Atomic St ructure

CHEMISTRY

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ATOMIC STRUCTURE

1. LAWS OF CHEMICAL COMBINATIONSLaw of conservation of mass: Laws of conservation of mass states that in a chemical reaction the weight ofproducts is always equal to the weight of reactants.Law of definite proportions: Law of definite proportions states that the elemental composition of a compoundalways remains same if it is analysed form various sources e.g. water from a river or ditch or pond either in Indiaor in USA would always give H : O ratio as 2 : 1.Law of Multiple Proportions: Law of multiple proportion states that elements combine in simple whole numberratios to form various types of compounds e.g. The ratio of N : O is 1: 1, 1 : 2 and 2 : 1 in NO, NO2 and N2O.respectively.

2. DALTON’S THEORY OF ATOMJohn Dalton developed his famous theory of atom is 1803. The main postulates of his theory were : Atom was considered as a hard, dense and smallest indivisible particle of matter. Each element consists of a particular kind of atoms. The properties of elements differ because of differences in the kinds of atoms contained in them. This theory provides a satisfactory basis for the law of chemical combination. Atom is indestructible i.e. it cannot be destroyed or created.Drawbacks It fails to explain why atoms of different kinds should differ in mass and valency etc. The discovery of isotopes and isobars showed that atoms of same elements may have different atomic

masses (isotopes) and atoms of different kinds may have same atomic masses (isobars). The discovery of various sub-atomic particles like X-rays, electrons, protons etc. during the 19th

century lead to the idea that the atom was no longer an indivisible and smallest particle of the matter..3. DISCOVERY OF ELECTRON

William Crookes found that certain rays come out of cathode and gain lots of energy because of high accelerationpotential before colliding with a gas molecule. This happens especially when the gas pressure is low. These raysare known as cathode rays.

Cathode

High Voltage

+

Gas at very low pressure

Cathode AnodeSuction pump

Fig. Cathode ray Tube experiment

Detailed study of cathode rays by J.J. Thomson led to the discovery of electrons. He observed that1.Cathode rays always travel in straight lines2.They are negatively charged. Cathode rays turn towards positively charged plates.3.Charge on particles constituting rays was determined by Oil

Drop experiment by Millikan as 191.6 10 .coulomb

4.Specific charge (e/m) does not change when the gas inside discharge tube was changed indicating that electronis fundamental particle.

e/m for electron = 111.758 10 /Coulombs kg .

5.Value of charge on electron = 191.6 10 Coulombs

6.Mass of the electron = 311.9 10 kg .

4. ANODE RAYS (DISCOVERY OF PROTON)It is well-known fact that the atom is electrically neutral. The presence of negatively-charged electrons in theatom amphasized the presence of positively-charged particles.To detect the presence of positively-charged particles, the discharged tube experiment was carried out, in whicha perforated cathode was used. Gas at low pressure was kept inside the tube. On passing high voltage betweenthe electrodes it was observed that some rays were emitted from the side of the anode. These rays passed through

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the holes in the cathode and produced green fluorescence on the opposite glass was coated with ZnS. These raysconsist of positively-charged particles known as protons.

+

ZnSCoating

H2 gas insideat low pressure

Perforatedcathode

For anode rays, e/m is not fundamental property as different gases used have different mass on C-12 scale.Highest e/m is for hydrogen gas.

5. DISCOVERY OF NEUTRONAfter the discovery of electrons and protons, Rutherford (1920) had predicted the existence of a neutral fundamentalparticle. In 1930, Bethe and Becker reported from Germany that if certain light elements, like beryllium, wereexpoesd to alpha rays from the naturally radioactive polonium, a very highly penetrating radiation was obtained.Similar results were obtained by Irene Cureia dn F. Jolit (1932). Chadwick (1932) demonstrated that this mysteriousradiation was a stream of fast moving particles of about the same mass as a proton but having no electric charge.Because of their electrical neutrality, these particles were called neutron.The lack of charge on the neutron is responsible for its great penetraing power.

Thus, a neutron is a sub-atomic fundamental particle which has a mass 241.675 10 g (approximately 1 amu),

almost equal to that of a proton or a hydrogen atom but carrying no electric charge. The /e m value of a neutronis thus zero.

6. CONSTITUENTS OF ATOMSome of the well known fundamental particles present in an atom are-protons, electrons and neutrons. Manyothers were discovered later viz positron, neutrinos etc.Subatomic Symbol Unit Charge Unit Mass Charge in Mass in AMUparticles Coulomb

Proton 1p1 +1 1 191.60 10 1.007825

Neutron 0n1 0 1 0 1.008665

Electron –1e0 –1 Negligible 191.602 10 45.489 10

Do you Know?Chandwick in 1932 discovered the neutron by bombarding elements like beryllium with fast moving -particles.He observed that some new particles were emitted which carried no charged and had mass equal to that ofproton.

7. THEORIES OF ATOMAfter the discovery of electrons, protons and neutrons several theories were proposed by various scientists suchas1.Plum Pudding Model (Thomson Model of Atom)2.Rutherford’s Model of Atom3.Planck’s quantum theory4.Bohr’s Atomic ModelThomson Model of AtomAccording to this model, electrons are embedded in uniform sphere of positive charge to confer electrical neutrality.This model was satisfactory to the extent that the electrostatic forces of repulsion among the electron cloud isbalanced by the attractive forces between the positively charged mass and the electron. How ever, this modelfails to explain the results of ionisation and scattering experiment and is, therefore discarded.Rutherford Model of Atom

When -particles are bombarded on thin foils ( 54 10 cm thick) of metals like gold, silver, platinum or copper,,

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most of them are passed through the metal foil with little or no deviation. However a small proportion of -particles are scattered through large angles and even bounced back i.e. deflected through 1800. From theseobservations, Rutherford drew the following conclusions:

Gold foil Zinc sulphidescreen

Reflected -particles

Lead plateLead block

Fig. Rutherford's Experiment

(a) As most of the -particles passed through the foil without undergoing any deflection, there must be sufficientempty space within the atom.(b) As -particles are positively charged, deflected by large angle there must be heavy small positively chargedbody present in the atom, which is called Nucleus.From these observations, Rutherford proposed the following model of atom:(a) Atom is composed of a positively charged nucleus where the whole mass of the atom is concentrated and theelectrons are present in relatively large volume around the nucleus. The total positive charge carried by nucleusmust be equal to the total negative charge carried by electrons and electroneutrality of atom is maintained.

Fig: Thomson Model

electron

positive sphere

(b) Electron are constantly moving around the nucleus in different orbits.The centrifugal force arising from this motion balances the electrostatic attraction between the nucleus and theelectron. Therefore the electron don’t fall into the nucleus.Drawbacks in Rutherford Model(a) The most fundamental objection arises from the electromagnetic theory of radiation which predicts that whena charged body moves in a circular path, it should radiate energy continuously. As the electron is a negativelycharged particle revolving around the nucleus, it should radiate energy continuously. As a result, the electronshould fall into the nucleus.(b) As the electron is continuously radiating energy, the spectra should be continuous. Actually, the spectra ofatom is a line spectra.

