1.

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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