= Solution
{c}
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,
$$
\boxed{\Delta E_{1s}\simeq\frac{2\pi}{5}k_ZR^2|\psi_{1s}(0)|^2.}
$$
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.
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