 Solution

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.