Explanation

(c) The potential energy of hydrogen atom
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{n}\ =$ $\frac{13.6}{n^{2}}$ $eV$
So, the potential energy in second orbit is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2}\ = \ -$ $\frac{13.6}{2^{2}}$ $eV$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2}\ = \ - \ \frac{13.6}{4}eV\ = \ - 3.4\ eV$
Now, the energy required to remove an electron from second orbit to infinity is
$U = E_{\infty} - E_{2}$ [From work-energy theorem and $E_{\infty} = 0$]
$\Rightarrow \ \ \ \ \ \ \ \ \ U = 0 - ( - 3.4)\ eV$
Or $\ \ \ \ \ U = 3.4\ eV$
Hence, the required energy is 3.4 eV.