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Expert-verified Found in: Page 409 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # The electromagnetic intensity thermally radiated by a body of temperature ${\mathbf{T}}$ is given by ${\mathbf{I}}{\mathbf{=}}{{\mathbf{\sigma T}}}^{{\mathbf{4}}}$ where ${\mathbf{\sigma }}{\mathbf{=}}{\mathbf{5}}{\mathbf{.67}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{8}}{\mathbf{}}{\mathbf{\text{W}}}{\mathbf{/}}{{\mathbf{\text{m}}}}^{{\mathbf{2}}}{\mathbf{\cdot }}{{\mathbf{\text{K}}}}^{{\mathbf{4}}}$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 $\frac{\mathbf{1}}{\mathbf{4}}{\mathbf{\text{c}}}$ .)

The intensity is $\left(5.64×{10}^{-8}\frac{\mathrm{W}}{{\mathrm{m}}^{2}\cdot {\mathrm{K}}^{4}}\right){\mathrm{T}}^{4}$

See the step by step solution

## Energy of photons in container (U)

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)

## Find the intensity.

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

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

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

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

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