How does the frequency of revolution of the electron in Bohr’s theory…

The correct answer is Option (3) → \( r^{-3/2} \)

In the Bohr model of the hydrogen atom, the electrostatic attraction of the nucleus gives the centripetal force for the electron. The frequency of revolution is the number of orbits the electron makes per second.

For an electron of mass \( m \) and charge \( e \) in an orbit of radius \( r \):

    \( \displaystyle \frac{mv^{2}}{r}=\frac{e^{2}}{4\pi\varepsilon_0\,r^{2}} \)

    \( \displaystyle \Rightarrow\; v=\frac{e}{\sqrt{4\pi\varepsilon_0\,m\,r}} \)

So \( v\propto r^{-1/2} \). The frequency of revolution is then:

    \( \displaystyle \nu=\frac{v}{2\pi r} \)

    \( \begin{aligned} \nu &\propto \frac{r^{-1/2}}{r} \\ &= r^{-3/2} \end{aligned} \)