Past exam of the atomic and laser physics course of the University of Oxford / 2021 / 4 / e / i / Solution

A nonrelativistic $2p$ wavefunction vanishes linearly at the origin. Its probability density is therefore proportional to $r^2$, so its first-order finite nuclear size correction shift scales as $R^4$, rather than the $R^2$ result for $1s$. It is positive but much smaller.

More explicitly, the normalized hydrogen-like ion radial function obeys $R_{21}^2(r)\simeq Z^5{r}^2/(24a_0^5)$ near the origin. Since the spherical harmonic has unit angular norm,ThusThis is about $2.4\mu \mathrm{eV}$ in the nonrelativistic model. At large $Z$, the relativistic $2p_{1/2}$ Dirac spinor has an $s$-type small component and a much stronger nuclear sensitivity than this nonrelativistic argument predicts; $2p_{3/2}$ remains much more suppressed. Neglecting spin-orbit interaction corrections would miss that distinction.