1.

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

Answer» For `lambda_(1) = 589 nm`
Frequency of transition `(v_(1)) = (c)/(lambda_(1))`
`= (3.0 xx 10^(8)ms^(-1))/(589 xx 10^(-9)m)`
Frequency of transition `(v_(1)) = 5.093 xx 10^(14) s^(-1)`
Similarly, for `lambda_(2) = 589.6 nm`
Frequency of transition `(v_(2)) = (c )/(lambda_(2))`
`= (3.0 xx 10^(8) ms^(-1))/(589.6 xx 10^(-9) m)`
Frequency of transition `(v_(2)) = 5.088 xx 10^(14) s^(-1)`
Energy difference `(DeltaE)` between excited states `= E_(1) - E_(2)`
where,
`E_(2) =` energy associated with `lambda_(2)`
`E_(1) =` energy associated with `lambda_(2)`
`DeltaE = hv_(1) - hv_(2)`
`= h(v_(1) - v_(2))`
`= (6.626 xx 10^(-34) Js) (5.093 xx 10^(14) - 5.088 xx 10^(14))s^(-1)`
`= (6.626 xx 10^(-34)J)(5.0 xx 10^(-3) xx 10^(14))`
`DeltaE = 3.31 xx 10^(-22) J`


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