JEE Advanced levelSingle correct optionDifficulty 3 of 3pt146
The first ionisation energy of He is 24.6 eV and that of hydrogen is 13.6 eV. Take 1 eV per atom = 96.485kJmol−1. The total energy to remove both electrons from one mole of gaseous He atoms is closest to
A7622kJmol−1
B5249kJmol−1
C2373kJmol−1
D3811kJmol−1
Answer: A
The second step is the ionisation of HeX+: 13.6 x 4 = 54.4 eV. Total = 24.6 + 54.4 = 79.0 eV per atom; 79.0 x 96.485 = 7622 kJ/mol. 5249 is only the second step and 2373 only the first.