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PHYSICS PHY523: ATOMIC,MOLECULAR & LASER SPECTROSCOPY

PG SEMESTER:- II, PAPER: - VII

M.Sc Physics:Second Semester

ATOMIC,MOLECULAR & LASER SPECTROSCOPY

Paper: VII Paper Code: PHY 523

UNIT-I

Spectroscopic terms and their notations Quantum states Atomic orbital Parity of the wave function Angular and radial distribution functions Spin orbit interaction Quantum mechanical relativity correction Lamb shift Zeeman effect Normal and anomalous Zeeman effect Paschen-Back effect Stark effect.

UNIT-II

Weak fields and strong field effects

Quantum mechanical treatment of stark effect

Hyperfine structure of spectral lines

Nuclear spin and hyperfine splitting

Intensity ratio and determination of nuclear spin

Breadth of spectral lines

Natural breadth

Doppler effect.

UNIT-III

Independent particle model

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He atom as an example of central field approximation Central field approximation for many electron atom Slater determiant L-S and j-j coupling Equivalent and non equivalent electrons Energy levels and spectra Spectroscopic terms: Hunds rule Lande interval rule Alkali spectra.

UNIT-IV

Rotational spectra of a diatomic molecule as a rigid rotator Vibrational energy of a diatomic molecule Diatomic molecule as a simple harmonic oscillator Energy levels and spectrums Morse potential energy curve PQR branches.

UNIT-V

Basic elements of a laser Threshold condition Population inversion Pumping mechanism – optical pumping Rate Equations for three level and four level Laser-system Ruby Laser Nd – YAG Laser Semiconductor laser carbon-dioxide laser Optical fibers – light wave communication Einstein’s A and B coefficients.

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UNIT - I

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Dr. Sachida Nand Singh

Associate Professor

Department of Physics

C.M.Sc. college, darbhanga

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(Q.) What is Zeeman effect? Deduce the classical theory of Zeeman shift (Ans.) Zeeman effect of is a magneto optical phenomenon. A source of light such as sodium vaporlamp is placed between the two pole pieces of an electromagnet. The emitted light can be analyzed along the parallel direction of applied field For these observation there are holes made in the pole – places. the light can also be analyzed along perpendicular direction to the applied field. To analyze the emitted radiation, we use high resolving power instrument such as L.G. plate and also we use Nickel prism arrangement. When magnetic field is on the parallel direction of B- field, we find that a single line is split into two component line the frequency of one component is greater than original line, and the frequency of another components is smaller than the original line by same amount. The side components are called 𝜎 component . The middle component is called 𝜋 component which is hidden in parallel observation. The other component is circularly polarized in opposite directions.

S

Pole

pipes

View parallel to field

View parallel to field

Pole pipes

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B – field electron

0

Y

When light is observed from a point which is perpendicular to the direction of B field the three component are seen. The three component are plane polarized. The 𝜎 component are polarized perpendicular to the -𝜋 components.

This effect when a single line is split into three components, was first of all observed by Zeeman so it is called Zeeman effect. If the strength of B field is weak; so that other effect can also be taken into account, then a single line split up into a large number of lines this effect is called anomalous Zeeman effect called normal because its treatment is classical. Classical theory of normal Zeeman effect the normal Zeeman effect is :- the orbital motion of electron can be resolve into three simple harmonic motion components directed along X – axis, Y – axis and Z – axis. If we assume that the B – field is applied along vertically upward direction on the plane of paper, then the Lorentz force on the electron Z axis will be zero. The simple harmonic motion along Y axis may assumed as made up of two circular motion of angular velocity 𝜔. Y+ is moving along anti-clock wise and Y- is due to clock wise angular motion. Similarly, the simple harmonic motion along X-axis may be assumed to the made up of two circular motion X+ and with X- angular velocity w. X- is due

Y+

Y_

X - axis

y- axis

Z

X

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to clockwise motion and X+ is due to anti-clockwise motion. The resultant anti-clockwise motion of electron will be X++Y+ and resultant clockwise motion will be X-+Y-. Now the centripetal force

on electron due to coulombs force will be F=−1

4𝜋∈0 𝑍𝑒2

𝑟2 towards the

Centre of circular orbit. If W be angular velocity of electron, then F = mw2r. where r is radius of circular path

Lorentz force = —e( �̅� × �̅� ) = —e( 𝑉𝑗̂ × 𝐵�̂�) = evB𝑖̂ = evB(-𝑖̂) Thus on the motion of anticlockwise the Lorentz force will act towards the Centre of the circle. Hence F + e𝑣𝛽 = m( w + ∆w)2r

Or, F + e𝑣𝛽 = mw2r+ 2mw∆wr (1)

But F = mw2r (2)

Hence, e𝑣𝛽 = 2mw∆wr Or, e𝑣𝛽 = 2mrw∆w Or, e𝑣𝛽 = 2m𝑣∆w

Or, ∆w = 𝑒𝐵

2𝑚 ------------------------------------------------------------------(3)

Therefore, change in frequency ∆𝜐= ∆𝑤

2𝜋

Hence, 2𝜋∆𝜐 = 𝑒𝛽

2𝑚 Or, ∆𝜐 =

𝑒𝛽

4𝜋𝑚 -------------------------------------(4)

Due to B field the frequency of anticlockwise motion will increase by ∆𝜐 and the frequency of clockwise motion will decrease by ∆𝜐. Due to transvers nature of light, the Z- component of motion is hidden, so we see only two components in parallel observations. Explanation of normal Zeeman effect by vector model, effect by vector model to explain normal Zeeman we leave the spin motion of electron. The angular momentum of electron due to its orbital motion is given by Pl= 𝑚𝑣𝑟 due to orbital motion of electron , the

circular electric current I = 𝑒

𝑇 where T is time period so I =

𝑒𝜐

2𝜋𝑟 . This

circular current is equivalent to electric dipole moment 𝜇𝑙 = iA = i𝜋r2. Where r = radius of the orbit so

O W

B

Z

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𝜇𝑙 = 𝑒𝜐𝜋𝑟2

2𝜋𝑟 =

𝑒𝜐𝑟

2 (3)

Hence, 𝜇𝑙

𝑃𝑙 =

𝑒𝜐𝑟

2𝑚𝜐𝑟 =

𝑒𝜐𝑟

2𝑚𝜐𝑟

or 𝜇𝑙 = 𝑒

2𝑚𝑝𝑙 (4)

The direction of 𝜇𝑙 is in opposite direction on of 𝑝𝑙 . Because charge

on the electron is negative . we also known that 𝑝𝑙 = 𝑙ℎ

2𝜋 where 𝑙 =

subsidiary quantum number. When B field is applied, the 𝑙 vector begins to precess around �̅� with angular velocity

𝜔𝑙 = 𝐵 𝜇𝑙

𝑃𝑙=

𝛽𝑒

2𝑚 (5)

The addition energy of electron due to this precession is given by ∆E = 𝜔𝑙𝑃𝑙cos𝜃 where 𝜃 is the angle between 𝑃𝑙 and 𝐵 field.

So, ∆∈ = 𝑒𝐵

2𝑚 ,

ℎ

2𝜋 lcos𝜃 =

𝑒𝛽ℎ

4𝜋𝑚 lcos𝜃

Or, ∆∈ = 𝑒𝛽ℎ

4𝜋𝑚𝑚I (6)

When 𝑚𝑙 = lcos𝜃 . when ml can have (2𝑙 + 1) value ranging from +l to –l . therefore the effect of B field is to split up each energy level into 2𝑙 +1 value Now let us suppose that in the presence of B field, there are two energy levels E1 and E2. in the absence of B field energy levels are E

(1) and E (2).

Hence E1 = E(1) + 𝑒𝛽ℎ

4𝜋𝑚𝑚𝑙1-----------------------------(7)

And E2 = E(2) + 𝑒𝛽ℎ

4𝜋𝑚𝑚𝑙2 ----------------------------------------------(8)

The quantity of energy radiated in the presence of B field will be given by

E2 – E1 = E(2) – E(1) + 𝑒𝐵ℎ

4𝜋𝑚(𝑚𝑙2- 𝑚𝑙1)

Or h𝜐 = h𝜐0 + 𝑒𝐵ℎ

4𝜋𝑚Δ𝑚𝑙 ----------------------------------------------------------(9)

Or, 𝜐 = 𝜐0 + 𝑒𝐵

4𝜋𝑚Δ𝑚𝑙 --------------------------------------------------------------(10)

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𝑚𝑙 = (2𝑙 + 𝑙)

2

𝜐 = 𝜐0 - 𝑒𝐵

4𝜋𝑚 𝜐0 , 𝜐 =𝜐0 +

𝑒𝐵

4𝜋𝑚

𝑒𝐵

4𝜋𝑚

Normal Zeeman effect

Here 𝜐0 = frequency of spectral line in absence of B field , 𝜐 = Frequency of line in presence of B field , Δ𝑚𝑙 = 0,or ±1. Thus there will be three possible lines 𝜐 = 𝜐0 for Δ𝑚l = 0.

𝜐 = 𝜐0 + 𝑒𝐵

4𝜋𝑚 for Δ𝑚l = 1

𝜐 = 𝜐0 - 𝑒𝐵

4𝜋𝑚 for Δ𝑚l = -1

The change of frequency is by 𝑒𝐵

4𝜋𝑚 it is called Lorentz unit.

Anomalous Zeeman effect: -

To explain anomalous Zeeman effect, we introduce the spin motion in addition to its angular motion. Thus the spin angular momentum vector of electron = S* and orbital angular momentum vector of electron =ɩ* therefore, total angular momentum vector of electron j*=ɩ*+S* (1)

l=2

l=1

1

0

-1

-2

1

0

-1

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The magnetic momentum due to orbital motion of electron

𝜇l = 𝑙∗ 𝑒ℎ

4𝜋𝑚 ----------------------------------- (2)

Where 𝜇𝑙 is directed in opposite to the direction of 𝑙∗. The magnetic moment due to spin of electron is given as

𝜇s = 2S*𝑒ℎ

4𝜋𝑚 -----------------------------------------(3)

𝜇s is also directed in opposite direction of s* since l* and S*precess around J*.

So 𝜇I and 𝜇s will also precess around 𝜇j . Here 𝜇j is not in line with the resultant of 𝜇I and 𝜇s.

To find the resultant magnetic

moment of electron ,we resolve 𝜇I

in the opposite direction of J*

and along normal to it. The average

of normal component will

be zero and the component along

opposite to J* will be 𝜇𝑙 cos (l*J*).

similarly the component of 𝜇s along

normal to J* will average out be

zero, and the component along

Opposite to J* will be 𝜇𝑠 𝑐𝑜𝑠 (S*J*)

Hence 𝜇j = 𝜇l cos (l*J*) +𝜇𝑠 𝑐𝑜𝑠 (S*J*)

Or, 𝜇j = 𝑒ℎ

4𝜋𝑚 [𝑙 ∗ cos(𝑙∗𝑗∗) + 2𝑠∗cos (𝑠∗𝑗∗)]

𝜇j

𝜇S

𝜇lS

0

𝜇l

L*

J*

S*

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But 𝜇j = J*g. In the term of bohr magneton

J*g = l*cos(l*J*) + 2S*cos (S*J*)

Or, J*g = L*[J*2+L*2-S*2] + 2S*[J*2+S*2-L*2]

Or, j*g = J*2+L*2-S*2 + J*2+S*2-L*2

Or, g = J*2+L*2-S*2/2J*2 + J*2+S*2-L*2/J*2

Or, g = J*2+S*2- L*2

Or, g = 1+𝐽(𝐽+1)+𝑆(𝑆+1)−𝑙(𝑙+1)

2𝐽(𝐽+1)

g Is called lange g splitting factor now 𝜇j= J*g.

If the atom is placed in weak magnetic field, then J*begins to precess around 𝐵.

Due to this precession the change in energy of electron = ∆𝐸 takes place .

When ∆𝐸 = 𝜇j Bcos( J*B ) = J*g B cos (J*B)

Or, ∆𝐸 = g𝐵𝑀j where 𝑀j = J*cos (J*B)

In terms of joule ∆𝐸 =g B𝑀j 𝑒ℎ

4𝜋𝑚 joule

∆𝜐 = ∆𝑀jgO

Where O = 𝑒𝐵

4𝜋𝑚𝑐 Lorentz unit.

Since ∆𝑀j has 2J+1 values ranging from +J to – J. so a given level splits up into 2J+1 sublevels . when this ∆𝑀j is subjected to selection rule ∆𝑀j = 0, ±1

1+

2J*2

2j* j*

2J*L* 2J*S*

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σ π π σ

We get the transition shown below for sodium D lines.

