Answer to Question #169519 in Molecular Physics | Thermodynamics for Khushi

Question

Derive Planck’s law of black body radiation.

Answer 1

Answer

The probability that a single mode

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

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

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

So average energy

$U=\sum E P(n)$

$U=kT$

The energy density of radiation range is

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

This is planck radiation distribution law.