8. ATOMIC NUMBER OF AN ELEMENT= Total number of protons present in the nucleus= Total number of electrons present in the atom* Atomic number is also known as proton number because the charge on the nucleus depends upon the number

of protons.* Since the electrons have negligible mass, the entire mass o the atom is mainly due to protons and neutrons only.

Since these particles are present in the nucleus, therefore they are collectively called nucleons.* As each of these particles has one unit mass on the atomic mass scale, therefore the sum of the number of

protons and neutrons will be nearly equal to the mass of the atom.Mass number of an element = No. of protons + No. of neutrons.* The mass number of an element is nearly equal to the atomic mass of that element. However, the main difference

between the two is that mass number is always a whole number whereas atomic mass is usually not a wholenumber.

* The atomic number (Z) and mass number (A) of an element ‘X’ are usually represented alongwith the symbolof the element as

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A

Z

Mass Number

Atomic Number

Symbol of theelementx

e.g. 23 3511 17,Na Cl and so on.

1.Isotopes:Such atoms of the same element having same atomic number but different mass numbers are called isotopes.1 2 31 1 1,H H and H and named as protium, deuterium (D) and tritium (T) respectively. Ordinary hydrogen isprotium.2. Isobars:Such atoms of different elements which have same mass numbers (and of course different atomic numbers) arecalled isobars.

e.g. 40 40 4018 19 20, ,Ar K Ca .

3. Isotones:Such atoms of different elements which contain the same number of neutrons are called isotones.

e.g. 14 15 166 7 8, ,C K O .

4. Isoelectronics:The species (atoms or ions) containing the same number of electrons are called isoelectronic.

For Example, 2 2 3, , , , ,O F Na Mg Al Ne all contain 10 electrons each and hence they are isoelectronic.

Illustration.Complete the following table:Particle Mass No. Atomic No. Protons Neutrons ElectronsNitrogen atom – – – 7 7Calcium ion – 20 – 20 –Oxygen atom 16 8 – – –Bromide ion – – – 45 36

Solution.For nitrogen atom.No. of electron = 7 (given)No. of neutrons = 7 (given) No. of protons = Z = 7 (atom is electrically neutral)

Atomic number = Z = 7Mass No. (A) = No. of protons + No. of neutrons = 7 + 7 = 14

For calcium ion.No. of neutrons = 20 (given)Atomic No. (Z) = 20 (given) No. of protons = Z = 20;No. of electrons in calcium atom = Z = 20But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the equation

2 2Ca Ca e but the composition of the nucleus remains unchanged.

No. of electrons in calcium ion = 20 – 2 = 18Mass number (A) = No. of protons + No. of neutrons = 20 + 20 = 40.For oxygen atom.Mass number (A) = No. of protons + No. of neutrons = 16 (Given)Atomic No. (Z) = 8 (Given)No. of protons = Z = 8,

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No. of electrons = Z = 8No. of neutrons = A – Z = 16 – 8 = 8

Some important characteristics of a wave:

Crest Crest

Trough Trough

a

a

Wavelength of a wave is defined as the distance between any two consecutive crests or troughs. It is representedby and is expressed in Å or m or cm or nm (nanometer) or pm (picometer).

8 101 Å 10 10cm m

9 121 10 , 1 10nm m pm m

Frequency of a wave is defined as the number of waves passing through a point in one second. It is representedby v (nu) and is expressed in Hertz (Hz) or cycles/sec or simply sec–1 or s–1.

1 Hz = 1 cycle/secVelocity of a wave is defined as the linear distance travelled by the wave in one second. It is represented by v andis expressed in cm/sec or m/sec (ms–1).Amplitude of a wave is the height of the crest or the depth of the trough. It is represented by ‘a’ and is expressedin the units of length.Wave number is defined as the number of waves present in 1 cm length. Evidently, it will be equal to thereciprocal of the wavelength. It is represented by v (read as nu bar).

1v

If is expressed in cm, v will have the units cm–1.Relationship between velocity, wavelength and frequency of a wave. As frequency is the number of wavespassing through a point per second and is the length of each wave, hence their product will give the velocityof the wave. Thus,

v v

Cosmic rays < - rays < X-rays < Ultraviolet rays < Visible < Infrared < Micro waves < Radio waves.9. PLANCK’S QUANTUM THEORY (1901)

It states1.Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy called Quanta.2.Each quantum is associated with a definite amount of energy E which is proportional to frequency of radiation.

E hv

Where, h = Planck’s constant = 346.626 10 secJoule .

v = Frequency of the light radiation

3.A body can emit or absorb radiations only in whole multiples of quantum i.e. E nhv

where 1,2,3......n

cv

where c = velocity of light

= wavelength

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hcE

9(a). Some Important Formulae1.A = Z + N (Number of neutrons)

2.dynamic mass of particle 2 1/ 20 /[1 ( / ) ]m m v c

3.Radius of nucleus 1/ 3 150 0( ) , 1.2 10R R A R m

4. c v

5.wave number = 1/v

6. /E hv hc hcv

7.9 2 21 2

20

1; 9.0 10 /4

q qF K K Nm C

r

8. 2 1E hv E E

Illustration 1.Calculate number of photon coming out per sec. from the bulb of 100 watt. If it is 50% efficient and wavelengthcoming out is 600 nm.

Solution.energy = 100 J

energy of one photon = 34 8

199

6.625 10 3 10 6.625 102600 10

hc

no. of photon = 19 19100 10 15.09 106.625

Illustration 2.Certain sun glasses having small of AgCl incorporated in the lenses, on exposure to light of appropriate wavelengthturns to gray colour to reduce the glare following the reactions:

( )hvAgCl Ag Gray Cl

If the heat of reaction for the decomposition of AgCl is 248 kJ mol–1, what maximum wavelength is needed toinduce the desired process?

Solution.

Energy needed to change = 3248 10 /J mol

If photon is used for this purpose, then according to Einstein law one molecule absorbs one photon. Therefore,

3248 10A

hcN

34 8 237

3

6.625 10 3.0 10 6.023 10 4.83 10248 10

m

.

10. BOHR’S ATOMIC MODELThe postulates of Bohr’s atomic theory regarding stability of electrons of an atom are as follows:(i) The electrons is an atom revolve around the nucleus only in certain selected circular orbits. These orbits areknown as energy levels or stationary states. An electron can be excited from a lower state to higher state with theabsorption of a quantum of energy, or can come down from a higher to lower state with emission of a radiationof energy (as shown in figure) equal to energy to quantum 2 1E E E hv . 2E and 1E are energies of theelectron associated with stationary orbits.

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+

Absorption of E

Emission ofradiation with energyE(E2 E1)

E1

Fig. Bohr's Atomic Model

(ii) The stability of the circular motion of an electron requires that the electrostatic force (due to the attractionbetween the nucleus and the electron) provides the necessary centripetal force for the motion of electron.

22

0

1 ( ). /4

Ze e mv rr

... (i)

where Z – atomic numbere – charge on electron

0 – permittivity of the charge in vacuum

r – distance between positive charge & electron(iii) Angular momentum of electron is quantised i.e. electron can revolve only in those orbits where its angularmomentum is an integral multiple of / 2h

/ 2mvr nh ... (ii)

where, v – velocity of electronm – mass of electron

h – Planck’s constant

1, 2,3........n are known as Principal quantum number..

