 Solution

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

P (r) d r = 4 π r^{2} ∣ ψ (r) ∣^{2} d r

, so

P (r) = 4 β^{3} r^{2} e^{- 2 β r} .

This radial probability density, rather than the density per unit volume, determines the most probable radius. Its expectation value is

⟨ r ⟩ = 4 β^{3} \int_{0}^{\infty} r^{3} e^{- 2 β r} d r = \frac{3}{2 β} = \frac{3 a _{0}}{2 Z} .

For its interior maximum, differentiate

lo g P = constant + 2 lo g r - 2 β r

. The stationary condition

2/ r - 2 β = 0

gives

r_{mode} = \frac{1}{β} = \frac{a _{0}}{Z} .

∣ ψ ∣^{2}

is instead largest at the nucleus; including the spherical-shell volume factor is essential to the requested radial mode.

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