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Preface

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Atomic and Molecular Spectroscopy

according to CSIR-NET/JRF, GATE and JEST examination. The whole book has been

divided into topic-wise sections.

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Summary of Physical Constants

speed of light in vacuuma 2.99 8 110 ms

permeability of vacuumb 7 1

7 110 Hm

permittivity of vacuum 201 ( )c

12 110 Fm

constant of gravitationc 6.673 11 3 1 210 m kg s

planck constant 6.626

(2 )h 1.054 34

elementary charge 1910 C

magnetic flux quantum, 0 2.067 15

electron volt 1.602 19

electron mass 9.109 3110 kg

proton mass p 2710 kg

proton/electron mass ratio p em m 1836.152

unified atomic mass unit u 1.66 2710 kg

find-structure constant, 7.297

inverse 137.035

Rydberg constant, R 1.097 7 110 m

Avogadro constant 6.022 23 1

Faraday constant, F 9.648 4 1

molar gas constant R 8.314 1 1J mol K

Boltzmann constant, A k 1.380

Stefan-Boltzmann constant, 5.670 8 2 410 Wm K

Bohr magneton, 9.274 24 110 JT


Atomic constants

Electron constants

fine-structure constant, 7.297

inverse 137.035

Rydberg constant, R 1.097 7 110 m

1510 Hz

2.179 18

13.605 eV

Bohr radiusb, 5.291 11

electron mass 9.109 3110 kg

in MeV 0.510 MeV

electron / proton mass ratio /e pm m 5.446

electron charge e 1910 C

electron specific charge / ee m 11 110 C kg

electron molar mass, e 5.485 7 110 kg mol

compton wavelength, 2.426 1210 m

classical electron radius, 20 2.817 1510 m

Thomson cross section, 28 / 3 er 6.652 29 210 m

electron magnetic moment 24 1

in Bohr magnetons,

in nuclear magnetons,

electron gyromagnetic ratio, 2 e e 1.760

electron g-factor,


Protons constants

Neutron constants

proton mass p 1.672 2710 kg

in MeV 938.271 MeV proton/electron mass ratio /p em m 1.836

proton charge e 1.602

proton specific charge / pe m 9.578 7 110 C kg

proton molar mass, A pN m 1.007 3 110 kg mol

proton compton wavelength, / ( )ph m c 1.321

proton magnetic moment 1.410 26 110 J T

in Bohr magnetons, p B 1.521

in nuclear magnetons, /p N 2.792

proton gyromagnetic ratio, 2.675 8 1 110 s T

Neutron mass 1.674 2710 kg

in MeV 939.565 MeV neutron/electron mass ratio 1838.683

neutron/proton mass ratio /n pm m 1.001

neutron molar mass, A n n 1.008 3 110 kg mol

neutron compton wavelength, 1.319

neutron magnetic moment 27 110 JT

in Bohr magnetons

in nuclear magetons

neutron gyromagnetic ratio, 1.832 8 1 110 s T


Muon and tau constants

Muon mass 1.883 2810 kg

in MeV 105.65 MeV

tau mass 3.167 2710 kg

in MeV 1.777 muon/electron mass ratio / em m 206.768

muon charge

muon magnetic moment 26 110 JT

in Bohr magnetons, B 4.841

in nuclear magnetons, / N 8.890

muon g-factor


Formula List

1. The empirical formula for the line spectrum of Hydrogen atom 2

8

2 2

where n = 3,4,5,... for H , H , H & H

2. Total energy in nth orbit 2

2n

3. For an electronic transition from n2th level to n1th level

2

2 21 2

1 1 1Rz

n n

4. Spin-orbit interaction

Interaction between electron spin magnetic moment and the internal

magnetic field

5. The relativistic term shift is

2 41

3

1 3

1 42

rE R zTr cm

hc nn

6. Lande-g-factor

7. Larmor frequency

8. Transverse deflection suffered by the atom is given by Stern-Gerlach experiment

9. Half-intensity line breadth in terms of frequency

10. In terms of wavelength

11. Wavelength of X-ray, for K

2

2 2


12. Rotatiaonl wave number 1F J BJ J

13. Vibrational Wave number

where osc

c

Vibrational constant

14. Dissociation energy is given by max

*0 maxe

2 1

4 2e

e e

e

x where

15. 3

213

21

321

21

16. 1 2N N N

17. Cavity life time

1 2

2( )

1c

cL

nLt

c nR R e

absorption coefficient of active medium

and = reflectivity of mirrors


Atomic-Molecular Spectra & Laser Physics

Contents # Introduction of Atomic and Molecualr Spectroscopy

1. a) b) Sommerfeld atomic model

2.

3.

4. Spectroscopic terms of L-S & J-J coupl

5. Spectra of Alkali-

6.

7.

8.

9. Rotational-

10. X-

11. Electronic Band

12. The Raman-Spectra

13.


Introduction to Atomic and Molecular Physics

Atomic and molecular physics is an interesting and important subject to understand and invent the world. Here, we start with a brief introduction of atomic and molecular physics

The atomic and Molecular physics is the world of atoms, molecules, Ions, Clusters and photons It is one of the most fundamental and important field in development of knowledge in physical and

chemical sciences It is the oldest branch of Quantum physics. The field contributed so much to the development of

quantum mechanics that until thirties of last century atomic and molecular physics was inseparable from quantum physics

It remains one of the most important subjects for the test ing grounds of the quantum theory Inspite of being the oldest field, it has retained its freshness all along and is continuing to contribute to

the understanding of the fundamental laws of nature and also to the development of new technologies

Research in this area today encompasses the study of structure and interaction of atoms, molecules, ions, biomolecules, clusters at wide ranging scales

Providing newer insights into practically all field of science and technology such as: Spectroscopy, Laser Physics & Technology, Plasma Physics, Nuclear Physics, Non-accelerator particle Physics, Astrophysics, Condensed Matter Physics and Material Sciences, Metrology, Biosciences, Atmospheric Sciences

Chemical sciences mainly analysis and reactions, biological physics, energy research and fusion studies.

Varieties of applications:

Laser , X-ray technology, Nuclear Magnetic Resonance (NMR), Magnetic Resonance Imaging(MRI), Electron Paramagnetic Resonance (EPR), Mossbauer spectroscopy, Laser atom cooling, Bose Einstein condensation (BEC), GPS system for navigation, Pollution detection, Various medical applications etc.

Wide ranging scales make the field of atomic & Molecular Physics more interesting today than it was ever We also see that the study of atomic and Molecular Physics is at the heart of every interdisciplinary research An understanding of the structure of atoms and Molecules is therefore a prerequistite of entering into several of the areas discussed earlier

Various scales in atomic Molecular Physics

In the following, we will come to know the characteristics of the atoms and molecules in various conditions in terms of temperature, photon intensities and frequencies

(Temperature range from to )

Fusion Plasma and stellar environment, Spectroscopy an d dynamics of highly charged ions, diagnostics techniques for plasma

Molecular interactions in astrophysical ices, formation of complex molecules in interstellar environment, fundamental question on formation of molecules which have relevance to origin of life


110 K

Atomic and Molecular beams, Clusters

Laser cooled atoms, atomic clock based on Cs fountain, Nano-fabrication using cold atoms

Quantum degenerate atomic gases, Bose-Einstein Condensation, Fermi Sea, Quantum many body physics, Matter wave optics, Atom interferometer, Atomic Physics as basis of condensed matter physics

Photon intensities used for interaction: 9 3 210 10 ( / )W cm obtained from normal light sources

High resolution spectroscopy for structure of atoms and molecules

3 9 210 10 ( / )W cm obtained from laser sources

Multi-photon processes, multi-step excitation processes, nonlinear optical processes, Rydberg and autoionizing levels, Ultra-trace analysis (single atom detection)

