Q 7.39 - An Introduction to Thermal Physics

Question

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).

Step-by-Step Solution

Answer

Hence,  formula for the photon spectrum as a function of wavelength is $u\left(l\right)=\frac{8\pi (kT{)}^4}{(hc{)}^3}\frac{1}{l^5}\frac{1}{e^{1/l}-1}$

Step 1: Given information

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

Step 2: Explanation

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)

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

 

Step 3: Calculations

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

style="width:30%" $$\begin{gathered} \frac{U}{V}=8\pi hc{\int }_{\infty }^0\frac{1}{{\lambda }^5}\frac{1}{e^{hc/kT\lambda }-1}d\lambda \\ \frac{U}{V}=8\pi hc{\int }_0^{\infty }\frac{1}{{\lambda }^5}\frac{1}{e^{hc/kT\lambda }-1}d\lambda \end{gathered}$$

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.