∆𝑀j =0 gives 𝜋 components . ∆𝑀j = ±1 gives rise to 𝜎 component.

For sodium N11 1S22S22P63S1

3/2

1/2

-1/2

-3/2

1/2

-1/2

2P3/2

2S1/2

D2 line- splitting

D1 line- splitting

2s1/2

2P1/2

+1/2

-1/2

+1/2

-1/2

mj

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Anomalous Zeeman effect.

Polarization rules for Zeeman lines : - The quantum transitions which obey ∆𝑀j = 0, or ±1 are permitted when we view in a direction perpendicular to the field : -

∆𝑀j = ± 1 ∴ plane polarized perpendicular to the magnetic field direction = 𝜎 component ∆𝑀j = 0, are plane polarized parallel to the magnetic field direction = 𝜋 component

When we view parallel to the field: -

∆𝑀j = ± 1 circularly polarized 𝜎 component ∆𝑚j = 0 , forbidden 𝜋 component

Paschen back effect: -

In the explanation of Zeeman effect, we assume that the external magnetic field is weak in comparison to the magnetic field produced due to orbital and spin motion of electron. In this case the precession of L*

and S* vector around J*is much faster than that of J*

around 𝐵.

But when the strength of B is

Increased the coupling between L* and S*

Breaks down and there is no existence

of J*. Now L* and S* precess around

B field independently. This is known as

Paschen back effect. Due to this short of splitting whatever be the pattern of anomalous Zeeman effect in weak magnetic field, it is converted into normal pattern in strong B field. The angular velocity of precession of L* around B is

B

ml

ms

𝑙*

S*

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𝜔l = 𝑒

2𝑚 β and the angular velocity of precession of S* around B is

𝜔s = 2𝑒

2𝑚 β. Therefore the change in interaction energy due to the

two motion is given by ∆𝐸= ∆𝐸lB + ∆𝐸sB (1)

Where ∆𝐸lB = 𝐵𝑒

2𝑚 l*

ℎ

2𝜋cos (l*B) (2)

And ∆𝐸sB = 2𝑒

2𝑚

𝑠∗ℎ

2𝜋 cos (s*B) (3)

Or,∆𝐸= 𝐵𝑒ℎ

4𝜋𝑚 [𝑙∗ cos(𝑙 ∗ 𝐵) + 2𝑆∗cos (𝑆∗𝐵)]

Or, ∆𝐸 =𝐵𝑒ℎ

4𝜋𝑚(𝑚l + 2𝑚s) (4)

(ml + 2ms) is called strong field quantum number. In term of frequency change

∆℧ =𝐵𝑒

4𝜋𝑚 (ml + 2ms)

In terms of wave number

∆℧ =𝐵𝑒

4𝑚𝑐 (ml + 2ms)

Since 𝑒𝐵

4𝜋𝑚𝑐 = Lorentz unit . so in the term of Lorentz unit

∆𝜐 = ∆ (𝑚𝑙+ 2ms ) in Lorentz unit.

Since ∆𝑚𝑙 = 0 or , ±1 but ∆𝑚𝑠= 0, so ∆ (𝑚𝑙 + 2𝑚s) = O, ± 1

now We get three different frequencies. It means the result is normal Zeeman effect now let us consider a principal series doublet.

2P 3/2 2S ½ and 2P ½ 2S ½

In the strong field P level is split up into six levels. For a particular value of l(l+1), ml has 2l+1 values hare 1, 0, -1. For each value of ml, ms has two values +1/2 and -1/2 levels with same value of ml + 2ms coincide, so on the whole we have five sub levels of P level and two sublevel of S level the value of different quantum number are given below.

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Term L S ml ms ml+2ms amlms

2P3/2

1 1 1

1/2 1/2 1/2

1 0 -1

1/2 1/2 1/2

2 1 0

a/2 0 -a/2

2P1/2

1 1 1

1/2 1/2 1/2

1 0 -1

- 1/2 -1/2 -1/2

0 -1 -2

-a/2 0 a/2

2S1/2

0 0

1/2 1/2

0 0

1/2 -1/2

+1 -1

0 0

2

1

0

-1

-2

+1

-1

Paschen back effect

ml + 2ms

2P3/2, 1/2

2S1/2

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UNIT-II

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STRAK APPANATUS

C

STARK EFFECT

The effect of electric field on the spectrum of the hydrogen was discovered by stark in 1913. He found that Balmer lines are splitted due to application of electric field. Later on it was found that this effect is exhibited by spectra of all elements. it is called stark effect. Stark arranged on auxiliary electrode F closed behind the cathode C at a distance of few millimetres is which is perforated as shown in figure. A very strong electric field of several thousand volts per cm is maintained between F and C. positive ions are axiliareted toward c. some Positive ions will pick up electrons on the way forming neutral M atoms. The fast atoms which pass through the perforation and emit light in FC are called canal rays.

These canal rays were made to pass into the space between the plates of highly charged condenser. The main features of the phenomenon are the following: -

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(1). All hydrogen lines form symmetrical pattern but the pattern depends on the quantum number n of the term involved. The number of lines and total width of the pattern increases with n. Thus the number of components of Hβ lines is greater than those of the Hα lines. similarly , the components of H𝛾 lines is greater than those of Hβ.

(2). The wave number difference is integral multiplies of a unit which is proportional to the field strength F. It is same for all hydrogen lines.

(3). The observations perpendicular to the direction of their electric field shows that the components are polarised in part parallel to the field – 𝜋 components and in part perpendicular to the field 𝜎 components .

(4) Up to the field of about 100000 volts per cm the resolution increases in proportion to the field strength. In this region we have linear stark effect. In case of more intense fields, more complicated effects are observed which is called quadrate stark effect.

(Q). Explain the linear stark effect of hydrogen atom.

(Answer). We have to explain the effect of external electric field on the energy levels of hydrogen atoms. The Schrodinger wave equation may be written as: -

∇2𝜓 + 2𝑚

ℏ2 [𝐸 +

𝑧𝑒2

4𝜋 0𝑟− 𝑒𝐹𝑧] Ψ = 0 (1)

The unperturbed Hamiltonian for hydrogen atom is given as:

H0 = [ℏ2

2𝑚 ∇2 −

𝑧𝑒2

4𝜋ℇ0𝑟] ----------------------2

The perturbation H’ which is the extra energy of the nucleus and electrons due to external field F is given by

𝜆H (1) = + eFZ = eFr cos𝜃

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Where 𝜆 =eF, H(1) = r cos𝜃

H(1)nlm , n’l’m’ = ∭Ψ*

nlmr cos𝜃Ψn’l’m’ r2sin𝜃drd𝜃d𝜙

Where Ψnlm are hydrogen atoms wave function Ground state of Ψ gives

n=1, l=0 and m=0

The wave function of H atom is spherically symmetric so, there is no degeneracy. The first order perturbation is given as

𝜆H(1)1,0,0,1,0,0 = eF∭𝑟𝑐𝑜𝑠𝜃

1

𝜋𝑎03 𝑒−2 (

𝑟

𝑎0) r2dr sin𝜃𝑑𝜙 (2)

a0 = Radius of first Bohr’s orbit

and 𝜓1,0,0 = 1

√(𝜋𝑎03)

𝑒−𝑟

𝑎0

The value of the integral is zero. Therefore, for ground state there is no first order stark effect.

For excited state n = 2 for hydrogen atom the value of l = 0, 1, and m = 1,0,-1

Hence the quantum number l and m have the following combination

(0,0), (1,0), (1,1), and (1, -1)

The wave function have a fourfold degeneracy, 𝜓200 , 𝜓211 , 𝜓210 and 𝜓2,1,-1. All have the same energy E2. The secular determinant is formed and solved. The simplified matrix becomes.

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−E(1) 0 −𝑎0 0

0 −E (1) 0 0 =0

−𝑎0 0 −E(1) 0

0 0 0 −E(1)

This equation has four roots

E (1) = 0, 0, +a0, -a0

The corresponding perturbed energy values are

𝜆E(1) = 0, 0, +3𝐹ℎ2 (4𝜋∈0)

𝑚𝑧𝑒2 , −3𝐹ℎ2

𝑚𝑧𝑒2 (4𝜋𝑒0)

This result agree with the result obtained for the sate n=2. the change in energy given in equation (2) gives rise to a term shift of the order of 12cm-1 which is large in comparison with fine structure splitting of n = 2 level of hydrogen. It is found that in modest electric field the condition of the linear stark effect is satisfied

The four roots of equation (2) imply that the two stats 𝜓2,1,1 and 𝜓2,1,-1. Are not shifted the remaining two states are.

𝜓+ = 1

√2 (𝜓2,0,0 + 𝜓2,1,0 ) , 𝜆E(1) =

3𝐹ℏ24π𝜖0

𝑚𝑧𝑒2 (3)

𝜓_ = 1

√2 (𝜓2,0,0 − 𝜓2,1,0 ), 𝜆E(1) =

3𝐹ℏ24π𝜖0

𝑚𝑧𝑒2 (4)

The function 𝜓+ corresponds of electron charge distribution whose centre of charge in the presence of electric field is displaced by the amount 3a0 along positive z- direction while converse is true for 𝜓-

. Therefore, linear stark effect degeneracy in L is removed. The function 𝜓+ and 𝜓_ do not have a definite orbital angular momentum. Thus when states of opposite parity are degenerate linear stark effect is observed.

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The energy of atom up to second order perturbation term is given as:

E = −𝑅𝑧2𝑐ℎ

𝑛2 + F 3𝑎0𝑒

2𝑍 n (n2 –n1) – F2 𝑎0

3

16𝑧4 n4 (17n2 – 3(n2 – n1)2 – 9ml2+

19) 5

The electric quantum number n1 n2 and m1 are related by equation

n = n1 + n2 + ml +1 6

Here n = Total quantum number n1 and n2 are new parabolic quantum numbers. The different quantum numbers take on the following values:

n = 1, 2, 3 …………………………..…, ∞

n1 = 0, 1, 2, 3,…………………………,(n −1)

n2 = 0, 1,2,3 …………….., (n −1)

𝑚𝑙 = 0,±1,±2,………… ± (𝑛 − 1)

the term shift is given by

∆𝑇=−∆𝐸

𝑐ℎ =

−3𝑎0𝑒

2𝑍𝑐ℎ Fn (n2−n1) (7)

Where F is in volt/cm

So, ∆𝑇 = F𝜐(𝑛2−𝑛1)𝑛

𝑍 × 6.0 × 10-5cm-1 (8)

equation (6), (7), (8) agrees will expectable result.

we conclude that due to stark effect levels are shifted by the amount which is a multiple of fundamental amount the stark shift for level n = 1 is zero.

The transitions are subjected to selection rule

∆ml = 0 , 𝜋 components

∆ml = ±1 ,𝜎 components

The ∆ml = +1 and -1 are polarised in opposite sense due to superposition principle . These two transitions give rise to linearly

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polarised transitions for transvers view and for longitudinal view it gives unpolarised transition.

𝜆 is written for ml and nn for n1 n6 for n2

It is found that other lines of alkali like

n n2 n1 (n2 –n1 ) n(n2 –n1 ) ml = n- n1-n2 -1

1

0

0

0

0

0

2

0 0 1 1

0 1 0 1

0 -1 1 0

0 -2 2 0

1 0 0 -1

3 0 0 0 1 1 1 2 2 2

0 1 2 0 1 2 0 1 2

0 -1 -2 1 0 -1 2 1 0

0 -3 -6 3 0 -3 6 3 0

2 1 0 1 0 -1 0 -1 -2

The Stark shift

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(Q). What are the hyperfine structures of spectral lines?

(Answer). When yellow sodium lines are observed with high dispersion instruments two components of this line is observed. one is of wavelength 5890.12 Å, it is called D1 lines and another is wavelength 5896.16 Å, it is called D2 line. It is found that other lines of alkali like spectra show the doublet characters. This is called doublet fine structure. The single line spectra of heavier atoms contain more than two lines. This structure is called multiplet structure. When these spectra are analysed with high resolving power spectrometers such as fabric – Perrot instrument, it is fond that there are some spectral lines separated by less than 1 Å. Such spectra are very closely packed. They are called hyperfine lines or hyperfine structure.

(Q). What are the causes of fine structure?

(Answer):- The fine structures of spectral lines are attributed to external nuclear effects. If we consider the spin orbit interaction and relativistic variation of mas of electron, then the phenomenon of fine structure of spectral lines could be easily explained.

(Q). Explain the hyperfine structure of spectral lines?

(Answer). According to Pauli suggestion the nucleus of atom also spins about an axis. The angular momentum of nucleus also gives rise to magnetic moment of nucleus. The interaction of magnetic moment of nucleus with magnetic field of extra – nuclear electrons gives rise to hyperfine structure of spectral lines.

