Explanation

(c) Given, ground state energy of hydrogen atom
$E_{1}\ = \ - 13.6\ eV$
Energy of electron in first excited state ($ie,$ $n$=2)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2}\ $= $- \frac{13.6}{(2)^{2}}$ $eV$
Therefore ,excitation energy
$\ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}E = E_{2}\ –\ E_{1}$
$\ \ = - \frac{13.6}{4}$ $- \ ( -$13.6) $= \ -$3.4 + 13.6 = 10.2 $eV$