• # question_answer In a hydrogen atom, the binding energy of the electron in the ground state is ${{E}_{1}}.$Then the frequency of revolution of the electron in the nth orbit is A) $\frac{2{{E}_{1}}}{{{n}^{3}}h}$                                  B) $\frac{2{{E}_{1}}{{n}^{3}}}{h}$C) $\sqrt{\frac{2m{{E}_{1}}}{{{n}^{3}}h}}$                    D) $\frac{{{E}_{1}}{{n}^{2}}}{h}$

Binding energy of electron in the nth orbit is ${{E}_{n}}={{E}_{1}}/{{n}^{2}}.$ Now, $m{{v}_{n}}{{r}_{n}}=\frac{nh}{2\pi }$and $\frac{1}{2}mv_{n}^{2}={{E}_{n}}$ $\Rightarrow$            $\frac{\left( \frac{1}{2} \right)mv_{n}^{2}}{m{{v}_{n}}{{r}_{n}}}=\frac{{{E}_{n}}}{\frac{nh}{2\pi }}$ $\Rightarrow$$\frac{{{v}_{n}}}{{{r}_{n}}}=\frac{4\pi {{E}_{n}}}{nh}=\frac{4\pi {{E}_{1}}}{{{n}^{3}}h}\Rightarrow \frac{4\pi {{E}_{1}}}{{{n}^{3}}h}$ $\therefore$    ${{f}_{n}}=\frac{{{\omega }_{n}}}{2\pi }=\frac{2{{E}_{1}}}{{{n}^{3}}h}$ Hence, the correction option is [a].