24



ABSORPTION CELL

(Q). Describe an experiment to demonstrate the hyperfine structure of spectral lines.

(Answer):- The hyperfine structure of spectral lines can be demonstrated by the Jackson and Kuhn experiment. The apparatus consists of an evacuated tube at the bottom of which a heating arrangement is provided. When sodium is heated in this tube, it is vaporised. At the upperpart of the tube there are cooling arrangement, where the sodium vapour is condensed radiations from sodium arc are allowed to pass through this vapour in horizontal direction. The emergent radiations are analysed with fabry – Perot instrument. The spectrum shows sharp absorption lines which are narrowly spaced in the back ground of broad emission lines. These sharp absorption lines are due to the fact that hyperfine structure exists.

(Q). Deduce the theory of hyperfine structure of spectral lines?

(Answer):- The nucleons in the nucleus of an atom also has orbital angular momentum as well as spin angular momentum. if the resultant angular momenta of all the nucleons within the nucleus

is Iℎ

2𝜋 or I*

ℎ

2𝜋 where I is called nuclear spin quantum number.

25



The nuclear spin vectors I* combines vectorically with J* which represent the total angular momentum of all extra – nuclear electrons. This coupling gives rise to a new vector F or F* which represents the total mechanical moment of the atom. here F is called Hyperfine quantum number. It is restricted to integral number or half integral number.

Now for simplicity suppose that the atom is of single valence electron. There will be interaction of nuclear moment I* with orbital motion of electron L* again there will be interaction of I* with S*. Here we can say that electron moves in the electric field of nucleus with angular

momentum = m (�̅� × �̅�) = l* ℎ

2𝜋

The value of electric field due to electron at nucleus at distance r from it

E= −𝑒

𝑟3

�̅�

(4𝜋𝜖𝜊)

The value of magnetic field generated due to orbit motion is given by

�̅� = 𝐸×�̅�

𝐶2

or �̅�= −𝑒

(4𝜋∈0)

�̅� × �̅�

𝑟3𝑐2

or �̅�= −𝑒ℎ

2𝜋𝑚0𝑐2 l*(

1

𝑟3)(1

4𝜋 0)

Since the value of r does not remain constant, so its average value is taken Due to this magnetic field �̅�the nucleus has the larmor precession around the field

direction with angular velocity 𝜔l given by 𝜔l = Bgi𝑒

2𝑀 where

𝑔𝑖e

2𝑀

= Ratio of magnetic moment to mechanical moment = 𝜇

𝐼∗ℎ

2𝜋

26



gi Is called nuclear g factor. it is 1

1840 times the g factor for an orbital

electron .

The interaction energy is given by

∆𝐸i= (Angular frequency of larmor precession ) × (Projection of nuclear mechanical moment on 𝑙*)

Or ∆𝐸I = gI 𝑒2

2𝑀𝑚𝜊 𝐼∗

ℎ

2𝜋 𝑙∗

ℎ

2𝜋 cos(𝐼∗𝑙∗) (

1

4𝜋𝜖0𝑟3)

As l* precesses around J* and J* and I* precess around there resultant F*. the cosine term is also averaged; which is given by

<cos(I*𝑙*)> = cos (I*𝐽*) cos (I*𝐽*)

So ∆𝐸𝐼𝑙 = 𝑔𝑖𝑒2

2𝑀𝑚𝜃𝑐2 𝑙∗ℎ

2𝜋

𝐼∗ℎ

2𝜋 cos (I*𝐽*) cos (I*𝐽*)

The international energy between two magnetic dipoles 𝜇𝑠 and 𝜇𝐼 is given by

∆𝐸𝐼𝑙 = 𝜇𝑠𝜇𝐼

4𝜋 0𝑟3 [cos(𝜇𝐼𝜇𝑆) − 3cos (𝜇𝐼𝑟)cos (𝜇𝑆𝑟)(𝑟)]

In the expression r is the distance between two magnetic dipoles.

The total interaction energy is given

As: ∆𝐸 = ∆𝐸𝐼𝑙 + ∆𝐸𝐼𝑠

Note

∆𝐸𝐼𝑙 = 𝑔𝑖 𝑒2

2𝑀𝑚𝜃𝑐2 𝑙∗ℎ

2𝜋

𝐼∗ℎ

2𝜋 cos (I*𝐽*) cos (I*𝐽*) <

1

4𝜋 0𝑟3 >

or ∆𝐸 = 𝑒2ℎ2

8𝜋2𝑀𝑚0 𝑔𝐼(I*𝐽*) cos (I*𝐽*) [

𝑙∗

𝐽∗ cos (𝑙∗𝑆∗)] +

𝑆∗

2𝐽∗ cos(𝑙∗𝑆∗ ) −

3𝑆∗

2𝐽∗ cos(𝐽∗𝑙∗ ) cos(𝑙∗𝑆∗ )] <1

𝑟3 > 1

4𝜋 0

27



according to Goldsmith, the expression with in the bracket must

bracket must be replaced by 𝑙

𝐽∗2

∗2

and hence

∆𝐸 = 𝑒2ℏ2

2𝑀𝑚0 𝑔𝐼(I*𝐽*) cos (I*𝐽*)

𝑙∗2

𝐽∗2 <

1

𝑟3 > 1

4𝜋 0

The averaged value of <1

𝑟3 > AND FOR I* J* cos (I* J*) give the

expression

∆𝐸 = 𝑔𝐼𝑅𝑐ℎ𝛼3𝑍2𝑚𝑜

𝑥3𝑀

1

𝐽(𝐽+1)(𝐿+1

2) 𝐹(𝐹+1)−𝐼(𝐼+1)−𝐽

2

In terms of wave number

∆𝑇 = 𝑔𝐼𝑅𝛼3𝑍2

𝑥3

𝑚𝑜

𝑀

1

𝐽(𝐽+1)(𝐿+1

2) 𝐹(𝐹+1)−𝐼(𝐼+1)−𝐽(𝐽+1)

2

The selection rule is ∆𝐹 = 0,±1, which explains the hyperfine structure. The hyperfine structure of 4722 Å line of Bismuth the

initial level of this line has got the value of I and J as 9

2 and

1

2

respectively. Hence it splits up into levels corresponding to F = 4

and F = 5, the final level has got I = 9

2 and J =

3

2 . The level divides

itself into four levels having F = 3,4,5,6 respectively. The six transitions are allowed following the selection rules.

28



152

197

255

3

4

5

6

I = 9/2

6P27s1(1

2 1

2 3

2)1

2

830

4

Bismuth ʎ = 4722Å

F=5

6P2(1

2 3

2 3

2 )3

2

29



UNIT-III

30



Alkali spectra

In alkali metals the outermost shell contains a single electron. The spectra emitted by alkali metals are characterised by single valence electron the man example of alkali metals are lithium, sodium, potassium, rubidium, and cesium the spectra of these elements are identical with the spectrum of hydrogen atoms. The H atom also contains a single valence electron there are other elements which show alkali spectra. They are singly ionised atoms of beryllium magnesium, calcium, doubly, ionised atoms of boron are aluminium and so on.

The alkali spectra can be grouped into four chief series as given below: -

1. Principal series: - This series arises due to transition between various, p levels and lowest s levels. The lowest s –level represents the ground state of the atom. The wave number of principal series given by the relation

�̅�p = 𝑅

(1+𝜇𝑠) 2 -

𝑅

(𝑚+𝜇𝑝)2 when m≥2

2. Sharp series: - The sharp series comes from transition from s-level to the lowest of p-level. the involve number is expressed as

�̅�s = 𝑅

(2+𝜇𝑝) 2 -

𝑅

(𝑚+𝜇𝑠)2 when m≥2

3. The Diffuse series: - it arises from the transition between the various D-levels and the lowest p-level. The wave number Is expressed by the relation:

�̅� D = 𝑅

(2+𝜇𝑝) 2 -

𝑅

(𝑚+𝜇𝐷)2 when m≥3

31



4. fundamental series or Bergmann series: - the fundamental series arises from to transition from various F levels to the lowest D- level. The wave no us expressed by the relation.

�̅� F = 𝑅

(3+𝜇𝐷) 2 -

𝑅

(𝑚+𝜇𝐹)2 when m≥4

Here 𝜇𝑆 𝜇𝑝 𝜇𝐷 𝜇𝑓 are the characteristics constriction of sharp,

principal, diffuse, and fundamental series respectively. The series name is given due to nature of current term. The first term is bigger than second term𝜇𝑆 > 𝜇𝑃 > 𝜇𝐷 > 𝜇𝐹 . The lowest possible values of m is 2 in case of principle and sharp series .

Orbitals magnetic moment: - the electron always moves about the nucleus in an orbit. Thus the electron may be considered as small circular current about the nucleus. The magnetic moment of electron due to its orbital motion is 𝜇. = 𝐼𝐴 where I = electric

current = 𝑒

𝑇 =

𝑒𝑣

2𝜋𝑟

Where r= radius of the orbit and 𝜈 = velocity of electron in the

orbit. A= area of circular orbit = 𝜋𝑟2, therefore, 𝜇 = 𝑒𝑣

2𝜋𝑟 𝜋𝑟2

or 𝜇= 𝑒𝑣𝑟2

2𝑟=

𝑒𝑣𝑟

2 --------------------(1)

The angular momentum of electron = 𝐿 = 𝑚𝜈𝑟. -----------------------(2)

Where m= mass of electron, 𝜈 = velocity of electrons, r= radius of orbit.

So 𝜇

𝐿 =

𝑒𝑣𝑟

2𝑚𝑣𝑟 =

𝑒

2𝑚

Therefore, 𝜇=𝑒

2𝑚 x L

32



or 𝜇 ̅= - 𝑒

2𝑚 �⃗� where �̅� is a vector and �̅� is also a vector. Since charge

on electron is negative hence direction of �̅� and that of �⃗� are opposite to each other. In quantum mechanism angular

momentum of electrons is determined by h/2𝜋 √𝑙(𝑙 + 1) where

l=0,1,2,3,4,etc. here l is called orbital quantum number. When l=1, then L=0 i.e. orbital angular momentum is zero. It is called s- orbit electron. The electrons in s-orbital have zero angular momentum, therefore its magnetic moment is zero.

For l=l , L = h/2𝜋 √2(𝑙) = √2 h/2𝜋

This is called p electron. For l=2, l= h/2𝜋√6

D electron and for f electron, L=ℎ

2𝜋 √3(4)

Therefore, 𝜇 l = -𝑒

2𝑚𝑐 ℎ

2𝜋 √𝑙(𝑙 + 1)

The possible component of angular momentum along any specified direction such as the direction of are external magnetic field which is always along z-axis. It is determined by quantum number 𝑚𝑙 as

𝑚𝑙ℎ

2𝜋 where 𝑚𝑙= 0, ±1, ±2 etc.

Here magnetic momentum component

( 𝑚𝑙)z = - 𝑒

2𝑚 𝑚𝑙

ℎ

2𝜋 =

𝑒ℎ

4 𝜋𝑚 𝑚𝑙

For s- electron 𝑚𝑙=0, So (𝑚𝑙)𝑠 = 0

For p orbit electron 𝑚𝑙 =0,, , ±1.

So 𝑚𝑙 = 𝑒ℎ

2 𝜋𝑚 , 0 and = -

𝑒ℎ

2 𝜋𝑚.

𝑒ℎ

4 𝜋𝑚 = 𝜇β = Bohr magneton.

33



Spin magnetic moment: -

Since, the electron has an intrinsic spin angular momentum of

magnitude √𝑠(𝑠 + 1) ℎ

2𝜋

Where s= spin quantum number. its value is 1

2 . The spin magnetic

moment associated with spin angular momentum = 𝜇s = -

gs√𝑠(𝑠 + 1)= 𝑒

4𝜋

Here gs coefficient is inserted, because the ratio of magnetic moment to the angular momentum differs from classical value. the value of 𝑔𝑠 = 2. The component of angular momentum along. Z-

axis is determined by 𝜇𝑆 and it is expressed by 𝜇𝑆 ℎ

2𝜋 with 𝜇𝑆 =

1

2 or

−1

2. Thus, (𝜇s)z= -gs 𝜇𝑆

𝑒ℎ

4 𝜋𝑚 =±

𝑒ℎ

4 𝜋𝑚= ± 𝜇 The electrons spin gives

wise to nearly one Bohr magnetic in the direction or opposite to the external field although the spin is ½ integral values.

Total magnetic moment

If we neglect the nuclear magnetic moment, then the vector sum of orbital angular momentum and the spin angular momentum gives the total angular momentum. The total angular quantum number is given by J = (l + s). the total angular momentum of

electron is given by √𝑗(𝑗 + 1) ℎ

2𝜋 .