Bohr’s Atomic Radiusfrom (i) &(ii) we have -

from (i) 2 20/(4 )v ze mr ... (iii)

& from (ii) 2 2 2 2 2 2/ 4v n h m r ... (iv)

Equating (iii) and (iv), we have2 2 2 2 2 2

0/(4 ) / 4Ze mr n h m r

or,2 2

02 2

(4 )4

n hr

mZe

r is called Bohr’s radius.

Where 0, , ,h e m and are constants, Thus, 2( / )r K n Z

Putting the values of 0, , ,h e m and , 11 2(5.297 10 ) ( / )r m n Z

Illustration 1.

For hydrogen atom 1Z , therefore radius of the first orbit = 11 2(5.297 10 )(1 /1)m

For He ion 2Z , therefore radius of the first orbit = 11 2(5.297 10 ) (1 / 2)m

112.649 10 m

Solution.Velocity of electrons in various orbits

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Substituting value of r in equation no. (ii)

2

0

2 ( / )4

Zev K Z nnh

The energy of an electron in an orbit

E = Kinetic energy + Potential energy = 2

2

0

12 (4 )r

Zemv

Putting 2( )mv from equation (i), we have

2 2 2

0 0 02(4 ) (4 ) 2(4 )Ze Ze ZeE

r r r

Now, putting the value of ' 'r , we have

2 2 42 2 18 2 2

2 2 20

2 ( / ) (2.18 10 ) ( / )(4 )

mZ eE K Z n J Z nn h

The energy difference between two energy levels n2 and n1 is given by

2 1

2 2 42 21 22 2

0

1 2 1/ 1/(4 )n n

mZ eE E E n nh

18 22 21 2

1 1(2.18 10 )J Zn n

It terms of wave-number, we have

2 2 21 21/ 1/Hv R Z n n

10(a). Bohr’s Model for Hydrogen like Atoms:

1. 2hmvr n

2.2 2

2 1812 2 22.178 10 / 13.6n

E z zE z J atom eVn n n

2 4

1 2

2 meEn

3.2 2 2

2 2

0.529 Å4n

n h nrZ Ze m

4.2 62 2.18 10 /ze zv m s

nh n

5.Revolutions per sec = 2 16

3

0.657 10/ 2 Zv rn

6.Time for one revolution = 16 3

2

1.52 102 / nr vZ

7. . . . .( . . 2 . .)nE K E P E P E K E

8.2

2

0

1. . 1/ 2 , . .4

zeK E mv P Er

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Illustration 1.What is the principal quantum number of H atom orbital is the electron energy is –3.4 eV? Also report theangular momentum of electron.

Solution.E1 for H = –13.6 eV

Now, 12n

EEn

2

13.63.4n

n = 2

Now, Angular momentum (mur) = 34

34 12 6.626 10. 2.1 10 sec2 2 3.14hn J

.

Illustration 2.Calculate the energy, velocity and radius of electron in Li2+ ion.

Solution.For Li2+ ion, z = 3, n = 1, then

radius = 2 10.529Å 0.529 Å 0.1763Å

3nz

velocity = 8 82.18 10 / sec 3 2.18 10 / sec.zcm cmn

86.54 10 / sec.cm

Energy2

2

913.6 13.61

zev evn

122.4 ev .

11. DEFINITION VALID FOR SINGLE ELECTRON SYSTEM :(i) Ground state:lowest energy state of any atom or ion is called ground state of the atom.Ground state energy of H-atom = –13.6 eVGround state energy of He+ ion = –54.4 eV(ii) Excited state:State of atom other than the ground state are called excited states:

n = 2 first exited staten = 3 second exited staten = 4 third exited staten = n + 1 nth exited state

(iii) Ionisation energy (IE) :Minimum energy required to move an electron from ground state ton is called ionisation energy of the atom or ion.Ionisation energy of H-atom = 13.6 eVIonisation energy of He+ ion = 54.4 eVIonisation energy of Li+2 ion = 122.4 eV(iv) Ionisation Potential (I.P):Potential difference through which a free electron must be accelerated from rest, such that its kinetic energybecomes equal to ionisation energy of the atom is called ionisation potential of the atom.I.P. of H atom = 13.6 VI.P of He+ Ion = 54.4 V

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(v) Excitation Energy:Energy required to move an electron from ground state of the atom to any other state of the atom is calledexcitation energy of that state.excitation energy of 2nd state = excitation energy of 1st state = 1st excitation energy = 10.2 eV.(vi) Excitation Potential:Potential difference through which an electron must be accelerated from rest to so that its kinetic energy becomeequal to excitation energy of any state is called excitation potential of that state.excitation potential of third state = excitation potential of second excitate state = seconds excitations potential =12.09 v.(vii) Binding Energy ‘or’ Separation Energy:Energy required to move an electron from any state to n is called binding energy of that state.Binding energy of ground state = I.E. of atom or Ion.

Illustration.A single electron system has ionisation energy 11180 kJ mol–1. Find the number of protons in the nucleus of thesystem.

Solution.

I.E. = 2

192 21.69 10Z J

n

3 219

23 2

11180 10 21.69 106.023 10

ZI

Z = 312. HYDROGEN SPECTRUM:

1. Study of Emission and Absorption Spectra:An instrument used to separate the radiation of different wavelengths (or frequencies) is called spectroscope ora spectrograph. Photograph (or the pattern) of the emergent radiation recorded on the film is called a spectrogramof simply a spectrum of the given radiation. The branch or science dealing with the study of spectra is calledspectroscopy.Emission SpectraWhen the radiation emitted from some source e.g. from the sun or by passing electric discharge through a gas atlow pressure or by heating some substance to high temperature etc, is passed directly through the prism and thenreceived on the photographic plate, the spectrum obtained is called ‘Emission spectrum’.Depending upon the sources of radiation, the emission spectra are mainly of two types:(i) Continuous spectra:When white light from any source such as sun, a bulb or any hot glowing body is analysed by passing through aprism it is observed that it splits up into seven different wide bound of colours from violet to red. These coloursare so continuous that each of them merges into the next. Hence the spectrum is called continuous spectrum.

PhotographicPlate

Prism

Beam

White light

Slit VIBGYOR

(ii) Line Spectra:When some volatile salt (e.g., sodium chloride) is placed in the Bunsen flame or an electric discharge is passedthrough a gas at low pressure light emitted depends upon the nature of substance.

PhotographicPlate

Prism

Beam

Slit

Platinum wire5896 Å5890 Å}Two yellow lines

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It is found that no continuous spectrum is obtained but some isolated coloured lines are obtained on thephotographic plate separated from each other by dark spaces. This spectrum is called ‘Line emission spectrum’or simply Line spectrum.2. Absorption spectra:When white light from any source is first passed through the solution or vapours of a chemical substance andthen analysed by the spectroscope, it is observed that some dark lines are obtained in the otherwise continuousspectrum. These dark lines are supposed to result from the fact that when white light (containing radiations ofmany wavelengths) is passed through the chemical substance, radiation. of certain wavelengths are absorbed,depending upon the nature of the element.

