The binding energy of an electron in the ground state of the He atom is equal to 24 eV. The energy required to remove both the electrons from the atom will be:

Question

The binding energy of an electron in the ground state of the He atom is equal to 24 eV. The energy required to remove both the electrons from the atom will be: (A) 59 eV (B) 81 eV (C) 79 eV (D) None of these

Tardigrade Admin · Accepted Answer

Ionization energy of He =(z2/n2) × 13.6 =(22/12) × 13.6 =54.4 eV Energy required to remove both the electrons = binding energy + ionisation energy = 24.6 + 54.4 = 79 eV

Q. The binding energy of an electron in the ground state of the He atom is equal to $24 e V$ . The energy required to remove both the electrons from the atom will be:

59 eV

81 eV

79 eV

None of these

Ionization energy of $He$

$= \frac{z ^{2}}{n ^{2}} \times 13.6$

$= \frac{2 ^{2}}{1 ^{2}} \times 13.6$

$= 54.4 e V$

Energy required to remove both the electrons

$=$ binding energy + ionisation energy

$= 24.6 + 54.4 = 79 e V$

Q. The binding energy of an electron in the ground state of the He atom is equal to 24eV. The energy required to remove both the electrons from the atom will be: