the entrance aperture of an optical fiber at anangleθ as shown in Fig. 8.14.

θ

Cladding ( )

Core ( )1n

n2

i

AIR(n = 1)

r

Figure8.14Cone of acceptance for the optical fiber.

This ray will enter the core at an angle of refractionr and propagate through thecore by total internal reflection. Leti be the angle of incidence (and reflection) at thecore-cladding interface. Then, we have by Snell’s law,

n0 sinθ = n1 sinr

= n1 sin(90− i)

246 Applied Optics

= n1 cosi

= n1(1− sin2 i)1/2 (8.10)

Whenθ corresponds to the maximum angleθm of entrance at which the ray is capableof propagating inside the fiber, the angle of incidence at thecore-cladding interface willbe equal to the critical angleic. The ray will be propagated by total internal reflectionfor all θ < θm. Thus,θm defines thesemi-angle of the cone of acceptance. Substitutingfor sinic from equation (8.9), we get

n0 sinθm = n1(1− n22/n

21)

1/2

= (n21 − n2

2)1/2

If the external medium is air, then,n0 = 1.

sinθm = (n21 − n2

2)1/2 (8.11)

The quantity sinθm is callednumerical aperture of the optical fiber. The numericalaperture may also be expressed in terms ofrelative refractive index differencewhich is defined as

= (n1 − n2)(n1 + n2)/2

=(n1 − n2)(n1 + n2)

(n1 + n2)2/2

(n2

1 − n22)

2n21

(8.12)

Substituting the value of (n21 − n2

2)1/2 from this equation in equation (8.11), we get,

sinθm = n1(2)1/2 (8.13)

The numerical aperture is an important parameter and is a measure of the light collect-ing ability of the optical fiber. It is interesting to note that the parameter is independentof the dimensions of the core. Typical values of the semi-angle of the cone of accep-tance and the numerical aperture of optical fibers are 10 and 0.2 respectively.

We may define a quantity callednormalized frequencyV given by

V = ka(n21 − n2

2)1/2 (8.14)

wherek is the propagation vector for the incident radiation,a is the core radius andn1

andn2 are the refractive indices of the core and cladding respectively. This can also be

Applied Optics 247

expressed in terms of numerical aperture and the relative refractive index differenceas

V = (2π/λ) · a · (NA) (8.15)

= (2π/λ) · a · n1 · (2)1/2 (8.16)

The normalized frequency is a dimensionless quantity and isalso called theV numberof the fiber. It gives the combined information about the three design variables forthe fiber namely, operating wavelength, core radius and the relative refractive indexdifference.

The number of modes that can be propagated through a multi mode step index fiberis given by

M = 1/2(2πa/λ)2 · (n21 − n2

2)

= 1/2(2πa/λ)2 · (NA)2 = V2/2. (8.17)

8.6.3 Modes of propagation in a fiber

There are various modes by which light waves can be made to propagate in an opticalfiber. The mode of propagation depends on the construction ofthe optical fiber. Wemay broadly classify the optical fibers as

(i) Single mode step index fiber,

(ii) Multi mode step index fiber, and

(iii) Multi mode graded index fiber.

In a step index fiber, the refractive index of the core is higher than that of the claddingand is constant (Fig. 8.15).

Hence, the refractive index changes abruptly at the interface between the core andthe cladding resulting in total internal reflection. In a single mode fiber, the core is sosmall that there is a single path for the light to propagate inthe fiber (Fig. 8.16).

248 Applied Optics

(a) (b) (c)

n n n

Figure 8.15Variation of refractive index of core and cladding in (a) single mode stepindex fiber, (b) multimode step index fiber and (c) multimode graded index fiber.

(a)

(b)

(c)

Figure8.16Propagation of light in (a) single mode step index fiber, (b) multimode stepindex fiber and (c) multimode graded index fiber.

Applied Optics 249

In a multimode fiber, there are many such paths possible for the propagation. Thepropagation in step index fibers is also referred to as due to reflective confinement.

In a graded index fiber, the refractive index of the core is maximum at the centreof the core and decreases gradually and becomes equal to thatof the cladding at theinterface between core and cladding. The propagation in such fibers is by refractiveconfinement.

8.6.4 Signal distortion in optical fibers

Propagation of a signal through the optical fiber involves total internal reflection oflight rays many times. Further, the rays are reflected at various angles. The raysreflected at higher angles travel greater distances than therays reflected at low angles.As a result, all the rays do not arrive at the end of the fiber simultaneously and the lightpulse broadens as it travels through the fiber. Since the output pulse does not matchwith the input pulse, the signal is said to be distorted.

When white light source is used instead of a monochromatic radiation, another kindof distortion occurs. Since radiation of different wavelengths has different velocities,they do not arrive at the output simultaneously. This distortion is calledchromaticdispersion.

The signal distortion is quite considerable in multimode step index fibers. In gradedindex fibers, the light travels with different velocities in different parts of the core asthe refractive index varies radially along the core. The rays travel faster near the outeredge of the core(since the refractive index is smaller) thannear the centre. Hence, allthe rays arrive at the output almost at the same time and the signal distortion is reduced.In a single mode step index fiber also the distortion is less than that in multi mode stepindex fibers.

8.6.5 Signal attenuation in optical fibers

Attenuation is the loss of optical power as light travels down a fiber. It is measuresin decibels (dB/km). Over a set distance, a fiber with a lower attenuation willallowmore power to reach its receiver than a fiber with higher attenuation. IfPin is the inputpower andPout is the output power after passing through a fiber of lengthL, the meanattenuation constantα of the fiber, in units ofdB/km, is given by

α = (10/L)log10(Pin/Pout) (8.18)

Attenuation can be caused by several factors, but is generally placed in one of twocategories: intrinsic or extrinsic.

250 Applied Optics

Intrinsic attenuation occurs due to inherent characteristics of the fiber. It is causedby impurities incorporated into the glass during the manufacturing process. When alight signal hits an impurity atom in the fiber, one of two things will occur: it willget scattered or it will be absorbed. Rayleigh scattering accounts for the majority(about 96%) of attenuation in optical fiber. If the scatteredlight is incident on the core-cladding interface at an angle greater than the critical angle for total internal reflection,the light will be totally internally reflected and no attenuation occurs. If the angle ofincidence is less than the critical angle, the light will be diverted out of the core andattenuation results. Intrinsic attenuation can also occurdue to absorption of signal inthe fiber. The radiation may be absorbed by the impurity atomsand re-emitted byspontaneous emission or converted into some other form of energy.

Extrinsic attenuation can be caused by two mechanisms, namely, macro-bendingand micro-bending of the fiber. All optical fibers have a minimum bend radius spec-ification that should not be exceeded. This is a restriction on how much bend a fibercan withstand before experiencing problems in optical performance or mechanical re-liability. A bend in the fiber may result in the modification ofthe angle of incidenceon the core-cladding interface and hence may lead to signal loss. A bend may alsoinduce strain in the fiber which affects the refractive index locally. Microbending isa small scale distortion of the core-cladding interface in alocalized region. This maybe related to pressure, temperature, tensile stress or crushing force to which the fibermight have subjected to.

8.7 Applications of optical fibers

Optical fibers find extensive applications in communicationof optical signals. Trans-mission using optical frequencies has the advantage of higher speed of transmissionand hence greater volume of information carried. They find application in all experi-mental arrangements where light is transmitted from one point to another without loss.A few important applications of fiber optics are mentioned here.

8.7.1 Fiber optic communication

A conventional communication system consists of a transmitter which is used to trans-mit a carrier wave, usually at radio or microwave frequency,modulated by the signal.The medium of transmission may be wires or free space. The signal is recovered bydemodulation at the receiving end.

In a conventional optical fiber communication system, the input signals, audio,video or other digital data, are used to modulate light from asource like a light emitting

Applied Optics 251

diode or a semiconductor laser and is transmitted through optical fiber(Fig. 8.17). Atthe receiving end, the signal is demodulated to reproduce the input signal.

SOURCE

INFORMATION

OF

TRANSDUCER

CONVERTER

A / D MODULATOR

(LASER DIODE)(MICROPHONE)

DEMODULATOR

(PHOTO DIODE)

D / A

CONVERTOR (RECEIVER)

TRANSDUCER

OF

DESTINATION

INFORMATION

OPTICAL FIBER CABLE

Figure8.17Optical communication system.

Communication using optical fibers has many advantages.

(i) The optical carrier frequency used for communication isin the range from 1013

to 1015 Hz which is many orders of magnitude higher than the frequency of thecarrier waves used in conventional communication. This means that informationcan be transmitted at a higher rate than in the case of radio ormicrowave com-munication. In other words, a greater volume of informationcan be carried overthe fiber optic system.

(ii) Optical fibers, because of their flexibility and lightweight can be handled moreeasily.

(iii) Optical fibers are usually fabricated from electrically insulating materials andhence are safe.

(iv) In conventional communication systems, electrical interference leads to signalfrom a parallel line being picked up. Such cross talk is absent in optical fiberseven when they are bunched together in large numbers.

(v) Optical fibers being insulators are immune to interference as there is no inductionof electromagnetic noise.

(vi) Attenuation of signal or transmission loss is very low and optical fiber cableswith losses as low as 0.01dBkm−1 have been fabricated.

252 Applied Optics

(vii) The life span of optical fibers is considerably high at 25-30 years as compared toaround 15 years of life of copper cables.

(viii) Optical fiber links are more economical compared to copper cables because ofthe longer life span, higher resistance to temperature and corrosion and ease ofmaintenance of optical fibers.

8.7.2 Applications in medicine and industry

Optical fibers are also useful for medical applications for visualization of internal por-tions of the human body. They can also be used for the examination of visually inacces-sible regions for engineering applications. A typical example of a flexible fiberscope(endoscope) is shown in Fig. 8.18.

PROBE

OBSERVATION

SOURCE OF

ILLUMINATION

Figure8.18Fiber optic endoscope.

Use of laser in combination with optical fibers is being exploited not only for theobservation of internal portions but also in the treatment of malignant tissues. A sim-ilar equipment will also be useful to examine parts of machinery which are otherwiseinaccessible to observation.

Optical fibers also find application in the fabrication of sensors which are devicesused to measure and monitor physical quantities such as displacement, pressure, tem-perature, flow rate etc.

Numerical Examples

8.1 Calculate the numerical aperture and hence the acceptance angle for an opticalfiber whose core and clading has refractive index of 1.45 and 1.40 respectively.

Solution: If n1 andn2 are the refractive indices of core and clading respec-tively, then, numerical aperture is given by,

sinθm = (n21 − n2

2)1/2

= (1.452 − 1.402)1/2

Applied Optics 253

= 0.38 (Ans.).

Acceptance angleθm = sin−1(0.38)

= 22.3 (Ans.).

8.2 A He-Ne laser emits light at a wavelength of 632.8nmand has an output powerof 5mW. How many photons are emitted each second by this laser?

Solution:

Wavelength of emitted lightλ = 632.8× 10−9m.

Energy of each photonhc/λ =6.62× 10−34× 3× 108

632.8× 10−9

= 3.138× 10−19J

Energy emitted by the laser= 5mW= 5× 10−3Js−1

Number of photons emitted=5× 10−3

3.138× 10−19

= 1.59× 1016photons/sec (Ans.).8.3 Find the ratio of population of the two energy states of the Ruby laser the tran-

sition between which is responsible for the emission of photons of wavelength694.3nm. Assume the ambient temperature as 27C.

Solution:

Energy of the emitted radiation= (E2 − E1) = hc/λ

=6.63× 10−34 × 3× 108

694.3× 10−9

= 2.86× 10−19J

The ratio of population (N2/N1) between two energy statesE2 andE1 is givenby the Boltzman relation,

(N2/N1) = exp− [(E2 − E1)/kT]

= exp−2.86× 10−19

1.38× 10−23 × 300= exp− 69.08

= 9.98× 10−31 (Ans.).

254 Applied Optics

Note: If this is the ratio of population, then there is no population inversionand there will be no laser emission. It should be noted that inruby laser, theenergy stateE2 is a meta stable state and hence Boltzman relation cannot beused to compute the population in that state.

Exercise

8.1 What are the advantages of optical communication over the other conventionaltypes of communication? (March 1999).

8.2 Explain the terms spontaneous, stimulated emissions and population inversion.Briefly explain the construction and working of a ruby laser with energy leveldiagram. (March 1999).

