Be³⁺ has lost three electrons, so it is a one-electron (hydrogen-like) ion and the Bohr formula applies directly:

\(r_n = \frac{n^{2}}{Z}\,a_0\)

where \(a_0\) is the Bohr radius. The radius grows as \(n^{2}\) but shrinks as \(Z\), because a larger nuclear charge pulls the electron in more tightly.

Set the two radii equal. For hydrogen in its ground state, \(Z = 1\) and \(n = 1\), so \(r = a_0\). For beryllium, \(Z = 4\). Therefore

\(\frac{n^{2}}{4}a_0 = a_0 \;\Rightarrow\; n^{2} = 4 \;\Rightarrow\; n = 2\).

The physical reading is worth stating: beryllium's nucleus has four times hydrogen's charge, so to sit at the same distance the electron must be promoted to a shell whose \(n^{2}\) is four times larger — that is, from n = 1 to n = 2. The stronger pull is exactly offset by the higher principal quantum number.

The option n = 0 is not a permitted value at all, since \(n\) must be a positive integer and n = 0 would give zero radius.

The same scaling explains a related fact: the energy of a hydrogen-like ion is \(E_n = -13.6\frac{Z^{2}}{n^{2}}\) eV, so although Be³⁺ at n = 2 has the same radius as ground-state hydrogen, it is bound four times more strongly.

Hence Be³⁺ has the same orbital radius in the n = 2 state.