The electromagnetic intensity thermally radiated by a body of temperature $\mathbf{T}$is given by $\mathbf{I}\mathbf{=}{\mathbf{\sigma T}}^{\mathbf{4}}$where$\mathbf{\sigma }\mathbf{=}\mathbf{5}\mathbf{.67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{\text{W}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{2}}\mathbf{\cdot }{\mathbf{\text{K}}}^{\mathbf{4}}$

This is known as the Stefan-Boltzmann law. Show that this law follows from equation (9-46). (Note: Intensity, or power per unit area, is the product of the energy per unit volume and distance per unit time. But because intensity is a flow in a given direction away from the blackbody, the correct speed is not $\mathbf{c}$. For radiation moving uniformly in all directions, the average component of velocity in a given direction is$\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\text{c}}$ .)

Expression for energy of photons in container (U) is given by-

$\mathbf{U}\mathbf{=}\frac{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{Vk}}_{\mathbf{B}}^{\mathbf{4}}{\mathbf{T}}^{\mathbf{4}}}{\mathbf{15}{\mathbf{h}}^{\mathbf{3}}{\mathbf{c}}^{\mathbf{3}}}$ ……. (1)

Where;

${\mathbf{k}}_{\mathbf{B}}\mathbf{\to }$Boltzmann's constant

$\mathbf{h}\mathbf{\to }$Planck's constant

$\mathbf{c}\mathbf{\to }$Speed of light in vacuum

$\mathbf{T}\mathbf{\to }$Temperature

$\mathbf{V}\mathbf{\to }$Volume of container

Alternate expression for intensity -

$\mathbf{I}\mathbf{=}\left(\frac{\text{energy}}{\text{volume}}\right)\left(\frac{\text{distance}}{\text{time}}\right)$ ……. (2)

To start, equation (2) can be rewritten slightly:

$\begin{array}{c}\mathrm{I}=\left(\frac{\text{energy}}{\text{volume}}\right)\left(\frac{\text{distance}}{\text{time}}\right)\\ \left.=\left(\frac{\text{energy}}{\text{volume}}\right)\text{(velocity}\right)\\ \mathrm{I}=\frac{\mathrm{U}}{\mathrm{V}}\mathrm{v}\dots \dots ..\left(3\right)\end{array}$ ….. (3)

Use that the average component of velocity in any particular direction is $\mathrm{c}/4$, set v equal to $\mathrm{c}/4$, and use equation (1) for the energy $\mathrm{U}$ in equation (3):

$\begin{array}{c}I=\left(\frac{U}{V}\right)v\\ =\left[\frac{\left(\frac{8{\pi }^{5}V{k}_B^4{T}^4}{15h^3{c}^3}\right)}{V}\right]\left[\frac{c}{4}\right]\\ =\left(\frac{2{\pi }^{5}V{k}_B^4}{15h^3{c}^3}\right)T^4\end{array}$

Substitute $1.38\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}$ for ${\mathrm{k}}_{\mathrm{B}},6.63\times {10}^{-34}\text{J}$.s and $3\times {10}^3\text{m}/\text{s}$ for $\text{c}$ in the above equation and obtain the equation as given below.

$\begin{array}{c}\mathrm{I}=\left(\frac{2{\mathrm{\pi }}^5{\mathrm{Vk}}_{\mathrm{B}}^4}{15{\mathrm{h}}^3{\mathrm{c}}^3}\right){\mathrm{T}}^4\\ =\left[\frac{2{\mathrm{\pi }}^5{\left(1.38\times {10}^{-23}\frac{\mathrm{J}}{\mathrm{K}}\right)}^4}{15{\left(6.63\times {10}^{-34}\text{J}.\text{s}\right)}^3\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}}\right)}\right]{\mathrm{T}}^4\\ =\left(5.64\times {10}^{-8}\frac{\mathrm{W}}{{\mathrm{m}}^2\cdot {\mathrm{K}}^4}\right){\mathrm{T}}^4\end{array}$