8.3 With a neat diagram, explain numerical aperture, and ray propagation in an op-tical fiber. Describe the types of optical fibers and modes of transmission.

(March 1999).

8.4 Describe the principle on which optical fiber works. Mentiontheir applications.

(August 1999).

8.5 What is lasing action? Describe the working of He-Ne laser with the help ofenergy level diagram. Mention industrial applications of lasers. (August 1999).

8.6 Distinguish between luminescence, fluorescence and phosphorescence. How arethese phenomena explained on the basis of energy band picture? (March 2000).

8.7 Explain the terms spontaneous emission, stimulated emission and populationinversion. With energy level diagram explain the working mechanism of a He-Ne gas laser. (August 2000).

8.8 Explain types of optical fibers. Mention advantages and limitations of opticalcommunication system. (August 2000).

8.9 Explain how transmission of light takes place in optical fibers. Discuss differenttypes of optical fibers. (March 2001).

8.10 Explain how lasing action takes place in He-Ne laser. (August 2001).

8.11 Explain the principle of optical fibers. Derive an expression for the numericalaperture. Calculate the numerical aperture of a given optical fiber if the refractiveindices of core and cladding are 1.623 and 1.522 respectively. (August 2001).

Applied Optics 255

8.12 Explain the conditions required for laser action. Describethe construction andworking of a He-Ne laser with necessary energy level diagram. (March 2002).

8.13 Explain the terms modes of propagation, cone of acceptance and numerical aper-ture. (March 2002).

8.14 Calculate the numerical aperture and angle of acceptance for an optical fiberhaving refractive indices of 1.565 and 1.498 for core and cladding respectively.

(March 2002).

8.15 Discuss the point to point optical fiber communication system. Mention theadvantages of optical fiber communication over the conventional communicationsystem. (Feb 2003).

8.16 Obtain an expression for energy density of photons in terms of Einstein?s coef-ficients. (Aug 2003).

8.17 Explain the construction and working of He-Ne gas laser. (Aug 2003).

8.18 Write a note on holography. (Aug 2003).

8.19 An optical glass fiber of refractive index 1.450 is to be clad with another glass toensure total internal reflection that will contain light travelling within 5o of thefiber axis. What maximum index of refraction is allowed for the cladding?

(Feb 2004).

8.20 Discuss the applications of laser. (Aug 2004).

8.21 Find the ratio of population of two energy levels out of whichone correspondsto metastable state if the wavelength of light emitted at 330K is 632.8 nm.

(Aug 2004).

[Note: It is not possible to solve this example if we assume a level tobe metasta-ble since we cannot use Boltzman relation].

Chapter 9

Modern Materials And Methods

Materials play a very important role in every sphere of life.The nomenclature for dif-ferent stages of development of human civilization as stoneage, bronze age, iron age,etc., indicate the importance of materials. The ever increasing demand for materialswith specific characteristics for diverse applications andthe constant search for betterand more efficient alternatives has made the study of materials important. Apart fromthe conventional materials that have been in use for over ages, new materials are beingdesigned and synthesized with attractive properties. Thischapter gives a glimpse ofthe current trends in modern engineering materials and alsoa few modern techniquesof analysis.

9.1 Ceramics

Ceramics are compounds between metallic and non-metallic elements. The term ce-ramics comes from the greek word “Keramikos”, which means burnt stuff. Thesematerials are usually bad conductors of heat and electricity. They are resistant to hightemperature and corrosive environments, but are usually brittle. Chemically, they areusually oxides, nitrides or carbides.

Depending on the application for which they are used, ceramics are classified asglasses, clay products, refractories and cements. The properties and applications of thevarious categories of ceramics are discussed below.

9.1.1 Glasses

Glasses are non-crystalline silicates. Oxides likeNa2O, CaO, Al2O3, B2O3, etc., areadded to modify their properties to suit various applications.They are optically trans-parent and easy to fabricate. Glasses are produced by heating the raw materials to atemperature above the melting point of all the constituents. In order to ensure transper-ancy, the melt should be mixed homogeneously and made free from bubbles. Uponcooling, the melt becomes more and more viscous, passes through a state of super-cooled liquid and then solidifies into glass.

The process of formation involves inclusion of internal stress, called thermal stresses.These defects are eliminated by heat treatments like annealing and tempering.Anneal-

256

Modern Materials And Methods 257

ing is a process in which the glass is heated to a temperature called annealing point.At this temperature, atomic diffusion is rapid and any residual stress may be removed.Then, it is slowly cooled to room temperature. The temperature at which transitionoccurs from supercooled melt into glass is called the glass transition temperature. Intempering, the glass is heated to a temperature above glass transitiontemperature butbelow the softening point. Then it is cooled to room temperature in a jet of air or inan oil bath. Tempered glasses are used for applications where high strength is needed.This includes doors, automobile windshields, eye lenses, etc.

9.1.2 Clay products

Clay is one of the most widely used raw material. They are usedin building bricks,tiles, porcelain tableware and sanitary ware. They are alumino silicates containingchemically bound water. For applications, they are mixed with minerals like flint,quartz and a flux like feldspar. For example, porcelain used for tableware may have acomposition of 50% clay, 25% quartz and 25% feldspar.

The formation of clay products includes grinding the raw materials into a fine pow-der and mixing with water to a desirable consistency. They are drawn into requiredshape by extrusion technique or moulded using a porous mouldwhen water evaporatesto leave solid material. The product so obtained may containsome water and needsto be dried. The process of drying is assisted by controlled air flow at suitable tem-perature. It may also be subjected to firing by heating it at a temperature in the range900−14000C to enhance its mechanical strength. During the process of heat treatment,formation of liquid glass may lead to filling up of pores with an increase in the densityof the material. This process is known as vitrification.

9.1.3 Refractories and abrasives

Refractories are materials capable of withstanding high temperature without meltingor decomposing. They also remain inert to the atmosphere. They are useful in the man-ufacture of bricks and as lining material for furnaces. Chemically, they are mixturesof oxides likeS iO2, Al2O3, MgO, etc. For commercial applications, the raw materialwith the required composition is taken in the form of a powder. Upon firing, the parti-cles undergo a bonding phase thereby increasing the strength. Porosity is an importantparameter and must be controlled to obtain suitable material properties.

Abrasive ceramics are materials which are known for their hardness or wear resis-tance. They possess a high degree of toughness. Diamond is the best known abrasivewith a high value for its hardness, but is relatively expensive. Other common ceramic

258 Modern Materials And Methods

abrasives include silicon carbide, tungsten carbide, aluminum oxide, silica, etc. Theyare used in several forms depending on the application as a powder, as a coating bondedto grinding wheels, etc. Grinding, lapping and polishing wheels make use of abrasivepowders in a liquid medium.

9.1.4 Cements

There are some inorganic ceramic materials grouped as cements. They have a char-acteristic property of forming a paste when mixed with water, which subsequentlyhardens. Examples are Portland cement, plaster of paris, lime, etc. They are useful insolid and rigid structures as construction material.

9.1.5 Cermets

A cermet is a composite material composed of ceramic (cer) and metallic (met) mate-rials. A cermet is ideally designed to have the optimal properties of both a ceramic andthose of a metal. Ceramics possess basic physical properties such as a high meltingpoint, chemical stability, and especially oxidation resistance. The basic physical prop-erties of metals include ductility, high strength and high thermal conductivity. Themetal is used as a binder for an oxide, boride, carbide, or alumina. Generally, themetallic elements used are nickel, molybdenum, and cobalt.Depending on the physi-cal structure of the material, cermets can also be metal matrix composites, but cermetsare usually less than 20% metal by volume.

Cermets are used in the manufacture of resistors (especially potentiometers), ca-pacitors, and other electronic components which may experience high temperatures.They are used for the high-temperature sections of jet engines as well as high temper-ature turbine blades. Spark plug is also another example of acermet. It is an electricaldevice used in some internal combustion engines to ignite the fuel by means of anelectric spark. Ceramic parts have been used in conjunctionwith metal parts as fric-tion materials for brakes and clutches. Bioceramics are materials that can be used inthe human body. They can be in the form of thin layers on metallic implants, compos-ites with a polymer component, or even just porous networks.These materials workwell within the human body for several reasons. Cermets are also used in dentistry asa material for fillings and prostheses.


9.2 Composite Materials

A composite may be defined as a combination of two or more dis-similar materialsthe properties of which are superior to those of the individual components. A majorityof composites consist of two phases, the matrix which is the continuous medium inwhich a second phase, namely the reinforcement is uniformlydistributed. A familiarexample is fiber glass which consists of glass fibers in a resinmatrix. Resin is light,durable and easy to mould but does not have sufficient strength and stiffness. On theother hand, glass fibers possess high strength and stiffness. By combining these twomaterials, it is possible to produce a new material with all the useful properties of boththe components. Composites are also referred to as engineered materials since they aredesigned and produced as per requirement.

Based on the type of matrix material used in the composite, they are classified intothree categories:

a) Polymer matrix composites

b) Metal matrix composites

c) Ceramic matrix composites.

Based on the nature and shape of the reinforcement used, theyare further classified asfiber reinforced composites, particle reinforced composites and flake reinforced com-posites.

The resultant properties of the composites are dependent onthe properties of thematrix as well as reinforcement materials. The matrix performs the important func-tion of binding and holding the reinforcement. It also protects the reinforcement frommechanical damage. The matrix material bears the load and transfers it to the rein-forcement. The reinforcement phase should be distributed uniformly through out thematrix. It should not react chemically with the matrix material even at elevated tem-peratures of use. Generally, the reinforcement material should have higher modulusof elasticity than the matrix material. But, both matrix material and the reinforcementshould have similar coefficients of thermal expansion.

Fiber reinforced compositesconsist of very thin fibers of reinforcement material,usually 5 to 20 microns thick, distributed in a suitable matrix material. The fibers maybe continuous, extending through the size of the sample or discontinuous, with suitablelengths for specific requirements. Depending on the manner in which the fibers arepacked within the matrix, they are further classified as oriented fiber composites andrandom fiber composites. The randomness of distribution of the fiber may be in two


dimensions (planar randomness) or in three dimensions (volume randomness). Thesecomposites are often anisotropic depending on the nature ofalignment of the fiberphase in the material.

In Particle reinforced composites, the reinforcement is in the form of fine parti-cles, usually smaller than 50 microns in size and constitutes 10 to 50 percent of thetotal volume of the composite. The properties are strongly dependent on the size of theparticulate matter and the inter-particle spacing. The load is carried by both the matrixand reinforcement materials, and more effectively in the case of finer particles. Exam-ples are refractory oxides (silica or alumina) in metallic matrices, metals in polymersand ceramics.

Flake reinforced compositescontain reinforcement material in the form of thinflakes or platelets. For example, layer materials like mica,graphite etc. distributed insuitable matrix. They are very limited in applications.

Modern composite materials have made a significant impact onthe present techno-logical development of mankind. It is now possible to designmaterials with specificproperties as per the requirements of the application. Composites may be designed topossess high strength and load bearing capacity. They can belightweight materialswith considerable resistance to corrosion and wear. They are therefore widely used incivil construction, automobiles and aeronautical engineering applications. However,they are relatively costly because of the technology neededin their synthesis. Furtherwork is currently in progress to evolve procedures for theirsynthesis and also to collectdesign data on the materials.

9.3 Smart Materials

Smart materials have one or more properties that can be dramatically altered by varyingthe condition to which they are subjected. Most everyday materials have physical prop-erties which cannot be significantly altered. A variety of smart materials already existand are being studied extensively. These include piezoelectric materials, magneto-rheostatic materials, electro-rheostatic materials and shape memory alloys.

Each individual type of smart material has a different property which can be sig-nificantly altered, such as viscosity, volume, and conductivity. The property that canbe altered influences what types of applications the smart material can be used for.

Piezoelectric materials have a unique property which makesthem useful as a smartmaterial. When a piezoelectric material is deformed, it produces a small but measur-able electrical charge on the sample. Conversely, when an electrical signal is applied toa piezoelectric material it experiences a significant change in size. Hence, piezoelectricmaterials are most widely used as sensors.


