A gas consists of atoms of mass `m` being in thermodynamic equilibrium at temperature `T`. Suppose `omega_(0)` is the natural frequency of light emitted by the stoms.
(a) Demonstrate that the spectral distribution of the emitted light is defined by the formula
`I_(omega) = I_(0)e^(-a(1-omega//omega_(0))^(2)`,
`(I_(0)` is the spectral intensity corresponding to the frequency `omega_(0), a = mc^(2)//2kT)`.
(b) Find the relative width `Delta omega// omega_(0)` of a given spectral line. i.e. the width of the line between the frequencies at which `I_(omega) = I_(0)//2`.

Explanation

Let `v_(x)` be the projection of the velocity vector of the radiating atom on the observation direction. The number of atoms with projections falling within the interval `v_(x)` and `v_(x) + dv_(x)` is
`n(v_(x))dv_(x ~) exp(-mv_(x)^(2)//2kT) dv_(x)`
The frequency of light emitted by the atoms mocing with velocity `v_(x)` is `omega = omega_(0)(1+(v_(x))/(c ))`. From teh expressions the frequency of atoms can be found `:n(omega) d omega = n(v_(x)) dv_(x)`. Now using
`v_(x) = c(omega - omega_(0))/(omega_(0))`
we get `n(omega) domega_(~) exp (-(mc^(2))/(2kT)(1-(omega)/(omega_(0)))^(2))(cd omega)/(omega_(0))`
Now the spectral radiation density `I_(omega) prop n_(omega)`
Hence `I_(omega)=I_(0)e^(-a(1-(omega)/(omega_(0)))^(2)), a = (mc^(2))/(2kT)`
(The constant of proportinality is fixed by `I_(0)`.)
(b) On putting `omega = omega_(0)+- (1)/(2) Delta omega` and demanding
`I_(omega) = I_(0)//2`
we get `(1)/(2) = e^(-a((Delta omega)/(2omega_(0)))^(2))`
so `a((Delta omega)/(2omega_(0)))^(2) = In2`
Hence `(Delta omega)/(2omega_(0)) = sqrt((2In2)kT//mc^(2))`
and `(Delta omega)/(omega) = sqrt((2In2)kT//mc^(2))`