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

Put $k_Z=Z{e}^2/(4\pi {ϵ}_0)$. Gauss's law for a uniform spherical nuclear charge distribution, with the electrostatic potential matched at $R$, gives the electronic potential energySubtracting the point-nucleus Coulomb potential leaves the nonnegative perturbationThe normalized nonrelativistic hydrogen-like ion $1s$ wavefunction has $|\psi (r){|}^2=Z^3{e}^{-2Zr/a_0}/(\pi a_0^3)$. Since $ZR/a_0\ll 1$, approximate it inside the atomic nucleus by $|\psi (0){|}^2$. First-order time-independent perturbation theory then givesDefine $E_H=-Z^2\mathrm{Ry}$, as the paper specifies for the hydrogen-like ion, rather than the neutral-hydrogen value. Using $\mathrm{Ry}=e^2/(8\pi {ϵ}_0{a}_0)$ givesto leading order in $ZR/a_0$. For $A=208$, $R\simeq 7.11\mathrm{fm}$, $ZR/a_0\simeq 0.0110$, andThe finite nuclear charge radius makes the energy level less bound. The quoted frequency is a level shift divided by $h$: a line frequency shifts by $(\Delta E_u-\Delta E_l)/h$, so a decay into this shifted $1s$ level alone has the opposite sign.

The formula is a nonrelativistic, small-nuclear charge radius estimate. Keeping the exponential in the first-order integral gives $\Delta E_{1s}=\Delta E_{\mathrm{leading}}[1-5ZR/(6a_0)+O((ZR/a_0{)}^2)]$, about a one-percent correction here. More seriously, $Z{\alpha }_{\mathrm{fs}}\simeq 0.598$ for lead, so relativistic contraction is substantial. A precise physical prediction needs a finite-nucleus Dirac equation calculation; the boxed estimate is the model requested, not precision heavy-ion spectroscopy.

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.

The finite nuclear size correction perturbation is unchanged for the same lead atomic nucleus. Its leading effect on one occupied $1s$ orbital is still proportional to that orbital's contact density,Other electrons screen the nuclear attraction and alter the self-consistent $1s$ wavefunction, usually reducing its contact density and hence its positive shift relative to the bare hydrogen-like ion. The reduction is modest for a deep $1s$ orbital: it penetrates well inside most other shells and still samples nearly the full $Z=82$ attraction. Replacing $Z$ by the neutral atom's asymptotic charge one would be inappropriate for this core orbital.

A neutral lead atom has two $1s$ electrons. Their contributions add in a frozen-orbital estimate of total atomic energy, roughly twice the per-electron shift, with relaxation and correlation corrections. An experimentally observed transition frequency again depends on the difference between the finite-size shifts of both participating many-electron quantum states. A quantitative neutral-lead result requires a screened relativistic many-electron calculation, not the one-electron formula with a single universally valid effective charge.