Explanation

(b) The minimum energy needed to ionise an atom is called ionisation energy. The potential difference through which an electron should be accelerated to acquire this much energy is called ionisation potential.
$\ \ \ \ \ \ \ \ \ \left( E_{2} \right)_{H}\ –\ \left( E_{1} \right)_{H} = \ 10.2\ eV$
or$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\left( E_{1} \right)_{H}}{4}$ $- \ \left( E_{1} \right)_{H}\ = \ 10.2\ eV$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( E_{1} \right)_{H}\ = \ - \ 13.6\ eV$
Hence ,ionisation potential energy is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \left( E_{\infty} \right)_{H} - \left( E_{1} \right)_{H}\ = \ 13.6\ eV$
$\therefore$ Ionisation potential = 13.6 V