1000 Solved Problems in Modern Physics

2.3 Solutions 117

Further from (2) d(Δω)=dω. The relative distribution of Doppler shift is

dN

N

=

exp

[

−

(

Δω

ΔωD

) 2 ]

√

π

dω

ΔωD

(6)

Thus a Gaussian distribution is produced in the Doppler shift due to the

random thermal motion of the source (Fig. 2.4).

The intensity of radiation is

I(ω)=

exp

[

−

(

Δω

ΔωD

) 2 ]

ΔωD

√

π

(7)

centered around the unshifted frequencyω 0. The width of the distribution at

the frequencies whereI(ω) falls to half the central intensityI(ω 0 ) is known as

the half width

Doppler half width=2(ln2)^1 /^2 ΔωD

= 2 ω 0

(

2 kT ln 2

Mc^2

) 1 / 2

(8)

Thermal broadening is most pronounced for light atoms such as hydrogen and

high temperatures, for example the Hαline (6,563A) has a Doppler width of ̊

0.6A at 400 K. ̊

2.49 Lande’ g-factor is

g= 1 +j(j+1)+s(s+1)−l(l+1)/ 2 j(j+1)

For the term^2 P 3 / 2 ,l= 1 ,J=^32 ,s=^12 andg=^43

For^2 S 1 / 2 ,l= 0 ,j=^12 ,s=^12 andg= 2

Fig. 2.5Anamolous Zeeman
effect in an alkali atom. The
lines are not equidistant