Find the quantum number `n` corresponding to the excited state of `He^(+)` ion, if on transition to the ground state that ion emits two photons in succession with wave lengths `108.5` and `30.4 nm`.

Explanation

Suppose the electron in the excited state is present in the shell `n_(2)`. First it falls from `n_(1) " to " n_(1)` and then from `n_(1)` to ground state (for which n =1). Thus, the two transitions involved and the corresponding wavelengths emitted are
(i) `n_(2) rarr n_(1), lamda_(1) = 108.5 nm = 108.5 xx 10^(-7)cm`
(ii) `n_(1) rarr 1, lamda_(2) = 30.4 nm = 30.4 xx 10^(-7) cm`
Applying Rydberg's formula first to case (ii), we get
`bar(v) = (1)/(lamda) = RZ^(2) ((1)/(1^(2)) - (1)/(n_(1)^(2))), " i.e., " (1)/(30.4 xx 10^(-7)) = 109677 xx 2^(2) xx ((1)/(1^(2)) - (1)/(n_(1)^(2)))`
or `(1)/(n_(1)^(2)) = 1 - (1)/(30.4 xx 10^(-7) xx 4 xx 109677) = 1 - 0.75 = 0.25 or n_(1)^(2) = (1)/(0.25) = 4 or n_(1) = 2`
Applying Rydberg's formula now to case (i), we get
`(1)/(108.5 xx 10^(-7)) = RZ^(2) ((1)/(2^(2)) - (1)/(n_(2)^(2))) = 109677 xx 2^(2) xx ((1)/(4) - (1)/(n_(2)^(2)))`
or `(1)/(n_(2)^(2)) = (1)/(4) - (1)/(108.5 xx 10^(-7) xx 4 xx 109677) = 0.25 - 0.21 = 0.04`
or `n_(2)^(2) = (1)/(0.04) = 25 or n_(2) =5`