Electro-rheostatic (ER) and magneto-rheostatic (MR) materials are fluids, whichcan experience a dramatic change in their viscosity. These fluids can change fromliquid state to a solid substance when subjected to an electric field or magnetic field.The effect is completely reversible. They regain their original viscosity when the fieldis removed. MR fluids experience a viscosity change when exposed to a magneticfield, while ER fluids experience similar changes in an electric field. The compositionof each type of smart fluid varies widely. The most common example of a magneto-rheostatic material is an oil consisting of suspended tiny iron particles.

MR fluids are being developed for use in car shocks, damping washing machine vi-bration, prosthetic limbs, exercise equipment, and surface polishing of machine parts.ER fluids have mainly been developed for use in clutches and valves, as well as enginemounts designed to reduce noise and vibration in vehicles.

9.4 Shape Memory Alloys

Shape memory alloys are metals, which exhibit two very unique properties, namely,elasticity and shape memory effect. Arne Olander first observed these unusual prop-erties in 1938, but no were any serious research work was madein the field of shapememory alloys until the 1960’s. The most effective and widely used shape memoryalloys include Nitinol (NiT i), CuZnAl, andCuAlNi.

The two unique properties mentioned above are made possiblethrough a solid statephase change which occurs in the shape memory alloy. A solid state phase change isone in which a molecular rearrangement occurs with the material remaining a solid. Inmost shape memory alloys, a temperature change of only about10C is sufficient toinitiate the phase change. The two phases, which occur in shape memory alloys, areMartensite, and Austenite.

Martensite is the low temperature phase of the material. It is the relatively soft andmay be easily deformed. The molecular structure in this phase is twinned. Austenitehas a cubic structure and occurs at higher temperatures. Theun-deformed Martensitephase is the same size and shape as the cubic Austenite phase on a macroscopic scale,so that no change in size or shape is visible in shape memory alloys until the Martensiteis deformed.

The shape memory effect is observed when a piece of shape memory alloy is cooledbelow a critical temperature (Fig.9.1). At this stage the alloy is composed of Martensitein its twinned form. This sample can be easily deformed by applying a suitable load.The deformed Martensite is transformed to the cubic Austenite phase by heating itto above a critical temperature. By cooling the alloy, the original twinned martensitestructure may be regained.


Austenite

Deformed martensite

Cooling

Heating

Loading

Load

Tem

par

atu

re

Twinned martensite

Figure9.1The two phases of shape memory alloys.

Pseudo-elasticity occurs in shape memory alloys, without achange in tempera-ture,when the alloy is completely composed of Austenite. Ifthe load on the shapememory alloy is increased, the Austenite gets transformed into Martensite simply dueto the loading. When the loading is decreased, the Martensite begins to transform backto Austenite and the sample springs back to its original shape.

Some of the main advantages of shape memory alloys include their attractive me-chanical strength and corrosion resistant properties. These alloys are expensive com-pared to other materials such as steel and aluminum. They have poor fatigue propertiesand a steel component may be hundred times more durable than aSMA element.

Shape memory alloys are being used for a variety of applications. They have beenused for military, medical and robotics applications. Nitinol couplers are used in F-14fighter planes to join hydraulic lines tightly. It is used in robotics actuators and ma-


nipulators to simulate human muscle motion. Some examples of applications in whichpseudo-elasticity is used are eyeglass frames, medical tools, cellular phone antennae,orthodontic arches etc.

9.5 Microelectromechanical Systems

Microelectromechanical systems(MEMS) are systems or devices constructed out ofmaterials that have strong inter-relation between their electrical and mechanical be-havior. For example, application of an electric signal leading to a mechanical changeor a mechanical change leading to the development of an electrical signal, as in thecase of piezoelectric materials, is referred to as an electro-mechanical effect. Whensuch devices are constructed on a micro-scale by processes similar to those employedfor fabrication of integrated circuits, the resulting devices are called MEMS. Depend-ing on the parameter that causes the change (Stimuli) and theparameter that changes(response), MEMS are classified as Sensors and Actuators.

9.5.1 Sensors

A sensor is one in which a mechanical stimuli results in an electrical response. Forexample, piezoelectric materials have the property of developing electrical charges ontheir surfaces when subjected to a mechanical pressure. Hence, a piezoelectric materialcan be used in the construction of a sensor.

Sensors can be classified based on the nature of the stimuli used in the device.Thermal sensors are devices that sense a change in the temperature and respond with achange in its electrical behavior. A thermo-resistive sensor has a device element whichhas its electrical resistance a strong function of temperature. A change in temperatureresults in a corresponding change in its electrical resistance. Hence, measurement oftemperature is possible through a measurement of the electrical resistance. Similarly,a thermocouple consisting of a junction of two dissimilar metals generates an e.m.fwhich depends on the temperature of the junction. Thus, a thermocouple can be usedas a thermal sensor.

Mechanical sensors respond to application of a force leading to changes in its shapeor size resulting in a corresponding change in its electrical behavior. Piezoelectricmaterials are examples of mechanical sensors (Fig.9.2).

In these materials, application of a stress, compressive ortensile, results in thedevelopment of electric charges on the surface leading to a potential difference. Theelectric field so generated is a measure of the stress to whichthe sample is subjected.Strain gauges are also examples of mechanical sensors (Fig.9.3).


(a) (b)

Figure 9.2 Working of a piezoelectric sensor crystal sample (a) in the absence of ap-plied force and (b) in presence of applied force.

−

G

P Q

RS

(b)(a)

+

Figure9.3Schematic of a strain gauge (a) and the experimental set up (b).

A strain gauge consists of a very thin, long metal wire, arranged in the form of ameandering pattern and bonded to the surface under study. Itcan also be in the form ofa thin film deposited on the surface. Any mechanical deformation of the surface resultsin a change in the dimensions of the metal wire leading to a corresponding change inits electrical resistance. The strain gauge may form one of the arms of a Wheatstone’snet to determine accurately the change in the resistance. The change in the resistancewill be proportional to the stress to which the surface is subjected.


Semiconductor materials, when used as Hall probes, will work as magnetic sensors.There are a few materials which show a large magneto-resistance, a phenomenon inwhich application of a magnetic field results in changes in the electrical resistance ofthe materials. Such materials can also be used as magnetic sensors.

Radiation sensors are very common and are used in various instruments as detec-tors of radiation. They are referred to as photo detectors and can be made sensitiveto radiations in the range from infrared to x-rays. Photo conducting materials andphotodiodes are used in these sensors.

9.5.2 Actuators

Actuators are devices which produce mechanical effects as a response to an appliedelectrical signal. Reverse piezoelectric effect, i.e., deformation of a solid when sub-jected to an electric field, is an example.

Bi-metallic strip is an example of a thermal actuator. It consists of two metalshaving widely different values for the coefficient of thermal expansion. The metals aretaken in the form of strips and are bonded to each other. When this bi-metallic strip isheated, because of the unequal expansion of the two metals, it bends into an arc. Thedevice can be used to make or break circuits.

In piezoelectric actuators, application of an electric field to the opposite faces ofa single crystalline piezoelectric crystal like quartz results in a change in the dimen-sions of the crystal. The crystal regains its original shapeand size when the appliedfield is removed. Further, the change in size may be positive or negative(increase ordecrease) depending on the direction of the applied electric field. This is an exampleof a mechanical actuator.

9.6 Nano Materials

Nano technology deals with structures of matter having dimensions of the order of anano meter. Even though the term nano technology is relatively new, structures anddevices of nanometer dimensions are known for a long time. Roman glass makersare known to have used nano sized metal particles to produce coloured glasses. Photo-graphic films contain silver halide which decomposes when exposed to light producingnano particles of silver. Richard Feynman, in a lecture at a meeting of the AmericanPhysical Society, predicted the potential applications ofnano materials.

Nano materials exhibit properties strikingly different from those of bulk materi-als. This is because every property of a material has a characteristic length associatedwith it. For example, the electrical resistance of a material is due to the scattering of


conduction electrons away from the direction of flow by collision with lattice atoms,impurities, etc. The effect of scattering on the resistance depends on how many suchcollisions takes place per unit distance of travel or the average distance travelled by theelectron before getting scattered. This characteristic length is called the mean free pathof conduction electron. When the dimension of the solid sample becomes comparableto this characteristic length, the fundamental property ofelectrical resistance changesand becomes a function of the dimension of the sample. This iscalled quantum sizeeffect and is responsible for the characteristic properties exhibited by the nano phaseof materials.

Nano materials are classified as quantum wells, quantum wires and quantum dots.In a three dimensional structure, if one dimension, say thickness, is of nano size, thenthe structure is called quantum well. If two dimensions are of nano size, then it is calleda quantum wire and if all the three dimensions are of nano size, then it is called a quan-tum dot. The word quantum is associated with the structures because the propertiesexhibited by them are described by quantum mechanics.

In a bulk metal, for example, the conduction electrons are free to move throughoutthe entire conducting medium. In other words, the electronsare completely delocal-ized. When one or more dimensions of the sample become small,say of the order ofa few atomic spacings, the delocalization of the electrons is restricted. The electronsexperience confinement as their movement is restricted in those dimensions which aresmall. The number of electrons with a particular energy becomes a function of the sizeof the sample. The density of states for a bulk metal, which isa measure of the numberof allowed energy states available for occupation at various energy values, is shown inFig.3.5. In the case of nano structures, the dependence of density of states on energywill also get modified depending on the degree of confinement.

The dependence of density of states on energy for different quantum structures isshown in the figure. Density of statesg(E) versus EnergyE in the case of (a) bulk ma-terial, (b) a quantum well, (c) a quantum wire, (d) a quantum dot and (e) an individualatom.

In the case of a bulk sample, since the number of electrons interacting is very large,the density of states as a function of energy is a continuum ofavailable states at variousenergies. As the size of the sample reduces in one or more dimensions, there will berestrictions on the number of available states at different energies. As we go from bulkto quantum well to quantum wire to quantum dot, the density ofstates approaches thatfor an individual atom. This results in drastic modifications in the electrical, thermaland other properties of nano structures.


g(E)

E E E EE

g(E) g(E) g(E) g(E)

(a) bulk material

(b) a quantum well

(c) a quantum wire

(d) a quantum dot

(e) an incividual atom

9.6.1 Synthesis of nano materials

There are two main approaches to nano technology. One is a top-down approach wherenano objects are constructed from larger samples without any need for control at atomiclevel. The other is the bottom-up approach where materials and devices are built fromatomic or molecular components. This approach of designingand manufacturing nanosystems with the necessary control at the molecular level iscalled molecular manufac-turing.

Gas condensationwas the first technique used to synthesize nanocrystalline metalsand alloys. In this technique, a material is vaporized usingthermal evaporation sourcesor electron beam heating in a reduced pressure. The formation of ultra fine particlesis achieved by collision of evaporated atoms with residual gas molecules. Variouskinds of vaporization methods like resistive heating, heating with high energy electronbeams, laser heating and induction heating may be used. The evaporated atoms loosekinetic energy by colliding with air molecules and condensein the form of small crys-tallites. Sputtering may also be used instead of thermal evaporation. Sputtering is anon-thermal process in which surface atoms are physically ejected from the surfaceby momentum transfer from an energetic bombarding ion beam in a glow discharge.Sputtering has been used in low pressure environment to produce a nanophase materi-als including silver, iron and silicon.

In vacuum deposition process, elements, alloys or compounds are vaporized anddeposited in a vacuum. The vaporization source is the one that vaporizes materials bythermal processes. The process is carried out at pressure much lower than that usedin gas condensation. The substrate may also be heated to a temperature ranging from


ambient to 500C. These deposits have particles or grains in the range of 1 to100 nmsize. The advantages associated with vacuum deposition process are high depositionrates and economy. However, the deposition of many compounds is difficult.

Chemical Vapour Deposition(CVD) is a well known process in which a solid isdeposited on a heated surface via a chemical reaction from the vapour or gas phase.The energy necessary for the chemical reaction to take placecan be provided by severalmethods. In thermal CVD the reaction is activated by a high temperature above 900

C. A typical apparatus comprises of gas supply system, deposition chamber and anexhaust system. The reaction may be activated by plasma at elevated temperatures. Thechemical reaction may also be induced by laser radiation which has sufficient photonenergy to break the chemical bond in the reactant molecules.Typical nanocrystallinematerials like SiC, Si3N, Al2O3, TiO2, SiO2, ZrO2 with average particle size of a fewnanometre have been synthesized by this technique.

