If we are talking fluorescence here then the relaxation of a vibrationally excited molecule is very fast with an average lifetime of 10^-12 seconds or less. This means that it is significantly faster than the loss of energy from an electronically excited state. As a result, fluorescence when it occurs always involves a transition from the lowest vibrational state (because any species that were in the higher vibrational states will already have dropped to the lowest state long before electronic relaxation starts).
This means that the energy of electronic relaxation is generally less than the energy of excitation. As a result, fluorescence emissions will generally be of lower frequency (or longer wavelength) than their corresponding absorption’s. This is known as a “Stokes Shift”.
2007-03-01 21:18:54
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answer #1
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answered by Anonymous
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When a photon is absorbed by something and then re-emitted, some of the energy of the photon is also absorbed by the material.
This energy reduction allows us to re-calculate the frequency from-: Energy = Plancks Constant x Frequency
This shows a reduction in energy, means a reduction in frequency.
The corresponding wavelength (being re-emitted) can then be calculated by-:
Speed of Light = Frequency x Wavelength
We can easily see from the latter equation that if the frequency is reduced, the wavelength is made longer (as the speed of light remains constant).
Hope this helps.....
2007-03-01 21:18:58
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answer #2
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answered by Doctor Q 6
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By conservation of energy, the photons emitted cannot have higher energy than the photons absorbed. The energy of a radiation photon is directly proportional to the frequency (is the Planck's constant times the frequency), and hence is inversely proportional to the wavelength.
2007-03-01 21:22:04
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answer #3
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answered by sciquest 4
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Doppler Effect?
2007-03-01 20:56:38
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answer #4
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answered by SS4 7
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