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

Ionization energy of  $H{e}^{+}=E_0-E_1=0-E_1$

$E_1\left(H{e}^{+}\right)=-19.6\times {10}^{-18}Jat{m}^{-1}$

$En=\frac{k{Z}^2}{n^2}$

$-19.6\times {10}^{-18}=\frac{-k\times 2^2}{1^2}$

$K=4.9\times {10}^{-18}$

$E_1\left(L{i}^{2+}\right)=\frac{4.9\times {10}^{-18}\times 9}{I^2}$

$=-44.1\times {10}^{18}Jato{m}^{-1}$

$=-4.410+{10}^{17}Jato{m}^{-1}$