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