 Solution

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

⟨ \frac{1}{r} ⟩ = 4 β^{3} \int_{0}^{\infty} r e^{- 2 β r} d r = β .

⟨ V ⟩ = - \frac{Z e ^{2} β}{4 π ϵ _{0}} = - \frac{ℏ ^{2} β ^{2}}{m _{e}} = 2 E .

Here the last equality uses

Z e^{2} / (4 π ϵ_{0}) = ℏ^{2} β / m_{e}

. Since this is an energy eigenstate,

E = ⟨ T ⟩ + ⟨ V ⟩

, and so

⟨ T ⟩ = - E

. The kinetic energy is positive because

E < 0

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