Explanation

(c)
$r_{H} = 0.529\mathring{\mathrm{A}}$
$r_{n} = r_{H} \times \frac{n^{2}}{Z}$
For $Li^{2 +}(n = 2),$
$r_{Li^{2 +}}{= r}_{H} \times \frac{(2)^{2}}{3} = \frac{r_{H} \times 4}{9}\$
For $Li^{2 +}(n = 3),$
$r_{Li^{2 +}} = r_{H} \times \frac{(3)^{3}}{3} = 3r_{H}$
For $Be^{3 +}(n = 2)$
$r_{Be^{3 +}} = r_{H} \times \frac{(2)^{2}}{4} = r_{H}$
For $He^{+}(n = 2)$
$r_{He^{+}} = r_{H} \times \frac{(2)^{2}}{2} = 2r_{H}$
Thus, $Be^{3 +}(n = 2)$ has same radius as that of the first Bohr’s orbit of $H$-atom