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

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

The normalized radial probability density givesConsequently the expectation value of the potential energy isHere the last equality uses $Z{e}^2/(4\pi {ϵ}_0)={\hslash }^2\beta /m_e$. Since this is an energy eigenstate, $E=⟨T⟩+⟨V⟩$, and so $\boxed{⟨T⟩=-E}$. The kinetic energy is positive because $E<0$, in agreement with the virial theorem for a Coulomb potential.

The probability of finding the electron in a spherical shell is $P(r)dr=4\pi r^2|\psi (r){|}^{2}dr$, soThis radial probability density, rather than the density per unit volume, determines the most probable radius. Its expectation value isFor its interior maximum, differentiate $\log P=\text{constant}+2\log r-2\beta r$. The stationary condition $2/r-2\beta =0$ givesThe density per unit volume $|\psi {|}^2$ is instead largest at the nucleus; including the spherical-shell volume factor is essential to the requested radial mode.

The nuclear beta decay is fast compared with the electronic orbital timescale. In the sudden approximation, the electronic wavefunction does not have time to change during the increase from $Z_i=1$ to $Z_f=2$. The subsequent probability of a particular daughter energy eigenstate is therefore its squared overlap with the initial state.

For the final $1s$ state, write ${\beta }_i=1/a_0$ and ${\beta }_f=2/a_0$. The sudden hydrogenic charge change gives the amplitudeThusThe final $2p$ states have spherical harmonics $Y_{1m}$, whereas the initial state is proportional to $Y_{00}$. Their angular overlap $\int Y_{1m}^{\ast }{Y}_{00}d\Omega$ vanishes by orthogonality. Accordingly $\boxed{P_{2p}=0}$ in the ideal spherically symmetric sudden-change model. Nuclear recoil or interactions with the emitted particles can produce small corrections. This is an overlap selection rule for a sudden change, not a photon-induced electric dipole transition.