Question

The ionization energy of hydrogen is $13.6\text{eV}$. The first ionization energy of $He$ is $24.6\text{eV}$. Calculate the energy given out (in eV) when $H{e}^{2+}$ nucleus combines with two electrons to form a helium atom in its ground state.

Answer 1

The first ionization energy of $H$ to $H^{+}$ is $13.6\text{eV}$
The first ionization energy of $He$ to $H{e}^{+}$ is $24.6\text{eV}\dots \dots \left(1\right)$
When a $H{e}^{2+}$ ion accepts an electron, the energy released is $Z^2=2^2=4$ times the first ionization energy of hydrogen atom since it forms a single electron, hydrogen like species.
Thus, the energy for the ionization of $H{e}^{2+}$ to $H{e}^{+}$ is $4\times 13.6=54.4\text{eV}\dots \dots \left(2\right)$
When (1) and (2) are added, we get the energy given out when $H{e}^{2+}$ ion gains 2 electrons and forms $He$ atom in its ground state.
It will be $24.6\text{eV}+54.4\text{eV}=79\text{eV}$