For a fluorescent molecule, the emission λ can be calculated by TDDFT with the excited state optimized geometry. But how can I tell from a calculation that a molecule is not fluorescent ? For non-fluorescent molecule, the standard calculation can also yield an emission wavelength ? Then how can I identify whether a molecule is fluorescent or not by calculation ? I am trying to explain substituent effect on the fluorescence properties of triphenylamine-based molecules. In solid state fluorescence experiment, all the molecules showed an fluorescence maxima except for first, one substituent which does not show any fluorescence peak and second, anotehr substituent for which experimental intensity was very weak. How to complement this experimental data with TDDFT?
I concur with Peter
What you ask for is not the fluorescence but its competition with other processes. It means that an excited molecule can also lose its energy by a non-radiative step to another electronic state with electronic energy converted to vibrational energy. TDDFT must thus be able to predict the rate of such a process.
What can be done in some instances is taking rules of thumb. 1. When the calculated/measured absorption to the fluorescent state is strong chances are that there is fluorescence. 2. The energy distance between the fluorescent state and the state to which it can transfer by a radiationless transition Is high. I save you the explanation
From simple TDDFT calculations you get also the electric transition dipole moments and oscillator strengths, so you can check whether the transition from the lowest (the Kasha rule) excited state is optically allowed or not.
Another issue is if the state order is correctly predicted by TDDFT for your system or not. This is or importance especially when there are doubly excited states, for which TDDFT gives too high energies, and the transitions from which are usually weak or forbidden.
Yours
Marcin
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