9 15 210 10 ( / )W cm High intense laser pulse sources

Intense field regime, above threshold ionisation, multiple ionisations, high harmonic generation (up to 300), Coulomb explosion, atomic stabilization

Spectral resolution used for spectroscopy

Ultra precision measurements, Atomic Clocks of accuracy 1 in 1610 and better, Ultra-precision magnetometry (

Tesla detection)

6100 10Hz Hz Parity violation experiments in atoms and molecules (non-accelerator particle physics), Study of hyperfine interactions, Hyperfine structure and Isotope shifts, Measurements of properties of unstable nuclei (Nuclear Physics via atomic physics), Cold atom physics

6 910 10 Hz Fine structure in atoms and molecules, atomic & molecular interactions, state selective processes in atoms and molecules

9 1510 10 Hz Spectrochemical analysis, Material characterization, Photoionization and photo-fragmentation, supercontinuum sources and their interactions

Excitation source used for modern research: Table-top instruments to high energy particle accelerator

Laser: to photons, used for rotational, vibrational and electronic spectroscopy

Synchrotron radiation sources: 5eV to 1000 eV photons: used to study valence to inner-shell processes

Ion traps and Atom trap: This is used to study the interactions of charged particles of MeV energies with atoms and molecules


Molecular spectrum is the signature and the fingerprint to identify the molecules. This has a deeper and direct application to our human society.

Now-a-days several spectroscopic methods are used in medical sciences for diagnosis the disease. In the following a glimpse of these methodologies is given

X-rays: the most common instrument used almost everywhere

The first X-ray device was discovered accidentally by the German scientist Wilhelm Roentgen (1845-1923) in 1895 He found that invisible rays emitted from cathode-ray tube that penetrate paper and wood but could not penetrate metal and bone. Roentgen used his device to examine the bone structure of the human hand

X-ray produced when a target metal (such as tungsten) is bombarded by highly accelerated electrons. These high energy electrons knock out the electrons from the deep energy levels such as K and L shells, the other electrons in higher energy levels occupy these vacant spaces by radiating extra energy in the form of X-rays

X-rays penetrate the body and the image is stored on silver halide films Bones and metals absorb or reflect X-rays-leaving the impression on the film, whereas the soft tissues transmit X-rays. This contrast produces and image of bone or metal deep inside the body. The helps to identify medical problems inside the body such as fractures of bones etc

Nuclear Magnetic Resonance (NMR) / Magnetic Resonance Imaging (MRI) A nucleus with non-zero spin has angular momentum and a magnetic moment such as 1H, 31P, 13C, 19F. This nuclear magnetic moment interacts with the external magnetic field. In resonance condition, this absorbs the energy To obtain the image: vary the external field as a function of position to localize signals

They speculated the earth, air, fire sky and water might from the basic elements from which the physical world is constricted.

They also developed various schools of thought about the ultimate nature of matter. Perhaps the most remarkable was the atomist school founded by the ancient Greeks Leucippus of Miletus and Democritus of Thrace about 440 BC.

For purely philosophical reason, and without benefit of experimental evidence, they developed the notion that matter consists of indivisible and indestructible atoms.

Today, we have come a long way developing many thoughts and ideas, carrying out several experimentations and producing lot of evidence.

An understanding of the structure of atoms and molecules is therefore a prerequisite for entering into several of the areas discussed earlier

The basic atomic and molecular physics is the scientific study of the structure of the atoms and molecules, its energy states, and its interactions with other particles and with electric and magnetic fields.


Chapter 1 Atomic Model

Bohr Model of the Hydrogen Atom

in diameter, surrounded by moving electrons in a region about

in diameter. If this model is studied under classical mechanics, the atom must be unstable because accelerating electron must continuously radiate energy and there may only be continuous spectrum, no spectral lines since atom is stable it emits a line spectrum originating from the structure and link between the three known concepts at that time The empirical formula for the line spectrum of the Hydrogen atom discovered by Balmer and Rydberg # Balmer while studying emission spectrum of Hydrogen has found a formula that correctly predicted the four visible lines of Hydrogen atom viz, and .

Balmer formula

where

, for line at

, for line at

, for line at

, for line at

# Rydberg had proposed a formula for line spectrum of H-atom

Where R is Rydberg constant =

# # Quantum nature of the emission and absorption of light Bohr

assumptions

H

H

H


:

Niels Bohr, in 1913, developed a model of atomic structure which was in accurate quantitative agreement with the observed hydrogen and hydrogen-like spectra. This model is based on certain postulates which are.

(i) An electron in an atom moves in a circular orbit the nucleus under the influence of the Coulomb attraction between the electron and the nucleus, according to the laws of classical mechanics.

and embodies some of the ideas concerning the stability of the nuclear atom.

(ii) Out of the infinite number of orbits which would be possible in classical mechanics, it is only

possible for an electron to move in an orbit for which the magnitude of its orbital angular momentum is an integral multiple of

This postulate introduces quantisation. We shall see that the quantisation of the orbital angular momentum of the atomic of the atomic electron leads to the quantisation of its total energy.

(iii) The electron, in spite of its accelerated motion, does not radiate electromagnetic energy while moving in an allowed orbit. Thus its total energy remains stationary.

This postulate removes the problem of the stability of the electron moving in a circular orbit, due to the emission of electromagnetic energy as demanded by classical theory

(iv) Electromagnetic radiation is emitted if an electron, initially moving in an orbit of total energy

discontinuously changes its motion so that it moves in a lower orbit of total energy

of the emitted radiation is given by

-atom Hydrogen | H-like atom Hydrogen)

Let us consider an atom consisting of a nucleus of charge and a single electron of charge and m

The condition of mechanical stability of the electron

Hydrogen like atom

#

L

fE

2

20

1

4

ze e mv

rr

He Li

2P


# Alkali atoms Na, K, Li, Rb etc

Outer most shell

Electrostatic force between nucleus and must be equal to the centripetal force. where, v is the speed of electron in its orbit

Now,Orbital angular momentum of electron is

nd postulates

where and is called principal quantum number

(4)

and therefore velocity of electron is

Radius of nth Bohr Orbit

where

Putting the value of quantities in we get

Velocity of Electron

we know that

or

4P

2

nhL

2

nhv

mr

2

02

zev

nh

0r

2 0.53Athnr n

2

0

1

4n

eV

n


The value of , which is less than 1% of the speed of light. We define the

dimension less quantity ratio of to C by

This ratio is called the fine structure constant

Total Energy of moving electron

We know that moving electron has two kind of energy one is electrostatic potential energy and other its kinetic energy

and

Total energy (E) = K.E + P.E

r We get total energy of moving electron in nth orbit is

(6)

Now, put

and for H-atom

we get energies of the various allowed orbits of H-atom

The energies are n total energy of the quantum state becomes less negative, approaching to zero as . The lowest energy level is called the ground state and the higher level excited state

The ground-state energy of the hydrogen atom is a convenient energy unit. It is called the

Rydberg (ry) and it value is

1V2

1

0 04

V e

C mr C C

2

0

1. (u)

4

zeP E

r

2

0

1. .

8

zeK E

r

2 2 2

0 0 08 4 8

ze ze zeE

r r r

2 4

2 2 20

11,2,3........

8n

mz eE n

h n

191

192

193

194

21.7 10 13.6

5.42 10 3.39

2.41 10 1.51

1.36 10 0.85

E J eV

E J eV

E J eV

E J eV

2 3


Note: In order to remove the electron from its ground state to infinity, a minimum of 13.6eV of energy is required. Thus 13.6eV is the binding energy of the Hydrogen atom

Term Value

The quantity for an infinitely heavy nucleus

It is constant for all atom

The lowest and the most stable energy level has largest term value

The ground state energy E1 of the hydrogen (Z =1) atom is a convenient energy unit. It is called the ry), and its numerical value is

In order to remove the electron from its ground state to infinity, a minimum of 13.6 eV of energy is required. Thus, 13.6 eV is the binding energy of the hydrogen atom, the energy which binds the electron to the nucleus. The binding energy (or ionisation energy) is numerically equal to the energy of the ground state of the atom.

