Solution
= Solution
For a $2p$ <atomic orbital>, $l=1$. The radial factor obeys
$$
R'=Ne^{-r/2}(1-r/2),\qquad R''=Ne^{-r/2}(r/4-1).
$$
Substitution into the <radial Schrödinger equation for a central potential> gives
$$
-\frac12R''-\frac{R'}r+\frac{R}{r^2}-\frac Rr=-\frac18R.
$$
Thus the <energy eigenvalue> is
$$
\boxed{E=-\frac18\,E_{\rm h}\simeq-3.40\ \mathrm{eV}.}
$$
The <normalization constant> cancels; $E_{\rm h}$ is the <Hartree energy>.