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Emission transitions in the Paschen series end at orbit `n=3` and start from orbit `n` and can be represented as `v=3.29xx10^(15) (Hz)[1//3^(2)-1//n^(2)]` Calculate the value f `n` if the transition is observed at `1285 nm`. Find the region of the spectrum. |
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Answer» `v=(3.29xx10^(15)Hz)((1)/(3^(2))-(1)/(n^(2)))` `lambda=1285 nm=1285xx10^(-19)m=1.285xx10^(-16)m` `v=(c)/(lambda)=((3xx10^(8)ms^(-1)))/((1.285xx10^(-6)m))=2.3346xx10^(14)s^(-1)` `2.3346xx10^(14)=3.29xx10^(15)[(1)/(3^(2))-(1)/(n^(2))]` `(2.3346)/(32.9)=(1)/(3^(2))-(1)/(n^(2)) or 0.071=(1)/(9)-(1)/(n^(2))` `(1)/(n^(2))=(1)/(9)-0.071=0.111-0.171=0.04` `n^(2)=(1)/(0.04)=25 or n=5` Paschen series lies in infrared of the spectrum. |
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