(a) Write the Planck distribution law in terms of the frequency f, rather than the wavelength $\mathit{\lambda }$, to obtain I(f). (b) Show that

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{ }\mathit{I}\mathbf{\left(}\mathit{\lambda }\mathbf{\right)}\mathit{d}\mathit{\lambda }\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi }}^{\mathbf{5}}{\mathbf{k}}^{\mathbf{4}}}{\mathbf{15}{\mathbf{c}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{3}}}{\mathit{T}}^{\mathbf{4}}$

where I($\mathit{\lambda }$) is the Planck distribution formula of Eq. (39.24). Hint: Change the integration variable from$\mathit{\lambda }$to f. You will need to use the following tabulated integral:

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{ }\frac{{\mathbf{x}}^{\mathbf{3}}}{{\mathbf{e}}^{\mathbf{\alpha }\mathbf{x}}\mathbf{-}\mathbf{1}}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{240}}{\left(\frac{2\pi }{\alpha }\right)}^{\mathbf{4}}$

(c) The result of part (b) is I and has the form of the Stefan– Boltzmann law, $\mathit{l}\mathbf{=}\mathit{\sigma }{\mathit{T}}^{\mathbf{4}}$(Eq. 39.19). Evaluate the constants in part (b) to show that$\mathit{\sigma }$has the value given in Section 39.5.

$I\left(\lambda \right)=\frac{2\pi h{c}^2}{{\lambda }^5\left(e^{hc/\lambda KT}-1\right)}$

$I\left(\lambda \right)=\frac{2\pi h{c}^2}{{\lambda }^5\left(e^{\frac{hc}{\lambda kT}}-1\right)}$

using the following relationship to calculate the speed of light;

$\begin{array}{r}c=v\lambda \\ \lambda =\frac{c}{v}\end{array}$

through substitution;

$$\begin{gathered} I=\frac{2\pi h{c}^2}{{\left(\frac{c}{v}\right)}^5\left(e^{\left(\frac{hc}{v}\right)kT}-1\right)} \\ =\frac{2\pi h{v}^5}{c^3\left(\frac{hv}{kT}-1\right)} \end{gathered}$$

Hence, in terms of frequency, the Planck's distribution law is;

$v=\frac{2\pi h{v}^5}{c^3\left(e^{\frac{hc}{kT}}-1\right)}$

$$\begin{gathered} I={\int }_0^{\mathrm{\infty }} I\left(\lambda \right)d\lambda \\ ={\int }_0^{\mathrm{\infty }} I\left(v\right)\frac{-c}{v^2}d \\ ={\int }_0^{\mathrm{\infty }} \frac{2\pi h{v}^5}{c^3\left(\frac{hv}{kT}-1\right)}\frac{-c}{v^2}dv \\ ={\int }_0^{\mathrm{\infty }} \frac{2\pi h{v}^3}{c^2\left(\frac{hv}{kT}-1\right)}dv \end{gathered}$$

Let $x=\frac{hv}{kT}$, and then differentiate both sides by $dx=\frac{h}{kT}dv$. The wavelength will be $v=\frac{kTx}{h}$. substituting as well.

$$\begin{gathered} I={\int }_0^{\mathrm{\infty }} \frac{2\pi h{v}^3}{c^2\left(\frac{hv}{kT}-1\right)}dv \\ ={\int }_0^{\mathrm{\infty }} \frac{2\pi h{\left(\frac{kT}{h}x\right)}^3}{c^2\left(e^x-1\right)}\frac{kT}{h}dx \end{gathered}$$

therefore;

$\begin{array}{r}I={\int }_0^{\mathrm{\infty }} \frac{2\pi h{\left(\frac{kT}{h}x\right)}^3}{c^2\left(e^x-1\right)}\frac{kT}{h}dx\\ =\frac{2\pi (kT{)}^4}{c^2{h}^3}{\int }_0^{\mathrm{\infty }} \frac{x^3}{e^x-1}dx\end{array}$

using the tabulated integral value as a substitute,

$$\begin{gathered} I={\int }_0^{\mathrm{\infty }} \frac{2\pi h{\left(\frac{kT}{h}x\right)}^2}{c^2\left(e^x-1\right)}\frac{kT}{h}dx \\ =\frac{2\pi (kT{)}^4}{c^2{h}^3}{\int }_0^{\mathrm{\infty }} \frac{x^3}{e^x-1}dx \\ =\frac{2\pi (kT{)}^4}{c^2{h}^3}\frac{1}{240}(2\pi {)}^4 \\ =\frac{2{\pi }^2{k}^4{T}^4}{15c^2{h}^3} \end{gathered}$$

Hence, proved $l=\frac{2{\mathrm{\pi }}^2{\mathrm{k}}^4{\mathrm{T}}^4}{15c^2{h}^3}$

$$\begin{gathered} \sigma =\frac{2\pi }{15c^2{h}^3} \\ =\frac{2{\pi }^2{\left(1.381\times {10}^{-23}\right)}^4(5800\mathrm{k}{)}^4}{15{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2{\left(6.626\times {10}^{-34}\right)}^3} \\ =5.67\times {10}^{-8}\mathrm{W}/{\mathrm{m}}^2\cdot {\mathrm{K}}^4 \end{gathered}$$

Hence, the constant is$\sigma =5.67\times {10}^{-8}\mathrm{W}/{\mathrm{m}}^2\cdot {\mathrm{K}}^4$