Show that ground state energy of an electron in H-atom is equal to the first excited state energy of electron in `He^+` ion (assuming their Rydberg's constants to be equal ).

Explanation

`E_(n) = - R_(H) (Z^(2))/(n^(2)) = - 13.60 (Z^(2))/(n^(2)) eV`
For H-atom in the ground state, `Z = 1, n =1, :. E_(1) = -13.60eV`
For `He^(+)` ion in the first excited state, `Z = 2, n = 2, E_(2) = - 13.60 xx (2^(2))/(2^(2)) = - 13.6 eV`
Hence, the desired result is proved.