Energy required for the electron excitation in Li^{++} from the first to the third Bohr orbit is

- 12.1 eV
- 12.4 eV
- 36.3 eV
- 108.8 eV

**Answer**

E_{1} = -[(13.6 x 3^{2})/1^{2}]

E_{3} = -[(13.6 x 3^{2})/3^{2}]

ΔE = E_{3} - E_{1}

= 13.6 x 3^{2} x [1 - (1/9)]

= 108.8 eV

**The correct option is D.**