Sol-gel processingis a wet chemical synthesis approach that can be used to gen-erate nanoparticles. It involves formation of colloidal suspension (sol) in continuousliquid phase (gel). A catalyst is used to start reaction and control pH. Sol-gel formationoccurs in four stages namely, hydrolysis, condensation, growth and agglomeration ofparticles. By controlling the growth parameters, it is possible to vary the structure andproperties of sol-gel derived inorganic networks. The significant potential of nanoma-terial synthesis and their applications is yet to be explored. Understanding more ofsynthesis would help in designing better materials.

Unlike many of the methods mentioned above,mechanical attrition produces itsnanostructures not by cluster assembly but by the structural decomposition of coarsergrained structures (top down approach). The ball milling and rod milling techniquesbelong to the mechanical alloying process which has received much attention as apowerful tool for the fabrication of several advanced materials. Mechanical alloying isa unique process, which can be carried out at room temperature.

A ball mill, a type of grinder, is a cylindrical device used ingrinding (or mixing)materials like ores, chemicals, ceramic raw materials and paints. Ball mills rotatearound a horizontal axis, partially filled with the materialto be ground plus the grindingmedium. Different materials are used as media, including ceramic balls,flint pebblesand stainless steel balls. An internal cascading effect reduces the material to a finepowder. Industrial ball mills can operate continuously, fed at one end and dischargedat the other end. High-quality ball mills are potentially expensive and can grind mixtureparticles to as small as 5 nm, enormously increasing surfacearea and reaction rates.The grinding works on principle of critical speed. The critical speed is the speed afterwhich the steel balls (which are responsible for the grinding of particles) start rotating


along the direction of the cylindrical device thus causing no further grinding.

9.6.2 Applications of nano materials

Although there has been much hype about the potential applications of nanotechnol-ogy, the current applications are limited to the ones developed in yesteryears. Theseinclude nanoparticles of titanium dioxide, silver, zinc oxide etc in sunscreen, cosmet-ics, food packaging, clothing, disinfectants, household appliances, paints and outdoorfurniture varnishes. One proposed application is the development of so-called smartmaterials. This term refers to any sort of material designedand engineered at thenanometer scale to perform a specific task, and encompasses awide variety of possiblecommercial applications. A nanosensor would be a smart material, involving a smallcomponent within a larger machine that would react to its environment and change insome fundamental, intentional way.

A promising field for the applications of nanotechnology is in the practice ofnanomedicine. This involves the creation of nanoscale devices for improved ther-apy and diagnostics. Such nanoscale devices are known as nanorobots or nanobots.These nanobots have the potential to serve as vehicles for delivery of medicine to re-pair metabolic or genetic defects. Similar to the conventional or macroscopic robots,nanobots would be programmed to perform specific functions and be remotely con-trolled, but possess a much smaller size, so that they can travel and perform desiredfunctions inside the human body.

Band gap engineering involves tailoring of band gaps with the intent to create un-usual electronic transport and optical effects, and novel devices. Most of the devicesbased on semiconductor nanostructures are band gap engineered quantum devices.Lasers fabricated using single or multiple quantum wells asthe active region have beenextensively studied over the last two decades. Quantum welllasers offer improved per-formance with lower threshold current and lower spectra width as compared to thatof regular double heterostructure lasers. Quantum dots have been used in lasers anddetectors. Nanostructures have been used for making photo-electrochemical cells forhigh efficiency conversion of light to electrical power due to its large surface area atwhich photo-electrochemical processes occur.

Applications of molecular nanotechnology to mechanical engineering will be aimedat realizing some mechanical systems on nano scale. The simplest mechanical systemwe can think of is a mechanical bearing. A conventional bearing consists of a shaft anda sleeve, with the relative motion of these components beingfacilitated by a suitablelubricant. The efficiency of such mechanical bearing lies in ensuring minimum wearand tear of the components along with minimum friction between them. A design of


(a) shaft (b) sleeve as seen along their axis

Figure9.4Components of a nano mechanical bearing showing (a) shaft and (b) sleeveas seen along their axis.Each circle represents a group of atoms.

nano mechanical bearing is shown in Fig.9.4.The components, namely the shaft and the sleeve, are polycyclic ring structures

consisting predominantly of atoms of carbon, hydrogen and nitrogen. The shaft hasa six fold symmetry about its axis and the sleeve has a fourteen fold symmetry. Thedimensions are designed such that the essential requirements for a satisfactory func-tioning of the bearing are met. Theoretical calculations show that the design yieldslow energy barrier to the rotation of the shaft within the sleeve. The advantages of thedesign include low static friction between the moving parts. The repulsive interactionsresist the movement of the shaft away from its axial alignment and displacement alongthe axis. These characteristics suggest the possibility ofextending nano technology topractical mechanical systems.

The existing and well understood conventional mechanical manufacturing tech-niques are useful in the top-down approach. However, molecular manufacturing in-volves chemical synthesis with precise placement of atoms and molecules which re-quires development of new tools and techniques. It is predicted that with the devel-opments in molecular manufacturing, various existing devices and their capabilitieswill improve by several orders of magnitude. A switch over from micro-devices tonano-devices is expected to reduce device sizes drastically and improve their speed. Inthe field of mechanical engineering, the conventional machinery may be replaced bymolecular machinery.

The future applications of nano materials will include next-generation computerchips, better insulation materials, phosphors for high-definition tv, low-cost flat-paneldisplays, tougher and harder cutting tools, high energy density batteries, high-power


magnets, high-sensitivity sensors, automobiles with greater fuel efficiency, aerospacecomponents with enhanced performance characteristics, longer-lasting satellites, longer-lasting medical implants, ductile, machinable ceramics, large electrochromic displaydevices, etc.

Due to the far-ranging claims that have been made about potential applicationsof nanotechnology, a number of concerns have been raised about what effects thesewill have on our society. Immediate issues include the effects of nanomaterials onhuman health and the environment. There is scientific evidence which demonstratesthe potential dangers posed by some toxic nanomaterials to humans or the environment.The extremely small size of nanomaterials means that they are much more readily takenup by the human body than larger sized particles. Nanomaterials are able to crossbiological membranes and access cells, tissues and organs that larger-sized particlesnormally cannot. Size is therefore a key factor in determining the potential toxicity of aparticle. However it is not the only important factor. Otherproperties of nanomaterialsthat influence toxicity include: chemical composition, shape, surface structure, surfacecharge, aggregation and solubility.

9.6.3 Scaling laws

The magnitudes of most of the physical parameters when expressed in nano scale dif-fer to a great extent from their macro scale values. The magnitudes for the nano scalesystems can be computed applying scaling laws to the values for macro systems. How-ever, the validity of scaling laws needs to be examined carefully. This is because macroscale systems are more or less defined by classical models anda transition to nano scaleusing scaling laws involves assumptions of the validity of these classical models. Nanoscale systems are atomic size structures where mean free path effects and quantum ef-fects are important. These effects may contribute differently to different physical prop-erties. It is convenient to study the nano systems separately under mechanical systems,electrical systems, thermal systems, etc., by applying classical continuum model.

Nano mechanical systems are useful for many applications. If we assume that themechanical strength of the material is a constant for a givenmaterial irrespective of itsdimensions, the total strength will be proportional to the area of the sample.

i.e., total strengthα L2 (9.1)

whereL represents a linear dimension. Expressed in nano scale, a stress of 1010Nm−2

will be equal to (1010/10−18)Nnm−2, i.e., 10nNnm−2. Similarly, shear stiffness whichincreases with sample area but decreases with increasing length, is proportional toL.


A sheer stiffness of 1012Nm−1 can be expressed as 103Nnm−1. Hence, deformation canbe written as

de f ormationα f orce/stiffnessα L. (9.2)

Assuming the density to be constant,

Massα volumeα L3. (9.3)

The mass of a cubic nanometre block of a material of density 5× 103kgm−3 will be5× 10−24kg. Hence,

Accelerationα (force/mass)α (L2/L3) α L−1 (9.4)

≈ 10−8Nnm2/5× 10−24kg

≈ 2× 1015ms−2. (9.5)

Thus, a cubic nano metre sample will experience large acceleration as compared tomacroscopic systems. Also, the effect of gravitational acceleration will be negligibleon nano mechanical systems.

These calculations are based on the assumption that scalinglaws are applicableeven when we consider nano systems. It is to be realized that atransition from macroto micro and nano scales are associated with major changes inthe conditions of con-struction and operation. In nano systems, for example, the surface becomes moreimportant than the volume. However, the influence of the surfaces on the propertiesare neglected.

In electromagnetic systems, classical scaling laws have tobe applied more carefullybecause quantum effects become dominant at small dimensions. If we assume theelectrical resistivity as a material constant,

Resistanceα (length/area)α L−1 (9.6)

Assuming an electrical resistivity of 1.5×10−7ohmm, the resistance of a cubic nanome-tre of copper would be 150 ohms. This result has to be examinedcarefully for its va-lidity since the calculation has ignored the effect of size on the electrical resistivity ofthe material.

Similarly, the models for thermal conductivity in solids also breakdown for nanosystems since the thermal phonons are associated with a meanfree path much largerthan the dimensions of the structure itself.

It may be concluded that the scaling laws based on classical models are not suitablefor describing the behaviour of nano systems.


9.6.4 Carbon nano clusters

Carbon has two stable allotropic forms, namely diamond and graphite. In diamond,the angle between the carbon-carbon bonds is 109 and in graphite, it is 120. It wasgenerally believed till recent times that no other carbon bond angles are possible. In1964, Phil Eaton synthesized a square carbon molecule,C8H8 called cubane. In 1983,L.Paquette synthesizedC20H20, a molecule having dodecahedron structure. Carbonnano clusters were discovered in 1985 by researchers at University of Sussex and RiceUniversity. Among these nano clusters containing various number of carbon atoms, thecluster containing 60 carbon atoms is the most prominent andis widely studied. Thiscluster is made up of 20 hexagonal and 12 pentagonal faces symmetrically arranged toform a molecular ball of carbon atoms (Fig.9.5). The clusteris named ‘Fullerene or‘Buckyball, after a noted architect, Richard Buckminster Fuller who popularized thegeodesic dome. Carbon nano clusters may be obtained by laserevaporation of carbon.The diameter ofC60 is about one nanometre. Chemically, they are quite stable and re-quire very high temperature to break them into atoms. However, they sublime at lowertemperature. This property is used in growing crystals and thin films of fullerenes.

Figure9.5Molecular structure of fullerene(C60)

C60 is highly electronegative and readily forms compounds withatoms that willdonate electrons to it. It is electrically a non conductor but become conducting whendoped with electropositive alkali metals.


C60 is a yellow powder which turns pink when dissolved in certainsolvents suchas toluene. When exposed to strong ultraviolet light, it polymerizes forming bonds be-tween adjacent balls. In the polymerized state, the C60 no longer dissolves in toluene.This property makes it useful as a photoresist in photolithographic processes. Otherphysical and chemical properties are being studied to evaluate the material for furtherapplications.

9.6.5 Carbon nano tubes

Carbon nano tubes are cylindrical fullerenes. They are tubes of carbon with a diameterof a few nanometer and several millimeters in length. They may be obtained as singlewalled tube or multi-walled tube with open or closed ends (Fig.9.6).

Figure9.6A single walled carbon nano tube

By virtue of their unique molecular structure, they are characterized by high tensilestrength, high ductility, high resistance to heat and chemically inert. They are metallicor semiconducting in nature depending on the diameter of thetube. Nano tubes withlarge diameters have conductivities higher than that of copper. Smaller tubes are semi-conducting with the band gap increasing with decrease in diameter. Typically, a tubeof diameter 2 nm has an energy gap of about 1eV. Carbon nano tubes show negativemagneto-resistance. There is a great interest in using these nano tubes for construct-ing electronic devices. There are several areas where carbon nano tubes are currentlybeing used. Some of the applications include flat-panel displays, scanning probe mi-croscopes, etc. They are also used to store lithium or hydrogen for fuel cells. They areuseful as catalysts in chemical reactions and are also used as chemical sensors.

