91影视

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

The equation 7.83 is:

$\frac{U}{V}=\frac{8蟺}{(hc{)}^3}{鈭�}_0^{鈭�}\frac{{系}^3}{e^{系/kT}-1}d系\left(1\right)$

Change the variables to the wavelength using,

$系=\frac{hc}{位}鈫�d系=-\frac{hc}{{位}^2}d位$

Substitute this into (1)

$\frac{U}{V}=-\frac{8蟺}{(hc{)}^3}{鈭�}_0^{鈭�}\frac{(hc{)}^4}{{位}^5}\frac{1}{e^{hc/kT位}-1}d位$

By changing the integration boundaries, at $系=0$the wavelength is$鈭�$and at$系=鈭�$the wavelength is zero.

localid="1647752119716">UV=8蟺hc鈭�鈭�01位51ehc/kT位-1d位UV=8蟺hc鈭�0鈭�1位51ehc/kT位-1d位

Changing the function to be dimensionless variables,

$l=\left(\frac{kT}{hc}\right)位$

Hence,

$\frac{U}{V}=\frac{8蟺(kT{)}^4}{(hc{)}^3}{鈭�}_0^{鈭�}\frac{1}{l^5}\frac{1}{e^{1/l}-1}d位$

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

Using python to solve this function. The code is:

The graph is:

The peak occurs at $l=0.2014$

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

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.