Explanation

(d) For an atom of atomic number $Z$, radius of $n$th orbit is given by
$r_{n} = \frac{kn^{2}}{Z}\ldots(i)$ where $k =$constant
For ground state of hydrogen, $Z = 1,\ n = 1$, so that
$r_{1} = \frac{k1^{2}}{1} = k$
Let $n$ be the energy state of $Be\ ^{+ + +}$ for which orbital radius is $r_{1}$. Put
$Z = 4\ $and$\ r_{n} = r_{1} = k$ in Eq.(i)
$r_{1} = \frac{r_{1}n^{2}}{4}\ or\ \ n^{2} = 4;n = 2$