Past exam of the physics A course of the University of Cambridge / 2019 / Paper 1 / Section B / B7 / b

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- Solution b

Solution

The normalized radial probability density gives

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

Consequently the expectation value of the potential energy is

⟨ 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

, in agreement with the virial theorem for a Coulomb potential.

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