Using Bohr’s formula for energy quantization, the ionisation potential of the ground state of \[L{i^{ + + }}\] atoms is
1. \[122\;{\rm{V}}\]
2. \[13.6\;{\rm{V}}\]
3. \[3.4\;{\rm{V}}\]
4. \[10.2\;{\rm{V}}\]

Hint: The above problem can be resolved using the fundamental concepts and applications of the ionization potential. The ionization potential is also known as the ionization energy, and it is analytically equal to the energy released when some electrons are being removed from the neutral atom. In this problem, we are given anionic species of lithium atoms. Then we have to apply the standard mathematical relation for the ionization energy of the ionic species. This mathematical relation includes the atomic number of lithium-ion and the energy at its first ionization state or the ground state. Then by substituting the values in the equation, we can obtain the final result.

Complete step by step answer:
The expression for the ionization potential of a Hydrogen like atom is given by,
\[\Rightarrow IP = \dfrac{{{E_1}}}{e} \times \dfrac{{{z^2}}}{{{n^2}}}\]
Here, e is the charge unit and its value in this case is unity, \[{E_1}\] is the energy of atom or ion at ground state and its standard value is \[13.6\;{\rm{eV}}\], z is the atomic number of lithium ion and its value is 3 and n is number of orbital and in this case, its value is taken as 1.
Solve by substituting the values in the above equation as,
\[
\Rightarrow IP = \dfrac{{{E_1}}}{e} \times \dfrac{{{z^2}}}{{{n^2}}}\\
\Rightarrow IP = \dfrac{{\left( {13.6\;{\rm{eV}}} \right)}}{{1\;{\rm{e}}}} \times \dfrac{{{{\left( 3 \right)}^2}}}{{{{\left( 1 \right)}^2}}}\\
\Rightarrow IP = 122.4\;{\rm{V}}
\]
Therefore, the ionisation potential of the ground state of \[L{i^{ + + }}\] atoms is \[122.4\;{\rm{V}}\] and option (1) is correct.

Note:To resolve the given problem, one must try to understand the basic phenomenon included in calculating ionization energy or the ionization potential. One should also try to remember the basic applications of ionization energy.