What is the energy (in J atom$$^{-1}$$) required for the following process? $$\text{Li}^{2+}(g) \rightarrow \text{Li}^{3+}(g) + e^-$$ (Take the ionization energy for the H atom in the ground state as $$2.18 \times 10^{-18}$$ J atom$$^{-1}$$)

The species $$\mathrm{Li^{2+}}$$ has only one electron, so it is a hydrogen-like ion with nuclear charge $$Z = 3$$.

For any hydrogen-like species, the energy of an electron in the ground state $$\left(n = 1\right)$$ is
$$E_1 = -\,Z^2 R_H$$
where $$R_H = 2.18 \times 10^{-18}\ \text{J atom}^{-1}$$ is the ionization energy of the hydrogen atom.

The ionization process
$$\mathrm{Li^{2+}(g) \;\rightarrow\; Li^{3+}(g) + e^-}$$
removes this electron from $$n = 1$$ to $$n = \infty$$, so the required energy equals the magnitude of $$E_1$$:

$$\text{Ionization energy of } \mathrm{Li^{2+}} = Z^2 R_H$$

Substituting $$Z = 3$$:

$$Z^2 R_H = 3^2 \times 2.18 \times 10^{-18}$$

$$= 9 \times 2.18 \times 10^{-18}$$

$$= 19.62 \times 10^{-18}$$

Rewriting with one non-zero digit before the decimal point:

$$19.62 \times 10^{-18} = 1.962 \times 10^{-17} \ \text{J atom}^{-1}$$

Therefore, the energy required is $$1.962 \times 10^{-17}\ \text{J atom}^{-1}$$.

Option C which is: $$1.962 \times 10^{-17}$$