PhotographicPlate

Prism

NaClSolutionSlit V

IBGY}ORWhite light

Dark lines in yellow region of continuous spectrum

PhotographicPlate

Prism

Beam

Slit

EMISSION SPECTRUM OF HYDROGEN:12(a). H-Atom Spectrum

When an electric discharge is passed through hydrogen gas at low pressure, a bluish light is emitted. When a rayof this light is passed through a prism, discontinuous line spectrum of several isolated sharp lines is obtained asshown in figure.

n=654

3

2

n=1

Energy

Lymann series

Balmer series

Paschen series

Bracket seriesPfund series

Energy levels of H-atom

All these lines observed in the hydrogen spectrum can be classified into the series as is tabulated in the table.The hydrogen Spectrum

Region Spectral lines n1 n2

UV Lyman series 1 2,3,4,.....Visible Balmer series 2 3,4,5,.....I R Paschen series 3 4,5,6,.....far-I.R Brackett series 4 5,6,7,.....

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far-I.R. P fund series 5 6,7,8,.....Illustration

Calculate the highest wavelength of line spectra of H-atom when the electron is situated in 3rd excited state.Solution.

Highest wavelength means lowest energy difference of electronic transition from one energy level to otherenergy level.

Hence, lowest energy transition will be 4 3n to n .

4113.6 0.85

16E ev ev

3113.6 1.549

E ev ev

4 3 ( 0.85 1.54)E E E ev

190.69 0.69 1.6 10hc ev J

34 87

19

6.626 10 3 10 18 100.69 1.6 10

m m

13. DE-BROGLIE RELATIONSHIPAccording to de-Broglie matter has dual character i.e. wave as well as particle, if the wavelength of matter be having mass m moving with velocity v , then,

hmv

where h is Planck’s constant 34(6.626 10 sec)Joul .

wave in phase wave out of phaseCase - I Case - II

For electron moving around a nucleus in a circular path, two possible waves of different wavelengths are possible.In case I, the circumference of the electron orbit is an integral multiple of wavelength.In case II, wave is destroyed by interference and hence, does not exist.Therefore, the necessary condition for a stable orbit of electron of radius r is, 2 r n

when 1,2,3,n etc .

As h

mv

2 ,2

nh nhr or mvrmv

This is simply the original Bohr condition for a stable orbit. Hence, the Bohr’s model of H-atom is justified byde-Broglie relationship.

13.(a)De-Broglie Relations:

h hmc p

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de-Broglie pointed out that the same equation might be applid to material particle by suing m for the mass of theparticle instead of the mass of photon and replacing c, the velocity of the photon, by v, the velocity of theparticle.

2 ( . .)h h

mv m K E

From the de-Broglie equation it follows that wavelength of a particle decreases with increase in velocity of theparticle. Moreover, lighter particles would have longer wavelengths than heavier paticles, provided the velocityis equal.If a charged particle Q is accelerated through potential difference V from rest then de-broglie wavelength is

2h

mQV

de-Broglie concept is more isgnificant for microscopic or sub-microscopic particles whose wavelength can bemeasured.The circumference of the nth orbit is equal to n times the wavelength of the electron.

2 nr n

Wavelength of electron is always calculated using De-broglie calculation.Illustration. 1

Calculate the wavelength of a body of mass 1 kg moving with a velocity of 10 m sec–1.Solution.

We know,h

mv

Substituting the values, m = 1, mg = 10–6 kg, v = 10 m sec–1.

and 34 2 16.625 10h kg m s

3429

6

6.625 10 6.625 1010 10

m

Illustration. 213.6 eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground states absorbs1.50 times as much energy as the minimum energy required for it to escape from the atom. What is the wavelengthof the emitted electron? ( 31 19 349.109 10 , 1.602 10 , 6.63 10 .em kg e coulomb h J s )

Solution.1.5 times of 13.6 eV, i.e., 20.4 eV, is absorbed by the hydrogen atom out of which 6.8 eV (20.4 – 13.6) isconverted to kinetic energy.

KE = 6.8 eV = 6.8 ( 191.602 10 coulomb ) (1 volt) = 181.09 10 J .

Now, 212

KE mv

or,18

631

2 2(1.09 10 ) 1.55 10 / .(9.109 10 )

KE Jv m sm kg

34

1031 6

(6.63 10 . ) 4.70 10 .(9.109 10 )(1.55 10 / )

h J s metresmv kg m s

14. HEISENBERG’S UNCERTAINTY PRINCIPLE (1927):If subatomic particles have wave nature then we can’t pinpoint where exactly a particle is. The idea was definedby Heinsenberg as - There is a limit to the precision to which the position and momentum of a particle may bedetermined simultaneously-

/ 4x p h

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Where x uncertainty in position of an electron.

p uncertainty in momentum of an electron, i.e., when we try to determine position for a subatomicparticle correctly, uncertainly in momentum will be very large or when we try to determine the momentumcorrectly then uncertainly in position will be large.ORIGIN OF QUANTUM THEORYWhen solid body heated it emit radiations in the forms of waves. The wave nature of light can be explained bydiffraction interference etc. But some other observable properties such as photoelectric effect, compton effectcould not be explained from wave nature. Hence a different theory is needed to explain these facts.

Illustration.Calculate the uncertainty in the velocity of a wagon of mass 2000 kg whose position is known to an accuracy of

10m .

Solution.

Uncertainty in position, 10x m

Mass of the wagon, 2000m kg

According to Heinsenberg’s principle,

.4hx m v

34 2 139 16.626 10 2.64 10

4 . 4 3.14 2000 10h kgm sv msm x kg m

.

14(a). Black Body RadiationWhen radiation falls on an object, a part of it is reflected, a part is absorbed and the remaining part is transmittedbecause no object is a perfect absorber. But the black body (e.g., a metallic hollow sphere with a small hole,blackened on the inside surface) absorbs completely all the radiations falls on it by successive reflections insidethe enclosure.

T1

T2

T3

Wave length

energy T1>T2>T3

The black body is not only a perfect absorber of radiation energy, but also an ideal radiation, i.e., when the blackbody is heated, it radiates the maximum amount of energy. The energy which is radiated is dependent on thetemperature of the black body and is independent of the nature of the interior material.The curves represent the distribution of radiation from a black body at different temperatures. The shape of thecurves couldn’t be explained on the basis of classical electromagnetic theory in which it was assumed that thebody radiates energy continuously. So, the intensity of radiation should increase continuously without limits asthe frequency increases. But the experimental observations are contrary to the classical view. For each temperature,there is a maximum in the curve corresponding to a particular wavelength, indicating the maximum radiation ofenergy. At higher temperature, the position of the maximum in the curve shifts towards shorter wavelength andbecomes more pronounced. To explain this black body radiation, Max Planck put forward new quantum theory.Planck Quantum Theory(a) Radiation energy is not emitted or absorbed continuously but discontinuously in the form of tiny bundles ofenergy, called quanta.(b) Each quanta is associated with a defined amount of energy (E) which proportional to the frequency ofradiation i.e, E v or, E hv where h is Planck’s constant 34(6.626 10 )J sec .

