 Solution

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.