The ionisation energy of  ${\mathrm{He}}^{+}$ is $19.6 \mathrm{x} 1018 \mathrm{J} {\mathrm{atom}}^{-1}$. The energy of the first stationary state of ${\mathrm{Li}}^{2+}$ is

$\mathrm{I}.\mathrm{E}.=+\mathrm{x}\times \frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{J}/\mathrm{atom}$

For He⁺, n = 1, Z = 2

$19.6\times {10}^{-18}=+\mathrm{x}\times \frac{2^2}{1^2}$

$19.6\times {10}^{-18}=\mathrm{x}\times 4$

$\frac{19.6\times {10}^{-18}}{4}=\mathrm{x}$

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

For ${\mathrm{Li}}^{2+}$atom,  $T.E.=-4.9\times {10}^{-18}\times \frac{Z}{n^2}J./atom$

($⸪ \mathrm{T}.\mathrm{E} \& \mathrm{I}.\mathrm{E}$. have opposite signs & rest formula is the same)

$-4.9\times {10}^{-18}\times \frac{3^2}{1^2}=-4.9\times {10}^{-18}\times 9$

$=-44.1\times {10}^{-18}\mathrm{J}{\mathrm{atom}}^{-1}\mathrm{or}-4.41\times {10}^{-17}{\mathrm{atom}}^{-1}$