(a) Potential energy $U = eV = eV_{0}\ln\frac{r}{r_{0}}$
$\therefore$ Force $F = - \left| \frac{dU}{dt} \right| = \frac{eV_{0}}{r}$
$\therefore$ The force will provide the necessary centripetal force. Hence $\frac{mv^{2}}{r} = \frac{eV_{0}}{r}$
$\therefore\ \Rightarrow v = \sqrt{\frac{eV_{0}}{m}}$ …(i)
and $mvr = \frac{nh}{2\pi}$ …(ii)
From equations (i) and (ii)
$mr = \left( \frac{nh}{2\pi} \right)\sqrt{\frac{m}{eV_{0}}}$ or $r \propto n$