Q. The binding energy of an electron in the ground state of He is equal to 24.6 eV. The energy required to remove both the electrons is
A) 49.2 eV
B) 24.6 eV
C) 38.2 eV
D) 79.0 eV
Explanation
(d) Helium atom has 2 electrons. When one electron is removed, the remaining atom is hydrogen like atom, whose energy in first orbit is
$E_{1} = - (2)^{2}(13.6\ eV) = - 54.4\ eV$
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 $= 24.6 + 54.4 = 79.0\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.