91Ó°ÊÓ

Changing variables in equation 7.83 to $\mathrm{λ}=hc/\mathrm{Ï\mu }$ and thus deriving a formula for the photon spectrum as a function of wavelength.

The equation 7.83 is:

$\frac{U}{V}=\frac{8\mathrm{Ï€}}{(hc{)}^3}{∫}_0^{∞}\frac{{\mathrm{Ï\mu }}^3}{e^{\mathrm{Ï\mu }/kT}-1}d\mathrm{Ï\mu }\left(1\right)$

Change the variables to the wavelength using,

$\mathrm{Ï\mu }=\frac{hc}{\mathrm{λ}}→d\mathrm{Ï\mu }=-\frac{hc}{{\mathrm{λ}}^2}d\mathrm{λ}$

Substitute this into (1)

$\frac{U}{V}=-\frac{8\mathrm{π}}{(hc{)}^3}{∫}_0^{∞}\frac{(hc{)}^4}{{\mathrm{λ}}^5}\frac{1}{e^{hc/kT\mathrm{λ}}-1}d\mathrm{λ}$

By changing the integration boundaries, at $\mathrm{Ï\mu }=0$the wavelength is$∞$and at$\mathrm{Ï\mu }=∞$the wavelength is zero.

localid="1647752119716">UV=8πhc∫∞01λ51ehc/kTλ-1dλUV=8πhc∫0∞1λ51ehc/kTλ-1dλ

Changing the function to be dimensionless variables,

$l=\left(\frac{kT}{hc}\right)\mathrm{λ}$

Hence,

$\frac{U}{V}=\frac{8\mathrm{π}(kT{)}^4}{(hc{)}^3}{∫}_0^{∞}\frac{1}{l^5}\frac{1}{e^{1/l}-1}d\mathrm{λ}$

$u\left(l\right)=\frac{8\mathrm{Ï€}(kT{)}^4}{(hc{)}^3}\frac{1}{l^5}\frac{1}{e^{1/l}-1}$

Using python to solve this function. The code is:

The graph is:

The peak occurs at $l=0.2014$

$$\begin{gathered} 0.2014=\left(\frac{kT}{hc}\right)\mathrm{λ} \\ \mathrm{λ}=0.2014\left(\frac{hc}{kT}\right) \\ \mathrm{λ}=\left(\frac{hc}{4.965kT}\right) \end{gathered}$$

This isn't the same as the solution to problem 7.37. This is due to the nonlinear relationship between energy and wavelength; for example, the energy difference between 1 eV and 2 eV is the same for 101 eV and 102 eV, but the wavelengths that correspond to these two intervals are not.