Suppose that the potential energy between elect and proton at a distance $r$ in a hypothetical hydrogen atom is given by $\frac{-k{e}^2}{3r^3}.$ Use Bohr's theory to obt energy levels of such a hypothetical atom.

For the hypothetical atom, the potential energy of electron revolving in the $n^{\text{th }}$ orbit is given by

     $U=-\frac{k{e}^2}{3r^3}$

The force on the electron in this potential field given by

      $F=-\frac{dU}{dr}=\frac{k{e}^2}{r^4}$

This force provides the necessary centripetal force the electron to revolve in a circle of radius $r$ in $n^{\text{th }}$ orbit with a speed $v$, so we have

      $$\begin{gathered} \frac{m{v}^2}{r}=\frac{k{e}^2}{r^4} \\ \Rightarrow m{v}^2=\frac{k{e}^2}{r^3} \end{gathered}$$

Applying Bohr's Quantisation Rule i.e.  $mvr=\frac{nh}{2\pi }$

     $\Rightarrow v=\frac{nh}{2\pi mr}$

Substituting equation (2) in (1), we get

$$\begin{gathered} m{\left(\frac{nh}{2\pi mr}\right)}^2=\frac{k{e}^2}{r^3} \\ \Rightarrow r=r_u=\frac{4{\pi }^{2}k{e}^{2}m}{n^2{h}^2}\text{ and }v=v_n=\frac{n^3{h}^3}{8{\pi }^{3}k{m}^2{e}^2}\backslash \mathrm{endarray} \end{gathered}$$

Now energy in $n^{\text{th }}$ orbit is $E=E_n=U+K$

$\Rightarrow E_n=-\frac{k{e}^2}{3r^3}+\frac{1}{2}m{v}^2$

From equation (1), we get $\frac{1}{2}m{v}^2=\frac{k{e}^2}{2r^3}$

Substituting this value of kinetic energy in equation $\left(3\right)$, we get

      $$\begin{gathered} E_n=-\frac{k{e}^2}{3r^3}+\frac{k{e}^2}{2r^3}=\frac{1}{6}\left(\frac{k{e}^2}{r^3}\right) \\ \Rightarrow E_n=\frac{1}{6}k{e}^2{\left(\frac{n^2{h}^2}{4{\pi }^{2}k{e}^{2}m}\right)}^3=\frac{n^6{h}^6}{384{\pi }^6{k}^2{e}^4{m}^3} \end{gathered}$$