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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.A. `3.31xx10^(-22)kJ`B. `3.31xx10^(-22)J`C. `2.98xx10^(-21)J`D. `3.0xx10^(-21)kJ` |
Answer» Correct Answer - B Step I : Wavelength `(lamda)` = 589 nm = `589xx10^(-9)m` Frequency (v) = `c/lamda=(3xx10^(8))/(589xx10^(-9)) =5.093xx10^(14)` cycles per sec Step II : Wavelength `(lamda)` = 589.6 nm = `589.6xx10^(-9)m therefore" "v=c/lamda=(3xx10^(8))/(589.6xx10^(-9))=5.088xx10^(14)` cycles per sec Energy difference between two excited states, `DeltaE=6.626xx10^(-34)(5.093-5.088)10^(14) =6.626xx10^(-34)xx5xx10^(-3)xx10^(14)=3.31xx10^(-22)J` |
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