The radius of the n^ th Bohr orbit of a Be^ 3+ ion is equal to the...

The radius of the $n^{\text{th}}$ Bohr orbit of a hydrogen-like species is given by:
$r_n = a_0 \dfrac{n^2}{Z}$

For a $\text{Be}^{3+}$ ion, the atomic number $Z = 4$.
Therefore, the radius of its $n^{\text{th}}$ orbit is:
$r_n = a_0 \dfrac{n^2}{4}$

The radius of the first Bohr orbit of a hydrogen atom ($Z = 1, n = 1$) is $a_0$.

According to the given condition:
$a_0 \dfrac{n^2}{4} = a_0$

$\dfrac{n^2}{4} = 1$

$n^2 = 4$

$n = 2$

Answer: 2