Find the quantum no: 'n' corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelengths.108.5 and 30.4 nm.

Options

A. 7
B. 5
C. 3
D. 4

Answer

5

Explanation

Given, `lamda_(2) =30.4 xx10^(-7) cm, lamda_(1)=108.5 xx10^(-7) cm`
Let excited state of `He^(+)` be `n_(2)`. It comes from `n_(2)" to "n_(1)` and then `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))=109678xx4[(1)/(1^(2))-(1)/(n_(1)^(2))]" "therefore n_(1)=2`
Now for `lamda_(1): n_(1)=2 and n_(2) : ` ?
`(1)/(lamda_(1)) =R_(H) Z^(2) [(1)/(2^(2))-(1)/(n_(2)^(2))], (1)/(108.5xx10^(-7))=109678xx4[(1)/(2^(2))-(1)/(n_(2)^(2))]" "therefore n_(2)=5`
Thus, excited state for He is `5^(th)` orbit.