Using Bohr's atomic model,derive an equation for the radius of the $n^{th}$ orbit of an electron.

(N/A) The atomic model of Bohr is shown in the figure.
Let the mass of an electron be $m$,its charge be $e$,its linear speed in the $n^{th}$ orbit be $v_n$,the radius of the orbit be $r_n$,and the charge on the nucleus be $Ze$,where $Z$ is the atomic number.
The necessary centripetal force is provided by the Coulombic attractive force between the electron and the nucleus. Thus,
$\frac{m v_n^2}{r_n} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(e)}{r_n^2} = \frac{Z e^2}{4 \pi \epsilon_0 r_n^2}$ ... $(1)$
From Bohr's second postulate,the angular momentum $L_n$ of an electron in the $n^{th}$ orbit is given by:
$L_n = m v_n r_n = \frac{n h}{2 \pi}$ ... $(2)$
From equation $(1)$,we have $m v_n^2 = \frac{Z e^2}{4 \pi \epsilon_0 r_n}$.
Multiplying both sides by $r_n^2$,we get $m v_n^2 r_n^2 = \frac{Z e^2 r_n}{4 \pi \epsilon_0}$.
Taking the square root,$m v_n r_n = \sqrt{\frac{Z e^2 m r_n}{4 \pi \epsilon_0}}$.
Equating this to equation $(2)$:
$\frac{n h}{2 \pi} = \sqrt{\frac{Z e^2 m r_n}{4 \pi \epsilon_0}}$
Squaring both sides:
$\frac{n^2 h^2}{4 \pi^2} = \frac{Z e^2 m r_n}{4 \pi \epsilon_0}$
Solving for $r_n$:
$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$