Question

The electron of a hydrogen atom revolves round the proton in $n^{\text{th }}$ circular orbit of radius $r_n=\frac{{\epsilon }_0{n}^2{h}^2}{\left(\pi m{e}^2\right)}$ with a speed $v_n=\frac{e^2}{2{\epsilon }_{0}nh}$. The current due to the circulating electron is proportional to $e^n$. Find $n$.

Answer 1

Answer: 5

Time period of revolution of electron in $n^{\text{th }}$ orbits is
$T_n=\frac{2\pi r_n}{v_n}=\frac{2\pi \left(\frac{{\epsilon }_0{n}^2{h}^2}{\pi m{e}^2}\right)}{\frac{e^2}{2{\epsilon }_{0}nh}}=\frac{4{\epsilon }_0^2{n}^3{h}^3}{m{e}^4}$
The current due to circulating electron in $n^{\text{th }}$ orbit is
$I_n=\frac{e}{T_n}=\frac{e}{\left(\frac{4{\epsilon }_0^2{n}^3{h}^3}{m{e}^4}\right)}=\frac{m{e}^5}{4{\epsilon }_0^2{n}^3{h}^3}$
$I_n\propto e^5$