Question

In a hydrogen atom,the binding energy of the electron in the $n^{th}$ state is $E_n$, then the frequency of revolution of the electron in the nth orbits is

• A. $2E_n/nh$
• B. $2E_{/}nh$
• C. $E_{/}nh$
• D. $E_n/nh$

Answer 1

The correct option is A $2E_n/nh$

By Bhor's theory

$mvr=\frac{nh}{2\pi }\Rightarrow 2nr=\frac{nh}{mv}$ ....... (1)$

Also, $\frac{m{v}^2}{2}=$ Binding energy $=E_n$ .......... $\left(2\right)$

We know.

$v=wr\Rightarrow v=2\pi fR$ ........... $\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

$f=\frac{v}{\left(\frac{nh}{mv}\right)}=\frac{m{v}^2}{nh}=\frac{26n}{nh}$

Option $=A$ is correct