(solved)Question 2.14 of NCERT Class XI Chemistry Chapter 2

Energy of a free electron at infinity = 0

Let principal quantum number = n

Let $\displaystyle E_n$ energy in the n-th orbit

Let $\displaystyle R_H $ = Rydberg's constant

by using Bohr's theory for H atom

$\displaystyle \therefore E_n = -R_H\bigg(\frac{1}{n^2}\bigg)$

-ve sign means energy is lowered

$\displaystyle \therefore$ energy required to remove = $\displaystyle (+)E_n$

$\displaystyle = R_H\bigg(\frac{1}{n^2}\bigg)$

$\displaystyle = 2.18 \times 10^{-18} \bigg(\frac{1}{5^2}\bigg)$

$\displaystyle = 8.72 \times 10^{-20} \text{ J }\underline{Ans}$

$\displaystyle \underline{\underline{\text{Comparison with IE}}}$

$\displaystyle E_1 = -\frac{R_H}{1^2}$

$\displaystyle E_5 = -\frac{R_H}{5^2}$

Dividing, $\displaystyle E_1 = 25 \times E_5$

IE = 25 times the answer