9.7 Liquid Crystals

The intermediate phase between the solid and the liquid is called the liquid crystalphase. Liquid crystal materials generally have some uniquecharacteristics. They usu-ally have a rod-like molecular structure, with the molecules showing a tendency to


point along a common axis. This common axis is called the director. This is in con-trast to molecules in the liquid phase, which have no intrinsic order. In the solid state,molecules are highly ordered and have little translationalfreedom (Fig.9.7). The char-acteristic ordering in the liquid crystal state is between the traditional solid and liquidphases. Substances that aren’t as ordered as a solid, but do have some degree of align-ment are called liquid crystals.

(a) (b) (c)

Figure 9.7 Arrangement of molecules in the three phases, (a) solid, (b)liquid crystaland (c) liquid.

To quantify the order present in a liquid crystalline material, an order parameterSis defined as

S = (1/2) < 3 cos2 θ − 1 > (9.7)

whereθ is the angle between the director and the axis of each individual molecule.In an isotropic liquid, the order parameter will be zero and for a perfect solid, it willbe equal to 1. Typical value for the order parameter of a liquid crystal depends ontemperature and may be anything between 0.3 and 0.9.

9.7.1 Classification of liquid crystals

There are many types of liquid crystal states, depending upon the nature and amountof order in the material. The prominent ones are :

(1) Nematic phase

(2) Smectic phase

(3) Cholesteric phase.


The nematic liquid crystal phase is characterized by molecules that have orien-tational order but no positional order (Fig. 9.8). The molecules can flow and theirposition is randomly distributed in the liquid. But, they have a tendency to point to-wards a particular direction leading to a finite orientational order. As a consequence,nematic liquid crystals are strongly anisotropic.

Figure 9.8 Orientational order in nematic liquid crystal. The arrow represents thedirector.

The smectic phase is another distinct phase of liquid crystal substances. Moleculesin this phase show some degree of translational order in addition to the orientationalorder (Fig.9.9). In the smectic state, the molecules maintain the general orientationalorder of nematics, but also tend to align themselves in layers or planes. The motionis restricted to within these planes but the layers themselves can move past each other.The increased order means that the smectic state is more “solid-like” than the nematic.There are many types of smectic phases possible with different kinds of ordering.In thesmectic-A mesophase, the director is perpendicular to the smectic plane, and there isno positional order within the layer. In smectic-B, the smectic plane is perpendicular tothe director, but the molecules are arranged in a network of hexagons within the layer.In the smectic-C phase, molecules are arranged similar to that in the smectic-A phase,but the director is not perpendicular to the smectic plane.

The cholesteric liquid crystal phase is composed of molecules containing a chiralcenter which produces intermolecular forces that favor alignment between moleculesat a slight angle to one another. This leads to the formation of a structure which canbe visualized as a stack of very thin 2-D nematic-like layerswith the director in eachlayer twisted with respect to those above and below (Fig.9.10).


(a)

(b)

(c)

Figure9.9Three major types of smectic liquid crystals. (a) Smectic A phase, (b) Smec-tic B phase as seen along the director axis and (c) Smectic C phase.

1

2P

Figure9.10Cholesteric liquid crystal showing the rotation of the director in a distanceequal to half the pitch.


An important characteristic of the cholesteric phase is thepitch. The pitch,p, isdefined as the distance it takes for the director to rotate onefull turn in the helix. Thisgives the cholesteric liquids the special property to selectively reflect light of wave-lengths equal to the pitch length, so that a particular colorwill be reflected when thepitch is equal to the corresponding wavelength of light in the visible spectrum. Thewavelength of the reflected light can be controlled by adjusting the chemical composi-tion of the cholesteric phase.

9.7.2 Applications of Liquid Crystals

Liquid crystal have made contribution in many areas of science and engineering, aswell as device technology. The most common application is liquid crystal displays(LCDs). They work on the effect of electric field on the optical properties of the liquidcrystal. A typical device is shown in Fig. 9.11.

6 5 4 3 2 1

Figure 9.11 Schematic of a liquid crystal display unit consisting of (1)polarizer,(2)electrode, (3) liquid crystal layer, (4) back electrode, (5) second polarizer crossedwith the first and (6) reflector.

It consists of a liquid crystal layer sandwiched between twopolarizers that are crossedwith each other. The liquid crystal selected is a twisted oneso that the light passingthrough the first polarizer is reoriented to pass through thesecond polarizer also. Whenan electric field is applied between the two transparent indium-tin oxide electrodes, allthe molecules in the liquid crystal align parallel to the electric field. Since the twopolarizers are crossed, the light does not get transmitted through the second polarizer.A reflector is placed after the second polarizer to reflect anylight incident on it. Thus,electric field can be used to switch a pixel on or off.

Liquid crystals have a many other uses. They can be used as thermal sensors sincemany of their properties are very sensitive to temperature.They are now being ex-plored for optical imaging and recording applications. They are used for nondestruc-tive mechanical testing of materials under stress. As new properties and types of liquid


crystals are being investigated, these materials are gaining increasing importance inindustrial and scientific applications.

9.8 Non Destructive Testing Of Materials

Non destructive testing(NDT)is the technique of determining the quality of a prod-uct without in any way affecting its properties. It is emerging as an important inter-disciplinary technique useful for meeting the requirements of reliability and safety.Based on the principle used for testing, the technique may beclassified as follows:

(1) Radiographic methods (2) Ultrasonic methods

(3) Magnetic methods (4) Electrical methods

(5) Optical methods (6) Thermal methods

The basic principle involved in non destructive testing is to allow energy in someform to pass through the test sample and measure the effects like changes in amplitude,intensity, velocity, frequency, phase, etc of the transmitted signal.

9.8.1 Radiographic methods

In radiographic methods of NDT, radiations like x-rays or gamma rays are used as theprobe. Recently, electron and neutron beams are also being used as probes for certainapplications. When a radiation of suitable energy and intensity is made to incidenton the specimen under test, it penetrates through the material. The intensity of thetransmitted radiation will be modified by the presence of defects. The defects may bein the form of deviations from the periodic arrangement of atoms in the lattice or maybe in the form of inclusion of foreign materials in the specimen. Such defects absorbor scatter the radiation differently from the other regions of the sample and henceresult in a modification of the transmitted intensity. Thesechanges in the intensityof transmitted radiation may be recorded and analyzed usingphotography or othermethods.

Radiation for a specific NDT application is selected on the basis of their suitabilityand efficiency. X-rays are often used for the analysis of defects in crystalline solids.Since the wavelength of x-rays can be of the order of magnitude of the inter-atomicdistances in crystalline solids, they produce significant diffraction contrast leading toeasy detection of crystal defects. X-rays are also used in the analysis of flaws, inclu-sions, pinholes, cracks etc generated in welding or castingprocesses. Gamma rays aremore energetic and can be used for thicker samples. In the case of materials of highatomic number, absorption poses a serious limitation to NDTtechniques using x-rays


or gamma rays. Alternatively, electron or neutron radiography may be suitable for suchapplications.

9.8.2 Ultrasonic methods

Ultrasonic waves are mechanical waves, like sound waves, with a frequency abovethe audible range. Similar to sound waves, they produce local changes in the densityof the medium in which they are travelling and propagate witha velocity determinedby the medium. Ultrasonic waves are generated by making use of piezoelectric effector magnetostriction. Piezoelectric oscillators are widely used since they give a widerange of frequencies from 20 kHz to 10 GHz, where as magnetostriction oscillatorscan be used normally upto 100 kHz. Piezoelectric transducers are made from materialsshowing piezoelectric effect, like quartz, tourmaline, lithium sulphate, etc. A typicalexperimental set up is shown in Fig.9.12.

HTQUARTZ

CRYSTAL

Figure9.12Set up for piezoelectric oscillator method.

for piezoelectric oscillator methodA properly cut quartz crystal is placed betweentwo metal plates forming a parallel plate capacitor with crystal as the dielectric medium.The plates are connected to the primary of a transformer which is coupled to an oscil-lator circuit. When the frequency of the circuit matches with the natural frequencyof the crystal, resonance will occur and the crystal is set into mechanical vibrations,producing ultrasonic waves.

The principle of magnetostriction can also be used to generate ultrasonic waves. Abar of ferromagnetic material like iron or nickel changes its length when subjected tostrong magnetic field. If the applied magnetic field is alternating, the rod will alter-nately expand and contract with twice the frequency of the applied field. This resultsin the generation of ultrasonic waves in the medium surrounding the rod.

The experimental set up (Fig.9.13) consists of a specimen rod placed inside asolenoid and a high frequency current is passed through the solenoid. Resonance oc-curs when the natural frequency of the rod matches with the applied frequency, result-ing in the generation of ultrasonic waves.


HTFerromagnetic

rod

Figure9.13Set up for Electrostriction method.

If f is the resonance frequency andL is the length of the sample, then the velocityv of the ultrasonic waves generated is given by

v = 2 f L

The velocity of ultrasonic waves in a solid may also be determined by Pulse echomethod. In this method, a short pulse of electrical signal ofabout 1 to 2µs durationis produced by a pulse generator (Fig.9.14). This pulse is applied to the piezoelectriccrystal which is attached to the solid under test. The electric pulse produces an ultra-sonic signal which travels into the sample. The pulse gets reflected from the other endof the solid sample. When the echo reaches the crystal, it generates an electrical signalin the piezoelectric crystal. The initial electrical pulse(A) applied to the crystal and theelectrical pulse generated by the echo (B) are recorded on a cathode ray oscilloscopescreen. By knowing the length of the sample and the time gap between the two pulses,the velocity of ultrasonic waves in the given sample can be calculated.

A

PULSE

GENERATOR

AMPLIFIER

HORIZONTAL

SWEEP

GENERATOR

SAMPLE

PIEZOELECTRIC CRYSTAL

B

Figure9.14A schematic diagram of a pulse echo detection system.


The velocity of ultrasonic waves in a solid medium depends onthe density of thematerial and also on the elastic constants. In the case of a thin rod like sample whosediameter is much smaller than the wavelength of ultrasonic waves, the velocity is givenby

v = (E/ρ)1/2 (9.8)

WhereE is the Young’s modulus andρ is the density of the solid sample.If the sample under consideration is not piezoelectric, then, the velocity of ultra-

sonic waves in that sample may be determined as follows:A quartz crystal in the form of a rod of cross sectional areaA and lengthLq is used

in an experimental set up shown in Fig. 9.15. The resonance frequencyfq correspond-ing to the maximum amplitude of the well-defined wave patternobserved on the CROis

Oscillator

circuitCRO Sample

SpecimanQuartz crystal

Metal

electrodes

Figure9.15Experimental set up to find the velocity of ultrasonic waves through a solidsample.

noted. The given sample is also taken in the rod form with the same area of crosssection A and lengthLs. The sample is attached to the quartz crystal using a glue andthe resonance frequencyfc of the composite (quartz+ sample) is determined as before.The natural frequencyfs of vibration of the sample is calculated as

fs = fc + (mq/ms)( fc − fq)

where fq, fs and fc are the resonance frequencies for the quartz crystal, sample understudy and the composite,mq andms are the mass of the quartz crystal and the sample.The velocity of ultrasonic waves in the sample can be calculated as

vs = 2 fsLs


and the Young’s modulus of the sample can be calculated as

E = v2sρ

Whereρ is the density of the simple.The pulse echo method can also be used to detect flaws in the given solid sample.

This is based on the fact that the ultrasonic waves are reflected from a crack or otherdefects which produce abrupt changes in the elastic properties of the material. Anecho will be produced when a defect or a flaw interrupts a beam of ultrasonic waves.In Fig. 9.16, peakA represents the transmitted pulse,B is the pulse reflected from theother end of the sample andC is the pulse reflected from the defect. The horizontaldistanceAC across the screen indicates the position of the defect. The height of thepeakC, which is determined

A

C

B

Figure9.16CRO display for a sample with a flaw.

by the intensity of the ultrasonic wave reflected from the defect, indicates to someextent the size of the defect. The advantages of the technique are that it is easy, reliableand fast. The method is particularly useful when the samplesare large and the x-raytechnique cannot be used because of the limited penetrationof x-rays in solids.

