 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s3/2/3/solution

Let

r

be the electron-nucleus separation and use the reduced mass

μ = m_{e} M / (m_{e} + M)

. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:

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

Eliminating

v

gives

r_{n} = \frac{4 π ϵ _{0} ℏ ^{2}}{μ Z e ^{2}} n^{2} = a_{0} \frac{m _{e}}{μ} \frac{n ^{2}}{Z} .

The total energy is

E = μ v^{2} /2 - Z e^{2} / (4 π ϵ_{0} r) = - Z e^{2} / (8 π ϵ_{0} r)

, hence

E_{n} = - \frac{μ Z ^{2} e ^{4}}{2 ( 4 π ϵ _{0} ) ^{2} ℏ ^{2} n ^{2}} = - h c R_{M} \frac{Z ^{2}}{n ^{2}}, R_{M} = R_{\infty} \frac{μ}{m _{e}} .

Here

a_{0}

is the ordinary Bohr radius and

R_{\infty}

the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is

M r_{n} / (M + m_{e})

; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

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