Calculate the energy required for the process $He_{(g)}^{+} \rightarrow He_{(g)}^{2+} + e^{-}$. The ionization energy for the $H$ atom in the ground state is $2.18 \times 10^{-18} \, J \, atom^{-1}$.

(D) The energy of an electron in a hydrogen-like species is given by the formula: $E_{n} = -2.18 \times 10^{-18} \left( \frac{Z^{2}}{n^{2}} \right) \, J \, atom^{-1}$.
For the process $He_{(g)}^{+} \rightarrow He_{(g)}^{2+} + e^{-}$,the electron is removed from the ground state $(n = 1)$ to infinity $(n = \infty)$.
The energy required is $\Delta E = E_{\infty} - E_{1}$.
Since $E_{\infty} = 0$,we have $\Delta E = -E_{1}$.
For $He^{+}$,the atomic number $Z = 2$. Substituting $Z = 2$ and $n = 1$ into the formula:
$E_{1} = -2.18 \times 10^{-18} \times \left( \frac{2^{2}}{1^{2}} \right) \, J = -2.18 \times 10^{-18} \times 4 \, J = -8.72 \times 10^{-18} \, J$.
Therefore,$\Delta E = 0 - (-8.72 \times 10^{-18} \, J) = 8.72 \times 10^{-18} \, J$.