Question

3rd separation energy of an electron in H-atom will be?

Answer 1

Solution:

The separation energy of an electron is expressed by the following equation:

$E_{n_2} - E_{n_1} = - 13.6 \times Z^2 \times [\dfrac{1}{{(n_2)}^2} - \dfrac{1}{{(n_1)}^2}]$      

For Hydrogen atom, Z = 1 and $n_2 = \infty$ and $n_1 = 4$ (3rd excited state)

Putting these values in the equation of separation energy we get:

$\Rightarrow E_{n_2} - E_{n_1} = - 13.6 \times 1^2 \times [\dfrac{1}{\infty}^2} - \dfrac{1}{4}^2}]\\\\\Rightarrow E_{n_2} - E_{n_1} = - 13.6 \times [0 - \dfrac{1}{16}]\\\\\Rightarrow E_{n_2} - E_{n_1} = - 13.6 \times [- \dfrac{1}{16}]\\\\\Rightarrow E_{n_2} - E_{n_1} = \frac{13.6}{16}\\\\\Rightarrow E_{n_2} - E_{n_1} = 0.85 \text eV$

Hence, 3rd separation energy of an electron in H-atom will be 0.85 eV.