Term Values: The energy values of the various quantum states (energy levels) divided by are called . Thus using equation

The quantity for an infinitely heavy

is a constant for all atoms, the last equation can now be written as

The lowest and the most stable energy level has the largest term value

origin of one-electron (Hydrogen-like) Atomic Spectra : The energy equation can e written, in term of as

419

2 20

1ry 21.7 10 13.68

mcJ eV

h

2 4

2 3 20

11,2,3...

8n

n

E mz eT n

hc h c n

( )R

4

2 308

meR

h c

419

2 20

1 21.7 10 13.68

mery J eV

h

nT2 4

2 3 20

1

8n

n

E mZ eT

hc h c n

R

4

2 308

meR

h c

R

R


Let us now calculate the frequency vmoving in an orbit of quantum number drops to a lower orbit of quantum number . If and be

or

but , where v is wave number, thus

We can now discuss the emission of electromagnetic radiation by a one-electron Bohr atom. The normal state of the atom is the state in which the electron has the lowest energy . When the atom receives

energy from outside by some means, say by collisions in an electric discharge, the electron makes a transition to a state of higher energy . Now, as a common tendency of all physical systems, the

atom emits its excess energy and returns to the ground state. This is accomplished by a series of transitions in which the electron drops to states of successively lower energy, finally reaching the ground state. In each transition, electromagnetic radiation is emitted with a wave number determined by equation (vii), for example if the electron is excited to the state and drops successively through the states and to the ground state ; then three lines are emitted with wave numbers given by equation (vii) for and and and and

for hydrogen so that equation (vii) is

This is identical with the series formula for the various series of the hydrogen spectrum, if

according to the Bohr model

Taking the known values of and c upto six significant figures, we obtain

which is in good agreement with the experimental value of

According to the Bohr model, each of the five known series of the hydrogen spectrum arises from a set of transitions in which the electron drops to a certain final state

Lyman series

Balmer series

Paschen series

Brackett series

Pfund series

These series are illustrated in terms of the energy-level diagram. The transition giving rise to a particular line of a series is indicated by an arrow going from an initial

in fn fE

2

2 2f i

' 1vv

c

2

2 2f i

f 3in

1,Z

2 2f i

4

2 308

meR

h c

0

fn


quantum state to the final quantum state . Since the distance between any two energy levesl is

proportional to the difference between the energy of the two levels, the length of any arrow is proportional to the wave number of the corresponding spectral line.

The theoretical explanation of Balmer particularly impressive because the Lyman, Brackett, Pfund series had not been discovered at the time the theory was developed by Bohr. The existence of these series was predicted by the theory, and the series were observed experimentally in 1916, 1922, 1924 respectively at the predicted positions

When a beam of continuous light is passed through a glass cell containing hydrogen gas in the atomic states and then sent into a spectrometer, a set of dark lines on a bright background is observed. This is the absorption spectrum of hydrogen. In terms of quantum theory the incident light is considered to be a beam of quanta (photons) of all possible ene are absorbed by the hydrogen atoms whose energies correspond to transitions between the allowed energy levels of the atoms. Hence discrete dark lines appear against the continuous background. The process of absorbing radiation is just the inverse of the emission process, and hence the absorption lines have exactly the same frequencies as the emission lines

The dark lines in the absorption spectrum are never completely black, but only appear so by Contrast with the bright background. The reason is that the atoms which are excited by absorbing radiation, re-radiate the absorbed energy almost atonce, but these photon come off in random direction with only a few in the same direction as the original beam of continuous light

No Balmer Lines in the absorption spectrum of Hydrogen: It is observed that for every line in the absorption spectrum there is a corresponding line in the emission spectrum of the substance. The reverse is, however not true. Only certain emission lines appear in absorption. Transitions can occur only from

to . Therefore, lines of only the Lyman series appear in the absorption spectrum.

However in absorption spectra of certain stars, lines of the Balmer series have been observed. This is due to the fact that, on account of the high temperature of the stellar atmosphere a sufficient number of hydrogen atoms are in the first excited state .

in fn


We can, however estimate the temperature of hydrogen gas in the atomic state at which the Balmer series will be observed in the absorption spectrum. According to Boltzmann probability distribution, the ratio of the number of atoms in the first excited state to the number of atoms in ground state, in a

sample in thermal equilibrium at temperature T is

Now, for hydrogen we have

and so

It is apparent that a significant fraction of the hydrogen atoms will initially be in the state provided

T is of the order of

Resonance Line: When an atom in the ground state is excited to a higher state by the absorption of monochromatic radiation, it can fall directly back to the ground state, or after passing through other lower states. Therefore the substance gives emission line of the same frequency as absorbed and also

If, however hydrogen gas is illuminated with monochromatic radiation corresponding to the first

line of the Lyman series this radiation will be absorbed and atoms will rise from and state. The only possibility in this case is the transition back to the ground state , with the emission of the same line as absorbed. A spectral line with this property is called a resonance wavelength) line of Lyman series is the resonance line of hydrogen

Concept of Reduced mass

Let us now drop the assumption that the nucleus is infinitely heavy compared to the electron and so Remains fixed in space. Infect, the nucleus has a finite mass, and the electron and the nucleus both revolve about their common centre of mass C which remain fixed in space. We must make correction in Bohr theory for the finite mass fo the nucleus

Let m be the mass of electron, M the mass of nucleus and r the distance between them. Let x be the distance of the electron from the centre of mass C. The distance of the nucleus from C will be .

Since the system is in equilibrium the, the moments of m and M about C will be equal that is

1n

2

1 2

1

//2

/1

E kTE E kT

E kT

n ee

n e

( )r x


This gives

and

Since both the electron and the nucleus revolve around C with the same angular velocity , the total orbital angular momentum of the atom, L, is the sum of the angular momenta of the electron and the nucleus. Thus

Let us write

Where m by a factor .

Then, the total angular momentum of the atom about the centre of mass is

To take nuclear motion into account, Bohr modified his second postulate and said that the electron can move only in those orbits of which the angular momentum of the atom is an integral multiple of . Thus, from this quantisation condition, we have

;

or

In the absence of nuclear motion, the corresponding equation is

A comparison of the last two equation shows that a replacement of m by

into account the finite mass of the nucleus. The energy of the electron in the nth orbit of a one-electron atom is now

Since is slightly less than m, the electron energies are slightly less negative than if the nucleus were at

Rest (that is infinitely heavy). The wavelength of the spectral lines computed on the basis of the above energy equation are slightly larger than those corresponding to an infinitely heavy nucleus, and agree more closely with the experimental value.

Mrx

m M

22L mx M r x

2 2 2 2

2 2

2

2

2

M r m rm M

m M m M

mM M m r

m M

mMr

m M

2

2

nhr

2

nhr

2

nhm r

2 4

2 2 20

1

8n

Z eE

h n


Variation in Rydberg constant due to Finite Nuclear Mass The finite nuclear mass casus a slight variation in the Rydberg constant from atom to atom. The

Rydberg constant for an atom having an infinitely heavy nucleus is

and that for an atom having a nucleus of mass M is

where = , therefore

Thus, is less than by a factor , and is different for different atoms, depending upon their

nuclear masses. Taking the known values of and c, upto six significant figures we obtain

For the most extreme case of hydrogen, . The Rydberg constant for hydrogen is worked out to

be Thus, is less than by about one part in 2000?

Then variation of Rydberg constant from element to element against the mass number is shown in

fig . The largest change occurs between and with increasing mass number, the Rydberg constant approaches more closely to .

