Since electrostatic force is a conservative force, therefore, $F=-\frac{dU}{dx}$

$\Rightarrow F=-\frac{dU}{dR}=\frac{K{e}^2}{R^4}$

This force provides the necessary centripetal force, $F_C=\frac{m{v}^2}{R}$

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

Using Bohr quantization of angular momentum theory,

$$\begin{gathered} \Rightarrow mvr=\frac{n\mathrm{h}}{2\mathrm{\pi }} \\ \Rightarrow r=\frac{4{\mathrm{\pi }}^{2}K{\mathrm{e}}^2\mathrm{m}}{n^2{\mathrm{h}}^2} \end{gathered}$$

Thus, Bohr's orbit is, $r=\frac{4{\mathrm{\pi }}^{2}K{\mathrm{e}}^2\mathrm{m}}{n^2{\mathrm{h}}^2}$