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. -122.4 eV
B. 122.4 eV
C. -13.6 eV
D. 13.6 eV

Answer

B

Explanation

For hydrogen and hydrogen like atoms,
`E_n = -13.6((Z)^2)/((n^2)))`
Therefore, ground state energy of doubly ionized
lithium atom (Z =3, n =1) will be
`E_1 = (-13.6) ((3)^2/((1))^2`
=- 122.4 eV
`:.` lonization energy of an electron in ground state
of doubly ionized lithium atom will be 122.4 eV.