This is a simple case. For an atom with z electron the rules for adding the orbital and spin angular momentum are needed in detail. This addition can be done several ways.

(i)law: - the separate orbital angular moments are combined vectorially into a resultant angular momentum determiner by �̅�.

√𝐿(𝐿 + 1) ℎ

2𝜋 .

In actual value of the number L depends on the direction in which the various angular momenta points with respect to one another.

34



The maximum value of L is ∑ 𝑙𝑧1 and the minimum is zero. I.e.

L=0,1,2,3,----- ∑ 𝑙𝑧1 .

A similar vector addition is made of the spin angular momenta

determined by 𝑆̅

√𝑠(𝑠 + 1) ℎ

2𝜋 , the value of S goes from a maximum ∑ 𝑠𝑧

1 =𝑧

2 by

internal of ½ to a minimum of 0 or ½ depending on whether z is even or odd.

Electron angular momentum: -

Suppose that the orbit in which electron is revolving

Is of radius r. the velocity of electron is �̅� and mass of electrons is m. then linear momentum of electron= �̅�= 𝑚𝑣̅̅ ̅̅ .

Its direction will be along, tangent line on the orbit the angular momentum of electron.

�⃗� = 𝑟 x𝑝 = 𝑟 x𝑚𝑣⃗⃗⃗⃗⃗⃗ = m( 𝑟 x𝜗 )

Or, �⃗� = mrv �̂�

|L|= mr2w------------(1)

The electric current i= = 𝑒

(2𝜋𝑟

𝑣)

= = 𝑒𝑣

2𝜋𝑟

The magnetic moment= I A = 𝑒𝑣

2𝜋𝑟 𝜋r2.

𝜇= 𝑒𝑣𝑟

2𝜋𝑟

𝜇= 𝑒𝑟2𝑤

2

�̅�

𝑚

35



𝜇/𝐿= 𝑒𝑟2𝑤

2𝑚𝑟2𝑤 = e/2m

𝜇/𝐿= (e/2m)

Or �̅� = -e/2m , �⃗� The direction of magnetic moment is opposite to that of angular momentum.

36



UNIT-IV

37



Q. Describe the diatomic molecule as a harmonic oscillator. Also describe the expression for frequency of Vibrate spectra.

Ans: -Let us suppose there is a diatomic

Molecule The masses of two atoms are

m1and m2, the separation between them

is re in equilibrium condition. Suppose that

centre of mass is at o point o1o=r1 and oo2=r2.

Then m1r1=m2re or m1r1=m2(re-r1)

Or (m1+m2) = m2re or r1= 𝑚2𝑟𝑒

𝑚1+𝑚2-----------(1)

Similarly, r2= 𝑚1𝑟𝑒

𝑚1+𝑚2---------------------------(2)

Suppose that due to external force, the length between m1and m2

becomes r.

Then restoring forces on ml atom is 𝑚1𝑑2𝑟1

𝑑𝑡2 = -K(r-re)

Where K is force constant.

Similarly, the force acting on m2 Particle is 𝑚2𝑑2𝑟1

𝑑𝑡2 = -K(r-re)

Here r1= 𝑚2𝑟

𝑚1+𝑚2 and r2

𝑚1𝑟

𝑚1+𝑚2

Thus we can write fence on m1 particle

𝑚1𝑚2

𝑚1+𝑚2

𝑑2𝑟

𝑑𝑡2= -K(r-re)

Since re is constant, hence 𝑑2𝑟

𝑑𝑡2 =𝑑2(𝑟−𝑟𝑒)

𝑑𝑡2

So ( 𝑚1+𝑚2

𝑚1+𝑚2)

𝑑2(𝑟−𝑟𝑒)

𝑑𝑡2 = -k(r-re)-------------(3)

Now, let us put 𝑚1𝑚2

𝑚1+𝑚2 𝑚1 (𝑟 − 𝑟𝑒) = 𝑥

38



Therefore, m1 𝑑2𝑥

𝑑𝑡2=-kx.

Or 𝑑2𝑥

𝑑𝑡2 + 𝑘

𝑚1 x=0

Or 𝑑2𝑥

𝑑𝑡2 + 𝜔2𝑥 = 0 where 𝜔2 = 𝑘

𝑚1----------(4)

Where w is the frequency of vibration.

𝜔=√𝑘

𝑚1 or 𝑛 =

1

2𝜋√

𝑘

𝑚1 or �̅� =

1

2𝜋𝑐√

𝑘

𝑚1

Here �̅� is the wave number of vibration, a vibrating diatomic molecule is approximated as a harmonic oscillator. The potential energy function under the influence of which. Nucleus vibrate is

given as 𝑣(𝑟) =1

2𝑘𝑥2. where x is the placement from the mean

position the Schrodinger wave equation can be written as.

𝑑2𝜓

𝑑𝑥2 + 8𝜋2𝑚1

ℎ2 (𝐸𝑣 −1

2 𝑘𝑥2)𝜓 = 0

When m1= reduced mass of the diatomic molecule of we put 𝛼= 8𝜋2𝑚′𝐸𝑣

ℎ2

𝛽=√4𝜋2𝑚′𝑘

ℎ2 = 2𝜋√𝑚′𝑘

ℎ2

Then 𝑑2𝜓

𝑑𝑥2 +( 𝛼-𝛽2𝑥2)𝜓 =0

Now let us introduce a dimension less independent variable 𝜉 =

√𝛽 𝑥 so that

𝑑2

𝑑𝑥2=𝛽𝑑2

𝑑𝜉2 . Thus above eqn becomes.

𝑑2𝜓

𝑑𝜉2 + (𝛼

𝛽+ 𝜉2)𝜓 = 0

It we try a solution of this equation.

In the form 𝜓(𝜉)= CU(𝜉) e

39



Them above equation becomes: - 𝑑2𝑈

𝑑𝜉2 -2𝜉𝑑2𝑈

𝑑𝜉2 +(𝛼

𝛽− 1)𝑈=0

This equation will become Hermit differential equation provided.

We put (𝛼

𝛽− 1) = 2𝜐 or

𝛼

𝛽= 2𝜐 + 1

Therefore, we can replace V(𝜉) by

Hermetic polynomial 𝐻𝜐(𝜉). Thus

𝑑2𝐻𝜐(𝜉)

𝑑𝜉2 -2𝜉 𝑑2𝐻𝜐(𝜉)

𝑑𝜉 +2𝜐 𝐻𝜐(𝜉) =0

Therefore, solution of Schrodinger equation becomes

𝜓= C𝐻𝜐 (𝜉) 𝑒−𝜉22. it is valid only for 𝜐= 0,1,2, ------ the restriction 𝜐

restricts energy value E.

Since 𝛼

𝛽 =2𝜐+1

Or 𝐸𝜐 = ℎ

2𝜋√

𝑘

𝑚′ (𝜐+1

2)

𝐻𝜐(𝜐+1

2) or 𝐸𝜐= hcw (𝜐+

1

2)

Here 𝜔 is the vibrational frequency of the vibrating diatomic molecule expressed in wave number. The above equation gives the allowed energy.

For the harmonic oscillator. If 𝜐=0. 𝐸𝜐=1

2hcw. If the energy is

transformed. Into term value, then G()= 𝐸𝜐/hc=w(ϑ+L2)

Putting 𝜐 =0,1,2,3, etc. we get G(𝜐)= 𝑤

2, 3𝑤

2, 5𝑤

2. Thus we have a

series f equispaced discrete vibrational levels. The common separation is w cm-1, Suppose that a transition accurses from an upper vibrational level in which the quantum number

40



Is 𝜐1 to a lower state with quantum number 𝜐”.

The change in vibrational energy will be Eϑ1-Eϑ”

=(𝜐’-𝜐”) w=G (𝜐’)- G(𝜐”)

The frequency of radiation in wave number will be

𝜐𝑢 =𝐸𝜐

′ − 𝐸𝜐"

ℎ𝑐= (𝜐′ − 𝜐")𝜔

= 𝐺(𝜐′) − 𝐺(𝜐")

Suppose 𝜐1=1 𝜐”=0 then (∆𝐸)1,0 = 𝐸 1’-E 0”

=hcw

The frequency of this transition will be

(𝜐𝑢)1,0=(∆𝐸)1,0

ℎ𝑐= 𝑤𝑐𝑚 − 1

Or w=1

2𝜋𝐸√

𝑘

𝑚′ cm-1

The vibrational spectrum is expected to consists of a single band at 𝜔cm-1. Therefore, an intense band in infrared spectrum is to be concluded as vibrational spectrum, owing to its origin to harmonic vibration of the nuclei along inter nuclear axis. However, infrared spectrum also consists some weak bands called overtones at frequencies slightly lesser than 2𝜔,3𝜔 etc. their appearance suggests that vibration deviate from being harmonic. thus analysis should be made by treating the diatomic, molecular vibrations as anharmonic.

−7

2

5

2

3

2

1

2

𝜐=

41



r1

CG

r

2 M2

Describe the origin of pure rotational spectra and their characteristics.

Ans:- the molecules which have permanent electric dipole moment can give rise to rotational spectra. Thus homo nuclear diatomic molecules such as O2, H2 N2 etc. do not exhibit pure rotational spectra. the heteor nuclear OH diatomic molecules such as HF1. Hcl, HBr, Nacl etc can exhibit rotation oral spectra. Such spectra appear in the far in frared region at wavelength 200x104Å or more. During the rotation of a hetero nuclear diatomic molecules, the components of the dipole moment in a fixed direction changes periodically with the frequency of rotation of molecule so it emits radiation of same frequency. Let us suppose that there are two masses M1 and M2 joined by a raged bar of ----- r =r1+r2.

The centre of mass of the system is at c point the system rotates about the centric of gravity. C here we assume that the bond length

ro does not change applying the Property of C.G

m1r1=m2r2=m2(r0-r1)

or r1= 𝑚2𝑟0

𝑚1+𝑚2 Similarly, r2=

𝑚2𝑟0

𝑚1+𝑚2

Now moment of eventer of the systems about the

CG point =I= m1r12+m2r2

2

or I= m1(𝑚2𝑟0

𝑚1+𝑚2)2+m2(

𝑚1𝑟0

𝑚1+𝑚2)2

or I= 𝑚1𝑚2𝑟02

(𝑚1+𝑚2)2 (𝑚1+𝑚2) = (

𝑚1𝑚2

𝑚1+𝑚2)𝑟0

2

or 𝐼 = 𝑚′𝑟02

If the molecule is rotating about c point with angular velocity 𝜔, then kinetic energy of the system if given by.

M1

42



𝑘 =1

2𝐼𝜔2. The angular moments of the system abut c point =9w

So k.E. =1

2

(9𝜔)2

9. But quantum mechanically we know that angular

momentum = 𝐽ℎ

2𝜋 where J= rotational quantum number. Hence

k.E= 1

2𝐽2

ℎ2

4𝜋29

Or Er= 𝐽2ℎ2

4𝜋29---------------------(1)

The Schrodinger wave equation in sheen’s polar coordinates can be written as: -

1

𝑟2

𝜕

𝜕𝑟(𝑟2𝜕𝜓

𝜕𝑟) +

1

𝑟2 sin𝜃

𝜕

𝜕𝜃(𝑠𝑖𝑛𝜃

𝜕𝜓

𝜕𝜃) +

1

𝑟2 𝑠𝑖𝑛2 𝜃

𝜕2𝜓

𝜕𝜃2+

8𝜋2𝑚

ℏ2(𝐸 − 𝑉)𝜓 = 0

Where symbol have their own significance. Here we have assumed that r0= constant. Hence eqn(ii) becomes.

1

sin𝜃 𝜕

𝜕𝜃 (𝛿𝑖𝑛𝜃

𝜕𝜓

𝜕𝜃)+

1

𝛿𝑖𝑛2𝜃 𝜕2

𝜕∅2 +

8𝜋2𝐼

ℎ2𝐸𝜓 = 0 -------(3)

From eqn (3) it is apparent that 𝜓 is a function of 𝜃 and ∅.

Let Ψ(𝜃,𝜙) = (𝜃)( 𝜙). Therefore, equation (3) may be written as:

Sin 𝜃 𝜕

𝜕𝜃 (sin

𝜕

𝜕𝜃) +

ꭍ

𝛷 𝜕²𝛷

𝜕𝜙 +

8𝜋²𝐼𝐸

ℎ² sin² 𝜃 = 0

On separation of variable we get.

