Solution
= Solution
Substitute the quantized <radius> in $mvr=n\hbar$ to get $v_n=e^2/(4\pi\epsilon_0\hbar n)$. <Coulomb potential energy> is $-e^2/(4\pi\epsilon_0r_n)$ and the stipulated virial relation makes the total one half of this value. Hence
$$
\boxed{v_n=\frac{\alpha c}{n},\qquad E_n=-\frac{me^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}=-\frac{\mathrm{Ry}}{n^2}.}
$$
The negative sign means a <bound state> relative to a separated stationary <proton> and <electron>.