1000 Solved Problems in Modern Physics

116 2 Quantum Mechanics – I

Term SPDF

l 0123

Parity=(−1)l + 1 − 1 + 1 − 1

2.47 J=l+s= 0 + 1 / 2 = 1 / 2
F=I+J,I+J− 1 ,...I−J
= 2 , 1 , 0

2.48 The observed frequency (ω) of radiation from an atom that moves with the
velocity v at an angleθto the line of sight is given by

ω=ω 0 (1+(v/c) cosθ)(1)

whereω 0 is the frequency that the atom radiates in its own frame of referenece.

The Doppler shift is then

Δω

ω 0

=

ω−ω 0

a 0

=

(v

c

)

cosθ (2)

As the radiating atoms are subject to random thermal motion, a variety of

Doppler shifts will be displayed. In equilibrium the Maxwellian distribution

gives the fractiondNNof atoms with x-component of velocity lying betweenvx

andvx+dvx

Fig. 2.4Thermal broadening
due to random thermal
motion

dN

N

=

exp

[

−

(vx

U

) 2 ]

√

π

dvx

U

(3)

whereu/

√

2 is the root-mean-square velocity for particles of massMat tem-

peratureT.Now

u=

(

2 kT

M

) 1 / 2

(4)

wherek= 1. 38 × 10 −^23 J/K is Boltzmann’s constant.

Introducing the Doppler widthsΔωDandΔλDin frequency and wavelength

ΔωD

ω 0

=

ΔλD

λ 0

=

U

c

=

(

2 kT

Mc^2

) 1 / 2

(5)