Suppose the potential energy between electron  and  proton at a distance $r$  is given by  $-\frac{K{e}^2}{3r^3}$ . If  $m$  be the mass of electron, then application of Bohr’s theory to hydrogen atom in this case shows that 
1) energy in the $n^{th}$  orbit is proportional to  $n^6$
2) energy is proportional to $m^{-3}$  
3) energy the  $n^{th}$  orbit is proportional to  $n^{-2}$
4) energy is proportional to  $m^3$

Since,  $\left|F\right|=\frac{dU}{dr}=\frac{K{e}^2}{r^4}\mathrm{...........}\left(1\right)$
 $\Rightarrow \frac{K{e}^2}{r^4}=\frac{m{\upsilon }^2}{r}\mathrm{............}\left(2\right)$
According to Bohr’s Quantisation Rule,  we have
$m\upsilon r=\frac{nh}{2\pi }$        ………….. (3)
From (2) and (3), we get 
$r=\left(\frac{4{\pi }^2{e}^{2}K}{h^2}\right)\frac{m}{n^2}=K_1\left(\frac{m}{n^2}\right)$       …………. (4)
Since total energy E is half the potential energy, so we have 
 $E=-\frac{K{e}^2}{6r^3}=-\frac{K{e}^2}{6{\left(\frac{K_{1}m}{n^2}\right)}^3}=-\frac{K{e}^2{n}^6}{6K_1^3{m}^3}$
So, total energy is $E\propto n^6$  and  $E\propto m^{-3}$