91Ó°ÊÓ

$I\left(\mathrm{λ}\right)=\frac{2\mathrm{π}h{c}^2}{{\mathrm{λ}}^5\left(e^{hc/\mathrm{λ}KT}−1\right)}$

$I\left(\mathrm{λ}\right)=\frac{2\mathrm{π}h{c}^2}{{\mathrm{λ}}^5\left(e^{\frac{hc}{\mathrm{λ}kT}}−1\right)}$

using the following relationship to calculate the speed of light;

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

through substitution;

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

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

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

$$\begin{gathered} I={∫}_0^{\mathrm{∞}} I\left(\mathrm{λ}\right)d\mathrm{λ} \\ ={∫}_0^{\mathrm{∞}} I\left(v\right)\frac{−c}{v^2}d \\ ={∫}_0^{\mathrm{∞}} \frac{2\mathrm{π}h{v}^5}{c^3\left(\frac{hv}{kT}−1\right)}\frac{−c}{v^2}dv \\ ={∫}_0^{\mathrm{∞}} \frac{2\mathrm{π}h{v}^3}{c^2\left(\frac{hv}{kT}−1\right)}dv \end{gathered}$$

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.

$$\begin{gathered} I={∫}_0^{\mathrm{∞}} \frac{2\mathrm{π}h{v}^3}{c^2\left(\frac{hv}{kT}−1\right)}dv \\ ={∫}_0^{\mathrm{∞}} \frac{2\mathrm{π}h{\left(\frac{kT}{h}x\right)}^3}{c^2\left(e^x−1\right)}\frac{kT}{h}dx \end{gathered}$$

therefore;

$\begin{array}{r}I={∫}_0^{\mathrm{∞}} \frac{2\mathrm{π}h{\left(\frac{kT}{h}x\right)}^3}{c^2\left(e^x−1\right)}\frac{kT}{h}dx\\ =\frac{2\mathrm{π}(kT{)}^4}{c^2{h}^3}{∫}_0^{\mathrm{∞}} \frac{x^3}{e^x−1}dx\end{array}$

using the tabulated integral value as a substitute,

$$\begin{gathered} I={∫}_0^{\mathrm{∞}} \frac{2\mathrm{π}h{\left(\frac{kT}{h}x\right)}^2}{c^2\left(e^x−1\right)}\frac{kT}{h}dx \\ =\frac{2\mathrm{π}(kT{)}^4}{c^2{h}^3}{∫}_0^{\mathrm{∞}} \frac{x^3}{e^x−1}dx \\ =\frac{2\mathrm{π}(kT{)}^4}{c^2{h}^3}\frac{1}{240}(2\mathrm{π}{)}^4 \\ =\frac{2{\mathrm{π}}^2{k}^4{T}^4}{15c^2{h}^3} \end{gathered}$$

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

$$\begin{gathered} \mathrm{σ}=\frac{2\mathrm{Ï€}}{15c^2{h}^3} \\ =\frac{2{\mathrm{Ï€}}^2{\left(1.381×{10}^{−23}\right)}^4(5800\mathrm{k}{)}^4}{15{\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^2{\left(6.626×{10}^{−34}\right)}^3} \\ =5.67×{10}^{−8}\mathrm{W}/{\mathrm{m}}^2â‹\dots {\mathrm{K}}^4 \end{gathered}$$

Hence, the constant is$\mathrm{σ}=5.67×{10}^{−8}\mathrm{W}/{\mathrm{m}}^2â‹\dots {\mathrm{K}}^4$