(c) A body can emit or absorb energy only in whole number multiples of quanta i.e. E nhv where 1,2,3 .n etc

15. PHOTOELECTRIC EFFECTSir J.J. Thomson has discovered this phenomenon of ejection of electron from the surface of a metal when lightof suitable frequency of strikes on it.

Atomic Structure

16

Only few metals show this effect under the action of visible light, but many more show it under the action ofmore energetic u.v. light. For every metal, there is a minimum frequency of incident radiation necessary to ejectelectron from that metal surface, is known as Threshold frequency ( 0v ). This 0v varies metal to metal.

The number of ejected electrons from the metal surface depends upon the intensity of the incident radiation.Greater the intensity, the larger is the number of ejected electrons.

Hence, according to quantum theory, when a photon of light of frequency 0( )v v strikes on an electron in ametal, it imparts it entire energy to the electron. Then some of its energy (equal to binding energy of electron withthe nucleus) is consumed to separate the electron from the metal and the remaining energy will be imparted to theejected electron.

20

12

hv hv mv where 0hv is the binding energy of work function of the electron and 212

mv is the kinetic

energy of electron. Alkali metals are mainly used for photoelectric effect. Cesium, amongst alkali metals, haslowest threshold energy and used largely in photoelectric cell.

16. SHAPES OF ORBITALSs-orbital: they do not have directional character. They are spherically symmetrical. The s-orbital of higherenergy levels are also spherically symmetrical. They are more diffused and have spherical shells within themwhere probability of finding the electron is zero.

z

x

yNode

2s orbital

In the s-orbital, number of nodes is (n – 1)p-orbital: ‘p’ orbital has a dumb-bell shape and it has a directional character.The two lobes of a p-orbital are separated by a plane that contains the nucleus and is perpendicular to thecorresponding axis. Such plane is called a nodal plane because there is no probability of finding the electron.

+

px

z

y

+

py

z

y

x x

+

pz

zy

x

In the absence of an external electric or magnetic field, the three p-orbitals of a particular energy level have sameenergy and are degenerate. In the presence of an external magnetic field or electric field this degeneracy isremoved.

d-orbitals: For d-orbitals five orientations are possible viz., dxy, dyz, dxz, 2 2x yd , 2z

d . All these five orbitals in theabsence of magnetic field are equivalent in energy and are degenerate.The shapes of the orbitals are as follows:

Atomic Structure

17

dxy

x

y

dxz

x

z

dyz

x

z

These three ‘d’ orbitals are similar. The maximum probability of finding the electron is in lobes which aredirected in between the axes. Nodal region is along the axes.

dz2

z

dx2-y2

y

x

These two d-orbitals are similar. Probability of finding the electron is maximum along the axes and the nodalregion is in between the axes.

17. QUANTUM NUMBERSThese are used to determine the region of probability of finding a particular electron in an atom.(a) Principal quantum number (n):This denotes the energy level or the principal or main shell to which an electron belongs. It can have onlyintegral values 1, 2, 3 etc. The letter K, L, M ..... are also used to designate the value of n. Thus, an electron in theK shell has n = 1, that is L shell has n = 2 and so on.

Illustration. 1The principal quantum number of 2s-electron is

Solution.n = 2(b) Azimuthal quantum numbers (l):This denotes the orbital (Sub-level) to which an electron belongs. It gives an idea about the shape of the orbital.l can have any value from 0 to (n – 1), for a given value of n,i.e. l = 0, 1, 2, ..... (n – 1)Value of l 0 1 2 3Sub Shell s p d f

The value of orbital angular momentum of the electron for a given value of ‘l’ is ( 1)2hl l

(c) Magnetic quantum number (m):It gives us the idea about the orientations an orbital can have in space in the presence of magnetic field. Thevalues of ‘m’ depends on ‘l’ orbital quantum number.Total value of m = (2l + 1) and it varies –l to +l.For example, for l = 0 the value of magnetic quantum number m is also equal to zero, i.e. ‘s’-orbital can haveonly one orientation in space in presence of magnetic field.(d) Spin quantum number (s):The electron while moving round the nucleus in an orbit also rotates or spins about its own axis either in a

Atomic Structure

18

clockwise direction or in an anticlockwise direction. Its value is 12

or 12

corresponding to clockwise or

anticlockwise spin.

(1) The value of spin angular momentum for a given value of s is ( 1)2hs s

(2) The spin magnetic moment of electron (excluding orbital magnetic momentum) is given by

( 2)effective n n BM (Where n = Number of unpaired electrons).

18. DISTRIBUTION OF ELECTRONS IN AN ATOMThe filling up of orbitals with electrons takes place according to certain rules which are given below:(i) The maximum number of electrons in a main shell is equal to 2n2, where n is the principal quantum number.(ii) The maximum number of electrons in a sub-shell like s, p, d, f is equal to (2l + 1), where l is the azimuthalquantum number for the respective orbitals. Thus s, p, d, f can have a maximum of 2, 6, 10 and 14 electronsrespectively.(a) Afbau PrincipleAccording to the principle, “Electrons are added progressively to the various orbitals in the order of increasingenergy”.What does the word ‘Afbau’ mean?Afbau is a German term which means building up or construction.The energy of various orbitals increase in the order given below:1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s < ......

(i) A new electron enters the orbitals for which (n + l) is minimum, e.g. if we consider 3d and 4s orbitals, theelectron will first enter 4s-orbitals in preference to 3d.This is because the value of (n + l) for 4s-orbitals is less (4 + 0 = 4) than that for 3d-orbital (3 + 2 = 5)(ii) In case where (n + l) values are the same, the new electron enters the orbitals for which ‘n’ is minimum, e.g.in a choice between 3d and 4p for which (n + l) values are same (3 + 2 = 5, 4 + 1 = 5), the electron will prefer togo to the 3d-orbital, since n is lower for this orbital.(b) Pauli’s Exclusion PrincipleIt states that it is impossible for two electrons in a given atom to have same set of quantum numbers.

Example:(a) n = 2, l = 0, m = 0, s = +1/2

n = 2, l = 0, m = 0, s = –1/2(b) n = 2, l = 1, m = 0, s = +1/2

n = 2, l = 1, m = 0, s = –1.2n = 2, l = 1, m = +1, s = +1/2n = 2, l = 1, m = +1, s = –1/2n = 2, l = 1, m = –1, s = +1/2n = 2, l = 1, m = –1, s = –1/2

(c) Hund’s Rule of Maximum MultiplicityAccording to this rule, electrons enter the orbitals (e.g. s, px, py, pz ...) in the same sub-level in such a way as to

Atomic Structure

19

give maximum number of unpaired electrons. In other words it means that pairing begins with the introduction ofthe second electron in the s-orbital, the fourth in p, etc.What is the electronic configuration of Cu (Z = 29)?1s2 2s2 2p6 3s2 3p6 3d10 4s1

Exceptional Electronic ConfigurationSome atoms such as copper and chromium exhibit exceptional electronic configuration.For example:Cr(Z = 24) has an electronic configuration1s2 2s2 2p6 3s2 3p6 3d5 4s1

It is because of the extra stability associated with the half-filled and completely filled orbitals.

