Derivation of Planck's Radiation Law

Derivation: Let $ N$ be the total number of Planck’s oscillators and $E$ be their total energy, then the average energy per Planck’s oscillator is

$ \overline{E}=\frac{E_{N}}{N} \qquad (1)$

Let there be $ N_{0}, N_{1} ,N_{2} ,N_{3},---N_{n}$ oscillator having energy $ E_{0}, E_{1}, E_{2}, -- E_{n}$ respectively.

According to Maxwell’s distribution, the number of oscillators in the $ n^{th}$ energy state is related to the number of oscillators in the ground state by

$ N_{n}=N _{0} e^{\tfrac{-nh\nu }{kt} }\qquad (2)$

Where $ n$ is a positive integer. So put $ n= 1,2,3,…….$. The above equation can be written for different energy states. i.e.

$ N_{1}= N _{0} e^{\tfrac{-h\nu }{kt} }$

$ N_{2}= N _{0} e^{\tfrac{-2h\nu }{kt} }$

$ N_{3}= N _{0} e^{\tfrac{-3h\nu }{kt} }$

$.............$

$.............$

So, the total number of Planck’s Oscillators –

$ N= N _{0} +N_{1}+N_{2}+N_{3}+.... N_{n}$

$ N= N _{0} + N_{0} e^{\tfrac{-h\nu }{kt}}+ N_{0} e^{\tfrac{-2h\nu }{kt}}+...+ N_{0}e^{\tfrac{-nh\nu }{kt}}$

$ N= N _{0}[1+e^{\tfrac{-h\nu }{kt}}+e^{\tfrac{-2h\nu }{kt}}+.... e^{\tfrac{-nh\nu }{kt}}] \qquad(3)$

Let $ x = e^{\frac{-h\nu }{kt}} \qquad (4) $

Then $ N = N_{0}[1+x+x^{2}+x^{3}+...+ x^{n}]$

$ N = \frac{N_{0}}{1-x} \qquad(5)$ [ from Binomial theorem]

Now the total energy of oscillators –

$ E_{N} = E_{0}N_{0}+ E_{1}N_{1}+ E_{2}N_{2}+...+ E_{n}N_{n}$

$ E_{N } = 0.N_{0}+ h\nu N_{0} e^{\tfrac{-h\nu }{kt}}+...+nh\nu N_{0} e^{\tfrac{-nh\nu }{kt}}$

$ E_{N } = N_{0}h\nu (e^{\tfrac{-h\nu }{kt}}+2e^{\tfrac{-2h\nu }{kt}}+...+ ne^{\tfrac{-nh\nu }{kt}})$

From equation $(4)$ put $ x = e^{\tfrac{-h\nu }{kt}}$ in above equation. i.e

$ E_{N } = N_{0}h\nu (x+2x^{2}+3x^{3}+...+nx^{n}) $

$ E_{N } = N_{0}h\nu x(1+2x+3x^{2}+....)$

$ E_{N } = \frac{N_{0}h\nu x}{(1-x)^{2}} \qquad(6)$ (from Bionomical theorem)

Now substituting the value of $N$ from equation $ (5)$ and $ E_{n}$ from equation $ (6)$  in equation $ (1)$ –

$ \overline{E}= \frac{\frac{N_{0}h\nu x}{(1-x)^{2}}}{\frac{N_{0}}{(1-x)}}$

$ \overline{E}= \frac{h\nu }{(\frac{1}{x}-1)}$

$ \overline{E}= \frac{h\nu }{e^{\tfrac{h\nu }{kt}}-1} \qquad(7)$

The number of oscillators per unit volume in wavelength range to $ ( \lambda + d\lambda )$ is $\frac{8\pi }{\lambda ^{4}} d\lambda$.

The energy per unit volume $ (E_{\lambda }d\lambda )$ in the wavelength range to $( \lambda +d\lambda ) $ is –

$ E_{\lambda}d\lambda = \frac{8\pi }{\lambda ^{4}}d\lambda \overline{E} \qquad(8)$

From equation $ (7)$ and $ (8)$ –

$ E_{\lambda}d\lambda = \frac{8\pi }{\lambda ^{4}}d\lambda \frac{h\nu }{(e^{\tfrac{h\nu}{kt}}-1)}$

$ E_{\lambda}d\lambda = \frac{8\pi hc}{\lambda ^{5}} \frac{d\lambda}{(e^{\tfrac{hc}{\lambda kt}}-1)}$

The above equation describes Planck’s radiation law and this law was able to thoroughly explain the black body radiation spectrum.

Wien’s Displacement law from Planck’s Radiation Law:

Planck’s radiation law gives the energy in wavelength region $ \lambda to \lambda +d\lambda $ as –

$ E_{\lambda}d \lambda = \frac{8\pi hc}{\lambda ^{5}}(\frac{1}{e^{\tfrac{hc}{\lambda kt}}-1})d\lambda \qquad(1)$

For shorter wavelength $ \lambda T$ will be small and hence

$ e^{\tfrac{hc}{\lambda kt}}> > 1$

Hence, for a small value of $\lambda T$ Planck’s formula reduces to -

$ E_{\lambda}d \lambda = \frac{8\pi hc}{\lambda ^{5}}(\frac{1}{e^{\tfrac{hc}{\lambda kt}}}) d\lambda$

$ E_{\lambda}d\lambda = \frac{8\pi hc}{\lambda ^{5}}e^{\tfrac{-hc}{\lambda kt}}d\lambda$

$E_{\lambda}d\lambda = A \lambda ^{-5} e^{\tfrac{-hc}{\lambda kt}}d\lambda$

Where $ A = 8\pi hc$

The above equation is Wien’s law of energy distribution verified by Planck radiation law.

Rayleigh-Jeans law from Planck’s Radiation Law:

According to Planck’s radiation law –

$ E_{\lambda}.d\lambda = \frac{8\pi hc}{\lambda ^{5}}\frac{1 }{e^{\tfrac{hc}{\lambda kt}}-1}.d\lambda$

For longer wavelength $ e^{\frac{hc}{\lambda kt}}$ is small and can be expanded as-

$ e^{\tfrac{hc}{\lambda kt}} = 1+\frac{hc}{\lambda kt}+\frac{1}{2!}(\frac{hc}{\lambda kt})^{2}+....$

Neglecting the higher-order term –

$ e^{\tfrac{hc}{\lambda kt}} = 1+\frac{hc}{\lambda kt}$

Hence for longer wavelength, Planck’s formula reduces to –

$ E_{\lambda}.d\lambda = \frac{8\pi kt}{\lambda ^{5}}[\frac{1}{1+\frac{hc}{\lambda kt}-1}]$

$ E_{\lambda}.d\lambda = \frac{8\pi kt}{\lambda ^{4}}.d\lambda$

This is Rayleigh Jean’s law verified by Planck Radiation Law.