Explanation

(d)
Bohr radius for $n$th orbit $= 0.53 \times \frac{n^{2}}{Z}$
Where, $Z =$atomic number
∴ Bohr radius of 2nd orbit of $Be^{3 +} = \frac{0.53 \times (2)^{3}}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.53\ \mathring{\mathrm{A}}$
(d) Bohr radius of 1st orbit of H$= \frac{0.53 \times (1)^{2}}{1}$
Hence, Bohr’s radius of 2nd orbit of $Be^{3 +}$is equal to that of first orbit of hydrogen.