Answer
The probability that a single mode
has energy En = nhν is given by the usual Boltzmann factor
P(n)=∑exp(−kThν)exp(−kThν)
Let x=exp(−kThν)=kThν For classicla limit
So average energy
U=∑EP(n)
U=kT
The energy density of radiation range is
u(ν)dν=c38πv2Udν=c3(e(kThν)−1)8πhv3dν
This is planck radiation distribution law.
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