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Find the quantum number n corresponding ...

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`.

Text Solution

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`lamda_(1)=30.4xx10^(-7)cm`
`lamda_(2)=108.5xx10^(-8)cm`
Let excited state of `He^(+)` be `n_(2)`, It comes from `n_(2)` to `n_(1)` and then from `n_(1)` to 1 to emit two successive photon.
`1/(lamda_(2))=R_(H)Z^(2)[1/((1)^(2))-1/(n_(1)^(2))]`
`:.1/(30.4xx10^(-7))=109678xx(2)^(2)[1/((1)^(2))-1/((n_(1))^(2))]`
`:.n_(1)=2`
Now for `lamda,n_(1)=2,n_(2)=?`
`1/(108.5xx10^(-7))=109678xx4[1/((2)^(2))-1/((n_(2))^(2))]`
`:.n_(2)=5`
Hence excited state for He is 5th orbit.
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