 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ib/paper-3/9d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

With the proton treated as fixed and electron mass

m

, the classical energy and angular momentum are

E = \frac{1}{2} m ∣ \dot{r} ∣^{2} - \frac{e ^{2}}{4 π ϵ _{0} r}, L = m r \times \dot{r} .

L = m r v

\frac{m v ^{2}}{r} = \frac{e ^{2}}{4 π ϵ _{0} r ^{2}}, m r v = n ℏ.

Eliminating

v

gives

r = 4 π ϵ_{0} n^{2} ℏ^{2} / (m e^{2})

a = \frac{4 π ϵ _{0} ℏ ^{2}}{m e ^{2}}, r_{n} = n^{2} a, n = 1, 2, \dots .

The kinetic energy is half the magnitude of the attractive potential energy. Hence

E_{n} = - \frac{e ^{2}}{8 π ϵ _{0} r _{n}} = - \frac{m e ^{4}}{2 ( 4 π ϵ _{0} ) ^{2} ℏ ^{2} n ^{2}} .

This is the circular-orbit Bohr model; including proton recoil replaces

m

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