Question

The wavelength of a resonat mercury line is `lambda==253.65nm`. The mean life time of mercury atoms in the state of resonance excition is `tau=.15 mu s`. Evaluate the ration of the Doppler line broadening to the the natural linewidth at a gas temperature `T=300K`

Answer 1

A short lived state of mean life `T` has an uncertainity in energy of `DeltaE~( ħ)/(T)` which is transmitted to the photon it emits as natural broadening. Then
`Delta omega_(nat)=(1)/(T)` so `Delta_(lambda)_(nat)=(lambda^(2))/(2pic tau)`
The Dopper broadening on the other hand arises from the thermal motion of radiating atoms. The effect is non-relativistic and the maximum brodening can be written as
`Delta(lambda_(Dop))/(lambda)=2 beta=(2v_(pr))/(c )`
Thus `(Deltalambda_(Dopp))/(Delta lambda_(nat))=(4piv_(pr)tau)/(lambda)`
substitution gives using `v_(pr)=sqrt((2RT)/(M))= 157 m//s`
`(Delta lambda_(Dopp))/(Delta lambda_(nat))~~1.2xx10^(3)`
Note:- Our formula is an order of magnitude estimate.