Q: As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is

(A) 1.51

(B) 13.6

(C) 40.8

(D) 122.4

Ans: (D)

Sol: For Hydrogen like atoms ;

$\large E_n = -\frac{13.6 Z^2}{n^2} eV$

Therefore , ground state Energy of doubly ionized Lithium atom (Z=3 , n=1) will be

$\large E_1 = -\frac{13.6 \times 3^2}{1^2}$

= 122.4 eV

Hence , ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4 eV