Question

In a hydrogen atom, the binding energy of the electron in the ground state is $E_1.$ Then the frequency of revolution of nth electron in the nth orbits is

• A. $\frac{2E_1}{nh}$
• B. $\frac{2E_1{n}^3}{h}$
• C. $\sqrt{\frac{2m{E}_1}{n^{3}h}}$
• D. $\frac{2E_1{n}^2}{h}$

Answer 1

The correct option is A $\frac{2E_1}{nh}$

$\begin{array}{c}\text{B.E of electron= K.E of electron (in that orbit)}\\ \frac{1}{2}m{v}^2=E_1-\left(i\right)\\ \therefore \text{angular momentum -}mvr=\frac{nh}{2\pi }-\left(ii\right)\\ \text{Now from (i) divided by (ii)}\\ \frac{v}{2r}=\frac{E_1\times 2\pi }{nh}\\ \frac{v}{2\pi r}=\frac{2E_1}{nh}\\ ⟹f=\frac{2E_1}{nh}\end{array}$