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 ${n^{th}}$ orbit is:
A.- $\dfrac{{2{E_n}}}{{nh}}$
B.- $\dfrac{{2{E_n}n}}{h}$
C.- $\dfrac{{{E_n}}}{{nh}}$
D.- $\dfrac{{{E_n}n}}{h}$

Answer 1

As we know that in an atom kinetic energy is equal to binding energy. Kinetic energy of an electron in any state can be given with the help of Bohr’s concepts.

Complete step by step solution:
According to Bohr’s statements, kinetic energy or binding energy ${E_n}$ of electron in ${n^{th}}$state is:
$\dfrac{1}{2}m{v^2} = {E_n}$ −−−− (i)
Bohr described the angular momentum of an electron of ${n^{th}}$state with following equation:
$mvr{\text{ }} = {\text{ }}\dfrac{{nh}}{{2\pi }}$ ​−−−− ( ii)
Here mvr is the angular momentum of the moving electron.
m= Mass of electron
v = Velocity of moving electron
r = Atomic radius i.e. distance of electron from nucleus
n = Principle quantum number
h= Plank’s constant
Now we want to calculate the frequency of revolution in ${n^{th}}$orbit i.e. $\dfrac{v}{{2\pi r}}$.
For this purpose we will divide equation (i) by equation (ii)
$\begin{aligned} \dfrac{{m{v^2}}}{{2mvr}} = \dfrac{{2\pi {E_n}}}{{nh}} \\\ \\\ \end{aligned}$
On making some rearrangements above relation become:
$\dfrac{v}{{2\pi r}} = \dfrac{{2{E_n}}}{{nh}}$
Therefore, the frequency of revolution of the electron in the nth orbit is = $\dfrac{{2{E_n}}}{{nh}}$

So the correct answer is option (A) according to the above explanation.

Note: Beware of what we need to find out in the question. If the question asked about calculating frequency only then we have to solve it in a different manner. But here we are asked for the frequency of revolution which is different from frequency.