1) 20.4 V

2) 13.6 V

3) 30.6 V

4) 40.8 V

Answer: 2) 13.6 V

Solution:

First excitation potential of a given atom is 10.2 V

Since first excitation potential is given, n = 2

Let the ionization potential = E

Total energy in ground state, (TE)_G = – E

For first excited state, (TE)_F = – E/n² = -E/4

E – (E/4) = 10.2

3E/4 = 10.2

E = 40.8/3

= 13.6 V

Ionization potential = 13.6 V