Which orbit of `Be^(3+)` ion has the same radius as that of the 1st orbit of H-atom?

Explanation

`r_(n)(H)=r_(1)(H)xxn^(2)`,
`r_(n)(Be^(3+))=r_(1)(H)xx(n^(2))/(z)=(r_(1)(H)xxn^(2))/(4)`
in n=2, then `n^(2)=4` and hence `r_(2)(Be^(3+))=r_(1)(H)`
thus the radius of the 2nd orbit of `Be^(3+)` ion is equal to that of the 1st orbit of H-atom.