The ionization energy of  $H{e}^{+}$ is $19.6\times {10}^{-18}Jato{m}^{-1}$. Calculate the energy of the first stationary state of  $L{i}^{2+}$

I.E of  $H{e}^{+}=E\times 2^2(asZforHe=2)$

I.E. of  $L{i}^{2+}=E\times 3^2$ (as Z for Li = 3)

Hence,  $\frac{I.E.\left(H{e}^{+}\right)}{I.E.\left(L{i}^{2+}\right)}=\frac{4}{9}$

Therefore, $I.E\left(L{i}^{2+}\right)=\frac{9}{4}\times I.E.\left(H{e}^{+}\right)$

$=\frac{9}{4}\times 19.6\times {10}^{-18}=4.41\times {10}^{-17}Jato{m}^{-1}$