Question

The binding energy of an electron in the ground state of $He$ is equal to $24.6eV$ . What is the amount of energy (in $eV$ ) required to remove both the electrons?

Answer 1

Answer: 79

Helium atom has 2 electrons. When one electron is removed, the remaining atom is hydrogen like atom, whose energy in first orbit is
$E_1=-{\left(2\right)}^2\left(13.6eV\right)=-54.4eV$
Therefore, to remove the second electron from the atom, the additional energy of 54.4 eV is required. Hence, total energy required to remove both the electrons $=24.6+54.4=79.0eV$