9.8.3 Magnetic methods

These methods are particularly useful for magnetic materials. The principle involved isto study the effect of defects on the magnetic behavior of the material. Whenthe givenspecimen is magnetized, the magnetic field will be distortedat the location of defects.The magnetic field distortion may be measured using a suitable method. For example,


the sample may be scanned with a search coil to detect changesin the induced voltageat the defect sites. Alternatively, magnetic particles such as iron oxide in the form offine powder or a colloidal solution are spread over the samplesurface. The magneticparticles arrange differently at the defects thereby revealing their presence.

9.8.4 Electrical methods

It is possible to gain insight into the quality of the sample by a study of differentelectrical properties of the material. However, for an analysis of a localized defect,some special techniques are used. For example, in eddy current testing, a test coilcarrying a high frequency alternating current is made use of. When the coil is broughtclose to a conductor, eddy currents are induced in the coil which in turn change theimpedance of the coil. The presence of a defect in the conductor modifies the wayin which the eddy currents are induced and hence result in corresponding changes inthe impedance of the test coil. The method is suitable for detection of surface cracks,variations in thickness and conductivity of coatings.

9.8.5 Optical methods

Optical microscopy is a widely used technique for surface examination and evalua-tion. Surface morphology and microstructures can be effectively studied by opticalmicroscopy. Modifications to the conventional microscopy includes interference mi-croscopy and phase contrast microscopy which are based on the optical phenomenaof interference and polarization respectively yield valuable information on the sam-ple surface. Holographic interferometry technique using laser helps in the study ofstrained surfaces and surfaces subjected to vibration. Optical examination of transpar-ent samples in transmission helps in the analysis of the internal features of the sample.

9.8.6 Thermal methods

The principle involved in thermal methods is to supply heat to a specimen and observethe resulting temperature distribution. The temperature at the location of defects willbe different from that at other regions. The temperature distribution may be measuredusing sensitive thermal detectors like thermocouples, bolometers or photo conductingmaterials. Thermography is a related technique in which thethermal radiation emittedby the test sample is analyzed. By virtue of the temperature at which the sample ismaintained, there will be emission of infra red radiation from the sample which canbe detected and measured using an infrared detector. At defect sites, the temperature


and hence the emission of infrared radiation will vary in intensity and wavelength. Animage may be constructed with the data to show the defects in the sample.

9.9 Quantum Computation

Quantum computation is the study of information processingusing quantum mechan-ical systems. Quantum mechanics was developed when the classical physics was notin a position to explain many phenomena. Quantum mechanics has been indispensablesince then and has been used in various fields with success. Any bulk sample may beconsidered to be made up of a large number of quantum systems and the behaviour ofthe bulk sample may be predicted by consolidating the behaviour of individual quan-tum systems. In an attempt to control the bulk behaviour, attempts are being made tocontrol individual quantum systems. Quantum computation is a result of the study ofquantum systems and its application to information processing.

9.9.1 Properties of quantum bits

In classical computation, the information is addressed andprocessed using ‘bits’. Bitis defined as a unit of information. On a similar concept, quantum information is builtusing ‘quantum bits’ or ‘qubits’. A classical bit can have two states- 0 and 1. Similarly,the qubit can be in any of the two states -|0 > and|1 >. This is referred to as ‘Diracnotation’. A classical bit can exist in any one of the two states where as the quantumbit can exist in states that are linear combination or superposition of the two states.This may be represented as

|Ψ>= α0|0> +α1|1> (9.10)

whereα0 andα1 are complex numbers. In other words, a qubit represents a vector intwo dimensional space. The states|0> and|1> are called basis states and are oppositeto each other in vector space.

Classical bit can be examined and its state may be determinedto be either 0 or 1.In the case of qubit, we can only find the probability that it exists in the state|0> or|1>. The probability that it exists in the state|0> is equal to|α0|2 and the probabilitythat it exists in the state|1> is equal to|α1|2. Since the total probability has to be 1, wehave

|α0|2 + |α1|2 = 1 (9.11)

This is called the normalization condition for the qubit. However, it is emphasized thatthe qubit exists in either|0> state or|1> state only, but with finite probabilities. For


example, if we represent a qubit by

|Ψ>=1√

2|0> +

1√

2|1> (9.12)

we only mean that a measurement gives a value|0> for the qubit in 50%[|1/√

2|2] ofthe trials and a value|1> in the remaining 50% of the trials.

We can find an example of a quantum mechanical system in an electron. Theelectron can exist in either ground state or an excited state. We can call these states|0> and|1> respectively. During the excitation process in presence ofa radiation field,the electron may be found in|0> state or|1> state. The qubit represented by a linearcombination of these states indicates the probability of the two states being occupied.

θ

Φ

X

Y

Z

Ο

ψ

1

Figure9.17Bloch sphere representation of a qubit.

Figure 9.17 shows a geometrical representation of a qubit. Equation (9.10) may berewritten as

|Ψ>= cos(θ/2) · |0> +eiφ sin(θ/2) · |1> (9.13)

whereθ andφ are real numbers representing the angular coordinates in the three di-mensional sphere of unit radius. This sphere is called the Bloch sphere and is used tovisualize the state of a single qubit.


9.9.2 Quantum gates

Classical computer circuits make use of logic gates. In an analogous way, quantumcomputation is carried out using quantum gates. Let us consider an example of asingle bit gate. A classical single bit logic gate is the NOT gate. The operation of thisgate is defined by the truth table in which the 0 and 1 states areinterchanged during anoperation; i.e., if the input to the NOT gate is 0, the output will be 1 and vise versa. Asingle bit quantum gate or a single qubit gate should not onlyinterchange the states|0 >and|1 > but also perform a similar operation on the qubit state whichis a superpositionof the two states. In other words, assuming the qubit NOT gateto be acting linearly, itshould take theα0|0 > +α1|1 > state to theα0|1 > +α1|0 > state. Quantum NOT gateis represented in matrix form as

X =

[

0 11 0

]

(9.14)

The quantum stateα0|0 > +α1|1 > is written in vector notation as

[

α0

α1

]

Here, the top entry corresponds to the amplitude of|0 > and the bottom entry tothe amplitude of|1 >. Then, the quantum NOT gate operation is represented by

X

[

α0

α1

]

=

[

α1

α0

]

(9.15)

The normalization condition for the quantum stateα0|0 > +α1|1 > requires that

|α0|2 + |α1|2 = 1 (9.16)

This must also be true for the quantum state after the action of the gate. It can be shownthat

X+X = I (9.17)

whereX+ is the adjoint ofX andI is the two by two identity matrix.While we have only one non-trivial single bit classical gate, there are many non-

trivial single qubit gates. Two important qubit gates are the Z-gate and the H-gate. TheZ-gate is represented as

Z =

[

1 00 −1

]

(9.18)


in which the operation results in the change in the sign of|1 > to− |1 >. The H-gate isrepresented as

H =1√

2

[

1 11 −1

]

(9.19)

in which the operation results in|0 > being changed into1√

2(|0 > +|1 >) and |1 >

being changed into1√

2(|0 > −|1 >).

Thus, the operation of three single qubit quantum gates are represented as follows:

α0|0 > +α1|1 > −X − α1|0 > +α0|1 > (9.20)

α0|0 > +α1|1 > −Z − α1|0 > −α0|1 > (9.21)

α0|0 > +α1|1 > −H − α0√2

(|0 > +|1 >) +α1√

2(|0 > |1 >) (9.22)

Since there are an infinite number of two by two unitary matrices, we can have aninfinite number of single qubit gates. But the properties of the complete set can beunderstood from the properties of a smaller set. In other words, quantum computationcan be generated on any number of qubits using a finite universal set of gates. Such auniversal set requires the use of quantum gates involving multiple qubits.

9.9.3 Multiple qubits

If we have two classical bits, then there would be four possible states, namely 00, 01,10 and 11. Similarly, a two qubit system has four states denoted by|00>, |01>, |10>and |11 >. Since the pair of qubits can also exist in super positional states, the statevector describing the qubit system may be represented as

|Ψ >= α00|00> +α01|01> +α10|10> +α11|11> (9.23)

whereα00, α01, α10 andα11 are the complex coefficients representing the amplitudes ofthe respective states. Similar to the case of a single qubit,the square of the amplituderepresents the probability of finding the system in that particular quantum state.

An important two qubit state is called ‘Bell state’ or ‘EPR pair’ named after thescientists Einstein, Podolsky and Rosen. It is representedby

|Ψ >= |00> +|11>√

2(9.24)


This state has the property that the measurement on the first qubit gives a result0 with probability (1/2) and 1 with probability (1/2). A measurement on the secondqubit also gives the same result. This indicates that the results of measurement arecorrelated. Such correlations have been a subject of intense study. It has been observedthat the measurement correlations are much stronger in the Bell state as compared towhat is observed in classical systems.

In general, if we consider a system of n qubits, the quantum state of the systemwill be specified by 2n amplitudes. This indicates the enormous potential of quantumcomputation.

Exercise

9.1 Write a note on nano-scale systems (Jan 2003)

9.2 What are composite materials and give a brief account of classification of com-posite materials. (Jan 2003)

9.3 Give a brief account of smart materials. (Jan 2003)

9.4 What is a bit and a quantum bit? (Aug 2003)

9.5 Describe different methods of fabrication of MEMS. (Aug 2003)

9.6 List the advantages and disadvantages of composite materials. (Aug 2003)

9.7 What are smart materials? Explain the functional properties of smart materials.

(Aug 2003, July 2006)

9.8 Explain the basic principles of quantum computation. (Feb 2004, July 2006)

9.9 Discuss the nanoscale systems. (Feb 2004)

9.10 Explain MEMS. Discuss its application. (Feb 2004)

9.11 Explain density of states for various quantum structures (Aug 2004)

9.12 Explain nano-tubes and its applications by giving their physical properties.

(Aug 2004)

9.13 Explain smart materials with two examples. (Aug 2004)

9.14 What are composite materials? Give their classifications. (Aug 2004)


9.15 Discuss nanoscale systems giving atleast one application in detail. (Feb 2005)

9.16 What are the advantages and disadvantages of composite materials? (Feb 2005)

9.17 Explain the term MEMS. Give a brief account of smart materials. (Feb 2005)

9.18 Explain the term MEMS. Discuss different materials used for MEMS.

(Aug 2005)

9.19 Explain the advantages and disadvantages of composite materials.

(Aug 2005, July 2006)

9.20 Discuss the different types of nano-scale systems. (Aug 2005)

9.21 What are composites? Discuss their merits in the context of modern applications.

(Jan 2006)

9.22 Write a note on nanotechnology and its importance. (Jan 2006)

9.23 What is a quantum bit? Explain. (Jan 2006)

Chapter 10

Special Theory of Relativity

10.1 Introduction

When such quantities as length, time interval, and mass are considered in elementaryphysics, no special point is made about how they are measured. Since a standard unitexists for each quantity, who makes a certain determinationwould not seem to matter –everybody ought to get the same result. For instance, there is no difficulty in finding thelength of a rocket when it is stationary and on earth. But whatif the rocket is in flightand we are on the ground? We have standard methods to determine the length of adistant object with knowledge of trigonometry. However, when we measure the lengthof a moving rocket from the ground, we find it to be shorter thanit is to somebody inthe rocket itself. In order to understand how this unexpected difference arises we mustanalyze the process of measurement when motion is involved.

10.1.1 Frames of Reference

When we say that something is moving, what we mean is that its position relative tosomething else is changing. A passenger moves relative to anairplane; the airplanemoves relative to the earth; the earth moves relative to the sun; the sun moves relativeto the galaxy of stars (the Milky Way) of which it is a member; and so on. In each casea frame of reference is part of the description of the motion.To say that something ismoving always implies a specific frame of reference. A frame of reference is usuallya Cartesian coordinate system and the position of any objectis defined with respect tothe frame. The choice of the frame of reference is determinedby our own convenience.

An inertial frame of reference is one in which Newton’s first law of motion holds.In such a frame, an object at rest remains at rest and an objectin motion continues tomove at constant velocity (constant speed and direction) ifno force acts on it. Anyframe of reference that moves at constant velocity relativeto an inertial frame is itselfan inertial frame.

