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General equation for$\mathbf{N}$particles of E energy is given by

role="math" localid="1660030263876" $\mathbf{N}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{鈭�}}\mathbf{N}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{dE}$鈥�..(1)

Where,

$\mathbf{N}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{鈫�}$Quantum distribution

$\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{鈫�}$ Density of states

$\mathbf{N}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\frac{\mathbf{E}}{{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}\mathbf{鈭�}\mathbf{1}}$ 鈥︹赌�(2)

Where,${\mathbf{K}}_{\mathbf{B}}$Boltzmann's constant.

Also,

$\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{8}\mathbf{蟺痴}}{{\mathbf{h}}^{\mathbf{3}}{\mathbf{c}}^{\mathbf{3}}}{\mathbf{E}}^{\mathbf{2}}$ 鈥︹赌�(3)

Where,

$\mathbf{V}\mathbf{鈫�}$Volume of container

$\mathbf{h}\mathbf{鈫�}$being Planck's constant

$\mathbf{c}\mathbf{鈫�}$Speed of light in vacuum.

Expression for energy of photons in a container is given by-

$\mathbf{U}\mathbf{=}\frac{\mathbf{8}{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{Vk}}_{\mathbf{B}}^{\mathbf{4}}{\mathbf{T}}^{\mathbf{4}}}{\mathbf{15}{\mathbf{h}}^{\mathbf{3}}{\mathbf{c}}^{\mathbf{3}}}$鈥�..(4)

Where,

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

$\mathbf{h}\mathbf{鈫�}$Planck's constant

$\mathbf{c}\mathbf{鈫�}$Speed of light in vacuum

$\mathbf{T}\mathbf{鈫�}$Temperature

$\mathbf{V}\mathbf{鈫�}$Volume of container

To get an expression for the number of photon per unit volume at some temperature R , equations (2) and (3) are inserted in equation (1), and simplified:

$\begin{array}{c}\mathrm{N}={鈭�}_0^{\mathrm{鈭�}}\mathrm{N}\left(\mathrm{E}\right)\mathrm{D}\left(\mathrm{E}\right)\mathrm{dE}\\ ={鈭�}_0^{\mathrm{鈭�}}\left[\frac{1}{{\mathrm{e}}^{\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}\right]\left[\frac{8\mathrm{蟺痴}}{{\mathrm{鈩�}}^3{\mathrm{c}}^3}{\mathrm{E}}^2\right]\mathrm{dEN}\\ =\frac{8\mathrm{蟺痴}}{{\mathrm{鈩�}}^3{\mathrm{c}}^3}{鈭�}_0^{\mathrm{鈭�}}\frac{\mathrm{E}}{{\mathrm{e}}^{\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}\mathrm{dE}\end{array}$

That can be simplified more by the use of the substitutions:

$\begin{array}{c}\mathrm{x}=\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ \mathrm{E}={\mathrm{xk}}_{\mathrm{B}}\mathrm{T}\\ \mathrm{dE}={\mathrm{k}}_{\mathrm{B}}\mathrm{Tdx}\end{array}$

So then substitute those into equation (6)

$\begin{array}{c}\mathrm{N}=\frac{8\mathrm{蟺痴}}{{\mathrm{h}}^3{\mathrm{c}}^3}{鈭�}_0^{\mathrm{鈭�}}\frac{\mathrm{E}}{{\mathrm{e}}^{\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}\mathrm{dE}\\ \mathrm{N}=\frac{8\mathrm{蟺痴}}{{\mathrm{h}}^3{\mathrm{c}}^3}{鈭�}_0^{\mathrm{鈭�}}\frac{{\left({\mathrm{xk}}_{\mathrm{B}}\mathrm{T}\right)}^2}{{\mathrm{e}}^{\mathrm{x}}鈭�1}\left({\mathrm{k}}_{\mathrm{B}}\mathrm{Tdx}\right)\end{array}$

$\mathrm{N}=\frac{8{\mathrm{蟺痴k}}_{\mathrm{B}}^3{\mathrm{T}}^3}{{\mathrm{h}}^3{\mathrm{c}}^3}{鈭�}_0^{\mathrm{鈭�}}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}鈭�1}\mathrm{dx}$

Equation (5) can then be used to fill in for the integral, $\mathrm{N}=\frac{8{\mathrm{蟺痴k}}_{\mathrm{B}}^3{\mathrm{T}}^3}{{\mathrm{鈩�}}^3{\mathrm{c}}^3}{鈭�}_0^{\mathrm{鈭�}}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}鈭�1}\mathrm{dx}$

Substitute $1.38脳{10}^{23}\text{J}/\text{K}$for ${\mathrm{k}}_{\mathrm{B}},6.63脳{10}^{鈭�34}\text{J}.\text{s}$ for $\mathrm{鈩�}$, and $3脳{10}^8\text{m}/\text{s}$ for c

$\begin{array}{l}\mathrm{N}=\frac{8\mathrm{蟺痴}{(1.38脳{10}^{鈭�23}\text{J}/\text{K})}^3{\mathrm{T}}^3}{{(6.63脳{10}^{鈭�34}\text{J}.\text{s})}^3{(3脳{10}^8\text{m}/\text{s})}^3}(2.40)\\ \mathrm{N}=\left(2脳{10}^7\frac{1}{{\mathrm{m}}^3.{\mathrm{K}}^3}\right){\mathrm{VT}}^3\end{array}$

To find the average energy of a photon isa cavity of volume $\mathrm{V}$, use that the number of photons$\mathrm{N}$times the average energy of the photons$E$is equal to the total energy of the photons$\mathrm{U}$:

$\begin{array}{l}\mathrm{U}=\mathrm{NE}\\ \mathrm{E}=\frac{\mathrm{U}}{\mathrm{N}}\end{array}$

Put the value and solve

$\begin{array}{c}\mathrm{E}=\frac{\left(\frac{8{\mathrm{蟺}}^5{\mathrm{v}}_{\mathrm{B}}^4{\mathrm{T}}^4}{15{\mathrm{h}}^3{\mathrm{c}}^3}\right)}{\left(\frac{8{\mathrm{蟺痴k}}_{\mathrm{B}}^3{\mathrm{T}}^3}{{\mathrm{鈩�}}^3{\mathrm{c}}^3}\right){\displaystyle {鈭�}_0^{\mathrm{鈭�}}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}鈭�1}}\mathrm{dx}}\\ \mathrm{E}=\frac{{\mathrm{蟺}}^4{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{15{\displaystyle {鈭�}_0^{\mathrm{鈭�}}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}鈭�1}}\mathrm{dx}}\end{array}$

That can be simplified with equation (5) again

$\begin{array}{l}\mathrm{E}=\frac{{\mathrm{蟺}}^4{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{15{\displaystyle {鈭�}_0^{\mathrm{鈭�}}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}鈭�1}}\mathrm{dx}}\mathrm{n}\\ \mathrm{E}=\frac{{\mathrm{蟺}}^4}{15\left(2.40\right)}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\\ \mathrm{E}鈮�2.7{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\end{array}$