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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