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

Explanation

For H-atom, radius of ground state is
`r_1 = (h^2)/(4 pi^2 m e^2) " " ….(i)`
For hydrogen like atom,
`r_n^. = (n^2h^2)/(4pi^2 m Ze^2) " "…..(ii)`
Dividing Eq. (ii) by Eq. (i)
`(r_n^.)/(r_1) = (n^2)/(Z)`
For `Be^(3+)` ion, Z = 4
`:. (r._H)/(r_1) = (n^2)/4`
Now, `r._n = r_i`
`:. n^2 = 4 " or " n = 2`.
Thus, the second orbit of `Be^(2+)` has the same radius as the Bohr.s radius of hydrogen atom.