For small z , ${\mathit{e}}^{\mathbf{z}}$ is approximately 1+z. (a) Use this to show that Planck's spectral energy density (3-1) agrees with the result of classical wave theory in the limit of small frequencies. (b) Show that, whereas the classical formula diverges at high frequencies-the so-called ultraviolet catastrophe of this theory - Planck's formula approaches 0.

The expression for Planck's spectral energy density is given as follows,

$\frac{\mathbf{d}\mathbf{U}}{\mathbf{d}\mathbf{f}}\mathbf{=}\frac{\mathbf{h}\mathbf{f}}{{\mathbf{e}}^{\mathbf{h}\mathbf{f}\mathbf{/}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}\mathbf{-}\mathbf{1}}\mathbf{\times }\frac{\mathbf{8}\mathbf{\pi }\mathbf{V}}{{\mathbf{c}}^{\mathbf{3}}}{\mathit{f}}^{\mathbf{2}}$

Here, dUis the electromagnetic energy, his the Planck’s constant, f is the frequency, ${\mathit{k}}_{\mathbf{B}}$ is the Boltzmann constant, c is the speed of light, T is the temperature, andVis the volume.

For black body radiation, determine the limiting cases of Planck's formula.

Use the linear approximation of Taylor expansion for the exponential function.

$\frac{dU}{df}=\frac{hf}{exp\left({\displaystyle \frac{hf}{k_{B}T}}\right)-1}\times \frac{8\pi V}{c^3}{f}^2$

It is known that $exp\left(\frac{hf}{k{}_{B}T}\right)\approx 1+\frac{hf}{k_{B}T}$. So, substitute it in the above expression.

$$\begin{gathered} \frac{dU}{df}=\frac{hf}{\left(1+{\displaystyle \frac{hf}{k_{B}T}}\right)}\times \frac{8V}{c^3}{f}^3 \\ =k_{B}T\times \frac{8\pi V}{c^3}{f}^3 \end{gathered}$$

Thus, Planck's spectral energy density (3-1) agrees with the result of classical wave theory in the limit of small frequencies

For f > >1 , the spectral density goes to zero at very high frequencies and as f becomes very large, classical formula diverges as it is proprional to $f^2$.

$\underset{f\to \infty }{\lim }\frac{hf}{exp\left({\displaystyle \frac{hf}{k_{B}T}}\right)-1}\times \frac{8\pi V}{c^3}{f}^3=\underset{f\to \infty }{\lim }\frac{C{f}^2}{exp\left({\displaystyle \frac{hf}{k_{B}T}}\right)}$

Here, $C=hf\frac{8\pi V}{c^3}$ and the value of exponential factor is must greater than 1 so,.

$\exp \left(\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\right)=1$

Apply all the conditions in the above expression.

$\frac{d\cup }{df}\to 0$

Thus, the Planck’s formula approaches 0.