Question

Using Bohr’s postulates for hydrogen atom, show that the total energy (E) of the electron in the stationary states tan be expressed as the sum of kinetic energy (K) and potential energy (U), where K = −2U. Hence deduce the expression for the total energy in the n^th energy level of hydrogen atom.

Answer 1

According to Bohr’s postulates for hydrogen atom electron revolves in a circular orbit around the heavy positively charged nucleus. These are the stationary (orbits) states of the atom.

For a particular orbit, electron is moves there, so it has kinetic energy.

Also, there is potential energy due to charge on electron and heavy positively charged nucleons.

Hence, total energy (E) of atom is sum of kinetic energy (K) and potential energy (U).

`i.e. E =K+U`

Let us assume that the nucleus has a positive charge Ze. An electron moving with a constant speed v along a circle of radius of r with centre at the nucleus.

Force acting on e⁻ due to nucleus is given by

`F =(Ze^2)/(4piin_0r^2)`

The acceleration of electron  `v^2/r`(towards the centre)

If, m = mass of an electron

Then, from Newton’s 2^nd Law.

`F =m(V_2/r)`

`=>Ze^2/(4piin_0r^2) = m(V^2/r)`

`=>r = (Ze^2)/(4piin_0 mv^2)   ..... (1)`

From Bohr’s quantization rules

`mvr = n h/2pi     ... (2)`

Where n = a positive integer

Plug-in the value of r from equation (i)

`mv*(Ze^2)/(4piin_0 (mv^2)) = n h/(2pi)`

`v=(Ze^2)/(2in_0hn)  ..... (3)`

So, kinetic energy `k = 1/2mv^2`

`= (mZ^2e^4)/(8in_0^2 h^2 n^2)   ..... (4)`

Potential energy of the atom

`U =-(Ze^2)/(4piin_0r)    .... (5)`

Using (iii) in equation (i),

`r = (Ze^2)/(4piin_0 m(Ze^2)^2/(2in_0 hn)^2`

`=(4in_0^2 h^2 n^2)/((4piin_0)mZe^2`

` r =(in_0h^2n^2)/(pimZe^2)`

Using value of r, in equation (v)

`U = (-Ze^2)/(4piin_0 ((in_0h^2 n^2)/(pimZe^2)))`

`U = (-Z^2e^4)/(4in_0^2 h^2n^2)`

So, the total energy,

E = K + U

`= +(Z^2e^4)/(8in_0^2 h^2 n^2) - (Z^2e^4)/(4in_0^2 h^2 n^2)`

`= - (Z^2e^2)/(8in_0^2 h^2 n^2)`