Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

The electromagnetic intensity thermally radiated by a body of temperature $T$ is given by $I = {σT}^{4}$ where $σ = 5 .67 \times 10^{- 8} W / m^{2} \cdot K^{4}$
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 $c$ . For radiation moving uniformly in all directions, the average component of velocity in a given direction is $\frac{1}{4} c$ .)

The intensity is $(5 .64 \times 10^{- 8} \frac{W}{m^{2} \cdot K^{4}}) T^{4}$

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Energy of photons in container (U)

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

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

Where;

$k_{B} \to$ Boltzmann's constant

$h \to$ Planck's constant

$c \to$ Speed of light in vacuum

$T \to$ Temperature

$V \to$ Volume of container

Alternate expression for intensity -

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

Find the intensity.

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

$\begin{matrix} I = (\frac{energy}{volume}) (\frac{distance}{time}) \\ = (\frac{energy}{volume}) (velocity) \\ I = \frac{U}{V} v \dots \dots . . (3) \end{matrix}$ ….. (3)

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

$\begin{matrix} I = (\frac{U}{V}) v \\ = [\frac{(\frac{8 π^{5} V k_{B}^{4} T^{4}}{15 h^{3} c^{3}})}{V}] [\frac{c}{4}] \\ = (\frac{2 π^{5} V k_{B}^{4}}{15 h^{3} c^{3}}) T^{4} \end{matrix}$

Substitute $1.38 \times 10^{- 23} m^{2} \cdot kg \cdot s^{- 2}$ for $k_{B}, 6 .63 \times 10^{- 34} J$ .s and $3 \times 10^{3} m / s$ for $c$ in the above equation and obtain the equation as given below.

$\begin{matrix} I = (\frac{2 π^{5} {Vk}_{B}^{4}}{15 h^{3} c^{3}}) T^{4} \\ = [\frac{2 π^{5} {(1 .38 \times 10^{- 23} \frac{J}{K})}^{4}}{15 {(6 .63 \times 10^{- 34} J . s)}^{3} (3 \times 10^{8} \frac{m}{s})}] T^{4} \\ = (5 .64 \times 10^{- 8} \frac{W}{m^{2} \cdot K^{4}}) T^{4} \end{matrix}$

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