Question

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. 54.4 eV
• C. 79 eV
• D. 108.8 eV

Answer 1

Answer: (C)

79 eV

Answer 2

The energy needed to remove one electron from the ground state of $He = 24.6ev$

As the $H^{+}$ is now hydrogen like ionization energy

$= | - 13.6|\frac{2^{2}}{1^{2}}eV \Rightarrow E = 54.4eV$

$\therefore$ To remove both the electrons, energy needed

$= (54.4 + 24.6)eV = 79eV$