All inertial frame are equally valid. Suppose we see something changing its posi-tion with respect to us at constant velocity. Is it moving or are we moving? Suppose weare in a closed laboratory in which Newton’s first law holds. Is the laboratory movingor it is at rest? These questions are meaningless because allconstant – velocity motion

291

292 Special Theory of Relativity

is relative. There is no universal frame of reference that can be used everywhere, nosuch thing as “absolute motion”.

The theory of relativity deals with the consequences of the lack of a universal frameof reference. Special relativity, which is what Einstein published in 1905, treats prob-lems that involve inertial frame of reference. General relativity, published by Einsteina decade later, treats problems that involve frames of reference accelerated with re-spect to one another. An observer in an isolated laboratory can detect accelerations,as anybody who has been in an elevator or on a merry-go-round knows. The specialtheory has had an enormous impact on much of physics, and we shall concentrate onit here.

10.1.2 Galilean transformation

The transformation from one inertial frame of reference to another is called a Galileantransformation. Let an event occur in an inertial frame of referenceS at the locationP(x, y, z) at any instant of timet. Consider another frame of referenceS′ which movesalong positivex direction of reference frameS with a velocityv. Let the originsOandO′ of the two frames of reference coincide att = 0 and the pointP be at rest withrespect to the frame of referenceS. With respect to the frame of referenceS′, it moveswith a velocityv and its coordinates change with time. At any instant of timet = t′,we have

x′ = x− vt

y′ = y

z′ = z

And t′ = t.

This is Galilean transformation and it provides space-timerelation of an event in differ-ent inertial frames. In this transformation, we have assumed that the time of an eventfor an observer inS is the same as the time for the same event inS′. This assumptionholds good for all classical cases where the velocityv is much smaller than the velocityof light c. For example, consider the case of a person in a train moving with a speedv. If he throws a ball with a speedu in the direction of motion of the train, the speedof the ball to a stationary observer outside the train will be(v + u). What happens ifwe replace the ball with a flash of light? Will the stationary observer find the speed oflight to be (c+ v) ? Experiments were carried out by Michelson and Morley to verifythe addition rule for the velocities of light and of the earth. The speed of earth in itsorbit around the sun is about 30 km/sec which is about (1/10,000) of the velocity of

Special Theory of Relativity 293

light. With precision experimental set up, they tried to measure the differences in thevelocity of light along and perpendicular to the direction of earth’s motion.

10.1.3 Michelson-Morley experiment

Albert A. Michelson (1852-1931)was born in Germany but came to the United statesat a very tender age with his parents, who settled in Nevada. He attended the U.S.Naval Academy at Annapolis where he became a science instructor. To improve hisknowledge of optics, in which he wanted specialize, Michelson went to Europe andstudied in Berlin and Paris. Then he left the Navy to work firstat the Case school ofApplied Science in Ohio, then at Clark University in Massachusetts, and finally at theUniversity of Chicago, where he headed the physics department from 1892 to 1929.Michelson’s specialty was high-precision measurement, and for many decades his suc-cessive figures for the speed of light were the best available. He redefined the meterin terms of wavelengths of a particular spectral line and devised an interferometer thatcould determine the diameter of a star (stars appear as points of light in even the mostpowerful telescopes).

Michelson’s most significant achievement, carried out in 1887 in collaboration withEdward Morley, was an experiment to measure the motion of theearth through the“ether”, a hypothetical medium pervading the universe in which light waves were sup-posed to occur. The notion of the ether was a hangover from thedays before lightwaves were recognized as electromagnetic, but nobody at thetime seemed willing todiscard the idea that light propagates relative to some sortof universal frame of refer-ence.

T

S

P CP

Figure10.1Set up for Michelson-Morley experiment


A schematic diagram of the Michelson-Morley experiment is shown in Fig. 10.1.A beam of light from a source S is split into two parts by a semi-silvered glass plateP.A part of the beam travels to the mirrorM1, gets reflected to the plateP on the silveredside and is again reflected into a telescope. The other part ofthe beam travels to mirrorM2, gets reflected to the plateP and transmitted into the telescope. A compensatingglass plateCP is used to compensate for the difference in the optical paths travelledby the two beams before interfering. If the transit time for the two parts of the beamis same, they arrive at the telescope to produce constructive interference. If one ofthe beams is travelling along the direction of earth’s motion, there should be a changein the transit time for that path and this should lead to a change in the interferencecondition.

Although the experiment was sensitive enough to detect the expected ether drift,to everyone’s surprise none was found. The negative result had two consequences.First, it showed that the ether does not exist and so there is no such thing as “absolutemotion” relative to the ether: all motion is relative to a specified frame of reference,not to a universal one. Second, the result showed that the speed of light is the same forall observers, which is not true of waves that need a materialmedium to travel (such assound and water waves).

The Michelson-Morley experiment set the stage for Einstein’s 1905 special theoryof relativity, a theory that Michelson himself was reluctant to accept. Indeed, not longbefore the concepts of relativity and quantum theory revolutionized physics, Michelsonannounced that “physical discoveries in the future are a matter of sixth decimal place”.This was a common opinion of the time. Michelson received a Nobel Prize in 1907,the first American to do so.

10.2 Postulates of Special Theory of Relativity

Based on all the theoretical and experimental data available, Einstein put forward hisSpecial Theory of Relativity. Two postulates underlie the special theory of relativity:

The laws of physics are the same in all inertial frames of reference.This postulate follows from the absence of a universal frameof reference. If the

laws of physics were different for different observers in relative motion, the observerscould find from these differences which of them were “stationary” in space and whichwere “moving”. But such a distinction does not exist, and principle of relativity ex-presses this fact. The second postulate is based on the results of many experiments:

The speed of light in free space has the same value in all inertial frames ofreference.


The speed of light is 2.998× 108 m/s to four significant figures. This means thatthe velocity of light has the same value for all observers andis independent of theirmotion or of the motion of the light source.

To appreciate how remarkable these postulates are, let us look at a hypotheticalexperiment, basically no different from actual ones that have been carried out in anumber of ways. Suppose personA turns on a searchlight just as personB takes off ina spacecraft at a speed of 2× 108 m/s. Both of them measure the speed of light wavesfrom the searchlight using identical instruments. From theground personA finds theirspeed to be 3× 108 m/s as usual. “Common sense” tell us that personB ought to finda speed of (3− 2)× 108 m/s, or only 1× 108 m/s for the same light waves. But he alsofind the speed to be 3×108 m/s, even though to personA, personB seems to be movingparallel to the waves at 2× 108 m/s.

There is only one way to account for these results without violating the principleof relativity. It must be true that measurements of space andtime are not absolute butdepend on the relative motion between an observer and what isbeing observed. IfpersonA were to measure from the ground the rate at which personB’s clock clicksand the length of his meter stick, personA would find that the clock ticks more slowlythan it did at rest on the ground and that the meter stick is shorter in the direction ofmotion of the spacecraft. To personB, his clock and meter stick are the same as theywere on the ground before he took off. To personA they are different because of therelative motion, different in such a way that the speed of light personB measures isthe same 3× 108 m/s as personA measures. Time intervals and lengths are relativequantities, but the speed of light in free space is the same toall observers.

Thus, the Galilean transformation equations relating the space and time coordi-nates in one frame of reference to those in the other frame of reference are not validfor cases where the velocityv approaches the velocity of light. Transformation equa-tions that apply to all speeds and also incorporate the constancy of the velocity oflight were derived by the German physicist H.A. Lorentz. These equations, known asLorentz transformation equations, for the case consideredearlier of two inertial framesof reference moving relative to one another with a velocityv alongx direction are asfollows:

x′ = λ(x− vt)

y′ = y

z′ = z

and t′ = λ(t − vx/c2)

whereλ = 1/(1− v2/c2)1/2.


Before Einstein’s work, a conflict had existed between the principles of mechan-ics, which were then based on Newton’s laws of motion, and those of electricity andmagnetism, which had been developed into a unified theory by Maxwell. Newtonianmechanics had worked well for over two centuries. Maxwell’stheory not only coveredall that was then known about electric and magnetic phenomena but had also predictedthat electromagnetic waves exist and identified light as an example of them. How-ever, the equations of Newtonian mechanics and those of electromagnetism differ inthe way they relate measurements made in one inertial frame with those made in adifferent inertial frame.

Einstein showed that Maxwell’s theory is consistent with special relativity whereasNewtonian mechanics is not, and his modification of mechanics brought these branchesof Physics into accord. As we will find, relativistic and Newtonian mechanics agree forrelative speeds much lower than the speed of light, which is why Newtonian mechanicsseemed correct for so long. At higher speeds Newtonian mechanics fails and must bereplaced by the relativistic version.

10.3 Time Dilation

Measurements of time intervals are affected by relative motion between an observerand what is observed. As a result, a clock that moves with respect to an observer ticksmore slowly than it does without such motion and all processes (including those of life)occur more slowly to an observer when they take place in a different inertial frame.

If some one in a space craft finds that the time interval between two events in thespace craft ist0, which is determined by events that occur at the same place inanobserver’s frame of reference, is called the proper time of the interval between theevents. When witnessed from the ground, the events that markthe beginning and endof the time interval occur at different places, and in consequence the duration of theinterval appears longer than the proper time. This effect is calledtime dilation (todilate is to become larger).

To see how time dilation comes about, let us consider the following example. Apulse of light is reflected back and forth between two mirrorsL0 apart (Fig. 10.2). Letthe two mirrors and the clock be at rest. The total time taken by the light pulse for thereturn journey ist0 and is called the proper time.

The proper timet0 is given by

t0 =2LO

c(10.1)


0

Figure10.2Passage of a light pulse between two stationary mirrors.

0

Figure10.3Passage of a light pulse between two mirrors moving with a velocity v.

Now, let us consider the case of the two mirrors and the clock in motion with avelocity v in a direction perpendicular to the direction of motion of the light pulse(Fig. 10.3). The time taken by the pulse for the return journey is t. Because theclock is moving, the light pulse, as seen from the ground, follows a zigzag path. Onits way from the lower mirror to the upper one in the timet/2, the pulse travels ahorizontal distance ofv(t/2) and a total distance ofc(t/2). SinceL0 is the vertical


distance between the mirrors,(ct

2

)2

= L20 +

(vt2

)2

t2

4(c2 − v2) = L2

0

t2 =4L2

0

c2 − v2=

(2L0)2

c2(1− v2/c2)

t =2L0/c

√

1− v2/c2(10.2)

But 2L0/c is the time intervalt0 between ticks on the clock on the ground, as inEq. (10.1) and so

t =t0

√

1− v2/c2(10.3)

Here is a reminder of what the symbols in Eq. (10.3) represent:

t0 = time interval on clock at rest relative to an observer= proper time

t = time interval on clock in motion relative to an observer

v = speed of relative motion

c = speed of light

Because the quantity√

1− v2/c2 is always smaller than 1 for a moving object,t isalways greater thant0. The moving clock in the spacecraft appears to tick at a slowerrate than the stationary one on the ground, as seen by an observer on the ground.

Exactly the same analysis holds for measurements of the clock on the ground bythe pilot of the spacecraft. To him, the light pulse of the ground clock follows a zigzagpath that requires a total timet per round trip. His own clock, at rest in the spacecraft,ticks at intervals oft0. He too finds that

t =t0

√

1− v2/c2

So the effect is reciprocal: every observer finds that clocks in motionrelative tohim tick more slowly than clocks at rest relative to him.

Our discussion has been based on a somewhat unusual clock. Dothe same conclu-sions apply to ordinary clocks that use machinery – spring – controlled escapements,


tuning forks, vibrating quartz crystals, or whatever – to produce ticks at constant timeintervals? The answer must be yes, since if a mirror clock anda conventional clock inthe spacecraft agree with each other on the ground but not when in flight, the disagree-ment between them could be used to find the speed of spacecraftindependently of anyoutside frame of reference – which contradicts the principle that all motion is relative.

10.4 Length Contraction

Measurements of lengths as well as of time intervals are affected by relative motion.The lengthL of an object in motion with respect to an observer always appears to theobserver to be shorter than its lengthL0 when it is at rest with respect to him. Thiscontraction occurs only in the direction of the relative motion. The lengthL0 of anobject in its rest frame is called itsproper length. (We note that in Fig. 10.3 the clockis moving perpendicular tov, henceL = L0 there.)

