Question

Using Bohr's postulates, derive the expression for the orbital period of the electron moving in the n^th orbit of hydrogen atom ?

Answer 1

According to Bohr's model, the centripetal force required for the revolution of electrons around the nucleus is provided  by the electrostatic force of attraction between the electron and the nucleus.

If m is the mass of electron moving with a velocity v in a circular orbit of radius r, then the necessary centripetal force is \[F = \frac{m v^2}{r} . . . . . (i)\]

Also, the electrostatic force of attraction between the nucleus of charge (+Ze) and electron of charge (-e) is

\[F = \frac{1}{4\pi \epsilon_o}\frac{Z e^2}{r^2} = \frac{KZ e^2}{r^2} . . . . . (ii)\]

\[\text { where }, K = \frac{1}{4\pi \epsilon_o}\]

From (i) and (ii)

\[\frac{m v^2}{r} = \frac{KZ e^2}{r^2} \]

\[ \Rightarrow r = \frac{KZ e^2}{m v^2} . . . . . (iii)\]

Also one of the Bohr’s postulates states that the electron revolves in stationary orbits where the angular momentum of electron is an integral multiple of h/2π.  

\[mvr = \frac{nh}{2\pi}\]

Here, h is Planck's constant and n is any positive integer, 1, 2, 3...

\[\Rightarrow r = \frac{nh}{2\pi vm} . . . . . (iv)\]

From (iii) and (iv), we get 

\[\frac{KZ e^2}{m v^2} = \frac{nh}{2\pi mv}\]

\[or\]

\[v = \frac{2\pi KZ e^2}{nh} . . . . . (v)\]

Now, we know frequency (\[\nu\]) of electron in Bohr's stationary orbit is given by

\[v = r\omega = r(2\pi\nu)\]

\[\nu = \frac{v}{2\pi r}\]

\[\text { From (v), we get }\]

\[\nu = \frac{v}{2\pi r} = \frac{2\pi KZ e^2}{nh . 2\pi r} = \frac{KZ e^2}{nhr}\]

\[\nu = \frac{{KZe}^2}{nhr} \]

For hydrogen atom, Z = 1. Therefore

\[\nu = \frac{{Ke}^2}{nhr}\]

Above expression is the orbital frequency of electron moving in the n^th orbit of hydrogen atom. 
Now, we know the orbital period (T) of the electron moving in the n^th orbit of hydrogen atom is related to ​(\[\nu\]) as \[T = \frac{1}{\nu}\]

\[ \Rightarrow T = \frac{nhr}{{Ke}^2}\]