Which state of triply ionised Beryllium `(Be^(+++))` the same orbital radius as that of the ground state hydrogen ?

Explanation

Radius of `n^(th)` orbit is given by
`r_n=(n^2h^2)/(4pi^2mkZe^2)`
`therefore r_n prop n^2/Z`
Given , `r_n(Be^(3+))=r_1(H)`
`(n^2/Z)_(Be^(3+))=(n^2/Z)_H`
`therefore n^2/4=1^2/1 rArr n^2=4`
or n=2