CONCEPT BUILDING EXAMPLES

Example 1.Find out the energy of the electron in the first excited state in an H-atom.

Solution.Energy of an electron in H-like atom is given by

28

2(2.18 10 ) ;ZE Jn

where z is the number of protons and n is

number of shell in which electron is present.For the first excited state, 2n

218

2

1(2.18 10 )2

E J

195.45 10E J

Example 2.

The K.E. of a moving electron is 255 10 J ; Calculate its velocity and the wavelength.

Solution :

21. .2

K E mv

253 1

31

2( . ) 2 5 10 1.048 109.1 10

K E Jvelocity v msm kg

Now, wavelength h

mv

347

31 3 1

6.626 10 6.949 10(9.1 10 ) (1.048 10 )

Js mkg ms

Example 3.If the electron of the hydrogen atom has been excited to a level corresponding to 10.2 electron volts, what is thewavelength of the line emitted when the atom returns to its ground state?

Solution.

2 1E E E

E hv hcE

10.2E eV 1910.2 1.6 10E J

Atomic Structure

20

34 8

19

6.624 10 3 1010.2 1.6 10

m

34 8 9

19

6.624 10 3 10 1010.2 1.6 10

nm

121.8nm .

Example 4.An electron jumps from fourth excited state to the ground stable in Li+2 -ion and the energy released in the formof photon is allowed to strike a metal ( x ) surface whose work function ( ) is 181 10 J . What is the K.E. &velocity of the electron ejected.

Solution.The amount of energy released is given by

18 22 21 2

1 1(2.18 10 )E J zn n

where z is the atomic number of H-like atom.

18 2 181(2.18 10 ) (3) 1 1.23 1016

E J J

Now, . .hv K E

18 18. . (1.23 10 1 10 )K E J

2 181 2.3 102

mv J

191 5 1

31

2 2.3 10 7.11 109.1 10

v ms ms

Example 5.Ionisation energy of hydrogen atom is 13.6 eV. What will be the ionisation energy of He+ and Li++ ions?

Solution.As we know ionisation energy for one electron system is given by

2

2. . 13.6 zI P eVn

For hydrogen atom, z = 1, n = 1, So, I.P = 13.6 eVFor He+ ion z = 2 and n = 1.

So, I.P = 13.6 213.6 (2)

i.e., . . 54.4I P eV For Li2+, z = 3 & n = 1; I.P. = 213.6 (3)

i.e., . 122.4I P eV .

Example 6.A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n . This excited atomcan make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excitedstate by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values ofn and z (ionisation energy of hydrogen atom = 13.6 eV).

Solution.Total energy liberated during transition of electron from nth shell to first excited state (i.e., 2nd shell)

10.2 17 17.2 eV

1227.2 1.602 10 erg

Atomic Structure

21

22 2

1 12H

hc R Z hcn

12 2

2 2

1 127.2 1.602 102HR Z hc

n

... (i)

Similarly, total energy liberated during transition of electron from nth shell to second excited state (i.e. 3rd shell):

4.25 5.95 10.2eV

1210.2 1.602 10 erg

12 2

2 2

1 110.2 1.602 103HR Z hc

n

... (ii)

Dividing eq. (i) by eq. (ii) 6n On substituting the value of n in eqns. (i) or (ii) 3Z .

Example 7.1 mol of He+ ion excited. Spectral analysis showed the existence of 50% ions in 3rd level, 25% in 2nd level andremaining 25% in ground state. Ionisation energy of He+ is 54.4 eV; calculate total energy evolved when all theions return to ground state.

Solution.25% of He+ ions are already in ground state, hence energy emitted will be from the ions present in 3rd level and2nd level.

2 2 21 2

1 1( )E IPn n

per ion or atom.

02 23 1

1 1(54.4)2 1 3

NE

for 0

2N

ions falling to ground state

0454.49N eV

and 02 22 1

1 1(54.4)4 1 2

NE

for 0

4N

ions falling to ground state.

0354.416

N eV

Hence total energy = 04 354.49 16

N

23 9154.4 6.023 10144

eV

23 199154.4 6.023 10 1.6 10144

J

4331.13 10 J

Example 8.Estimate the difference in energy between the 1st and 2nd Bohr orbit for a hydrogen atom. At what minimumatomic number, would a transition from n = 2 to n = 1 energy level result in the emission of X-rays with

83.0 10 m . Which hydrogen atom like species does this atomic number correspond to?

Solution.For hydrogen atom, the expression for energy difference between two energy level is,

2 22 2

1 1HE R hc

n n

Atomic Structure

22

So, 7 34 8

2 111.09677 10 6.626 10 3 10 14

E E E

81.635 10 J

For hydrogen like species, the same expression is

22 22 2

1 1 1Hz R

n n

or,2

2 21 2

1 1 1Hz R

n n

So, 2 7

8

1 3(1.09677 10 )43 10

z

or, z = 2

So, the species is He+ because z = 2.Example 9.

What is the degeneracy of the level of the hydrogen atom that has the energy(a) –RH, (b) –RH/9

Solution.2/n HE R n

(a) ,HE R

1n when 1n , 0l , 0em

The level is degenerate.

(b) / 9HE R

3n when 3n , 0,1,2l

when 0l , 0em (3s-orbital)

when 1l , 1,0,1em (3p-orbital)

when 2l , 2, 1,0,1, 2me (3d-orbitals)This is 1 + 3 + 5 = 9 states in all. The degeneracy is 9.

Example 10.Calculate the uncertainty in the velocity of a ball of mass 150g if uncertainty in this position is

341Å ( 6.6 10 )h Js

Solution.From Heisenberg uncertainty principle we know that

4 4 4h h hx p or xm v or v

m x

Substituting value: 10150 0.15 , 1Å 10m g kg x m

34

10

6.6 104 3.14 10 0.15

v

24 13.499 10v ms .

Subjective Solved ExamplesExampe1. 11

Calculate the waelength and wave number of the spectral line when an electron in H-atom falls from higherenergy state 3n to a state 2n . Also determine the energy of a photon to ionise this atom by removing theelectron from 2nd Bohr’s orbit. Compare it with the energy of photon required to ionise the atom by removingthe electron from the ground state.

Solution.

Atomic Structure

23

+e n=2n=3

Photon emitted

+1e n=2

n=

Photon absorbed

n=1

First calculate the energy ( )E between the To ionise the atom from 2n , the

Bohr orbits 3n and 2n using: responsible transition will be 2n n .

18 22 21 2

1 12.18 10E Z Jn n

18 2

(2 ) 2 2

1 12.18 10 12

E J

18 2(3 2) 2 2

1 12.18 10 (1) ( 1)2 3

E J Z

195.45 10 J

193.03 10 J To ionise the atom from ground state

Now this energy difference is the energy of the ( 1)n , the transition is 1 .

photon emitted.18 2

2 2

1 12.18 10 11

E

PhotonhcE hv hcv

182.18 10 J

193.03 10 6560.3Åhc

and 6 11 1.52 10v m

Example 12A hydrogen atom in the ground state is hit by a photon by a photon exciting the electron to 3rd excited state.The electron then drops to 2nd Bohr orbit. What is the frequency of radiation emitted and absorbed in theprocess?