The length contraction can be derived in a number of ways. Perhaps the simplestis based on time dilation and the principle of relativity. Let us consider what happensto unstable particles called muons that are created at high altitudes by fast cosmic-ray particles (largely protons) from space when they collide with atomic nuclei in theearth’s atmosphere. A muon has a mass 207 times that of the electron and has a chargeof either+e or −e; it decays into an electron or a positron after an average lifetime of2.2µs (2.2× 10−6s).

Cosmic ray muons have speeds about 2.994×108 m/s (0.998c) and reach sea level inprofusion – one of them passes through each square centimeter of the earth’s surfaceon the average slightly more than once a minute. But int0 = 2.2µs, their averagelifetime, muons can travel a distance of only

vt0 = (2.994× 108 m/s)(2.2× 10−6s)

= 6.6× 102m= 0.66 km

before decaying, whereas they are actually created at altitudes of 6 km or more.To resolve the paradox, we note that the muon lifetime oft0 = 2.2µs is what an

observer at rest with respect to a muon would find. Because themuons are hurtlingtowards us at the considerable speed of 0.998c, their lifetimes are extended in ourframe of reference by time dilation to

t =t0

√

1− v2/c2=

2.2× 10−6s√

1− (0.998c)2/c2= 34.8× 10−6s= 34.8µs


The moving muons have lifetimes almost 16 times longer than those at rest. In atime interval of 34.8µs, a muon whose speed is 0.998c can cover the distance

v t = (2.994× 108 m/s)(34.8× 10−6s) = 1.04× 104m= 10.4 km

Although its life time is onlyt0 = 2.2µs in its own frame of reference, a muon canreach the ground from altitudes of as much as 10.4km because in the frame in whichthese altitudes are measured, the muon lifetime ist = 34.8µs.

What if somebody were to accompany a muon in its descent atv = 0.998c, so thatto him or her the muon is at rest? The observer and the muon are now in the same frameof reference, and in this frame the muon’s lifetime is only 2.2µs. To the observer, themuon can travel only 0.66km before decaying. The only way to account for the arrivalof the muon at ground level is if the distance it travels, fromthe point of view of anobserver in the moving frame, is shortened by virtue of its motion. The principle ofrelativity tells us the extent of shortening it must be by thesame factor of

√

1− v2/c2

that the muon lifetime is extended from the point of view of a stationary observer.We therefore conclude that an altitude on the ground find to beh0 must appear in

the muon’s frame of reference as the lower altitude

h = h0

√

1− v2/c2

In our frame of reference the muon can travelh0 = 10.4km because of time dila-tion. In the muon’s frame of reference, where there is no timedilation, this distanceabbreviated to

h = (10.4 km)√

1− (0.998c)2/c2 = 0.66 km

As we know, a muon traveling at 0.998c goes this far in 2.2µs.The relativistic shortening of distances is an example of the general contraction of

lengths in the direction of motion:

L = L0

√

1− v2/c2 (10.4)

Clearly the length contraction is most significant at speedsnear that of light. Aspeed of 1000km/s seems fast to us, but it only results in a shortening in the directionof motion to 99.9994 percent of the proper length of an objectmoving at this speed.On the other hand, something traveling at nine-tenths the speed of light is shortened to44 percent of its proper length, a significant change.

Like time dilation, the length contraction is reciprocal effect. To a person in aspacecraft, objects on the earth appear shorter than they did when he or she was on the


ground by the same factor of√

1− v2/c2 that the spacecraft appears shorter to some-body at rest. The proper lengthL0 found in the rest frame is the maximum length anyobserver will measure. As mentioned earlier, only lengths in the direction of motionundergo contraction. Thus to an outside observer a spacecraft is shorter in flight thanthe ground, but it is not narrower.

10.5 Twin Paradox

We are now in a position to understand the famous relativistic effect known as the twinparadox. This paradox involves a twin, one of which remains on the earth while theother goes on a voyage onto space at the speedv and returns. Dick is 20y old when hetakes off on a space voyage at a speed of 0.80c to a star 20 light years away. To Jane,who stays behind, the pace of Dick’s life is slower than hers by a factor of

√

1− v2/c2 =√

1− (0.80c)2/c2 = 0.60= 60%

To Jane, Dick’s heart beat only 3 times for every 5 beats of herheart; Dick takesonly 3 breaths for every 5 of hers; Dick thinks only 3 thoughtsfor every 5 of hers.Finally Dick returns after 50 years have gone by according toJane’s calendar, but toDick the trip has taken only 30y. Dick is therefore 50y old whereas Jane, the twin whostayed home, is 70y old.

To look at Dick’s voyage from his perspective, we must take into account that thedistanceL he covers shortened to

L = L0

√

1− v2/c2 = (20 light years)

√

1−(0.80c)2

c2= 12 light years

To Dick, time goes by at the usual rate, but his voyage to the star has takenL/v =15y and his return voyage another 15y, for a total of 30y. Of course, Dick’s lifespanhas not been extended tohim, because regardless of Jane’s 50-y wait, he has spent only30y on the roundtrip.

The twin paradox has been verified by experiments in which accurate clocks weretaken on an airplane trip around the world and then compared with identical clocks thathad been left behind. An observer who departs from an inertial system and then returnsafter moving relative to that system will always find his or her clocks slow comparedwith clocks that stayed in the system.

10.6 Relativity of Mass

When a force is applied to an object free to move, the force does work on the objectthat increases its kinetic energy. The object goes faster and faster as a result. Because


the speed of light is the speed limit of the universe, however, the object’s speed cannotkeep increasing in proportion as more work is done on it. But conservation of energyis still valid in the world of relativity. As the object’s speed increases, so does its mass,so that the work done continues to become kinetic energy eventhoughv never exceedsc.

To investigate what happens to the mass of an object as its speed increases, let usconsider an elastic collision (that is, a collision in whichkinetic energy is conserved)between two particlesA andB, as witnessed by observers in the reference framesSandS′ which are in uniform relative motion withS’ moving in +ve x-direction withrespect toS with a velocityv. The properties ofA andB are identical when determinedin reference frames in which they are at rest. Before collision, particleA had been atrest in frameS and particleB in frameS′. Then, at the same instant,A was thrown inthe+y direction at the speedVA while B was thrown in the−y′ direction at the speedVB, where

VA = VB (10.5)

Hence the behavior ofA as seen fromS is exactly the same as the behavior ofB asseen fromS′.

When the two particles collide,A rebounds in the−y direction at the speedVA,while B rebounds in the+y′ direction at the speedVB.

If linear momentum is conserved in theS frame, it must be true that

mAVA = mBVB (10.6)

Inserting these expressions forVA andVB in Eq. (10.6), we see that momentum isconserved provided that

mA = mB

√

1− v2/c2 (10.7)

BecauseA andB are identical when at rest with respect to an observer, the differ-ence betweenmA andmB means that measurements of mass, like those of space andtime, depend upon the relative speed between an observer andwhatever he or she isobserving.

In the example above bothA andB are moving inS. In order to obtain a formulathat gives the mass of a body measured while in motion in termsof its massm0 whenmeasured at rest, we need only consider a similar example in whichVA andVB are verysmall compared withν. In this case an observer inS will see B approachA with thevelocityv, make a glancing collision (sinceVB ≪ v), and then continue on. InS

mA = m0 and mB = m


and som=

m0√

1− ν2/c2(10.8)

The mass of a body moving at the speedv relative to an observer is larger thanits mass when at rest relative to the observer by the factor 1/

√

1− v2/c2. This massincrease is reciprocal; to an observer inS′, mA = m andmB = m0. Measured from theearth, a spacecraft in flight is shorter than its twin still onthe ground and its mass isgreater. To somebody on the spacecraft in flight the ship on the ground also appears tobe shorter and to have a greater mass.

Relativistic mass increases are significant only at speeds approaching that of light.At a speed one - tenth that of light the mass increase amounts to only 05 percent,but this increase is over 100 percent at a speed nine - tenths that of light. Only atomicparticles such as electrons, protons, mesons, and so on havesufficiently high speeds forrelativistic effects to be measurable, and in dealing with these particles, the “ordinary”laws of physics cannot to used. Historically, the first confirmation of Eq. (10.8) was thediscovery by Bucherer in 1908 that the ratioe/mof the electron’s charge to its mass issmaller for fast electrons than for slow ones. This equation, like the others of specialrelativity, has been verified by so many experiments that it is now recognized as one ofthe basic formulas of physics.

As v approachesc,√

1− v2/c2 in Eq. (10.8) approaches 0, and the massm ap-proaches infinity. Ifv = c, m = ∞, from which we conclude thatv can never equalto c: no material object can travel as fast as light. But what if a spacecraft moving atv1 = 0.5c relative to the earth fires a projectile atv2 = 0.5c in the same direction? Weon earth might expect to observe the projectile’s speed asv1 + v2 = c. Actually, veloc-ity addition in relativity is not so simple a process, and we would find the projectile’sspeed to be only 0.8c in such a case.

10.7 Massless Particles

Can a massless particle exist? To be more precise, can a particle exist which has norest mass but which nevertheless exhibits such particle like properties as energy andmomentum? In classical mechanics, a particle must have restmass in order to haveenergy and momentum, but in relativistic mechanics this requirement does not hold.

Let us see what we can learn from the relativistic formulas for total energy andlinear momentum: Totaly energy,

E =m0c2

√

1− ν2/c2(10.9)


Relativistic momentump =

m0v√

1− v2/c2(10.10)

Whenm0 = 0 andv < c, it is clear thatE = p = 0. A massless particle with aspeed less than that of light can have neither energy nor momentum. However, whenm0 = 0 andv = c, E = 0/0, which are indeterminate:E and p can have any values.Thus Eqs. (10.9) and (10.10) are consistent with the existence of massless particlesthat possess energy and momentum provided that they travel with the speed of light.

There is another restriction on massless particles. From Eq. (10.9),

E2 =m2

0c4

1− v2/c2

And from Eq. (10.10)

p2 =m2

0v2

1− v2/c2

p2c2 =m2

0v2c2

1− v2/c2

Subtractingp2c2 from E2 yields

E2 − p2c2 =m2

0c4 −m2

0ν2c2

1− ν2/c2=

m20c

4(1− ν2/c2)

1− ν2/c2= m2

0c4

i.e.,E2 = m20c

4 + p2c2 and

E =√

m20c

4 + p2c2 =

√

E20 + p2c2 (10.11)

According to this formula, if a particle exists withm0 = 0, the relationship betweenits energy and momentum must be given by

For massless particles,E = pc (10.12)

This does not mean that massless particles necessarily occur, only that the laws ofmechanics do not exclude the possibility provided thatν = c andE = pc for them.In fact, massless particles of two different kinds-the photon and the neutrino – haveindeed been discovered and their behavior is as expected.


Numerical examples

1. A spacecraft is moving relative to the earth. An observer on earth finds that,according to his clock, 3601 seconds elapse in a period of onehour as per theclock on the spacecraft. What is the spacecraft’s speed relative to earth?

Here,t0 = 3600s, t = 3601s

t =t0

√

1− v2/c2

Or v = c√

1− (t0/t)= 7.0696× 106 m/s

2. Solar energy reaches the earth at the rate of about 1.4 kWm−2 of the surfaceperpendicular to the direction of the sun. By how much does the mass of the sundecrease per second due to this energy loss? The mean radius of the earth’s orbitround the sun is 1.5× 1011m.

Energy lost per second= (1.4× 103) × (4Π)(1.5× 1011)2 = 4× 1026J.

Mass lost per second,M0 = E0/c2 = 4.4× 109 kg.

3. An electron and a photon have a momenta of 2 MeV/c. Find the total energy ofeach.

Electron energyEe =

√

m20c

4 + p2c2 = 2.064 MeV

Photon energyEp = pc= 2 MeV.


Physical Constants

Avogadro number,NA = 6.025× 1023 per gram mole

Boltzmann constant,k = 1.38× 10−23 JK−1

Electron charge,e= 1.60× 10−19 coulomb

Electron mass,m= 9.11× 10−31 kg

Permeability of free space,µ0 = 4π × 10−7(1.257× 10−6) Hm−1

Permittivity of free space,∈0= 8.85× 10−12 Fm−1

Planck’s constant,h = 6.62× 10−34 Js

Proton rest mass,mp = 1.67× 10−27 kg

Velocity of light, c = 3× 108 ms−1

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