91Ó°ÊÓ

Average component velocity in the given direction $=\frac{1}{4}c$

Useful integration, ${∫}_0^{∞}\frac{x^3}{e^x-1}dx=\frac{{\mathrm{π}}^4}{15}$

Some useful values:

$\mathrm{π}=3.14,k_B=1.38×{10}^{-23}J·{\text{K}}^{-1},h=6.63×{10}^{-34}\text{J}·\text{s, and}c=3×{10}^8\text{m}·{\text{s}}^{-1}$

Planck's formula is given as

$\frac{dU}{df}=\frac{hf}{e^{hf/k_{B}T}-1}×\frac{8\mathrm{π}V}{c^3}{f}^2$

By simplifying Planck's formula:

$\begin{array}{rcl}dU& =& \frac{hf}{e^{hf/k_{B}T}-1}×\frac{8\mathrm{π}V}{c^3}{f}^{2}df\\ & =& \frac{8\mathrm{π}hfc}{4c^3}×\frac{f^2}{e^{h//k_{B}T}-1}df\\ & =& \frac{2\mathrm{π}h}{c^2}×\frac{f^3}{e^{h//k_{B}T}-1}df\\ & & \end{array}$

Let,

$$\begin{gathered} \frac{hf}{k_{B}T}=x \\ df=\frac{k_{B}T}{h}dx \end{gathered}$$

Substituting and simplifying further,

$\begin{array}{rcl}dU& =& \frac{2\mathrm{π}h}{c^2}×\frac{f^3}{e^{h//k_{B}T-1}}df\\ & =& \frac{2\mathrm{π}h}{c^2}×\frac{x^3{k}_B^3{T}^3}{\left(e^x-1\right)h^3}×\frac{k_{B}T}{h}dx\\ & =& \frac{2\mathrm{π}{k}_B^4{T}^4}{h^3{c}^2}×\frac{x^3}{\left(e^x-1\right)}dx\\ & & \end{array}$

The total energy of the radiation in volume V at temperature T can be calculated as follows:

${∫}_0^{∞}dU=\frac{2\mathrm{π}{k}_B^4{T}^4}{h^3{c}^2}{∫}_0^{∞}\frac{x^3}{\left(e^x-1\right)}dx$

Using the given information,

$\begin{array}{rcl}U& =& \frac{2\mathrm{π}{k}_B^4{T}^4}{h^3{c}^2}×\frac{{\mathrm{π}}^4}{15}\\ & =& \frac{2×{\left(3.14\right)}^5×{\left(1.38×{10}^{-23}J·{\text{K}}^{-1}\right)}^4{T}^4}{\left(6.63×{10}^{-34}\text{J}·\text{s}×{\left(3×{10}^8\text{m}·{\text{s}}^{-1}\right)}^2\right)}×\frac{1}{15}\\ & =& 5.66×{10}^{-8}×T^4\text{W}·{\text{m}}^2·{\text{K}}^4\\ & & \end{array}$

Therefore, the intensity is given by

$I=\mathrm{σ}{T}^4$

Where, $\mathrm{σ}=5.66×{10}^{-8}{\text{Wm}}^2{\text{K}}^4$

The Stefan-Boltzmann law:

$I=\mathrm{σ}{T}^4$

Where, $\mathrm{σ}=5.66×{10}^{-8}{\text{Wm}}^2{\text{K}}^4$