NTA Abhyas JEE Main2020ChemistryStructure of AtomPractice
Ionisation energy of He + is 19 .6 × 10 − 18   J per atom. The energy of the first stationary state (n   =   1) of Li 2+ per atom is
Options
- A4.41 × 10 − 16 J
- B4.41 × 10 − 17   J
- C2.2 × 10 − 15   J
- D8.82 × 10 − 17 J
Correct answer
B. 4.41 × 10 − 17   J
Step-by-step solution
IE = Z 2 n 2 × 13 .6 eV …(i) IE 1 IE 2 = Z 1 2 n 1 2 × n 2 2 Z 2 2 …(ii) Given IE 1 = 19 .6 × 10 − 18 , Z 1 = 2, n 1 = 1, Z 2 = 3 and n 2 =1 . Substituting these values in equation (ii) 19 .6 × 10 − 18 IE 2 = 4 1 × 1 9 Or IE 2   =   19 .6   ×   10 − 18   ×   9 4 =   4 .41   ×   10 − 17   J/atom