Find the quantum number `n` corresponding to the excited state of `He^(o+)` ion if on transition to the ground state that ion emits two photon in succession with wavelength `108.5` and `30.4 nm`

Explanation

Gives `lambda_(2) = 30.4 xx 10^(-7) cm, lambda_(2) = 108.5 xx 10^(-7) cm` Let the excited state of `He^(Theta)` be `n_(2)` if comes from `n_(2)` to`n_(1)` and then `n_(1)` to `1` to emit two successive photon
`(1)/(lambda_(2)) = R_(H)Z^(2)[(1)/(1^(2)) - (1)/(n_(1)^(2))]`
`(1)/(30.4 xx 10^(-7)) = 109678 xx 4[(1)/(1^(2)) - (1)/(n_(1)^(2))]`
`:. n_(1) = 2`
`Now for lambda_(1) : n_(1) = 2 and n_(2) ,`?
`(1)/(lambda_(1)) = R_(H)Z^(2)[(1)/(2^(2)) - (1)/(n_(2)^(2))]`
`(1)/(108.5 xx 10^(-7)) = 109678 xx 4 [(1)/(2^(2)) - (1)/(n_(2)^(2))]`
`:. n_(2) = 5`
Thus excited state for lies is fifth orbit