 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2021/paper-1/section-b/b6/b/solution

V (r) = - e^{2} / (4 π ϵ_{0} r)

, so

\frac{1}{r} \frac{d V}{d r} = \frac{e ^{2}}{4 π ϵ _{0} r ^{3}} .

Use precisely the radial normalization stated in the PDF: the radial probability measure is

4 π r^{2} ∣ ψ ∣^{2} d r

. The supplied function has

∣ ψ ∣^{2} = \frac{r ^{2}}{96 π a _{0}^{5}} e^{- r / a_{0}}, 4 π r^{2} ∣ ψ ∣^{2} d r = \frac{r ^{4} e ^{- r / a_{0}}}{24 a _{0}^{5}} d r .

Its integral is one, since

\int_{0}^{\infty} r^{4} e^{- r / a_{0}} d r = 24 a_{0}^{5}

. In particular,

⟨ r^{- 3} ⟩ = \frac{1}{24 a _{0}^{5}} \int_{0}^{\infty} r e^{- r / a_{0}} d r = \frac{1}{24 a _{0}^{3}} .

This calculation uses the supplied convention, rather than multiplying a separately normalized standard radial function by an extra

4 π

j (j + 1) - l (l + 1) - s (s + 1) = 3/4 - 2 - 3/4 = - 2

⟨ H_{1} ⟩ = - \frac{ℏ ^{2}}{2 m ^{2} c ^{2}} \frac{e ^{2}}{4 π ϵ _{0}} \frac{1}{24 a _{0}^{3}} = - \frac{ℏ ^{2} e ^{2}}{192 π ϵ _{0} m ^{2} c ^{2} a _{0}^{3}} ≃ - 3.02 \times 1 0^{- 5} eV .

Here

a_{0} = 5.29 \times 1 0^{- 11} m

is the supplied Bohr radius. The negative sign means the

j = 1/2

member of the

l = 1

pair is shifted downward by this spin-orbit interaction; it is not the total hydrogen binding energy.

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