-electron (hydrogen-like) atoms, assuming infinitely heavy nucleus, was obtained as

Taking into account the finite mass of the nucleus, this must be written as

4

2 308

meR

h c

4

2 30

M

1

m

m

M4

2 30

1

8 1 1M

RmeR

m mh cM M

MR R1

1m

M

0

1

1836

m

M

R

1H 2H

R

2

2 2f i


For hydrogen (H), Z=1. Hence writing RH instead of RH instead of the expression becomes

For singly-ionised helium and so

For doubly-ionised lithium (Li++), Z = 3, and so

For triply-ionised beryllium , and so

theory corrected for finite nuclear mass agrees with the experimental data to within 3 parts in 100,00 The difference in the Rydberg constant for hydrogen and (ionised) helium can be utilised to compute the ratio of the mass m of the electron to the mass MH of the hydrogen nucleus (that is proton).

Determination of m/MH : The Rydberg constant for hydrogen atom is given by

Where is the Rydberg constant for an infinitely heavy nucleus, m is the mass of electron and is

the mass of hydrogen nucleus (that is, mass of proton). Similarly, the Rydberg constant for He+ is given by

Where is the mass of a helium nucleus. Since the mass of helium nucleus is very nearly equal to

four times the mass of a hydrogen nucleus we write

Dividing equation (i) by (ii) we get

or

or

2

2 2M

f i

2 2H

f i

2 2Li

f i

2 2Be

f i

1 /H

H

RR

m M

R HM

eHHe

1 / 4He

H

RR

m M

1 / 4

1 /

HH

He H

m MR

R m M

H HeH H

H He He HH


or

and can be found by measurements of wavelengths in the series of helium and hydrogen spectra.

Hence , the ratio of the mass of electron to the mass of proton (that is , hydrogen nucleus) can be

determine.

Further if (specific charge of hydrogen ion or the charge required to liberate 1 gram-

equivalent of any substance. Known as Faradey constant) is given, we can evaluate the specific charge of

electron .

is 96500 coulomb/gram and is 1/1836

Discovery of Heavy Hydrogen The variation of Rydberg constant with the (finite) mass of the nucleus resulted in the discovery of

deuterium (heavy hydrogen). Deuterium is an isotope hydrogen; its nucleus has a mass almost exactly double that of ordinary hydrogen. Therefore, the Rydberg constant for deuterium is slightly greater than for hydrogen: as shown below

and

Consequently the wavelength of the spectral lines of deuterium are slightly shorter than those of the corresponding spectral lines of hydrogen. The line of deuterium, for example has a wavelength of

while that of hydrogen is a small difference but sufficient for the identification of deuterium. The general nature of the two spectra is however exactly same because atoms of both hydrogen and deuterium have the same electron structure.

Wilson-Sommerfeld Quantisation Rules Planck introduce his quantum theory by quantizing the energy E of an atomic

oscillator. According to Planck an atomic oscillator can have only discrete energies which are multiples of , where is the frequency of the oscillator. Bohr on the other hand, quantised the a

angular momentum of the electron moving in a circular orbit Wilson and Sommerfeld in 1916, enunciated a set of rules for the quantisation of any physical system for which the coordinates are

- the Planck and the Bohr quantisation as special cases. These rules can be stated as follows.

For any physical system in which the coordinates are periodic functions of time, there exists a quantum condition for each coordinate. These quantum conditions are

Where is one of the coordinates, is the momentum associated with that coordinate and is a

quantum number taking integral values.

4

He H

HeHH

R Rm

RMR

H

e

m

H

8 11

11109737

1096781

1 11836

H

H

R cmR cm

m

M

11109737

1097071

1 12 1836

D

D

R cmR cm

m

M

6561A

iq in


The symbol . The

Let us apply Wilson-Sommerfeld quantum conditions to a one-dimensional simple harmonic oscillator. Its total (potential + kinetic) energy in terms of position x and linear momentum is

or

The relation between and is the equation of an ellipse with semi-axes a and b given

and Any instantaneous state of motion of the oscillator is represented by some point on the ellipse, plotted on a two dimensional space having coordinates

and

travels once around the ellipse so that the value of the integral is just equal to the area of the ellipse .

That is

If be the frequency of oscillation, then . Thus

According to the Wilson-Sommerfeld quantisation rule, expressed by equation (i) is an integer

Comparing the last two expressions,

or

ellipses in phase space, the area enclosed between successive ellipses always being h, as shown in fig

iq

in

xp

221 1

2 2x

E U K

pkx

m

x xp

x xp

1' /

2k m

'

Enh


Derivatio angular momentum form the Wilson-Sommerfeld rule. The Positions of an electron moving in a circular orbit can be specified in terms of plane polar coordinates which are periodic functions of time. The

momentum associated with the radial coordinate is , while that associated with the angular

coordinate is which is constant. For a circular orbit, is constant so that is zero.

Hence we apply the Wilson-Sommerfeld rule to the angular coordinate only. The rule becomes in this case

or or

or

The angular momentum of an electron is usually designated by , so that we may write The last expression as

which is Bohr quantisation law

De- According to de Broglie conception of matter waves, an electron of mass m moving with velocity

has associated with it a wavelength given by

-character of electron can account for the limited number of permissible orbits in an atom, and also enables us to deduce the quantised angular momentum of an electron in an atom.

We know that according to Bohr, an electron in an atom revolves in circular non-radiating orbits around the nucleus. Let us fit this idea with the wave-nature of electron. Since the electro does not radiate energy while moving in its orbit, the wave associated with it must be a

around an orbit only when the circumference of the orbit can contain exactly an integral number of de- Broglie wavelengths, as show in fig, for the first three Bohr orbits. This means that only those orbits are permitted for which the radius r is given by

If this were not so, then the waves in each travel around the orbit will not be in phase, and in a large number of travels the waves would interfere with each other in such a way that their average intensity would be zero. This would mean that an electron cannot be found in such an orbit. Hence, the last expression is necessary condition for the electron to exist in an orbit.

Substituting in the last expression we get

or

r r

drp m

dt

2 ,d

p I mrdt

r rp

L

h

m

2nh

rm

2

nhr

m


Hence the magnitude of the angular momentum of the electron in its orbit must be

According to the classical theory, an oscillating charged system would emit radiation of a

Particular frequency, namely, the frequency of the oscillator itself. According to quantum theory, however radiation is emitted as a result of the system making a transition from a quantum state to

another quantum state and the frequency of radiation is determined by the difference in energy

between the two states:

Bohr, in 1923 enunciated with established a correspondence between the classical frequency and the quantum frequency. This principle consists of two parts:

(i) The predictions of the quantum theory for the behaviour of any physical system must correspond to the predictions of the classical theory in the limit in which the quantum numbers specifying the state of the system become very large

(ii) Any selection rules which are necessary to obtain the required correspondence in the limit of large quantum numbers also hold for small quantum numbers. Let us apply this principle to the hydrogen atom. Let be the velocity of electron (mass m, charge e) revolving in a Bohr orbit of radius r of the hydrogen atom. The condition of mechanical stability of the electron is

The quantum condition is

,

These two equations give

and

The classical frequency of revolution of the electron in the orbit is

Now, the Rydberg constant for infinitely heavy nucleus.

2

nhL m r

in

fn

2 2

20

1

4

m e

r r

2

nhm r

2

0

20

2 2 20

/ 21

2 /

e nh

n h me

4 4

2 3 3 2 3 30 0

4

2 30

3

2R cf

n


This is the frequency which must be radiated by the moving electron classically. Now, the frequency radiated on the basis of quantum theory, when the electron drops from an initial

Orbit is given by

Again introducing , we have

This may be written in the following form

Let us consider the transition and write and . Then the radiated

quantum frequency is

When n is very large we can write

Under this condition, the emitted quantum frequency is

Equations (i) and (ii) yield

If we consider transitions we shall have

Thus for very large quantum numbers n, the quantum theory frequency of the radiation is identical with the classical frequency of the revolution (or its harmonics) of the electron in the orbits. This is in

(1) When an electron de-excites to lower states a photon is emitted. When an electron absorbs a photon it is excited to higher energy levesl.

