Explanation

(b) Hydrogen atom normally stays in lowest energy state ($n$=1), where its energy is
$\ \ \ \ \ \ \ \ \ \ E_{1}\ = \ \frac{Rhc}{1^{2}}\ = \ - Rhc$
On being ionized its energy becomes zero. Thus, ionization of hydrogen atom is
= energy after ionisation – energy before ionisation
= $0 - ( - Rhc)\ = Rhc$
$= (1.097\ \times \ 10^{7}\ m^{- 1})\ (6.63\ \times \ 10 - 34\ J - s)(3\ \times \ 108\ ms^{- 1})$
=$21.8\ \times \ 10^{- 19}\ J$
=$\frac{21.8\ \times 10^{- 19}}{1.6\ \times 10^{- 19}}$ $= 13.6\ eV$