Question

Derive an expression for the potential and kinetic energy of an electron in an orbit of a hydrogen atom according to Bohr’s model.

Answer 1

Hint: According to Bohr’s model, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. for an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is given by the coulomb’s law of attraction between the electron and nucleus. The gravitational attraction may be neglected as the mass of electrons and protons is very small.

Complete step by step solution:
Since the electron is moving in a fixed orbit, therefore the centripetal force is equal to the attraction between electron and proton.
$\therefore {\displaystyle \frac{m{v}^2}{r}}=k{\displaystyle \frac{e^2}{r^2}}$
$\Rightarrow m{v}^2=k{\displaystyle \frac{e^2}{r}}$ , ….equation (1)
Where $m$ is the mass of the electron, r is the radius of the orbit, $e$ is the charge of the electron, and $k={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$ is a constant
Now express the formula for the angular momentum of the electron in orbit
$\therefore mvr={\displaystyle \frac{nh}{2\pi }}$ , where $h$ is planck's constant.
$$\Rightarrow v={\displaystyle \frac{nh}{2\pi mr}}$$
We can substitute this expression $v$ in equation 1
$\therefore m{\left({\displaystyle \frac{nh}{2\pi mr}}\right)}^2=k{\displaystyle \frac{e^2}{r}}$
$\Rightarrow r={\displaystyle \frac{n^2{h}^2}{4{\pi }^{2}km{e}^2}}$ ……equation (2)
From equation (1) we can write
$\therefore E_e={\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{1}{2}}k{\displaystyle \frac{e^2}{r}}$
Using equation (2) we can put $r$
$\therefore E_e={\displaystyle \frac{k{e}^2}{2}}{\displaystyle \frac{4{\pi }^{2}km{e}^2}{n^2{h}^2}}$
$\Rightarrow E_e={\displaystyle \frac{2{\pi }^2{k}^{2}m{e}^2}{n^2{h}^2}}$
Now for the potential energy we have
$E_p=-k{\displaystyle \frac{e^2}{r}}$
Write the expression $r$ from equation 1 in the above equation
$\therefore E_p=-k{e}^2{\displaystyle \frac{4{\pi }^{2}km{e}^2}{n^2{h}^2}}$
$\Rightarrow E_p=-{\displaystyle \frac{4{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}}$
Hence $E_e$ and $E_p$ are the kinetic energy and potential energy respectively.

Note: Here the kinetic energy and potential energy for the electron is different from the other objects having more mass. When we talk about electrons we neglect the gravitational attraction of the earth. But we don’t consider its mass zero.