Sin 𝜃 𝜕

𝜕𝜃 (sin 𝜃

𝜕

𝜕𝜃) + B sin² 𝜃 = -

𝐼

𝜙 𝜕²𝛷

𝜕𝜙2 =m²

Thus equation (4) may be separated as:

Sin 𝜃 𝜕

𝜕𝜃(sin 𝜃

𝜕

𝜕𝜃)+ B sin² 𝜃 = m²...........................(5)

- 𝐼

𝜙 𝜕²𝛷

𝜕𝜙2 =m²....................................................(6)

43



From equation (6) we get Ə²𝛷

Ə𝛷² + m²φ = 0........(7)

The solution of equation (7) may be written as: 𝛷 = c eᴵᵐᶲ .........(8)

For 𝛷to be single value function , it show have some value for 𝛷=0 and 𝛷 =2𝜋

So 𝛷= C e°=C and ϕ = C e±ᴵᵐᶲ = C e±²iᵐ

So C=C e±2iπm

Or e±2πim =1

Or cos 2πm+ I Sin 2πm=1

It is only possible when m=0 g an integer thus m=0,1,2,3,4 etc.

Now equation.....(5) may be written as:

1

sinƟ 𝜕

𝜕𝜃 (sin 𝜃

𝜕

𝜕𝜃)+(𝛽 −

𝑚²

𝑠𝑖𝑛²Ɵ) = 0.........(9)

Let us put x= cos Ɵ

then 𝜕

𝜕𝜃 = (

𝜕

𝜕𝑥 ) (

𝜕𝑥

𝜕𝜃 ) =-sin Ɵ

𝜕

𝜕𝜃

So 𝜕

𝜕𝜃 = - sin Ɵ (

𝜕

𝜕𝑥 ).............(10)

So 𝜕

𝜕𝜃 = - sin Ɵ (

𝜕

𝜕𝑥 ) .............(11)

Therefore , equation (9) because

1

sinƟ (- sin Ɵ

𝜕

𝜕𝜃 ) (-sin²Ɵ

𝜕

𝜕𝑛 ) =+ {b-

𝑚2

(1−𝑥2)}=0

So 𝜕

𝜕𝑥 {(1-x²)

𝜕

𝜕𝑥 } + (𝛽 −

𝑚2

(1−𝑥2))........(12)

equation (12) is ladenger differentials equation. the value of x can

very form +1 to -1. Now let us write ↔=(1-x²)𝑚

2 𝐺

44



From equation (13) we can get the solution of equation (12) only when B = l(l+1) where l=0, or 1,2,3 etc.

So 8𝜋²𝐼𝐸

ℎ² = l(l+1) or E =

ℎ²

8𝜋²𝐼 (l) (l+1) for rotational quantum number

if we write J in place of l,

then Eᵣ = ℎ²

8𝜋²𝐼 (J) (J + 1)..........(14)

Where J = 0,1,2,3 etc. This equation gives the allowed rotational energy to a molecules property I and quantum no. J. If thus

energy is converted to rotational term value, it becomes F(J)=𝐸𝑟

ℎ𝑐 =

ℎ

8𝜋²𝐼𝐶 J(J+1)...............................(15)

F (J) is called the rotational term value (m) thus F(J)m= BJ (J+1) m‾¹. Where J=0,1,2,3 etc. β is called the rotational constant. If rotational transition occurs from higher value of J to lower value, then frequency of spectral lines in term of wave number is given

as: √r= 𝐸𝑟′−𝐸𝑟′′

𝑛𝑐 =

ℎ

8𝜋²𝐼𝐶[Jᴵ(Jᴵ+1)-Jᴵᴵ(Jᴵᴵ+1)

Or √𝑟 = B{Jᴵ(J+1)-Jᴵᴵ(Jᴵᴵ+1)]..........(16)

If a molecules possess dipole moment it can interact with oscillating electro magmatic radiation. the molecular can with draw or give up energy to the radiation. hence a molecule must possess diplo moment to give rise rotational spectra. the solution rule for rotational transition is ΔJ=(Jᴵ-Jᴵᴵ)= ±1.

Thus 𝜐𝑟= F(Jᴵᴵ+1)-F(Jᴵᴵ)

=β[ Jᴵᴵ+1) (Jᴵᴵ+2)-Jᴵᴵ(Jᴵᴵ+1)]

=2 β (Jᴵᴵ+1) where Jᴵᴵ=0,1,2,3, etc.

Thus the frequency for consecutive lines in the pure rotational spectrum of a diatoms molecules are 2B, 4B, 6B etc. Thus on wave number scale, they are equal distance limes

45



Q. Give the theory of rotation vibration spectra of a diatomic molecule. Discuss the effect of isotopes on the spectra.

Ans. The near infra-red spectra of molecule consists of band having closed limes arranged in particular manner. This suggests that in a vibrational transition, the molecules also changed its rotational energy state, thus it may be treated as vibrating rotator to a first approx. motion we may assume that a diatomic molecule may executed rotation and vibration quit independently. We may assume that there is no integration between rotational energy and vibrational energy. E total= Erot+Evib

EUR={G(𝜐)+F(J)}ch

or EUR=( 𝜐+1

2 )ℎ𝑐𝜔 +

ℎ²

8𝜋²𝐼 J(J+1)

Now, suppose that a simultaneous transition from the vibrational level 𝜐′to the level 𝜐′′ and from the rotational level Jᴵ to the level JII occurs, then change in energy= EᴵUR-EᴵᴵUR= (𝜐ᴵ- 𝜐ᴵᴵ) ℎ𝑐𝜔+ ℎ²

8𝜋²𝐼{Jᴵ(Jᴵ+1)-Jᴵᴵ(Jᴵᴵ+1)}

The frequency of radiation +1)

The frequency of radiation

𝜐 = EIur - EII

ur

Or 𝜃= (𝜐I + 𝜐 II ) w + B { J I (JI + 1) – JII (JI + 1) }

Where B = ℎ

8𝜋²𝐼𝐶 =

ℎ

8𝜋²𝑚ˈ𝑟²𝐶

Although, we have assumed at that vibration and rotation are occurring without interaction which is not the reality. A molecule vibrates 10³ times during the course of a single motion.

A change in bond length cause a change in moment of interia and corresponding change in B. Therefore, interaction between two energy is takes place.

hC

46



An increase in vibrational energy is accompanied by an increase in vibrational amplitude. Since B, the rotational constant depend on

(1

𝑟²) where r represent inter nuclear distance or bond length. It will

depend on the vibrational quantum no v an increase in vibrational energy will lead to an increase in the average bond length raw. The rotational constant B is smaller in the upper vibrational state than in the lower an quantum of the form BU = Be-x

(u+1

2).............(1)

Gives the value of BU. The rotational of constant in vibrational level v, in term of the equilibrium value be and x, a small positive constant dv, which allows for nonregedidity of the molecule is related to the equilibrium value de by:

𝐷𝜐=Dₑ+Bₑ( 𝜐+1

2)+.........(2)

Where Be is small compared to de which is it solve small. Therefore, the rotational energy of a molecule can be modified to take the form:

Eᵣ = 𝐸𝜐 J(J+1)hc- 𝐷𝜐J²(J²+1)hc......(3)

Where 𝐵𝜐and 𝐷𝜐bear the same significant as explain in equation (1) and (2). In the lowest level v=0, these becomes B0 and D0. In equation (2) be is small compared to de, so we write de in steal of Dₑ is regarded as constant for all vibrational level.

The total vibrational and rotational energy is thus

EUR= (𝜐+1

2)hc we – (𝜐+

1

2)² hc x 𝜔e+...........

𝛽𝜐𝐽(𝐽 + 1)ℎ𝑐 − 𝐷𝑒𝐽2(𝐽2 + 1)ℎ𝑐 + ⋯………………….(5)

This expression gives the frequency of limes constituting P and R branches of vibrational bond.

Ie. J-1 to J, 𝜐 (P) = 𝜐O – ( Bi 𝜐 + BiI 𝜐) J + ( Bi 𝜐 + BiI 𝜐) J2 + 4 Dej3 +……6

J to J-1 , 𝜐(R) = 𝜐O – ( Bi 𝜐 + BiI 𝜐) J + ( BI𝜐 + BII𝜐) J2

47



Where J=1,2,3 etc.........4DeJ³+...........(7)

The constant Bᴵ𝜐 and Bᴵᴵ𝜐 are the value of bv in the invited upper and final lower states in the vibrational transition, 𝜐0 is the frequency of centre of the bond. For 𝜐=1 to 𝜐=0

If we ignore de then equation (6) and (7) can be written as

𝜐P,R= 𝜐0 + ( Bi 𝜐 + BiI 𝜐) m + ( Bi 𝜐 + BiI 𝜐) m2 cm-

Where m=+1, +2, +3 for R branch

And m=-1, -2, -3 for P branch

Since the average value of inter nuclear distance re increases with

vibrational energy the vibrational constant 𝛽𝜐=ℎ

8𝜋²𝐼𝐶=

ℎ

8𝜋²𝑚ᴵ𝑟²ₑ is

small in the upper vibrational state than in the lower vibrational state. Since Bᴵᴵ𝜐 denotes the lower vibrational state, so Bᴵ𝜐‹13 it means band head appear in R branch on the origin such a bond is said to be degraded toward the red. Thus, in vibrational-rotational band are degraded to red only. But in electronic band, both red and violet degraded bands have been observed.

(11) for both p and r branches, we have Bᴵ𝜐‹ Bᴵ𝜐 the separation between the lines of R branch will decrease with increasing J value whereas separation between lines of p branches will increase with increasing J.

APPLICATION: - With the help of rotation- vibration spectra the equilibrium moment of i.e. And corresponding nuclear separation rₑ can be evaluated.

The two isotopic form of the same molecule have different reduced masses but force constant will remain the same. Therefore, the equilibrium vibration frequency of the two form we for two isotopic forms.

48



Also suppose that m1’ and m2’ be the reduced mass therefore, from the rotation force constant

K=4π²w²ₑmᴵ it follows that 𝑤₂

𝑤₁ =(

𝑚ˈˌ

𝑚ˈ₂)½ = 𝜑

So w₂= 𝜑𝑤,

Since, the anharmonicity constant x is proportional to the

equilibrium frequency, so that we can write x₂= 𝜑xˌ. Thus frequency shift of isotopic molecular spectra can be calculated the experimental and theoretical shift are in good agreement when supports the validity of the theory.

49



UNIT- V

50



Dr. Arun Kumar Singh

Associate Professor

Department of Physics

C.M. Sc. College, Darbhanga

51



UNIT-V, LASER

A laser stands for light wave amplication by stimulated

emission of radiation. The most widely known and spectacular

properties of lasers are that they emit unidirectional and

monochromatic beams of light. These remarkable Properties

have an amazing numbers of applications in the field of pure

science, chemistry, Biology and medicine. It is also called

optical maser because it operates in the visible portion of the

spectrum. When emission form an atomic system occurs in the

radio frequency range , the term used is raser, and stimulated

emission is in the infrared range ,the term used is Iraser.

The transition from one energy state to another is followed by a

transfer of energy. If energy in whatever from is supplied to the

system is raised from a lower energy state E1 to a higher energy

state E2. such a transition is called absorption. The reverse

process i,e the downward transition E2→E1 is called emission

.In addition to the transition that occur in the process of

emission and absorption of radiation there is a third process

predicted by Einstein im1916,called stimulated emission.

Einstein’s equation and population inversion :-

52



Let us consider the number of atoms per unit volume ,that

exist in a given energy state . This number is called population

N, and is by Boltzmann’s equation .

N=e-E/KT i

Where E is the energy level of the system, K- Boltzmann

constant and T is the absolute temp.

In any process of absorption or emission at least two

energy states are involved. Population may be compared in to

states. If N1 and N2 be the population in two states and E1 and

E2 their respective energy level then,

N1= e-E1/KT

And N2= e-E2/KT

N2/N1= e-E2/KT/ e-E

1/KT

Or, 𝑁2= N1𝑒−(𝐸2−𝐸1)/𝐾𝑇 ii

A Plot of this equation yields an exponential known as a

normal or Boltzmann’s distribution of atomic population. The

term ‘normal’ means that the system is in thermal equilibrium.

If we substitute Bohr’s frequency condition {E2-E1=hv} in

equation (2)

We get

N2=N1e-hv/KT iii

53



Now Let us first assume that an ensemble of atoms in the

thermal equilibrium and that it is not subjected to any external

radiation field. At room temp or higher, a certain number of

atoms will be an excited state.

These atoms may loose energy in the form of radiation in units

of quanta hv, without any outside stimulus .This is called

spontaneous emission.

