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smmoh4 smmoh4
wrote...
Posts: 118
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11 years ago
This is about the fluorescence spectrophotometer. Please help!
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wrote...
11 years ago
Because of the Franck-Condon Principle, absorption will probably take you to a vibrationally excited state of the electronically excited state.

Usually, this relaxes to the vibrational ground state of the electronically excited state, dissipating energy as heat.

Finally, emission will largely be to a vibrationally excited level of the electronic ground state.

The 0-0 transition will correspond to the lowest energy component of the absorption, and the highest energy component of the emission. if there is a significant change in geometry between ground and excited states, this component can be extremely weak.

In phosphorescence, there is in addition relaxation to a lower (often triplet) electronic excited state, causing an even more pronounced energy difference.

Hope this helps.
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