Question

Sol. [Key Concept: Bohr radius $r = a_{0}\frac{n^{2}}{Z} = 70.53$ pm gives $\frac{n^{2}}{Z} = \frac{70.53}{52.9} = \frac{4}{3}$, satisfied by $n = 2,\ Z = 3$ (Li$2 +$).]

$${r_{\text{species}\text{~}} = a_{0}\frac{n^{2}}{Z} }{70.53 = 52.9\left( \frac{n^{2}}{Z} \right) }{\frac{n^{2}}{Z} = \frac{70.53}{52.9} = 1.33 = \frac{4}{3} }$$Data will satisfy if $n = 2$ and $z = 3$

$${Li}^{+ 2},2 \Rightarrow \frac{4}{3} = 1.33 $$