Explanation

(c) Energy of an electron in ground state of an atom (Bohr’s hydrogen like atom) is given as
$E = - 13.6Z^{2}eV$ [$Z =$ atomic number of the atom]
$\Rightarrow E_{ionisation} = 13.6\ Z^{2}$
$\Rightarrow \frac{\left( E_{ion} \right)_{H}}{\left( E_{ion} \right)_{Li}} = \left( \frac{Z_{H}}{Z_{Li}} \right)^{2} = \left( \frac{1}{3} \right)^{2} = \frac{1}{9}$