(2) Radius of nth Bohr orbit,

where,

Principal quantum number Z = Atomic number

in

4

2 3 2 20

i f

f i

R

2 2f i

fn n

22

2 1'

1

nR c

n n

22

2 1

1

n

n n2 2

3

2'

R cv

n

2 2

2 24n

n hr

kZe m 0

20

n

n


(3) Velocity of electron in nth orbit

where, c is velocity of light

(4) Angular frequency of electron in nth orbit

where

(5) Frequency of electron in nth orbit

(6) Period of revolution of electron in nth orbit

(7) Electric current due to electron motion in nth orbit for a charge moving in a circle with frequency v,

(8) A circulating charge constitutes a current which is a source of magnetic field. We assume nucleus to be

carrying coil is given by

Hence magnetic induction product at nucleus due to electron motion in nth orbit,

22 2 2 40

3 3 3n

3 2 4

0 3

0

16 2

3

4.459 10rad/secn

Z

n

22 2 2 4 15 20

3 3 3 3n

n

2 2 4

0 3

4where,

k e mv

h

( )BT

3 3

2 4 2 2

2

4

nn

n

rT

V

n hT

me k Z3

0 2n

nT T

Z

3

0 2 4 2

16 3

2

1.5 10secn

nT

Z

2 2 2 5

3 3

2

3

2

4

nn n

n

n

eI ev

k Z e mI

n h

Z mI

n

0

2

iB

R

4 3 3 7 2

0 5 5

3 2

5

8n

n

k Z e mB

n h

Z mB

n


(9) Magnetic moment produced due to electron motion in nth orbit.

Substitute for and

(10) Potential energy of electron in nth orbit

(11) Kinetic energy of electron in nth orbit

(12) Total energy of electron in nth orbit

(R = Rydberg constant)

(13) For an electronic transition from level to level

(14) Ionisation energy: Total energy zero of a hydrogen atom corresponds to infinite separation between electron and nucleus. Total positive energy implies that atom is ionised and electron is in unbound (isolated) state moving with certain kinetic energy. The minimum energy needed to ionise an atom is called ionisation energy

The potential difference through which an electron should be accelerated to acquire this energy is called ionisation potential Ionisation potential,

(15) Binding energy: Energy liberated when constituents of a system are brought from infinity to assemble the system. The binding energy is negative of ionisation energy

(16) Excitation energy: The energy required to excite an electron from low energy level to a high energy level. The potential required to accelerate an electron so that it acquires excitation energy is termed excitation potential

nI

22

2

27.2n

n

kZeU Z eV

r n

2 2

2

13.6

2n

n

kZe ZKE eV

r n

2 2 2 4 2

2 2

2

2

n n n

n

E U KE

kZe k me Z

r n h2

2

RchZ

n2

2

13.6ZeV

n

2thn 1

thn

2

2 21 2

2

2

13.6

ionisation n nE E E E

ZeV

n

2

2

13.6nionisation

E ZV volt

e n

excitation higher lower

excitationexcitation

E E E

EV

e


(17) Number of spectral lines obtained due to transition of electron from nth orbit to lower orbits is

Problem 1. -like lithium atom. Solution: -like atom is given by,

Where Z = atomic number of the atom.

The ionisation energy of this atom is equal in magnitude to energy of ground state

or

or Ans

Problem 2. Find the ratio of minimum to maximum wavelength of radiation emitted by electron in

Solution: Energy of radiation corresponding to transition between two energy levels and is given

by E . E is minimum when and

hence,

E is maximum when and (when the atom is ionised)

Hence

So,

or

or

Problem 3. Solution: Energy of electron in nth Bohr orbit of hydrogen atom is given by

Hence,

or or

The angular momentum of an electron in nth orbit is given as . Putting we obtain

Ans

2

2

13.6ZE eV

n

2

2

ion HH

ion Li Li

2ion H

ion Li

2 21 2

min

2

max

min

max

3

4

E

E

max

min

/ 3

/ 4

hc

hc

min

max

3

4

2

13.63.4

n

2

nhL 2,n


Problem 4. A proton strikes another proton at rest. Assume impact-parameter to be zero, i.e., head-on collision. How close will the incident proton go to other proton?

Solution: The protons approach each other till their relative velocity becomes equal to zero. In this process the projectile proton is retarded and the target proton accelerated due to mutual force of repulsion. At the closest distance of approach both the protons will be moving with same velocity.

Since Columbian repulsive force is internal for system of protons we can apply law of conservation of momentum.

We get,

Change in KE

This change in energy is equal to electrical potential energy

Hence, Ans

Problem 5. Consider ionisation energy of a hydrogen atom as . Which of the followings: electrons,

hydrogen ion and helium ion, is more effective in ionising the hydrogen atom? Solution: We assume that a perfectly inelastic collision occurs between incident particle and hydrogen atom. From conversion of momentum we can write

Initial kinetic energy

Final kinetic energy

The decrease in kinetic energy,

This decrease in kinetic energy must be equal to ionisation energy

Which shows that the greater the mass of the incident particle the smaller the fraction of the initial kinetic energy that can be used for ionisation. When an electron is used, , i.e., , hence the initial

kinetic energy of the electron is almost, completely used for ionisation. When a hydrogen atom is used, the initial kinetic energy must be double that of the electron. For a helium atom

i.e., the energy must be five time that of the electron Ans

2

2 00

1 12

2 2 2m m

22 20 0

0

2

20 0

er

m

iW

20 0

1

2W m

21

2HW m m u

2 20

H

2 22 0

0 0

0

1

2 2 H

H

H

mW W m

m m

mW

m m

iW 0W

00 1 0

H

H


Note: In gas-discharge plasma, ionisation is carried out almost by electrons, while ionisation by the ions plays insignificant role

Problem 6. Consider energy level diagram of a hydrogen atom. How will the kinetic energy and potential energy of electron vary if the electron moves from a lower level to a higher level? Solution: Kinetic energy of an electron in a hydrogen atom is given by,

Remember that KE is always

Positive, hence as electron is excited to higher states its KE decreases

Potential energy is given by

Remember that potential energy is always negative for a particle in bound state. Secondly, magnitude of potential energy twice that of the kinetic energy. As n increases (i.e., electron moves to higher level) potential energy becomes less negative, it grows although magnitude of potential energy decreases. Identical energy tends to zero for , which is maximum value of potential energy. It occurs at infinite separation between electron and ionised atom.

Problem 7. Determine the wavelength of the first Lyman line, the transition from to . In what region of the electromagnetic spectrum does this line lie?

Solution: In this case, . Since , we have

122 nm which lies in the UV region

Problem 8. Determine the maximum wavelength hydrogen in its ground state can absorb what would be next smaller wavelength that would work? Solution: Maximum corresponds to minimum energy thus the jump is from the ground state to the first excited for which the energy is 13.6-3.4 = 10.2eV, the required wavelength is 122 nm. The next possibility is to jump from ground state to the second excited state, which requires 0.6-1.5= 12.1eV and corresponds to a wavelength

4 2

2 2 20

2

2

8

me ZKE

n h

KZ

n

4

2 20

2

1as KE

n

4 2

2 2 204

me ZPE

n h

2 4

2 2 20

2 1

34 8

7

18

3 1

c hc

v E E


Ans

Problem 8. Estimate the average kinetic energy hydrogen atoms (or molecules) at room temperature and use result to explain why nearly all H atoms are in the ground state at room temperature and hence emit no light Solution: According to kinetic theory the average kinetic energy of atoms or molecules in a gas is given by

where T is temperature in Kelvin room temperature is about so

or, in electron volt

Ans

The average kinetic energy is thus very small compared to the energy between the ground state the next higher energy state (13.6-3.4 = 10.2eV) any atoms in excited states emit light and eventually fall to the ground state. Once in the ground state, once in the ground state, collisions with other atoms can transfer energy of 0.04eV on the average. A small fraction of atoms can have much more energy (in accordance with the distribution of molecular speeds), but even kinetic energy that is 10 times the average is not nearly enough to excite atoms above the ground state. Thus, at room temperature, nearly all atoms are in the ground state. Atoms can be excited to upper states at very high temperatures or by passing current of high energy electrons through the gas, as in a discharge tube.

