Change variables in equation 7.83 to $\lambda =hc/ϵ$ and thus derive a formula for the photon spectrum as a function of wavelength. Plot this spectrum, and find a numerical formula for the wavelength where the spectrum peaks, in terms of hc/kT. Explain why the peak does not occur at hc/(2.82kT).

Changing variables in equation 7.83 to $\lambda =hc/ϵ$ and thus deriving a formula for the photon spectrum as a function of wavelength.

The equation 7.83 is:

$\frac{U}{V}=\frac{8\pi }{(hc{)}^3}{\int }_0^{\infty }\frac{{ϵ}^3}{e^{ϵ/kT}-1}dϵ\left(1\right)$

Change the variables to the wavelength using,

$ϵ=\frac{hc}{\lambda }\to dϵ=-\frac{hc}{{\lambda }^2}d\lambda$

Substitute this into (1)

$\frac{U}{V}=-\frac{8\pi }{(hc{)}^3}{\int }_0^{\infty }\frac{(hc{)}^4}{{\lambda }^5}\frac{1}{e^{hc/kT\lambda }-1}d\lambda$

By changing the integration boundaries, at $ϵ=0$the wavelength is$\infty$and at$ϵ=\infty$the wavelength is zero.

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

Changing the function to be dimensionless variables,

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

Hence,

$\frac{U}{V}=\frac{8\pi (kT{)}^4}{(hc{)}^3}{\int }_0^{\infty }\frac{1}{l^5}\frac{1}{e^{1/l}-1}d\lambda$

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

Using python to solve this function. The code is:

The graph is:

The peak occurs at $l=0.2014$

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

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.