The binding energy of an electron in the ground state of a neutral Helium atom is known to be 24.6 e

Question

The binding energy of an electron in the ground state of a neutral Helium atom is known to be 24.6 eV. What is the minimum total energy required to completely remove both electrons from the neutral Helium atom in its ground state?

Professor PC Thomas Classes · Answer

The first ionization energy (removing the first electron) is given as $E_1 = 24.6 	ext{ eV}$.
After removing one electron, the system becomes a hydrogen-like $He^+$ ion ($Z=2$).
The energy required to remove the remaining electron from its ground state is given by the Bohr model:
$E_2 = 13.6 	imes Z^2 = 13.6 	imes 2^2 = 54.4 	ext{ eV}$.
Total energy required to remove both electrons = $E_1 + E_2 = 24.6 	ext{ eV} + 54.4 	ext{ eV} = 79.0 	ext{ eV}$.