Calculate the energy required for the process  ${\mathrm{He}}^{+}\left(\mathrm{g}\right)\to {\mathrm{He}}^{2+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}$. The ionization energy for the H-atom in the ground state is $2.18\times {10}^{-18} \mathrm{J}{\text{ atom }}^{-1}$

Energy of electron in unielectron atomic system,

${\mathrm{E}}_{\mathrm{n}}=\frac{-2{\mathrm{\pi }}^2{\mathrm{mZ}}^2{\mathrm{e}}^4}{{\mathrm{n}}^2{\mathrm{h}}^2}$

For H-atom, ionisation energy (IE) = ${\mathrm{E}}_{\infty }-{\mathrm{E}}_1$

$\mathrm{IE}=0-\left(-\frac{2{\mathrm{\pi }}^2{\mathrm{me}}^4{1}^2}{1^2{\mathrm{h}}^2}\right)\text{ (where, }\mathrm{Z}=1\text{ and }\mathrm{n}=1\text{ for }$H-atom)

$\mathrm{IE}=2.18\times {10}^{-18} \mathrm{J}{\text{ atom }}^{-1}$

$\text{ For }{\mathrm{He}}^{+},\mathrm{IE}={\mathrm{E}}_{\infty }-{\mathrm{E}}_1$

$=0-\left(\frac{-2{\mathrm{\pi }}^2{\mathrm{me}}^4{2}^2}{1^2{\mathrm{h}}^2}\right)=4\times \frac{2{\mathrm{\pi }}^2{\mathrm{me}}^4}{{\mathrm{h}}^2}$

$=4\times 2.18\times {10}^{-18} \mathrm{J}{\text{ atom }}^{-1}$

$=8.72\times {10}^{-18} \mathrm{J}{\text{ atom }}^{-1}$

$\therefore \text{ The energy required for the process }$

${\mathrm{He}}^{+}⟶{\mathrm{He}}^{2+}+{\mathrm{e}}^{-}\text{is }8.72\times {10}^{-18}{\text{ atoms }}^{-1}\text{. }$