= Solution
Let $r$ be the electron-nucleus separation and use the <reduced mass> $\mu=m_eM/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative <angular momentum>:
$$
\frac{\mu v^2}{r}=\frac{Ze^2}{4\pi\epsilon_0r^2},\qquad \mu vr=n\hbar.
$$
Eliminating $v$ gives
$$
\boxed{r_n=\frac{4\pi\epsilon_0\hbar^2}{\mu Ze^2}n^2
=a_0\frac{m_e}{\mu}\frac{n^2}{Z}.}
$$
The total <energy> is $E=\mu v^2/2-Ze^2/(4\pi\epsilon_0r)=-Ze^2/(8\pi\epsilon_0r)$, hence
$$
\boxed{E_n=-\frac{\mu Z^2e^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}
=-hcR_M\frac{Z^2}{n^2},\quad R_M=R_\infty\frac\mu{m_e}.}
$$
Here $a_0$ is the ordinary <Bohr radius> and $R_\infty$ the infinite-nuclear-mass <Rydberg constant>. The electron's radius about the centre of mass is $Mr_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.
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