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`

Find the quantum number `n` corresponding to the excited state of `He^

1.	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`
Answer» 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

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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`

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