Using the Bohr model,calculate the electric current created by the electron when the $H$-atom is in the ground state.

(A) The orbital radius of an electron in the $n^{\text{th}}$ orbit of an $H$-atom is given by $r_{n} = \frac{\epsilon_{0} n^{2} h^{2}}{\pi m Z e^{2}}$.
For the ground state,$n = 1$ and $Z = 1$,so $r_{1} = a_{0} = 0.53 \times 10^{-10} \text{ m}$.
The velocity of the electron in the $n^{\text{th}}$ orbit is $v_{n} = \frac{Z e^{2}}{2 \epsilon_{0} n h}$. For $n = 1$,$v_{1} = 2.187 \times 10^{6} \text{ m/s}$.
The frequency of revolution $f_{1}$ is given by $f_{1} = \frac{v_{1}}{2 \pi r_{1}}$.
The electric current $I$ is $I = f_{1} e = \frac{v_{1} e}{2 \pi r_{1}}$.
Substituting the values: $I = \frac{(2.187 \times 10^{6} \text{ m/s}) \times (1.6 \times 10^{-19} \text{ C})}{2 \times 3.1416 \times (0.53 \times 10^{-10} \text{ m})}$.
$I \approx 1.05 \times 10^{-3} \text{ A} = 1.05 \text{ mA}$.