= Solution
Write $C=e^2/(4\pi\epsilon_0)$. The <shell theorem> makes the <potential energy> of the <Electron> constant inside the charged <spherical shell>. Relative to the point-proton <Coulomb potential>, the perturbation is
$$
\Delta V(r)=\begin{cases}
C\left(\dfrac1r-\dfrac1b\right),&0<r<b,\\
0,&r\geq b.
\end{cases}
$$
It is positive: the finite charge distribution weakens the attraction close to the origin.
The <first-order energy correction> is the <expectation value> of this perturbation in the unperturbed <hydrogen atom> <ground state>. Integrating its <probability density> over spherical volume gives
$$
\Delta E=\frac{4C}{a_0^3}\int_0^b e^{-2r/a_0}\left(r-\frac{r^2}{b}\right)\,dr.
$$
For a <proton> radius much smaller than the <Bohr radius>, expand the exponential uniformly over this short interval. The leading integral is
$$
\int_0^b\left(r-\frac{r^2}{b}\right)\,dr=\frac{b^2}{6},\qquad
\Delta E=\frac{2Cb^2}{3a_0^3}+O\left(\frac{Cb^3}{a_0^4}\right).
$$
Since $E_{1s}=-C/(2a_0)$, the signed fractional change is
$$
\boxed{\frac{\Delta E}{E_{1s}}=-\frac43\left(\frac b{a_0}\right)^2+O\left(\frac{b^3}{a_0^3}\right)}.
$$
Equivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the <energy> increases and the binding-energy magnitude decreases. This is the <finite proton-shell correction to hydrogen ground energy>.
If no small-radius expansion is made, the same <time-independent perturbation theory> integral gives, with $\beta=b/a_0$,
$$
\frac{\Delta E}{|E_{1s}|}=2\left[1-\frac1\beta+\left(1+\frac1\beta\right)e^{-2\beta}\right].
$$
Its expansion starts with $4\beta^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed <Schrödinger equation>.
Back to article page
