The electric potential between a proton and an electron is given by $V=V_0\ln \frac{r_n}{r_0},$ where  $r_0$ is a constant and "$r_n$" is the distance of electron from the proton. Assuming Bohr’s model to be applicable, variation of $r_n$ with ‘n’. ‘n’ being the principal quantum number is  $r_n\propto n^x$. Then the value of ‘x’ is 

Given potential energy (U) between electron and proton 
$=e{V}_0\ln \frac{r}{r_0} \left[\because \left|U\right|=eV\right]$    
$\therefore \left|F\right|=\left|\frac{-du}{dr}\right|=\frac{d}{dr}\left[e{V}_0{\log }_e\frac{r}{r_0}\right]=\frac{e{V}_0}{r_0}\times \frac{1}{r}$ 
This force will provide the necessary centripetal force 
 $\therefore \frac{m{v}^2}{r}=\frac{e{V}_0}{r{r}_0}\Rightarrow m{v}^2=\frac{e{V}_0}{r_0}\mathrm{..........}\left(i\right)$
As per Bohr’s postulate,  $mvr=\frac{nh}{2\pi }\mathrm{...........}\left(ii\right)$
From eq. (i) and (ii) 
$\frac{m^2{v}^2{r}^2}{m{v}^2}=\frac{n^2{h}^2{r}_0}{4{\pi }^2\times V_0^{ }e}\Rightarrow r^2=\frac{n^2{h}^2{r}_0}{4\pi V_{0}me}\Rightarrow r\propto n$