Problem 9. Use the Bohr model to determine the ionisation energy of the He+ ion, which has a single electron. Also calculate the minimum wavelength a photon must have a cause ionisation Solution: We want to determine the minimum energy required to lift the electron from its ground state and to

barely reach the free state at . The ground state energy of is given by with and

. Since all the symbols are the same as for the calculation for hydrogen except that Z is 2 instead of 1, we

see that will be times the for hydrogen i.e.,

Thus to ionise the He+ ion should require 54.4 eV and this value agrees with experiment. The minimum wavelength photon that can cause ionisation will have energy and wavelength

Ans.

If the atom has absorbed a photon of greater energy (wavelength shorter than 22.8 nm) the atom could still be ionised and the freed electron would have kinetic energy of its own.

Problem 10. Electrons of energy 12.09 eV can excite hydrogen atoms to which orbit is the electron in the hydrogen atom raised and what are the wavelengths of the radiations emitted as it drops back to the ground state?Solution: The energies of the electron in different states are

for

for

3,

2K kT

23 21

21

19

2

2

13.6Z eV

n

2Z

E

c hc

v hv34 8

19

6.63 10 3 1022.8

54.4 1.6 10nm


and for

Evidently the energy needed by an electron to go to the level ( or M-level) is 13.6-1.51

= 12.09 eV. Thus the electron is raised to the third orbit of principal quantum number Now an electron in the level can return to the ground state by making the following

possible jumps (i) to and then from to (ii) to Thus the corresponding wavelengths emitted are

(a) For to

or

Ans

This wavelength belongs to the Balmer series and lies in the visible region (b) For to

or Ans

belongs to the Lyman series and lies in the ultraviolet region.

(c) For the direct jump to

or Ans

which also belongs to the Lyman series and lies in the ultraviolet region.

Problem 11. A doubly ionised lithium atom is hydrogen like with atomic number Z = 3. Find the wavelength of

radiation required to excite the electron in from the first to the third Bohr orbit given the ionisation energy of hydrogen atom as 13.6 eV. Solution: The energy of nth orbit of a hydrogen-like atom is given as,

Thus for atom, as Z = 3, the electron energies for the first and third Bohr orbits are

For

For

Thus the energy required to transfer an electron from level to level is,

Therefore, the radiation needed to cause this transition should have photons of this energy.

3E

2 21

1

36

5R

7

366563A

5 1.097 10

2 22

2 7

2 23

3 7

2Li

2Li

1,n

2

1 2

13.6 3122.4

1E eV

3,n

3E


The wavelength of this radiation is,

or

Ans

Problem 12. A particle of charge equal to that of an electron, -e and mass 208 times the mass of electron (called -meson) moves in a circular orbit around a nucleus of charge +3e. (Take the mass of the nucleus to be

infinite). Assuming that Bohr model of the atom is applicable to this system: (i) derive an expression for the radius of the nth Bohr orbit (ii) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr

orbit for the hydrogen atom. (iii) find the wavelength of the radiation emitted when the -meson jumps from the third orbit to the

) Solution: (i) Proceeding as in , we have the radius of the nth orbit given by equation (11)

Substituting and we get

Further writing for , we obtain

(ii) The radius of first Bohr orbit for hydrogen is

or

For ( -mesonic atom) = (hydrogen atom), we have

or or Ans

(iii) Proceeding again as in , the energy for the nth orbit is given by equation (15)

Substituting and ,we get

34 8

19

2 2 2 20 0

22n

ee

2

h

2 20

2624n

e

n hr

m e

2

1 2e

20

1 2e

nr 1r2 2 2

0 02 2624 e e

n h h

m e m e

n

2

h


where is Rydberg constant

when the meson jumps from the third orbit to the first orbit, the difference in energy is

radiated as a photon of frequency v given by

As we have

or

or

Ans.

Problem 13. A pi-meson hydrogen atom in a bound state of negatively charged pion (denoted by

) and a proton. Estimate the number of revolutions a -meson makes (averagely) in the ground

state of the atom before it decay (mean life of a -meson ) Solution: nth orbit is given as

Radius for of -meson atom (ground state) will e

or

velocity in nth orbit

for

Time of one revolution

Probable number of revolutions before -meson decay,

4

2 2 20

4

2 3 20

2

234

18728

1872

en

e

m eE

n h

m e hc

h c n

Rhc

n4

2 30

e

3 1

hcE E

10

7

8

2 2

2 2,

4n

e

h nr

m ke 0


Ans

not spectral lines into several close, distinct components when observed under equipment of high resolution.

ogen atom actually consists of several components which are close together in energy

the electron could move in elliptic orbits also, in addition to B and shape of the allowed elliptic orbits, as well as the total energy of an electron moving in such an orbit.

Can be described in terms of the plane polar coordinates and , which being periodic functions of time, must be quantised separately.

If and be the radial and the angular momenta of the electron then according to Wilson-

Sommerfeld quantisation rules, we have

and

where

The second integral is easily evaluated because according to classical mechanics, the angular momentum

of any isolated system is constant thus

or

This is restriction on the orbital angular momentum is the same as in Bohr circular orbit theory To evaluate the first integral, let us write

and

where and are the radial and the angular components of the velocity of the electron

Now, the polar equation of the ellipse is

810

19

102 10

5.6 10

Meanlife

Time period

r

rp p

rn

p

r r

2


Where (ratio of semi-minor to semi-major axis)

Taking logarithmic differentiation of r with respect to we get

or [from eq. (vi)]

or

Now, using equation (v) we have

Also

Using equation (vii) we have

Hence the integral in equation (i) becomes

The value of this definite integral is

But , by equation (iii)

or

or

or

but , a property of ellipse.

21b

a

2 2

1 1 sin

1

dr

d ar

1 sin

1 cos

dr

r d2 2

2

r

dr dr drp mr m m m

dt d d

2r

drdr d

d2

1r

drp dr p d

r d

2 2

2r

2 2 2

2

0

r

2

12 1

1

2r

2

khp

2r

2

r

2

r r

21b

a


Since and are integers, we may put

through the nucleus, Sommerfeld postulated that the azimuthal quantum number that is

The radial quantum number however takes the zero value for which the orbit is circular that is

circular. That is

Therefore n can take on the values

For a given value of n, the k can assume only the values

The equation (ix) now becomes

This is the quantum condition for the elliptic orbits, out of all the classical possible ellipses, the electron can move only in those orbits for which the ratio of the minor to the major axis is the ratio of two integers. This quantises the size and the shape of the allowed elliptic orbits Total Energy of Electron in allowed Orbit: Let us now calculate the total energy E of an electron in a quantised elliptic orbit. The energy is the sum of the kinetic energy K and the potential energy U. that is

Using equation (iv) and (v) we have

Again using equation (viii) we have

or xi)

Now, from equation (vi) we have

or

or

rn

r

rn

b k

a n

22 2 2

0

1 1

2 4

Zem r r

r

2 22

20

r

2

22

2 2

22 2 2

2

2


or

Substituting this in equation (vii) we have

This is identical with equation hence we can compare the coefficients of and in these two equations. Then we have

and

but . Therefore the last two equation become

xii)

and

dividing equation (xii) by the square of equation (xiii) we have

or

but by equation (iii) and by equation (x)

electron still depends only on the principal quantum number n, and is independent of the azimuthal quantum number k.

