InterviewSolution
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InterviewSolution
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The longest wavelength doublet absorption is observed at `589` and `589.6 nm`. Caiculate the frequency of each transition and energy difference between two excited states. |
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Answer» `lambda_(1)=589 nm = 589xx10^(-9)m` `v_(1)=(c)/(lambda_(1))=(3xx10^(8)ms^(-1))/(589xx10^(-9)m)=(3000)/(589)xx10^(14)s^(-1) " " =5.0934xx10^(14)s^(-1)` `lambda_(2)=589.6 nm = 589.6xx10^(-9)m` `v_(2)=(c)/(lambda_(2))=(3xx10^(8)m s^(-1))/(589.6xx10^(-9)m)=(3000)/(589.6)xx10^(14)s^(-1)=5.0882xx10^(14)s^(-1)` `DeltaE = E_(1)-E_(2)=(hc)/(lambda_(1))-(hc)/(lambda_(2))=hc[(1)/(lambda_(1))-(1)/(lambda_(2))]` `=(6.626xx10^(-34)Js xx3xx10^(8) ms^(-1))[(1)/(589xx10^(-9)m)-(1)/(589.6xx10^(-9)m)]` `=(19.878xx10^(-34)xx10^(8))/(10^(-9))[(589.6-589)/(589.6xx589)]J` `=(19.878xx10^(-17)xx0.6)/(589.6xx589)J=3.43xx10^(-22)J` |
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