Question: Q. 4. Taking the Bohr radius as $a_{0}=53 \mathrm{pm}$, the radius of $\mathrm{Li}^{++}$ion in its ground state, on the basis of Bohr’s model, will be about (a) $53 \mathrm{pm}$. (b) $27 \mathrm{pm}$. (c) $18 \mathrm{pm}$. (d) $13 \mathrm{pm}$.
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Solution:
Ans. Correct option : (c)
Explanation : According to Bohr’s model of an atom, radius of an atom in its ground state is
$$ r=\frac{r_{0}}{Z} $$
where, $r_{0}$ is Bohr’s radius and $\mathrm{Z}$ is atomic number. As given that,
$r_{0}=53 \mathrm{pm}$ and atomic number of Lithium atom is 3
so, $r=\frac{53}{3}=17.67 \mathrm{pm} \approx 18 \mathrm{pm}$