Size and Shape of Sommerf From equation (xiii) we have

Applying , and so that , we have

22 2 22 2 2

2

2 2 22 2

1 2 1sin1

1 cos 1

a ar

rra

2

2 2 2

r

2 2 2

2

2 20

2

2 20

2

2

mZe a

p b

2 4 2 2

2 2 20

2

khp

b k

a n

2 4 4

2 2 2 20

1

8

mZ e eE

h h n

2 2

204

mZe ba

p

2

k hp

b k

a n

akb

n2 2

2 20

mZe aa

n h


or

Now, (say), the radius of the smallest Bohr orbit of hydrogen atom hence, we may write

Further using , we have

Now the theory predicts that for a given value of can take n possible different values

. This means that for a given n, there are n there are n

Orbits of different eccentricities (according to the conditions , which can be occupied by the electron.

Let us consider the first few cases of hydrogen atom (Z=1) For ; we have

So that from equation (xvi) and equation (xvii) we have

and

this is a circular orbit of radius

for we have (i) so that

(i) so that

Thus, for , and an elliptic orbit with semi-major axis

and semi-minor axis

For , there are three types of possible orbits (i) so that

(ii) so that

(iii) so that ,

Thus out of the three possible orbits of radius , the second is an elliptic orbit of semi-major axis and

semi-minor axis and the third is an elliptic orbit with semi-major axis and semi-minor axis and

semi-minor axis and semi-minor axis

akb

n

0

nkb a

Z

2,n

0

09a 09a

06a 09a 09a

09a 03a


We see that corresponding to each value of the principal quantum number n, there are n different allowed orbits. One of these is circular, just the orbit described by the original Bohr theory. The others are elliptic, all having the same semi-major axis as the radius of the circular orbit, but different semi-minor axes. The orbit having the lowest k is most elliptic.

But despite the very different paths followed by the electron moving in different possible orbits for a given n, the total energy of the elecorbits adds no new energy levels, and hence fails to explain the fine structure. The several orbits

To designate an orbit of given n and k values the value of n is followed by one of the letters

respectively .For example the state is written as 3s; and that

as 3p. It is seen that the s-electron orbit (lowest k) is most elliptical in any family of orbits having

same major axis (same n)

S Sommerfeld removed the degeneracy in the total energy of the electron moving in different k-orbits for a

given n by introducing the relativistic variation of mass of the electron. For an electron in the innermost

orbit f the hydrogen atom, , or less. Although this would give a relativistic correction to the total

energy only of the order of , but it is just the order of the splitting of energy levels required to explain the observed find structure of hydrogen spectral lines

In an elliptic orbit, the velocity fo electron varies and is largest near the perihelion. The actual size of the relativistic correction depends upon the average velocity of the electron which, in turn depends on the ellipticity of the orbit. This means that the correction is different for different k-orbits of a given and thus the degeneracy is removed

Using the relativistic expression for the kinetic energy of the electron

where is the rest mass of the electron Somerfield obtained the following equation

for the path of the electron:

where is a constant. this shows that r does not return to a given value when

increases by but only when it increases by which is greater than . Thus

the orbit of the electron is not a uniform rotating ellipse, but the perihelion of the orbit precesses in the same direction as the revolution of the electron. The advance of the

210410

,n

0m

2 2


perihelion per revolution is , as shown in fig

Taking the effect of relativistic precession into account, Sommerfeld calculated the total energy of an electron in an orbit characterised by the quantum number n and k as

where the quantity is a pure number - . (

is equal to the ratio of the velocity of electron in the first Bohr orbit of hydrogen to the velocity of light c in vacuum)

The last expression may also be written as

where is the Rydberg constant for an infinitely heavy nucleus. The term values of the

hydrogen-like atom are therefore

or

the first term of this equation is the same as that derived by Bohr for circular orbits and gives the major part of the energy. The second term is the relativity correction such that

Now,

which is positive since . Since depends on k, it is different for different k-orbits of the same n.

2 4 202 2 20

2

0

2 2 2

2

402 308

m eR

h c

2 2 2

2

2 4 2

2 3

T4 2

3

22 7 1 1

1.097 10137

R m

41

3

T


SOLVED PROBLMES

1. Calculate the speed of electron in the nth orbit of hydrogen atom. If relativistic correction is important for > 0.005, find for which orbit this correction is necessary. Given:

and

Soln: Let be the velocity of electron (mass m, charge e) revolving in a Bohr orbit of hydrogen atom of

radius r. The condition of mechanical stability of the electron is

The quantum condition is

(ii)

Dividing equation (i) by equation (ii), we get

Substituting the given values:

This is the required value. Now

Thus, for we have

Hence the relativistic correction is necessary for orbit only ,

2. How many revolutions does an electron in the state of a hydrogen atom make before dropping to

the state? The average life time of an excited state is second

Soln: Let be the velocity of electron (mass m, charge e) revolving in a Bohr orbit of hydrogen atom of radius r. The condition of mechanical stability of the electron is

The quantum conditions is

,

These two equation give

and

The number of revolutions of the electron in the orbit per second is

For the state, the frequency of revolution is

2 2

20

1

4

m e

r r

2

nhm r

2

0

0.0073,0.0036,0.0024......c

810

2 2

20

1

4

m e

r r

2

nhm r

2

0

4 4

2 2 3 2 3 3 30 0

2 2

2 4 8

me me c cf R

r n h h c n n


Substituting the given value of and the known value of c, we get

Hence the number of revolutions of the electron in its life-time of second is

3. The series limit wavelength of the Balmer series in hydrogen spectrum is . Calculate Rydberg constant for hydrogen atom.

Soln: The wavelength of the lines of Balmer series are given by

For the series limit wavelength, . Thus

or

4. Calculate the wavelength of the eight lines of the Balmer series of hydrogen atom. Given:

Soln: The wavelength of the lines of Balmer series are given by

For the 8th line,

or

Now, the Rydberg constant for hydrogen is

where is Rydberg constant for infinitely heavy nucleus, m is the mass of electron, and is the

mass of hydrogen nucleus (that is proton). We know that is 1/1836

Hence, from equation (i) have

4

R cf

7 1 8 1

4 1

1.097 10 3.0 10

4

8.2 10

m msf

s810

n

H

H

H

R HM

7 17 1

H

7

7 1


5. A beam of electrons bombards a sample of hydrogen. Through what minimum potential difference must the electrons have been accelerated if the first line of the Balmer series is to be emitted? Explain how many possible spectral lines can be expected if the atom finally attains the normal state. Given:

and .

Soln: Normally the hydrogen atoms are in the ground state . The first Balmer line is emitted when the

atom returns from to state. Hence in order to emit this line the atom must be first raised to the state by electron bombardment. Therefore the energy of the bombarding electrons must be equal to the difference of

The energy of hydrogen atom in the nth state is

Substituting the given values

Hence the bombarding electrons must be accelerated by 12.1 volt. If the atom finally attains the ground state, the possible number of spectral lines is

6. Show that the ionisation potential of is four times the value for hydrogen atom. Soln: The ionisation potential of an atom corresponds to the energy of the atom in the ground state. For one-electron atom, the ground-state energy is (ignoring effect of finite nuclear mass)

or

for and H, we have and 1 respectively

Hence the ionisation potential of is 4 times that of H

7. A positronium atom is a system consisting of a positron and an electron revolving about their common centre of mass, which lies half way between them compare its ionisation energy and emission spectrum with those of the hydrogen atom (with infinitely heavy nucleus)

Soln: The positron has the same mass m as the electron and has equal but positive charge. The reduced mass of the electron-positron atom is therefore

While the reduced mass of electron in hydrogen is very nearly m.