Again we consider the rate of this transition, that is the number

of atom in the higher state that make the transition to the lower

state , per second. Since N2 is the number of atoms in the

higher state the rate of the spontaneous transition P21, is given

by

P21=N2A21 iv

Where A21 is a proportionality Constant.

Now Let us assume that the system is subjected to some

radiation field. In that case, One of two processes may occur .It

depends on the direction of the radiation field with respect to

the phase of the oscillator. If the phases of the two systems

coincide, a quantum of the radiation field may cause the

emission of another quantum.

This process is called stimulated emission. Its rate is

𝑃21 = 𝑁2 𝐵21 𝜇𝑣 v

54



Where B21 is the another constant of proportionality and is the

energy density as a function of frequency v.

On the other hand, if the field of radiation field is in

a direction opposite to that of the oscillator, the impulse

transferred counteracts the oscillation energy is concurred, and

the system is raised to a higher state. This is the case of

absorption. Its rate is

P12=N1B12 vi

Where N1 is the number of atoms in the lower state

and B12 is another constant of proportionality. The three

constants A21B12 and B21 are known as Einstein’s co-efficients.

This indicates

that N2 decreases

exponentially as a

function of energy

difference between two levels.

If a quantum of energy ℎ𝜐, of the right frequency is incident on

an atom, the transition is upward and hence the quantum will

→ E

N2

↑

55



E2E

1E2

E1E

2E1

ABSORPTION

hv

be absorbed as sown in fig 1. If the transition is downward a

quantum is emitted.

This emission may occurs either spontaneously, fig-2 or it may

be occurs with the atom being stimulated to return to the lower

state fig 3.

With the system in thermal equilibrium, the net rate of

downward transition must equal the net rate of upward

transition.

N2A21+N2B21=N1B12

Or 𝑁2

𝑁1 A21B21

𝜇𝑣 𝑁2

𝑁1=B12

Or 𝑁2

𝑁1 =

𝐵12 𝜇𝑣

𝐴21+ 𝐵21 vii

But form equation iii we have

𝑁2

𝑁1=e-hu/KT

(1)

(2)

(3)

SPONTANEOUS

EMISSION

STIMULATED

EMISSION

56



Thus from vii we have

𝐵12 𝜇𝑣

𝐴21+𝐵21 𝜇𝜐= e-hv/KT

viii

On solving ,this yields

𝜇𝑣= A21

B12

1

𝑒ℎ𝑣/𝑘𝑇 − 𝐵21𝐵12

ix

A21+B21

B12= ehv/KT

A21

B12+

B21

B12= = ehv/KT

A21

B12= (ehv/KT-

B21

B12)

Or 𝜇𝑣= A21

B12

1

𝑒ℎ𝑣/𝑘𝑇 − 𝐵21𝐵12

In order to maintain thermal equilibrium, the system must

release energy in the form of radiation. The spectral distribution

of this radiation follows Planck’s radiation law in terms of

energy density (as a function of wavelength)

𝜇𝜆 =8𝜋ℎ𝑒

𝜆3 1

𝑒−ℎ𝑐/𝜆𝐾𝑇−1 x

The energy density must be consistent with Planck’s law

of any value of T. This is possible only when

57



B21=B12 xi

And 𝐴21

𝐵12 =

8𝜋ℎ𝑣3

𝑐3 xii

Where h is Planck’s constant, 𝜐 the frequency ,and c

velocity of light. Equations xi & xii are called Einstein’s

equations.

Generally, when an atomic system is in equilibrium absorption

and spontaneous emission take place side by side but when

N2>N1 absorption dominates. An incident quantum is more

likely to be absorbed than to cause emission. But one can find a

material that could be included to have majority of atom in

higher state than in lower state i.e. N2>N1 than we have a

condition called population inversion.

THE THRESHOLD CONDITION

Let us consider an optical resonator which consists of two

mirrors, one fully reflecting, separated by distance ‘𝑙’. The gap

consists of active medium.as for example ruby rod. Then

optical length of the medium is

𝑙’=𝑙 i

The optical standing waves are set up in the medium if

𝑙’=𝑚𝜆/2 ii

58



Where m is an integer. If ‘r’ is reflection co-efficient and I be the

intensity of falling light on the mirror, then

Ir=r𝑙 iii

If one mirror is made partially transparent then useful output

can be obtained.

The co-efficient of reflection for such a mirror is always less

than 1. If Kv is the absorption co-efficient at a frequency then

intensity of laser beam after traversing a distance x in the

medium is given by

𝐼𝜐(x)=𝐼𝜐(𝜐)𝑒−𝐾𝑣𝑥 iv

If Kv is negative i’e instead of absorption of energy there is

transverse in the energy then.

𝐼𝜐(x)=𝐼𝜐(𝜐)𝑒𝛼𝑥 v

Where α= -𝑘𝜐’ called unsaturated gain co-efficient This is so

because it represents gain before amplification. If R1 and R2 be

the reflection co-efficients of the two mirrors, the intensity per

passes will be diminished by a factor of R1R2.

Let R1R2=𝑒−2𝛽 or, 𝛽 = −1

2ln 𝑅1 𝑅2 vi

𝛽 represent loss of radiation of per single passage. As a result,

intensity changes to I0 exp (α𝑙-β) per single passage

59



For oscillation to be sustained in a laser the

amplification must be sufficient enough to compensate for energy

loss, thus

exp (α𝑙-β)>1 vii

α𝑙>β viii

Thus the threshold is reached when

α𝑙=β ix

Hence a laser operates when

α ≥β

𝚤 x

Now a laser beam is considered of frequency and v + dv and

intensity 𝐼𝜐 per unit frequency interval travelling in x direction

through a layer of active medium bounded by a width dx. Then

energy incident per unit area per unit time on the slab in the

frequency interval dv is 𝐼𝜐dv

The energy absorbed per unit time is

-δ𝐼𝜐= 𝐼𝜐𝑑𝜐𝑘𝜐dx xi

Total energy absorbed per second.

-dI = ∫ 𝐼𝜐𝑑𝜐𝑘𝜐dx

60



=𝐼𝜐dx∫𝐾𝜐𝑑𝜐 xii

In the above expression slight variation in intensity due to

change in frequency has been neglected.

If ρ(𝜐) be the energy density then,

𝐼𝜐=Uρ(𝜐) xiii Thus equation (12) may be written as

-dI= 𝑈ρdx ∫𝐾𝜐𝑑𝜐 xiv

In term of Einstein co-efficient energy absorbed is given by.

-dI=ℎ𝜐ρ (𝜐)(B12N1-B21N2)dx xv

From equation (xiv) and (xv) we have

Uρ(𝜐)dx∫𝐾𝜐𝑑𝜐 = ℎ𝜐ρ (𝜐)(B12N1-B21N2)dx xvi

Or, ∫ 𝐾𝜐𝑑𝜐 =ℎ𝜐

𝑈 N1B12 1-

𝐵21 𝑁2

𝐵12 𝑁1

= ℎ𝜐

𝑐N1B12 1-

𝐵21𝑁2

𝐵12 𝑁1 xvii

But, 𝐵21

𝐵12=

𝑔1

𝑔2 xviii

And, 𝐴21

𝐵21 =

8𝜋𝑉2ℎ𝜐

𝑐3 xix

Form the expressions of Einstein’s A&B co-efficient

61



We have

𝑢𝜐= 𝐴21/𝐵21

(𝑔1𝑔2

) (𝐵12𝐵21

) 𝑒𝑥𝑝(ℎ𝜐21 𝐾𝑇

)

With this equation, eq (17) becomes

∫𝑘𝜐𝑑𝜐 = ℎ𝜐

𝑐

𝑐3

8𝜐2ℎ𝜐

𝐴21𝑔2

𝑔1 (1 −

𝑔1

𝑔2

𝑁2

𝑁1) xx

∫𝑘𝜐𝑑𝜐 = 𝜇𝑐2 𝑔2𝑁1𝐴21

8𝑔𝜋𝑣2 (1 −𝑔1

𝑔2

𝑁2

𝑁1) xxi

Thus the absorption at frequency v can be written as

𝑘𝜐 = 𝜇𝑐2 𝑔2𝑁1𝐴21

8𝑔𝜋𝜐02 (1 −

𝑔1

𝑔2

𝑁2

𝑁1) g(𝜐) xxii

Where g(𝜐) accounts for the frequency distribution in the line, 𝜐0 is

the central frequency other line.

Now the unsaturated gain co-efficient α can be written as

α= 𝜇𝑐2 𝑔2𝑁1𝐴21

8𝑔1𝜋𝜐02 (

𝑔1

𝑔2 𝑁2

𝑁1− 1)𝑔(𝜐) xxiii

using equation (xvii),A/B in terms of 𝜔 ,we have

α= 𝜇𝑐2 𝜋2𝑔2𝑁1𝐴21

𝑔1𝜔02 (

𝑔1

𝑔2 𝑁2

𝑁1− 1)𝑔(𝜔) xxiv

but, g(𝜔)= 𝛾

2 .

1

(𝜔−𝜔02)+

𝛾2

4

xxv

62



Here 𝛾 is damping co-efficient. The maximum value of g(𝜔)is

when

𝜔 = 𝜔 0 and it is 2

r

But damping co-efficient

𝛾 = 1

𝜏21 = A21 xxvi

Hence,

𝛼(𝜔0) = 2μπc2g2N1

g1ω02 (

g1N2

g2N1− 1) xxvii

The maximum gain decrease as the sequence of the frequency

For a laser to operate

𝜇𝑐2𝑔2𝐴21

𝑔1𝜔02 (

𝑔1

𝑔2𝑁2 − 𝑁1)𝑔(𝜔) ≥

𝛽

𝑙

At the threshold

𝑔1

𝑔2𝑁2 − 𝑁1 =

𝛽

𝑙

𝑔1 𝜔02

𝜇 𝑐2𝑔2𝐴21𝑔(𝜔) xxviii

For non-desecrate state

63



N2-N1= 𝛽𝜔0

2

𝜇𝑐2𝑙𝐴21𝑔(𝜔) xxix

SHAWLOW-TOWNES CONDITION

P0=Initial number of photons in laser resonator

P=number of Photons that remain after m passages.

Number change due to finite reflectivities of mirrors.

Thus

𝑃𝑜 = 𝑃𝑜𝑒−𝛽𝑚 i

𝛽 =1

2𝑙𝑛𝑅1𝑅2 ii

If 𝜏𝑜= time taken by the photon in a single passage then,

t = time taken for the m passage = m𝜏0

Hence, P=P0𝑒−𝛽𝑡/𝜏0 iii

If 𝜏p, the average list of photon

Then,

𝑝

𝑝𝑜=

1

𝑒= 𝑒−𝛽𝜏𝑝/𝜏𝑜 iv

Or, 𝑒𝛽𝜏𝑝/𝜏𝑜=e

64



𝜏𝑝 =𝜏𝑜

𝛽=

𝑙

𝑐𝛽 v

Now equation,

N2-N1=𝛽 𝜔0

2

𝜇𝑐2𝑙𝐴21𝑔(𝜔) becomes

N2-N1=𝑙′ 𝜔0

2 𝜏21

𝜇𝑐3𝜏𝑝𝑙𝜋2𝑔(𝜔) vi

But,

𝑙′ = 𝜇𝑙

∴N2-N1=𝜔0

2𝜏21

𝜋2𝑐3𝜏𝑝𝑔(𝜔) vii

But, 𝑔(𝜔)= 2

𝜔𝑜 viii

Hence,

N2-N1=𝜔0

2𝜏21𝜔𝑜

2𝑐3𝜋 𝜏𝑝 ix

This is Shallow Townes condition for laser oscillation thus the

population difference required for laser oscillations depends

upon the two life times, the spontaneous life time for upper

energy line and effective decay time of cavity.

Requirement for population inversion and laser action :-

65



For population inversion and to produce laser action the

following arrangement are required. (i) An active medium (ii) A

pumping arrangement.

I. Active medium: - An active medium is a medium in which

atoms are in metastable energy state. Such atoms produce

more stimulated emission than spontaneous emission and

when excited soon reach the state of population inversion

leading to laser action.

II. Pumping: - The procedure to achieve population inversion

is called pumping. For achieving and maintaining the

condition of population inversion, the atoms are raised

continuously from the lower energy level to the upper

energy level. This required energy to be supplied to the

system. There are a number of techniques used for

producing population inversion as optical pumping,

electrical discharge, direct conversion etc. However,

optical pumping is very important and convenient method

.

Optical pumping:- The procedure adopted to achieve

population inversion is called pumping .In this method the

active material is illuminated with light of suitable

frequency=E2-E1/h. As a result, an atom in the lower energy

66



No

n-R

adia

tive

Tran

siti

on

𝜐𝑜

Metastable

State

(Lasing Level)

𝜐

state E1 absorbs the incident photon of energy ℎ𝜐 and is

raised to higher energy level E2.

