If the first ionization energy of hydrogen atom is 13.6 e V, then the second ionization energy of He atom is 

$$\begin{gathered} {\mathrm{E}}_{\mathrm{H}} =-\frac{13.6}{{\mathrm{n}}^2} ; ( {\mathrm{E}}_1)\mathrm{H} =-13.6 \mathrm{eV} \\ {\mathrm{E}}_1( {\mathrm{He}}^{+ })=-13.6 \times \frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-13.6\times \frac{{\left(2\right)}^2}{1}=-54.4\mathrm{eV} \\ {\mathrm{IP}}_2 \mathrm{of} \mathrm{He}\hspace{0.17em}\mathrm{will} \mathrm{be} \mathrm{endothermic} \mathrm{with} \mathrm{same} \mathrm{magnitude}; \end{gathered}$$