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

Explanation

For hydrogen-like atoms or ions, the radius of the n-th orbit is given by, `r_(n)=0.529xx(n^(2))/(Z)`
In the ground state of H-atom, n=1 (1st orbit) and Z(at. No.)=1
`therefore`Radius of 1st orbit of H-atom, `r_(1)=0.529xx(1^(2))/(1)=0.529` Å
Let the n-th orbit of `Be^(3+)` ion has the same radius as that of the ground state of H-atom.
Now, `r_(n)(Be^(3+)" ion")=0.529xx(n^(2))/(4)" "[because" at. no of Be=4"]`
Thus, `0.529xx(n^(2))/(4)=0.529`
or, `n^(2)=4 or, n=2`
Hence, second orbit of `Be^(3+)` ion has same radius as that of ground state of H-atom.