Ground State

THREE LEVEL SCHEME

As the excited atoms lose their energy by spontaneous emission

and drop to the lower energy level in a very short time. Thus

the process fails to produce necessary population inversion

Therefore, in practice, population inversion is brought about by

following schemes.

(a) A three level scheme (b) a four level scheme

Three level scheme: - Let us consider an atom with energy

levels E1, E2 and E3 with increasing value of energy. E1 is the

Stim

ula

ted

Ab

sorp

tio

n

Short-lived

state

Stim

ula

ted

Emis

sio

n

Excited

Radiation

Amplified Laser

Radiation

𝜐

E3

E2

E1

67



ground state, E3 is short lived state and E2 is an intermediate

meta-stable state.

The transition from E3 to E1 is for bidden and transition from

E3 to E2 is allowed.

When these atoms are irradiated with an exciting radiation

of the right frequency 𝜐0=E3-E1/h, the atom are excited to

the E3 state by the process of stimulated absorption. Some

excited atom quickly drops to the intermediate level E2 by

spontaneous emission E2 or by non-radiative process there by

converting their excess energy into vibrational K.E of the

atoms forming the substance as shown in fig. As E2 is a

metastable state, the atoms remain in this excited state for

comparatively longer time of 10-3 sec as compared to 10-8, for

the short-lived state E3 and for this time, the population of

the state, E2 is more than that of state E1 thus resulting in

population inversion of the collection of atoms. If an atom in

the state E2 decays by spontaneous emission or stimulated

emmision.it emits a radiation of frequency.

𝜐=E2-E1/h. The photon may produce stimulated emission

from another atom there by giving two coherent photons,

moving in the same direction.

These two photons produce two more photons and so on,

producing an amplified beam of photons.

68



𝜐𝑜

Metastable State

𝜐′

Upper lasing level E2

E1

Four Level Scheme: - In four level scheme there is an additional

very short-lived energy state E11 between the ground state E1

and meta stable state E2, When exciting radiation of right

frequency 𝜐0=E3-E1/h is incident on the lasing medium the

atoms in the

ground state E1 are excited to level E3 by the process of

stimulated absorption. The atoms stay at the E3 level for a very

short time of about 10-8 sec and quickly dropdown to the

metastable state E2.

Spo

nta

neo

us

Ab

sorp

tio

n

Short-lived State

Stim

ula

ted

Emis

sio

n

Excited

Radiation Amplified Laser

Radiation 𝜐

E3

E2

69



E1

Very Shortlisted State

Lower lasing level

GROUND STATE

(FOUR LEVEL SCHEME)

Spontaneous transition from the metastable E2 to the level 𝐸1′

is forbidden and, therefore cannot take place. As a result, the

atoms accumulate at the level E2 and the population at this

level rapidly increases. Again the level 𝐸1′ is well above the

ground level E1 so that (𝐸1′ - E1)>>KT.

Therefore, at normal temperature atoms can’t jump to the

level E11 from the ground level 𝐸1 because of thermal energy.

As a result, the level 𝐸1′

is almost virtually empty. The

population inversion is thus maintained between the levels

E2 and 𝐸1′ .

Further, when the atom in the state E2 decays by

spontaneous emission or stimulated emission to the state 𝐸1′ .

it emits a photon of frequency 𝜐′= E3-E1/h.

This photon further produces stimulated emission

from the another atom and thus starts the laser action. The

level E11 being a very short lived state the atoms immediately

relax further to the ground state, ready for being pumped to

the level E3.

70



Comparison of four level pumping scheme with three level

pumping scheme: -

In three level pumping scheme the atoms finally return to

the ground state directly from the metastable states.

Therefore, in order to achieve population inversion pumping

has to go on until more than half of the ground state atoms

reach the metastable state, which is the actual lasing level, as

number of the atoms in the ground state is very large a very

high pumping power is required for the purpose. But in case

of four level scheme the terminal level is almost virtually

empty so that the condition of population inversion is readily

established even if a smaller

number of atoms arrive at the upper lasing level.

Consequently only a small pumping is required to establish

population inversion in four level scheme.

In three level pumping scheme, as soon as

stimulated emission starts. the population inversion

condition returns to normal population condition and lasing

stop as soon as the excited atoms drop to ground level. But in

case of the four level schemes, the populations inversion

71



continues without interruption and laser light is continuously

obtained. Semiconductor laser is a two level laser, ruby laser

is athree level laser and Nd Y A G is a four level laser.

RUBY LASER

72



Ruby rod is taken in the form of a cylindrical rod of 4cm in

length and 0.5cm in diameter. Its ends are grounded and

polished such that the end faces are exactly parallel and also

perpendicular to the axis of the rod. One face is silvered to

achieve 100% reflection while the other is silvered to give

10% transmission. The rod is surrounded by a helical

photographic flash lamp filled with Xenon. The lamp

produces flashes of white light whenever activated by the

power supply. The system is cooled with the help of a cool

out circulating around the ruby rod.

Working: - Ruby consist of a crystal of Al2O3 doped with 0.03

to 0.05% of Cr2O3 so that cr+3 ions replace some of the Al

atoms. The crystal field splits up the energy levels of cr+3 ions

in such a way that it has a short lived energy state E3=2.26ev

and an intermediate metastable energy state E2=1.79ev above

the ground state E1 .

73



2.26ev

1.79ev

Op

tica

l pu

mp

ing

The Xe flash tube as an optical pump and only a small fraction

of the energy emitted by it in the form of blue green radiation of

λ=5500A0 corresponding to energy level E3-E1 is absorbed by

the Ruby and used to excite the cr+3 ions the rest of the energy

is used up in heating the apparatus, which therefor, has to be

kept cold by the circulating liquid nitrogen through the glass

tube surrounding the ruby rod. The Cr+3 ions excited to the

short-lived 2.26ev level jump to the 1.79ev metastable level in

the first step by the losing 0.47ev of energy. As a result the level

E2 becomes more population than level E1.

The length of the Ruby rod is made an exact multiple

of half wavelength so that the light waves trapped in it form

an optical standing wave. Thus, the photons undergo

multiple reflection from the silvered end of the crystal till the

beam become sufficiently intense to emerge out of the

partially silvered end. Thus a highly intense,

E2

E1

E3

ʎ=5500Å

Lasertransition

ʎ=6943Å Metastable state

Radiation-less

transition 0.47ev

74



monochromatic, coherent and unidirectional beam is

obtained.

POPULATION INVERSION AND RUBY LASER

The phenomenon by which higher probability stimulated

emission is achieved is called population inversion.

The transition probability for induced emission

depend upon following factors as.

(i) The number of atoms in the excited state.

(ii) The energy density of incident radiation.

Let us consider a collection of large number of

atoms N0. in thermal equilibrium. If at TK temp

N1 be the number of atoms in the energy state E1

and N2 that in the energy state E2 then from

Maxwell =Boltzmann statistics, the distribution of

atoms in different energy state is given by

N1=N0e-E1/KT and N2=N0e

-E2/KT

Or,N2/N1=(e-E2/KT)/(e-E1/KT)=e-(E2-E1)/KT

Or, N2=N1e-(E2-E1)/KT=N1e

-hv/KT i

Where hv=(E2-E1)

As E2>E1, N2<N1 Thus in the normal distribution the number

of atoms in the higher energy state is less than the number of

75



atoms in the lower energy state. In other words, the population

of atoms in higher energy level is less than that in lower energy

levels.

When radiation of matching frequency v=E2-E1/h

is incident on such a collection the atom in excited due to

stimulated absorption. The excited atoms return to the normal

state by any of the two processes i.e spontaneous emission or

stimulated emission. In order to achieve higher probability of

stimulated emission two condition must be satisfied.

(i)The higher energy state should have a longer mean life i.e it

should be a metastable state.

(ii)The number of atom in the higher energy state E2 must be

greater than that in E1.

Thus to establish a situation in which the number of atom in

the higher energy level is greater than that in the lower energy

level is called population inversion. It is a non-equilibrium

condition The procedure adopted to achieve population

inversion is called pumping e.g. optical pumping electrical

discharge, direct conversion etc. However, optical pumping is

very important and convenient method for population

inversion.

In optical pumping, the active material is illuminated with light

of suitable frequency.

76



𝜐=E2-E1/h

Q :- Describe with a neat sketch the construction and working

of a He-Ne laser with special reference to energy level diagram

mention some industrial applications of laser. What is meant by

nonlinear effect?

Ans :- On the basis of the nature of used gas, we may

distinguish atomic and molecular gas lasers. An example of gas

laser is the helium-neon laser.

Construction: -

A schematic diagram of He-Ne laser is shown in diagram.

77



It consists of a pyrex tube of about 0.5m in length and

5mm in diameter. The tube is filled with He-Ne mixture in the

ratio 5:1at a total pressure of 1mm of Hg. The tube has parallel

mirrors at both ends of which one is partly transparent. The

spacing between the mirrors is equal to an integral number of

half wavelength of the laser light. The mixture in the tube is

ionized by passing a D.C current through the gas He-atoms are

exited very efficiently by electron

impact into the 2s levels while Ne atoms are much less readily

excited by the electrons as shown in energy-level diagram This

excited 2s state of helium is relatively long lived the energy of

this level is 20.61ev which is almost the same as the energy of

the 5s level in neon, which 20.66ev. Hence the energy of the

helium atoms is easily transferred to the neon atoms to the 55

state results in a population inversion between the 55 and 3p

states.

The purpose of the He-atoms 1s thus to help achieve

a population inversion in the Ne-atoms. The spontaneous

transition from the 5s state to the 3p state produce photons of

wavelength 632.8mm which in turn trigger stimulated

transitions.

2S

5S

78



*

*

ENERGY - LEVEL DIAGRAM

Photons travelling parallel to the tube are reflected back and

forth between the mirror placed at the ends and rapidly build up

in to intense beam which escapes through the end with the

lower reflectivity. The Brewster ‘s angle end windows allow

light of one polarization to pass through without any reflection

losses. The electrons impacts excite the He and Ne atoms

occurs all the time and a He-Ne laser operates continuously.

APPLICATION OF LASER

1.In Technical and industrial fields :-

20.66ev

2P

3P

18.70ev

3S

SPONTANEOUS

EMISSION

632.8mm

LASER

TRANSITIO

N

ELECTRON MPACT

1S2

20.61ev

RADIATIONLESS

TRANSITION

79



The laser beam is used for cutting fabrics for clothing in

one hand , and steel sheets on the other.

It can drill minute holes in paper clips, single hair strand and

hard materials including teeth and diamond, Laser welding

melts and joins metallic rods laser heat treatment can harden

surfaces of engine crankshafts and the cylinder walls. Laser

beam is used to vapourise unwanted materials during

manufacture of electronic circuits on semiconductor chips.

2.In the Medical field :-

The laser beam is used for spot welding of detached retina

in grafting cornea in drilling holes in bones destroying specific

cancerous areas with in tissues in bloodless surgery like vocal

cord operation, stomach ulcers, kidney stones in cutting and

sealing small blood vessels during brain operation and while

performing microsurgery on cell and chromosomes.

3.In science and Research :-

In science and research laser has become a very useful

tool. Experiments with Michelson Interferometer using laser

80



show that fringes are obtained even with path difference of 9m

against 3m with ordinary light. Laser is also bring used in

ionization and dissociation studies of gases It is also used in

spectrochemical analysis of solids in conjunction with mass and

emission spectrometer and in a investigation of energy transfer

by exciting a specific vibration level and absorbing its decay in

polyatomic molecules.

Non Linear Effect :-

The refractive index and the absorptivity of a material are

independent of light that passes through the material. some

laser have electric field strengths as high as 109v/m, more than

enough to cause the breakdown of air (3x106v/m) and a few

orders of magnitude less than electric field holding a crystal

together.

The coulomb force between the nucleus and the

electron in an hydrogen atom, is

F=8x10-8 newton.

Now electric field strength

�⃗� = 𝐹

𝑞 =8x10-8/1.6x10-19=5x1011v/m

The above value is nearly same as the local field inside

crystal. This causes a number of phenomena such as the

invention of high power lasers. Light form such laser can easily

81



change the refractive index as well as the absorptivity of a

material. This effect of changing the properties of materials by

laser is called non-linear effect.

SEMICONDUCTOR INJECTION LASER

In 1961Bernard and Duraffourg gave the idea that

stimulated emission of photons is possible in semiconductors

form transition between conduction band and valance band.

Science then semiconductors laser has been widely developed.

The first semiconductor laser has made from gallium arsenide,

GaAs. These lasers are very useful particularly in optical

communication.

