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Find the quantum number `n` corresponding to the excited state of `He^(Theta)` 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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Answer» `lambda_(1) = 30.4 xx 10^(-7) cm` `lambda_(2) = 108.5 xx 10^(-7) cm` Let the excited state of `He^(Theta)` be `n_(2)` it comes from `n_(2)` to `n_(1)` and then from `n_(1)` to `1` to emit two successive photons `(1)/(lambda_(2)) = R_(H)Z^(2)[(1)/((1)^(2)) - (1)/(n_(1))^(2)]` `:. n_(1) = 2` Now for `lambda_(2)n_(1) = 2,n_(2) = `? `(1)/(108.5 xx 10^(-7)) = 109678 xx (2)^(2)[(1)/((2)^(2))- (1)/((n_(2))^(2))]` `:. n_(2) = 3` Hence the excited state of `He^(Theta)` is fifth orbit |
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