The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each radiation and energy difference between two excited states.

Explanation

`lamda_(1)=589nm=589xx10^(-9)m`
`therefore`Frequency, `v_(1)=(c)/(lamda_(1))=(3.0xx10^(8)m*s^(-1))/(589xx10^(-9)m)`
`=5.093xx10^(14)s^(-1)`
`lamda_(2)=589.6nm=589.6xx10^(-9)m`
`therefore`Frequency `(v_(2))=(c)/(lamda_(2))=(3.0xx10^(8))/(589.6xx10^(-9))=5.088xx10^(14)s^(-1)`
The difference in energy `(DeltaE)=E_(1)-E_(2)=h(v_(1)-v_(2))`
`=6.626xx10^(-34)xx(5.093-5.088)xx10^(14)=3.313xx10^(-22)`J.