Question

Using Bohr's postulates, derive the expression for the total energy of the electron in the stationary states of the hydrogen atom ?

Answer 1

According to Bohr’s 2^nd postulate, we have : `L_n= mv_nr_n =(nh)/(2π)`

where,
n = principle quantum
v_n = speed of the moving electron in the n^th orbit
r_n  = radius of the n^thorbit
The electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force to the electron.

\[\frac{m v^2}{r} = \frac{1}{4 \pi\epsilon_0}\frac{e^2}{r^2}\]

`V_n = e/(sqrt4π∈_0 mr_n)`

`∴ V_n = 1/n e^2/(4π∈_0)1/((h/(2π))`

`r_n =(n^2/m)(h/(2π))^2(4π∈_0)/e^2 `

Total energy, E_n = K.E. + P.E. =

\[\frac{m v^2}{2} - \frac{e^2}{4 \pi\epsilon_0 r}\]

Substituting the values, we get:

`E_n =-e^2/( 8πe_0) (m/n^2)((2π)/h)^2`

⇒ `E_n = (me^4)/(8π^2∈_0^2 h^2)`

`∴ E_n = - (2.18 xx 10^-18)/n^2`j

`E_n = 13.6/n_2 eV`