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

Hint:The energy required to remove an electron from any state is equal to the total energy of the electron in the orbit. The total energy of the electron is due to the kinetic energy and the electrostatic potential energy.

Formula used:
\[{E_n} = - 13.6\left( {\dfrac{{{Z^2}}}{{{n^2}}}} \right)eV\]
where \[{E_n}\] is the total energy of the electron in a hydrogen-like atom of atomic number Z in the quantum state number n.

Complete step by step solution:
It is given that the Li atom is doubly ionized, that means two electrons from the Li atom are removed initially. As the total number of electrons in a neutral Li atom is equal to the atomic number, i.e. 3; so the total number of electrons in a doubly ionized Li atom is 1. Hence, the doubly ionized Li atom can be assumed as a hydrogen-like atom.

For the ground state, \[n = 1\]
Putting the values in the expression for the energy as per Bohr model, we get
\[{E_1} = - 13.6 \times \left( {\dfrac{{{3^2}}}{{{1^2}}}} \right)eV \\ \]
\[\Rightarrow {E_1} = - 122.4\,eV\]
Total energy of the electron after getting ionized is zero. So the final energy of the electron is zero.

The energy required is equal to the difference of the energy of electrons in the final state and in the initial state.
\[E = \Delta E\]
\[\Rightarrow E = 0.0\,eV - \left( { - 122.4\,eV} \right)\]
\[\therefore E = 122.4\,eV\]
Hence, 122.4 eV energy is required to remove an electron from the doubly ionized Li atom.

Therefore, the correct option is D.

Note: The energy required to remove a loosely bound electron from the influence of the nucleus of the atom. Closest the electron is from the nucleus of the atom, greater will be the required energy to remove the electron from the influence of the nucleus of the atom.