The electric potential between a proton and an electron is given by $V=V_0\ln \frac{r}{r_0}$, where $r_0$ is a constant. Assuming Bohr's model to be applicable, write the variation of $r_n$,
n being the principal quantum number

$$\begin{gathered} \text{ Potential energy }U=eV=e{V}_0\ln \frac{r}{r_0} \\ \text{ Force }F=-\left|\frac{dU}{dr}\right|=\frac{e{V}_0}{r{r}_0}\text{ . } \end{gathered}$$

Therefore the force will provide the necessary centripetal force. 

Hence,

 $$\begin{gathered} \frac{m{v}^2}{r}=\frac{e{V}_0}{r{r}_0} \\ \Rightarrow v=\sqrt{\frac{e{V}_0}{m{r}_0}} \end{gathered}$$………(i)

and

 $mvr=\frac{nh}{2\pi }$………(ii)

$$\begin{gathered} \text{ From equations (i) and (ii), } \\ m\sqrt{\frac{e{V}_0}{m{r}_0}}\cdot r=\frac{nh}{2\pi } \\ \Rightarrow r=\frac{nh}{2\pi }\sqrt{\frac{r_0}{me{V}_0}} \\ \Rightarrow r_n\propto n \end{gathered}$$