Explanation

(c) The excitation energy in the first excited state is
$\ \ \ \ E\ = \ {RhcZ}^{2}\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right)$ $=$ (13.6 eV) $\times$ $Z^{2} \times \frac{3}{4}$
$\therefore$ 40.8 = 13.6 $\times$ $Z^{2} \times \frac{3}{4}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Z = 2$
So, the ion in problem is $He^{+}$. The energy of the ion in the ground state is
$E\ = \ \frac{{RhcZ}^{2}}{1^{2}}\ =$ $13.6\ \times \ 4\ = \ 54.4\ eV$
Hence, $54.4\ eV$ is required to remove the electron from the ion.