Question

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.

Short answer

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

(b) The Planck’s formula approaches 0.

Step-by-step solution

Expression for the planck’s spectral energy density

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.

(a) Verification of the planck’s spectral energy density agrees with the result of classical wave

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

(b) Veification of the fact that Planck’s formula approaches 0 whereas the classical formula diverges at high 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.