= Solution
With the proton treated as fixed and electron mass $m$, the classical <energy> and <angular momentum> are
$$
E=\frac12m|\dot{\mathbf r}|^2-\frac{e^2}{4\pi\epsilon_0r},\qquad
\mathbf L=m\mathbf r\times\dot{\mathbf r}.
$$
For a circular orbit, $L=mrv$ and <Coulomb's law> determines its <centripetal acceleration>:
$$
\frac{mv^2}{r}=\frac{e^2}{4\pi\epsilon_0r^2},\qquad mrv=n\hbar.
$$
Eliminating $v$ gives $r=4\pi\epsilon_0n^2\hbar^2/(me^2)$. Thus the <Bohr radius> and orbit radii are
$$
\boxed{a=\frac{4\pi\epsilon_0\hbar^2}{me^2},\qquad r_n=n^2a,\quad n=1,2,\ldots.}
$$
The <kinetic energy> is half the magnitude of the attractive <potential energy>. Hence
$$
\boxed{E_n=-\frac{e^2}{8\pi\epsilon_0r_n}
=-\frac{me^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}.}
$$
This is the circular-orbit <Bohr model>; including proton recoil replaces $m$ by the reduced mass.
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