Question

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

Answer 1

Time period of revolution of electron in $n^{\text {th }}$ orbits is
$T_{n}=\frac{2 \pi r_{n}}{v_{n}}=\frac{2 \pi\left(\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}}\right)}{\frac{e^{2}}{2 \varepsilon_{0} n h}}=\frac{4 \varepsilon_{0}^{2} n^{3} h^{3}}{m e^{4}}$
The current due to circulating electron in $n^{\text {th }}$ orbit is
$I_{n}=\frac{e}{T_{n}}=\frac{e}{\left(\frac{4 \varepsilon_{0}^{2} n^{3} h^{3}}{m e^{4}}\right)}=\frac{m e^{5}}{4 \varepsilon_{0}^{2} n^{3} h^{3}}$
$I_{n} \propto e^{5}$