 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2021/s03/3/d/solution

Δ E = E_{3} - E_{2} = \frac{( 9 - 4 ) ℏ ^{2}}{2 m r ^{2}} = \frac{5 ℏ ^{2}}{2 m r ^{2}} .

Using

E_{γ} = h c / λ

gives

λ = \frac{2 h c m r ^{2}}{5 ℏ ^{2}} = \frac{4 π c m r ^{2}}{5ℏ} .

No numerical mass or radius is supplied, so this is the determinate result. It is the wavelength conditional on the transition occurring. An isolated particle on a circle with no coupling would remain in its energy eigenstate; a charged particle coupled to electromagnetic radiation can emit subject to its transition selection rules. The

3 \to 2

change has

Δ n = - 1

, compatible with an electric-dipole angular factor for a charged ring particle.

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