Calculate the ionization energy of $H$,$He^{+}$,and $Li^{2+}$ in $kJ \, mol^{-1}$.

The ionization energy $(IE)$ of a hydrogen-like species is given by the formula: $IE = 1312 \times Z^2 \, kJ \, mol^{-1}$.
For $H$ $(Z=1)$: $IE = 1312 \times (1)^2 = 1312 \, kJ \, mol^{-1}$.
For $He^{+}$ $(Z=2)$: $IE = 1312 \times (2)^2 = 1312 \times 4 = 5248 \, kJ \, mol^{-1}$.
For $Li^{2+}$ $(Z=3)$: $IE = 1312 \times (3)^2 = 1312 \times 9 = 11808 \, kJ \, mol^{-1}$.
Thus,the values are $1312 \, kJ \, mol^{-1}$,$5248 \, kJ \, mol^{-1}$,and $11808 \, kJ \, mol^{-1}$ respectively.