PRINCIPLE: - A semiconductor has small energy gap Eg

(=1ev) between valance band occupied by electrons and

82



The empty conduction band as shown in fig (i.a). At room

temperature however some of the valence electrons acquire

thermal energy greater then Eg and cross over in to the

conduction band leaving behind holes in the valence band. If a

light photon of energy grater than Eg happens to interact with

the electrons one of the two processes may occur (i) the photon

may be absorbed by the valence band electrons which would be

excited to conduction band leaving behind a hole in valance

band.

(ii)The photon may stimulate an already excited

conduction band electron which would drop to the

83



valance band electron emitting a fresh photon in

coherence with the stimulating photon. The probability of

the two processes depends upon whether most of the

electrons are in the valence band on in the conduction

band. If, however by some means a large number of

electron are made available in higher energy states

stimulation emission is promoted. This situation is called

population inversion. This can be achieved near a p-n

junction having high doping densities and forward

currents.

Basic structure: - The basic structure of a gallium arsenide p-n

junction used as an injection laser is shown in fig 2.

84



The pair of parallel planes perpendicular to the plane of the

junction is polished while the two remaining sides of the diodes

are roughened. when a forward bias is applied to the laser

diode, a current flow. The injected electrons move from the n-

side to the p-side and holes from the p-side to the n-side. As the

electron and holes recombine photons are emitted. These

photons are reabsorbed or radiated away. This is the

spontaneous emission which occurs in all directions at low

current. As the current increased eventually a threshold current

is reached at which the emitted photon stimulated the emission

of more photons. These photons are internally reflected several

times at the polished walls, stimulating more and more

photons, all coherent with them when the photons beam

becomes sufficiently intense, it emerges out from the junction.

The main difficulty with GaAs laser is the high

threshold current density(≈105A/cm2) at room temperature.

Hence this laser could be operated only at law temperature at

which the required current density is lower.

Semiconductor laser are similar to other lasers in that the

emitted radiation is intense, monochromatic, directional and

highly coherent, both spatially or temporally. However, in

some respects semiconductor laser differ from other lasers as.

85



(i) In semiconductor laser the electron transition is associated

with the band structure of the materials whereas in other laser

the transition take place between discrete energy levels.

Therefore, the radiation from a semiconductor laser is less

monochromatic than that form other lasers.

(ii) A semiconductor laser is very compact in size less than

1mm.

(iii) In the P-n junction laser, the laser action is produced by

passing a forward current through the function. Hence the

emitted laser light can be modulated in to light pulses simply by

modulating the current.

(iv) semiconductor lasers have short photon life times.

Therefore, modulation at higher frequency can be achieved.

(v) Due to its compact size and capability of high frequency

modulation, is the best light source for optical fiber

communication.

THE CARBONDIOXIDE (CO2)LASER

The CO2 laser is a molecular gas laser .A molecule is made up

of two or more atoms. Hence, in addition to the electronic

motions, atoms in the molecule, may vibrate in different modes

and the molecule as a whole may rotate about some axis.

86



The three modes of vibration of a co2 molecule is shown in

given figure.

These are

(i)symmetric stretching mode having frequency

𝜐1=1337 cm-1 (ii) Bending mode having

Frequency, 𝜐2= 667 cm-1 and (iii) Asymmetric stretching. As

such a molecule has discrete electronic levels (like atoms), each

electronic level has a number of vibrational levels associated

with it, and each

ANTISYMMETRIC STRETCHING

87



LASER 9.6µm

𝜐𝑁 =1

vibrational has a number of rotational levels associated with it.

A transition between two electronic levels gives radiation in the

visible and ultraviolet region and a transition between

vibrational levels of same electronic levels give radiation in the

infra-red region, while a transition between rotational levels of

the same vibrational levels gives radiation in the far infra-red or

microwave region.

Working:- The laser uses Co2 gas mixed with N2 gas. The active

centres are the Co2 molecules which lose when transition occur

between a rotational levels associated with a vibrational level,

and a rotational levels associated with a lower vibrational level

of the electronic ground state. The vibrational and rotational

levels of the electronic ground state of Co2 and those of N2 are

shown in given fig

LASER 10.6µm 𝜐1 𝜐2

𝜐3

A

CO2 N2

→EN

ERG

Y(C𝑚

1)

)

𝜐N=0

C

88



When a discharge is passed in a tube containing Co2 the

electrons collision excite the N2 and Co2 molecules to higher

vibrational-rotational levels. The collision cross-section for

excitation of co2 to levels A is very large. This level is also

populated by radiation-less transition from upper excited levels.

Further the level A of Co2 is nearly coincident with the highly

populated 𝜐𝑁=1 level of N2 Thus, there is a very efficient

transfer of energy from a N2 molecule to a Co2 molecule,

resulting in the excitation of Co2 molecule. This increases the

pumping efficiency of the Co2 laser.

As a result above processes, population inversion is

established between level A and the lower B and C. This

initiates laser action producing laser beams at 9.6𝜇m and

10.6𝜇m.

APPLICATIONS:- Output powers of several watts to several

hundred watts are obtained from CO2 laser. Hence CO2 laser

because of its high output power, used in industry for welding,

drilling, cutting etc. They are also used in open air

communication system and optical radar systems.

CONSTRUCTION: -

89



Experimental arrangement of CO2 laser is shown in diagram. It

consists of a discharge tube about 26cm long and of 1.5mm2 in

crossectional area. The discharge tube is filled with CO2, N2

and He in the ratio 1:4:5 and small amount of water vapour.

However, laser action only takes place in co2 which is the active

medium between the vibrational levels and rotational levels

accompanying the vibrational levels.

ND : YAG LASER

NEODYMIUM YTTRIUM ALUMINIUM GARNET

LASER

The neodymium ion is a rare earth ion. It produces emission at

about 1um when doped into a solid state proper host crystal.

Most commonly used host materials are yttrium aluminum

garnet (YAG) crystal and glass. When doped in YAG, Nd3+

90



UPPER LASER LAVEL

LOWER LASER LEVEL

GROUND LEVEL

ions take the place of yttrium ions. Doping Concentrations are

typically of the order of 0.725% by weight corresponds to about

1.4x1026 atoms per cubic meter.

PRINCIPLE AND WORKING :-

The basic principle

of operation of the

Nd : YAG laser is the

Same as that of a

Ruby laser. It is also

a four level system.

The energy level

Diagram is shown

in the adjacent fig.

The pumping of Nd3+ Ions to the upper state is achieved by

using intense flash of light

E3

E2

𝐸1′

E1

1.06𝜇m

91



from a xenon flash tube or a krypton are lamp giving light of

wavelength 5000A0 to 8000A0. The light excites the Nd3+ ions

from the ground state E1 to various higher energy state of E3.

These ions immediately relax down to the level E2 which is a

metastable level with a life times of 0.25m sec. Laser action

takes place between the upper level E2 and lower laser level E1

at a wavelength of about 1.06um in the infra-red region.

The active medium for the laser is the rare earth metal ion Nd3+

dopped with yttrium aluminium Garnet (Y3Al5O12). The YAG

itself does not take part in the laser process. It only provides a

host crystal lattice so that when Nd3+ ion is placed in the YAG

lattice, it is subjected to the electrostatics field of the

surrounding ions, called the crystal field. The crystal field

modifies the transition probabilities between various energy

level of Nd3+ ions so that some transition which are forbidden

become allowed to crystal field. The pumping flash lasts only

for a short times 1ms. The laser output, therefore, is in the form

of a pulse which starts about 0.5ns after the pumping flash is

produced. This represents the time, for the population to build

up. The stimulation process builds up even more rapidly. It,

therefore, depopulates the upper levels much faster than the

pumping can replace the excited atoms. Thus the laser action

92



momentarily stops until population inversion builds up again

the process repeats itself.

Q:- Establish Einstein’s Equation for Boltzmann’s ratio and

explain population inversion with reference to Boltzmann’s

distribution curve.

Ans:- The number of atom. Per unit volume that exits in a

given energy state is called population and is denoted by N. It is

represented by Boltzmann’s equation as,

N=e-E/KT i

Where E is energy level of the system ‘K’ Boltzmann constant

ant T, the absolute temp.

In any process of absorption or emission at least two energy

state are involved. Thus if N1 and N2 be the populations in the

two states and E1 and E2 their respective energy levels then.

N1= 𝑒−𝐸1/𝐾𝑇

And N2= 𝑒−𝐸2/𝐾𝑇

Or N2/N1= 𝑒−𝐸2/𝐾𝑇/𝑒−𝐸1/𝐾𝑇 ii

The ratio of the population in these two states. N2/N1 ia called

Boltzmann’s ratio or relative population. Form Bhor’s

frequency condition equation (ii) may be written as,

N2=N1 𝑒−ℎ𝜐/𝐾𝑇 iii

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Let us assume that an ensemble of atoms is in thermal

equilibrium. At room temp. or higher, a certain number of

atoms will be an excited state. These atoms may loose energy in

the form of radiation in units of quanta ℎ𝜐 without any outside

stimulus. This is called spontaneous emission.

The rate of this transition i,e the number of atoms in the higher

state that make the transition to lower state per second. Since

N2 is the number of atoms in the higher state, the rate of the

spontaneous transition is given by

P12=N1A21 iv

Where A21 is proportionality constant.

Now the system is supposed to be subjected to some

radiation field. If the phases of the two system coincide, a

quantum of the radiation field may cause the emission of

another quantum. This process as anticipated by Einstein is

now called stimulated emission. Its rate is represented as,

P21=N2B21𝜇𝜐 v

Where B21 is proportionality constant and 𝜇𝜐 is the energy

density as a function of frequency 𝜐.

If the phase of the radiation field is in a direction opposite

that of the oscillator and the system is raised to a higher state by

absorbing energy then rate is expressed as

P12= N1B21𝜇𝜐 vi

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ABSORPTION

ℎ𝜐 E2E1

E2E1 (a)

Where N1 is the number of atoms in the lower state and B12 is

proportionality constant.

The three constant A21, B21 and B12 are known as

Einstein’s co-efficient.

The phenomena of absorption and emission can be shown

by following diagrams

If a quantum of energy 𝜇𝜐 incident on an atom the transition is

upward and hence the quantum will be absorbed as fig(a).

Conversely. If the transition is downward a quantum is emitted.

This may occur either spontaneously as fig(b) or it may occur

with the atom being stimulated to return to lower state as fig(c).

If the system be in thermal equilibrium, then the net rate

of downward transition must equal the net rate of upward

transition.

SPONTANEOUS

EMISSION

STIMULATED

EMISSION

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N2A21+N2B21𝜇𝜐 =N1B12 𝜇𝜐

Or,𝑁2

𝑁1 A21+ N1B12𝜇𝜐 /N1=B12𝜇𝜐

Or, 𝑁2

𝑁1= B12𝜇𝜐/A21+ B21𝜇𝜐 vii

With the help of equation (iii) above equation is written as,

B12𝜇𝜐/A21+B21𝜇𝜐=𝑒−ℎ𝜐/𝐾𝑇 viii

On solving this equation, we get

𝑢𝜐=A21/B12{1/𝑒ℎ𝜐/𝐾𝑇-B21/B12} ix

In order to maintain thermal equilibrium. The system must

release energy in the form of radiation. The spectral distribution

of this radiation follows Planck’s radiation formula in term of

energy density as.

Uλ=8 hc/λ5 1/ehc/KTλ-1 x

The energy density must be consistent with Planck’s law for

any value of T. This is possible only when.

B21=B12 xi

and A21/B12=8hc/ λ5 xii

Above equation (xi)and (xii) are called Einstein’s equation

when an atomic system is in equilibrium, absorption and

emission(spontaneous) take place side by side. But N2<N1 so

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absorption dominates. But when N2>N1 I’e in a material in

which majotity of atoms in the higher state than in lower state

then this condition is called population inversion.

E

E3

E2

E1

N3

N2

N1 N

(POPULATIOPN OF THREE LEVEL SYSTEM UNDER CONDITION OF

THERMAL EQUILIBRIUM GRAPH 1)

N3

E

E3

E2

E1

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THREE LEVEL SYSTEM SHOWING POPULATIOPN

INVERSION OF E2 w.r.t E1 GRAPH 2

In graph-1 the three states of a three levels system may be

designated as E1, E2 and E3. The uppermost state E3 is

populated least and the lowest state E1 is populated most. The

curve shown represents the Boltzmann distribution. Since

the population in the various states is such that N3<N2<N1 the

system is absorptive rather than emissive. But when the system

is being excited by supplying extraneous energy then this may

cause either N3 or N2 or both exceed N1 as shown in graph-2

N2

N1 N

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