As per Bohr model , the minimum energy (in eV) required to remove electron from the ground state of doubly ioinized `Li` alom `(Z = 3)` is

Options

A. 1.51
B. 13.6
C. 40.8
D. 122.4

Answer

B

Explanation

For hydrogen and hydrogen-like atom ,
`E_(n) = - 13.6 ((z)^(2))/((n)^(2)) eV`
Therefore ground state energy of doubly ionized lithium atom `(z = 3, n = 1)` will be
`E_(n) = (- 13.6) ((3)^(2))/((2)^(2)) = - 122.4 eV`
Therefore , Lonization energy of an electron in ground state of doubly ionized lithium atom will be `122.4 eV`