Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of $$Li^{2+}$$ ion in its ground state is given by $$\frac{1}{X}a_0$$, where $$X =$$ ________. (Where $$a_0$$ is the first Bohr's radius.)

For a hydrogen-like ion (single electron around nucleus of charge $$+Ze$$), Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n = \frac{n^2 a_0}{Z}$$

where
$$n$$ = principal quantum number,
$$Z$$ = atomic number of the nucleus,
$$a_0$$ = Bohr radius for the hydrogen ground state.

For $$Li^{2+}$$ the atomic number is $$Z = 3$$ because lithium has three protons. The ground state corresponds to $$n = 1$$.

Substituting $$n = 1$$ and $$Z = 3$$ into the formula,

$$r_1 = \frac{(1)^2 a_0}{3} = \frac{a_0}{3}$$

This can be written as $$\frac{1}{X} a_0$$ with $$X = 3$$.

Hence, $$X = 3$$, which matches Option C.