Show that Planck’s radiation law resolves the ultraviolet catastrophe.

Strategy The ultraviolet catastrophe occurs because the number of configurations through which a standing wave can form inside the cavity becomes infinite as λ → 0. We want to find out what happens to $\ell(\lambda, T)$ if we let λ → 0. We also need to investigate the total energy of the system, especially for the large number of small-wavelength oscillators.

Question

Show that Planck’s radiation law resolves the ultraviolet catastrophe.

Strategy The ultraviolet catastrophe occurs because the number of configurations through which a standing wave can form inside the cavity becomes infinite as λ → 0. We want to find out what happens to [latex]\ell(\lambda, T)[/latex] if we let λ → 0. We also need to investigate the total energy of the system, especially for the large number of small-wavelength oscillators.

Accepted Answer

If we let λ → 0 in Equation (3.23), the value of [latex]e^{h c / \lambda k T} \longrightarrow \infty[/latex] The exponential term dominates the [latex]\lambda^{5}[/latex] term as λ → 0, so the denominator in Equation (3.23) is infinite, and the value of [latex]\ell(\lambda, T) ightarrow 0[/latex]. Note that as the wavelength decreases, the frequency increases [latex](f=c / \lambda), ext { and } h f \gg k T[/latex]. Few oscillators will be able to obtain such large energies, partly because of the large energy necessary to take the energy step from 0 to hf. The probability of occupying the states with small wavelengths (large frequency and high energy) is vanishingly small, so the total energy of the system remains finite. The ultraviolet catastrophe is avoided.

[latex]\ell(\lambda, T)=\frac{2 \pi c^{2} h}{\lambda^{5}} \frac{1}{e^{h c / \lambda k T}-1}[/latex] (3.23)

Question 3.9: Show that Planck’s radiation law resolves the ultraviolet ca...

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