Which state of the triply ionized `Be^(+++)` has the same orbital radius as that of the ground state of hydrogen? Compare the energies of two states.

Explanation

For H-like particles, `r_(n) = 0529 xx (n^(2))/(Z)`
For ground state of H-atom, `Z = 1, n = 1, r_(1) = 0.529 Å`
For `Be^(3+) (Z = 4)`, if nth orbit has the same radius, then `0.529 xx (n^(2))/(4) = 0.529 or n^(2) = 4 " i.e., " n = 2`
Thus, second orbit of `Be^(3+)` ion has same radius as that of ground state (1st orbit) of H-atoms