Question #169519

Derive Planck’s law of black body radiation.


1
Expert's answer
2021-03-12T07:14:45-0500

Answer

The probability that a single mode

has energy En = nhν is given by the usual Boltzmann factor

P(n)=exp(hνkT)exp(hνkT)P(n) =\frac{exp(-\frac{h\nu }{kT}) }{\sum exp(-\frac{h\nu }{kT}) }


Let x=exp(hνkT)=hνkT{exp(-\frac{h\nu }{kT}) }=\frac{h\nu }{kT} For classicla limit

So average energy

U=EP(n)U=\sum E P(n)


U=kTU=kT

The energy density of radiation range is

u(ν)dν=8πv2Udνc3=8πhv3dνc3(e(hνkT)1)u(\nu) d\nu =\frac{8\pi v^2 U d\nu}{c^3}\\ =\frac{8\pi hv^3 d\nu}{c^3(e^(\frac{h\nu}{kT}) -1)}

This is planck radiation distribution law.


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