 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1/section-b/b7/a/solution

Use the nonrelativistic Coulomb potential energy

V = - Z e^{2} / (4 π ϵ_{0} r)

, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian is

\nabla^{2} ψ = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d ψ}{d r}) = (β^{2} - \frac{2 β}{r}) ψ .

Substituting into the time-independent Schrödinger equation gives

\frac{H ψ}{ψ} = - \frac{ℏ ^{2} β ^{2}}{2 m _{e}} + \frac{1}{r} (\frac{ℏ ^{2} β}{m _{e}} - \frac{Z e ^{2}}{4 π ϵ _{0}}) .

1/ r

must vanish. Therefore

β = Z / a_{0}

, where

a_{0}

1 = 4 π ∣ α ∣^{2} \int_{0}^{\infty} r^{2} e^{- 2 β r} d r = \frac{π ∣ α ∣ ^{2}}{β ^{3}} .

Choosing the irrelevant overall phase so that

α

is real and positive gives

α^{2} = β^{3} / π

E = - \frac{ℏ ^{2} β ^{2}}{2 m _{e}} = - \frac{m _{e} Z ^{2} e ^{4}}{2 ( 4 π ϵ _{0} ) ^{2} ℏ ^{2}} ≃ - 13.6 Z^{2} eV .

Replacing

m_{e}

by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

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