Solution.

Energy is absorbed when electron moves from ground state ( 1)n to 3rd excited state ( 4)n .

+en=2

Photon absorbed

n=1

n=1

n=4

First calcuate the energy difference btween 1n and 4n .

Use: 18 2(1 4) 2 2

1 2

1 12.18 10E Zn n

.

Atomic Structure

24

Here, 1 21, 1, 4Z n n Put 1 2n and 2 4n in the expression of E , to get:

18 2

(1 4) 2 2

1 12.18 10 11 4

E J

18 2(4 2) 2 2

1 12.18 10 12 4

E J

182.04 10 J 194.08 10 J

This is the energy of the photon absorbed. This is the energy of the photon emitted.

Use: 182.04 10PhotonE hv J to get: Use: 194.08 10PhotonE hv J

153.08 10v Hz 146.16 10v Hz

Similarly, when electron jumps from

4n to 2n , energy is emitted and is given by the same relation.Example 13

A hydrogen like ion, ( 2)He Z is exposed to electromagnetic waves of 256.4 Å. The excited electron givesout induced radiations. Find the waelength of the indicated radiations, when electron de-excites back to theground state. R = 109677 cm–1.

Solution:He+ ion contains only one electron, so Bohr’s From 3n , the electron can fall back to themedhot is applicable here. It absorbs a photon of ground state in three possible ways (transitions)

wavelength 256.4Å . Assume the electron to 3 1, 3 2, 2 1

be in ground state initially. Let it jumps to an Hence three possible radiations are emitted.excited state n2. Find the wavelengths corresponding to these

22 21 2

1 1 1v R Zn n

transitions.

Substitute for 8256.4Å 256.4 10 cm , The wavelength ( ) for transition, 3 1 willbe same i.e., 256.4 Å. Find for 3 2 and2 1 using the same relation.

109677, 2R Z for He+ ion, (3 1) 256.4 , (3 2) 1641.3ÅÅ

1 1n and find 2n . (2 1) 303.9Å

7 28 2 2

1 2

1 1 1109677 10 (2)256.4 10 1 n

2 3n

n=3n=2

n=1

Example 14Hydrogen gas when subjected to photon-dissociation, yields one normal atom and one atom possessing 1.97eV more energy than normal atom. The bond dissociation energy of hydrogen moleucle into normal atoms is103 kcals mol–1. Compute the wave length of effective photon for photon dissociation of hydrogen molecule inthe given case.

Solution:

2H H H

where H is normal H -atom and *H is excited H-atom. So the energy requird to dissociate 2H in this

manner will be greater than the usual bond energy of 2H molecule.

E (absorbed) = dissociation energy of 2H extra energy of excited atom.

Energy required to dissociated in normal manner

Atomic Structure

25

3103 10 cal per mol (given)

319

23

103 10 4.18 7.17 10 /6 10

J atom

The extra energy possessed by excited atom is 1.97 eV19 191.97 1.6 10 3.15 10J J

E (absorbed) = 19 197.175 10 3.15 10 J

181.03 10 J

Now calculate the wavelength of photon corresponding to this energy.

181.03 10 1930 ÅPhotonhcE

Example 15An electron in the first excited state of H-atom absorbs a photon and is further excited. The de Broglie wavelengthof the electron in this state is found to 13.4 Å. find the wavelength of photon absorbed by the electron in Å.Also find the longest and shortest wavelength emitted when this electron de-excites back to ground state.

Solution:Note: The energy state 1n is known as Ground State

The energy state 2n is known as First Excited State

The energy state 3n is known as Second excitedState and so on.

+e

n=2

n=n

Photon

The electron from 2n absorbs a photon and is further excited to a higher energy level (let us say n ).

The electron in this energy level ( n ) has a de Broglie wavelength ( ) 13.4Å

ee e

hm v

and 6 12.18 10nZv msn

[vn is the velocity of e– in nth Bohr orbit]

6 12.18 10hv

m n

34

10 31

6.626 10(13.4 10 ) (9.1 10 )

6 12.18 10n

4n

Now find the wavelength of the photon responsible for the excitation from 2 4n to n .Using the relation :

18 22 2(2 4)1 2

1 12.18 10E Zn n

194.09 10 J [n1=2, n2=4, Z=2]

19

(2 4)4.09 10hcE

4863.1Å

The Longest wavelength emitted when this electron 19(4 3) 1.06 10E

Atomic Structure

26

(from 4n ) falls back to the ground state will corresponds 191.06 10PhotonhcE E J

to the minimum energy transition. 18752.8Å

The transition corresponding to minimum energy will Shortest wavelength : 4 1

be 4 3 . 18 2

(4 1) 2 2

1 12.18 10 11 4

E

Note: The transition corresponding to maximum energy 182.04 10 J

will be 4 1 . (4 1) PhotonhcE E

( .)Energy diff PhotonhcE E hv

973.2Å

1

PhotonPhoton

E or E v

Using the same relation :18 2

(4 3) 2 21 2

1 12.18 10E Zn n

1 2[ 3, 4, 2]n n Z

Example 16A single electorn orbits around a stationary nucleus of charge Ze , where Z is a constant and e is themagnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the thirdBohr. Find :(a) the value of Z(b) the energy required to excite the electron from 3 4n to n .(c) the wavelength of radiation required to remove electron from 2nd Bohr’s orbit to infinity(d) the kinetic energy, potential energy and angular momentum of the electron in the first orbit.(e) the ionisation energy of above one electron system in eV.

Solution.Since the nucleus has a charge Ze , theatomic number of the ion is ‘ Z ’.

(a) The transition is 1 22 3n n by absorbing a photon of energy 47.2 eV .

47.2E eV Using the relation:

22 21 2

1 113.6E Z eVn n

2

2 2

1 147.2 13.6 52 3

Z Z

(b) The required transition is 1 23 4n n by absorbing a photon of energy E .

Find E by using the relation: 2

2 21 2

1 113.6E Z eVn n

2

2 2

1 113.6(5)3 4

E eV

16.53E eV

(c) The required transition is 1 22n n by absorbing a photon of energy E .

Atomic Structure

27

Find E by using the relation:

22 2

1 113.6(5) 852

E E eV

Find of radiation corresponding to energy 85 eV..

34 8

19

6.626 10 3 1085 (1.6 10 )

hcE

146.16 Å

(d) If energy of electron be En, then KE = –En and PE = 2En

2 2

2 2

13.6 13.6 5 3401n

ZE eVn

(340 ) 340KE eV eV

2( 340 ) 680PE eV eV

Angular momentum ( )2hl n

346.626 101

2l

341.05 10 J s

(e) The ionisation energy (IE) is the energy required to remove the electron from ground state to infinity. Sothe required transition is 1 . The ionisation energy

21( ) 13.6( )IE E Z eV

213.6 5 340IE eV Example 17

With what velocity should an alpha ( ) particle travel towards the nucleus of a copper atom so as to arrive ata distance 10–13 m from the nucleus of the copper atom?