The Rydberg constant for positronium, is therefore half that for hydrogen (with

infinitely heavy nucleus). Thus

2H

n

R hcE

n

1 3

8

9HE E R hc

He

21

He 2Z

He

H

He

4

2 308

e

h cPR

1

2PR R


The energy states of the positronium atom would then be given by

The corresponding expression for hydrogen atom (with infinitely heavy nucleus) is

Now, for positronium atom as well as for hydrogen atom. Hence we conclude from equation (i) and (ii) that the ionisation energy, which is numerically equal to the energy of the lowest state of

positronium atom is half of that of hydrogen atom.

The wave lengths of the emitted spectral lines of positronium atom and hydrogen atom are given by

and

Thus

The wavelengths of positronium lines are double than those of hydrogen atom

Bohr-Sommerfeld Theory of Hydrogen Atom

1. The wavelength of the first line of Balmer series of hydrogen is . Calculate (i) the ionisation potential and (ii) the first excitation potential of the hydrogen atom.

Soln: (i) The wavelengths of the Balmer

For the first line and 106562.8A=6562.8 10 m

or 7 1

10H

The ionisation potential of an atom is numerically equal to the (ionisation) energy required to remove an electron completely from the atom in the ground state. In hydrogen atom, the electron stays in the first orbit . Hence the energy required to remove this electron to infinity (where the energy is

considered to be zero) is numerically equal to the energy of the electron in the first orbit. We know that electron energy in the nth orbit of hydrogen atom is given by

4

2 2 20

n

2

Ignoring the effect of finite nuclear mass, we can take

2

Hn

R hcE

n

For the ground orbit

2 2PP f i


1 H

7 1 34 8 1

19

19

19

1.0.97 10 6.63 10 3.0 10

21.8 10

21.8 1013.6

1.60 10 /

m Js ms

J

JeV

J eV

Hence the energy required to remove the electron from orbit to infinity is 13.6eV

Obviously, the ionisation potential of the hydrogen atom is 13.6 volt (ii) The first excitation potential of the atom is the energy required to shift the electron from to

orbit, that is now,

1 13.6E eV

and 12

2n

1 2 13.6 3.4 10.2E E eV

2. Calculate the ionisation potential of hydrogen atom. Given:

and

Soln: The ionisation potential refers to the binding energy of the hydrogen atom (the energy binding the electron to the nucleus). Which is numerically equal to the energy of the lowest state corresponding to

, this is with Z = 1

4

1 2 20

31 19 4

2 212 2 2 34

18

18

19

(9.11 10 )(1.60 10 )

8 8.85 10 / 6.63 10

2.17 10

2.17 1013.6

1.60 10 /

kg C

C N m Js

J

JeV

J eV

The ionisation potential is 13.6 V

3. Find the energy, momentum, and wavelength of a photon emitted by a hydrogen atom making a direct transition from an excited state with to the ground state. Also find the recoil speed of the hydrogen

atom in this process. Given: and

.

Soln: The energy of electron in hydrogen atom (Z =1) in a state n is given by

2

hn

The energy of the photon emitted during an electron transition from to is


10 1 2 2

7 1 34 8 1

19

19

19

1 1

1 10

991.097 10 6.63 10 3.0 10

100

21.6 10

21.6 1013.5

1.60 10 /

HE E R hc

m Js ms

J

JeV

J eV

The momentum of the photon is

to find the wavelength of the photon, we see that the atom drops to the ground state . Hence the

photon emitted belongs to the Lyman series. The wavelengths of the spectral lines in this series are given by

For we have

or 8

7 1H

finally, after the emission of photon, the hydrogen atom recoils with a momentum of

(same as that of the photon). Hence the recoil speed of the hydrogen atom of mass is

4. A photon ionises a hydrogen atom from the ground state. The liberated electron recombines with a

proton into the first excited state, emitting a photon. Find (i) the energy of the free electron and (ii) the energy of the original photon. Given:

.

Soln: (i) The energy of the electron in hydrogen atom (Z =1) in the nth state is given by

2

7 1 34 8 1

2

19

2

19

19 2 2

(1.097 10 )(6.63 10 )(3.0 10 )

21.8 10

21.8 10 1 13.6

1.60 10 /

Hn

R hcE

n

m Js ms

n

J

n

J eV

J eV n n

The energy of electron in the first excited state is

Further, the energy of the emitted photon is


34 8 1

10

19

Hence the total energy of the liberated electron is (ii) The energy required to ionise hydrogen atom, that is to remove the electron from the atom in the ground state to infinity (where the energy is zero) is numerically equal to the energy of the

electron in the state, that is which is

1 2

The required energy of ionisation is then13.6eV . Hence the energy of the original photon is the energy

of ionisation plus the energy of the liberated electron that is

5. A meson (charge , mass where m is mass of electron) can be captured by a proton to form

a hydrogen-energy, and the wavelength of the first line in the Lyman series of such an atom. The mass of the proton is 1836 times the mass of electron. The radius of first Bohr orbit and the binding energy of hydrogen are

and 13.6 eV respectively. .

Soln: The reduced mass of the system is

From Bohr Theory, the radius of the first orbit of a hydrogen like atom for is given by

(taking finite mass of nucleus in consideration)

2 2 2

0 0 01 2 2 2

The quantity in the bracket is the first Bohr orbit of hydrogen atom which is . Therefore

The binding energy of an atom is equal to the energy of the lowest state

theory, the ground state energy of the hydrogen-like mesic atom with is given by

Hence the binding energy is 2530 eV. The wavelengths of the Lyman lines are given by

2 2

1 1 1

1MR

n

where is the Rydberg constant for the mesic atom. For the first line, , so that

4

3 MR

Now, 4

2 30

M and 4

2 30


M

Hence 4 4

3 3 186MR R

Now, (given)

Lyman series is in the X-ray region of the muonic atom spectrum.

6. Calculate the radius of the first Bohr orbit in ( 82)Z for a atom. The masses of meson

and proton are 207 times and 1836 times respectively the mass of electron given:

and .

Soln: The reduced mass of -mesic atom (formed by a meson and a proton) is

where m is mass of electron ( 1)n of a hydrogen-like mesic atom taking

finite mass of nucleus in consideration is 2

01 2

Substituting the given values:

7. A mixture of ordinary hydrogen and tritium (a hydrogen isotope whose nucleus is approximately three times more massive than ordinary hydrogen) is excited and its spectrum studied. Calculate the shift in

wavelength for the H lines of the two kinds of hydrogen. (Given: )

Soln: The wavelength of the H lines ( 3 2)n n of hydrogen 11 and tritium are given by

and

where and TR are Rydberg constants for hydrogen and tritium respectively for both).

Taking reciprocals and subtracting equation (ii) from (i) we get

H TH T

H T H

But 3T HM M (given)

Now, and 1

1836H

m

M(we know)


8. Radiation from a helium ion is nearly equal in wavelength to the H line, first line of Balmer series

(a) Between what values of n does the transition in occur? (b) is the wavelength larger or smaller

than that of H line? (c) Calculate the wavelength difference. Given: and

Soln: (a) The wavelengths of the Balmer series hydrogen (H), and those of the Pickering series of helium ion

are given by

(i)

and ; (ii)

equation (ii) can be written as

2 2He

He

(iii)

A comparison of equation (i) and (iii) shows that alternate lines of Pickering series ( 6,8,10.....)n nearly

coincide with lines of Balmer series ( 3,4,5...)n ; nearly because He H (not exactly). Thus the

transition in corresponding to transition for line ( 3 2)n n in H occurs for the following

values

(b) Since He H , we conclude that He H

(c) The wavelength of the first line of H, and that of are given by

and

6 6

1 1

9 9 8

36 1 1 36 1 1 369.1177 10 cm 9.1139 10 cm

5 5 5109677cm 109722cm

363.8 10 cm 27.36 10 2.736 10 2.736A

5

H HeH HeR R

cm cm

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