The ionisation energy of an electron in the ground state of helium atom is 24.6eV. The energy required to remove both the electron is

Options

A. 51.8eV
B. 79 eV
C. 38.2 eV
D. 49.2eV

Answer

B

Explanation

First ionization energy of helium atom is `I_(1) = 24.6eV`
Second ionization energy of helium atom is
`I_(2) = (13.6(2)^(2))/((1)^(2)) eV= 54.4eV`
`:.` Energy required to remove both the electrons from helium atom is
`=I_(1) + I_(2) = 24.6eV + 54.4eV + 79eV`