Solution.As -particle appraoches towards the Cu nucleus, it decelerates due to repulsion from it and finally itsvelocity will become zero at point A (which is the turning point). After that, particle will move in the leftdirection (and accelerating)

AV=0

Cu Nucleus(+29e)

r0

-particleV

To arrive at a distance (r0) from the nucleus, the kineticc energy of alpha particle should be equal to theelectrostaic potential energy of it, i.e., KE = EPE

2

0

12

NKq qm v

r

:m mass of -particle = 274 (1.67 10 )kg

:v velocity of -particle = ?

9 29 10 /K N m C

Atomic Structure

28

q charge on -particle = 192 ( 1.6 10 )C

r0 = Minimum distance of approach

Note: 2 , 4 pq e m m [ -particle is He2+ ion or He Nucleus]

qN = charge in Cu nucleus = Ze = 29(+1.6 10–19C)d = distance from nucleus = 10–13 m

0

2 NKq qvm r

Substituting the given values, we get,66.325 10 / .v m s

Note: This is a simple cases where velocity of -particle is directed towards the centre of the Cu-nucleus.

A Cu

r0

-particle

V=0

Note: When there is a difference between the velocity vector of -particle and the Cu(target) nucleus, thetrajectory is more complicated.

Target-particle

Example 18Find the energy required to excite 1.22 litre of hydrogen atoms gas at 1.0 atm and 298 K to the first excitedstate of atomic hydrogen. The energy requied for the dissociatio of H-H bonds is 436 kJ/mol. Also calculatethe minimum frequency of a photon to break this bond.

Solution.Let us, first find the number of moles of hydrogne atoms.

2

1 1.22 0.050.0821 298H

PVnRT

Thus the energy required to break 0.05 moles of H2 (H-H bond) = 0.05 436 19.62 .kJ

Now calculate the energy needed to excite the H-atoms to first excited state i.e., to 2n (First excited state isreferred to 2n ).

18 22 2

1 12.18 10 (1) /1 2

E J atom

181.635 10 /J atom

No. o H atoms = (No. of H2 molecules) 223 22(0.05 6.02 10 ) 2 6.02 10

Atomic Structure

29

The energy required to excited the given number of H-atom = 22 186.02 10 1.635 10 98.43J kJ

So the total energy required

19.62 98.43 118.05kJ

Now the energy required to break to single

H-H bond = 3

1923

436 10 7.238 106.023 10

= Energy supplied by the photon

19 347.328 10 6.626 10 ( )hv v

151.09 10v Hz

Example 19Estimate the differnce in energy between 1st and 2nd Bohr’s orbit for a H-atom. At what minimum atomicnumber (Z), a transition from 2n to 1n energy level would result in the emission of radiatio with wavelength

83.0 10 ?m Which hydrogen atom like species this atomic number corresponds to? How much ionisation

potential is needed to ionise this species? 7 1( 1.097 10 )R m

Solution.The difference in energy is given by E :

18 22 2

1 12.18 10 (1) /1 2

E J atom

18 111.65 10 1.65 10 10.2J ergs eV

For a H-like atom, 83.0 10 m .

18 2

2 2(2 1)

1 12.18 101 2

E Z J

PhotonhcE

Solve to get : Z = 2Hence the H-like atom is He+ ion.To ionise, He+ ion, ionisation energy (IE) = –(E1)

2( 13.6 2 ) 54.4IE eV

The ionisation potential (IP) is the voltage difference required to generate this much energy.

( ) 54.4IE qV e IP eV

(required) 54.4IP Volt

Example 20

A stationary He+ ion emits a photon correspondings to the first line ( )H of Lyman series. The photon thusemitted, strikes a H-atom in the ground state. Find the velocity of the photoelectrons ejected out of the hydrogenatom. The value of 7 11.097 10 .R m

Solution.

The difference in energy ( )E will be equal to the energy of the photon emitted.

First line in Lyman series corresponds to the transition 2 1 .

18 22 2

1 12.18 10 (2) /1 2

E J atom

186.54 10 J

Atomic Structure

30

The photon of this much energy strikes a H-atom in the ground state. Note that the ionisation energy of H-atomis 182.18 10 J . This will be the work function of H-atom. Using the Einstein’s photoelectric equation:

20

12i e eKE E W m v [Ei = Incident energy]

02( )i

ee

E Wvm

18 18

31

2(6.54 10 2.18 10 )9.1 10ev

63.09 10 /ev m s

We can also calculate the wavelength of electron ejected out = 102.36 10 2.36 Åm

34

38 6

6.626 10 2.36 Å9.7 10 3.09 10e

e e

h mm v

Example 21An electron in a hydrogen like species, makes a transition from nth Bohr orbit to next outer Bohr ( 1n ).Find an approximate relation between the dependence of the frequency of the photon absorbed as a function of‘ n ’. Assume ‘ n ’ to be large value ( 1n ).

Solution.

( 1)

18 2( ) 2 2

1 12.18 10( 1)n n

energy differenceE hv Z Jn n

18 2

2 2

2 12.18 10( 1)nhv Z J

n n

.

Since 1n (given)

1 ~ ; 2 1 2n n n n

18 2

4

22.18 10 nhv Z Jn

3v n .

Atomic Structure

31

MIND MAP

1. According to the quantumtheory, the radiant energy isemitted by atoms & moleculesin small discrete amounts(quanta)m rather than over acontinuous rante. The energy ofeach quanta is given by E = hv.

2. According to Bohr model, theangular mometum of an electronis an integral multiple of / 2h .Bohr’s model is applicable singleelectron species (hydrogen likespecies).

3. The radius of an orbit is givenby 2 2 2 2/ 4 .r n h kZms Thevelocity of an electron in an orbitis given by / 2v nh mr and theenergy of an electron in an orbitis given by 2 2 2 22 4 /E pk Z ms n h .

9. In photoelectric effect,electrons are elected from thesurface of certain metalexposed to light of at least acertain minimum frequency.

0 . .hv hv K E

4. In bohr model, an electronemits a photon when it dropsfrom a higher energy state to alower energy state.

8. Four quantum numberscharacterise each electron in anatom. The principal quantumnumber(n) indentifies the mainenergy level, the angularquantum number (l) indicatesshapes of orbital, the magneticorientation of orbital in spaceand the spin quantumnumber(s) indicates thedirection of the electron’s spinon its axis.

5. The emission spectra ofhydrogen is obtained whenelectron from an ecited state isdeexcited to the ground state.The release of specific amountsof energy in the form of photonsaccounts for the lines in thehydrogen spectrum. v of eachline in the spectrum can begiven by

2 21 21/ 2(1/ ) (1/ )]Ryz n n

7. An orbital may be defined asa region in space around thenucleus where the probabilityof finding the electron ismaximum.

6. De Broglie exendedEinstein’s wave particledescrition of light to all mattersin motion. The wavelength ofa moving particle of mass mand velocity v is given by deBroglie equation, / .h mv

ATOMICSTRUCTURE