Question

Using Bohr’s postulates, obtain the expression for total energy of the electron in the n^th orbit of hydrogen atom.

Answer 1

According to Bohr’s postulates, 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 provided by Coulomb’s force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small.

`So, (mv^2)/r = (ke^2)/r^2 => mv^2 = (ke^2)/r    .................. (1)`

Where, m = mass of electron, r = radius of electronic orbit and v = velocity of electron.

`mvr = (nh)/(2pi)=> v = (nh)/(2pimr)`

From eq(1), we get that:

`m((nh)/(2pimr))^2 = (ke^2)/r => r = (n^2h^2)/(2pimr)................... (2)`

(i) Kinetic energy of electron:

`E_K = 1/2 mv^2 = (ke^2)/(2r)`

Using eq (2),we get:E_k`(ke^2)/2 (4pi^2kme^4)/(n^2h^2) = (2pi^2k^2me^4)/(n^2h^2)`

(ii) Potential energy:

`E_p = - (k(e)xx (e))/r = - (ke^2)/r`

Using eq (2), we get `E_p= -ke^2 xx (4pi^2kme^2)/(n^2h^2) = - (4pi^2k^2me^4)/(n^2h^2)`

Hence, total energy of the electron in the n^th orbit

`E = E_p +E_k = (4pi^2k^2me^4)/(n^2h^2) +  (2pi^2k^2me^4)/(n^2h^2) = - (2pi^2k^2me^4)/(n^2h^2) = -13.6/n^2  eV`