Question

Calculate the energy required (in $eV$ ) to remove both the electrons of $He$ atom, if the binding energy of an electron in the ground state of $He$ is equal to $24.6eV$ .

Answer 1

The energy needed to remove one electron from the ground state of $He=24.6eV$ As the $He^{+}$ is now hydrogen-like, Ionisation energy $=\frac{13 . 6 Z^{2}}{n^{2}}eV$
$=\frac{13 . 6 \times 2^{2}}{1^{2}}eV$
$=54.4eV$
$\therefore $ To remove both the electrons, energy needed $=\left(\right.54.4+24.6\left.\right)eV=79eV$