An expression for the frequency of revolution of the electron in the Bohr's orbit is :

$\frac{me^4}{4\epsilon^2_0h^3}\times \frac{1}{n^3}$

The correct answer is option (1) : $\frac{me^4}{4\epsilon^2_0h^3}\times \frac{1}{n^3}$

In Bohr's orbit

$mvr=\frac{nh}{2\pi}$ ............. (i)

$\frac{e^2z}{4\pi \epsilon r^2}=\frac{mv^2}{r}$

$mv^2r=\frac{e^2z}{4\pi \epsilon_0}$ ............ (ii)

On putting the value of mvr from equation (i)

$v\times \frac{nh}{2\pi }\times \frac{e^2z}{4\pi \epsilon_0}$

$v=\left(\frac{e^2z}{2nh}\right)$

$\frac{m\times e^2z}{2nh}z=\frac{nh}{2\pi}$

$r=\frac{n^2h^2}{\pi me^2z}$

Now, $f=\frac{v}{2\pi r}$

$=\frac{e^2z}{\frac{2nh}{\frac{2\pi n^2h^2}{\pi me^2z}}}$

$=\frac{me^4}{4\epsilon^2_0h^3}\times \frac{1}{n^3}$