Question

How much energy is required to ionise H atom if the electron occupies n = 5 orbits? Compare your answer with the ionization enthalpy of H atom

[ energy required to remove the electron from n =1 orbit].

Answer 1

The energy of an electron in nth orbit is $E=\frac{-21.8\times {10}^{-19}}{n^2}J/atom$.

The ionization energy for the electron in the fifth orbit is:

$\Delta E=E_2-E_1=-21.8\times {10}^{-19}\frac{1}{n_2^2}=\frac{1}{n_1^2}$

$\Delta E=E_2-E_1=-21.8\times {10}^{-19}\frac{1}{\infty }=\frac{1}{5^2}=8.72\times {10}^{-20}$ J.
The ionization energy for the electron in the first orbit is:

$\Delta E^{\prime }=E_2-E_1=-21.8\times {10}^{-19}\frac{1}{n_2^2}=\frac{1}{n_1^2}$

$\Delta E^{\prime }=E_2-E_1=-21.8\times {10}^{-19}\frac{1}{\infty }=\frac{1}{1^2}=21.8\times {10}^{-19}$.
The ratio of the ionization energy of electron in first orbit to the ionization energy of electron in fifth orbit is:

$\frac{\Delta E^{\prime }}{\Delta E}=\frac{21.8\times {10}^{-19}}{8.72\times {10}^{-20}}=25$.