L9 - Biophysical Chemistry Chemistry 24a Winter Term...

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Biophysical Chemistry Chemistry 24a Winter Term 2009-10 Instructor: Sunney I. Chan Lecture 9 January 29, 2010 Fluorescence Resonance Energy Transfer (FRET). Fluorescence Depolarization. Fluorescence Resonance Energy Transfer (FRET) Also called “Singlet-Singlet Energy Transfer”. Quenching of fluorescence of a donor via migration of donor excitation to a suitable donor. Consider two chromophores: Donor D Acceptor A Assume that they are sufficiently far apart that their charge clouds do not overlap. That is, there is no exciton interaction between them, or the exciton interaction is negligible. = = resonance energy transfer D b D a A b A a rapid vibrational relaxation Vibrational relaxation Donor Acceptor Donor de-excitation and acceptor excitation coupled in resonant interaction lead to energy transfer from donor to acceptor: D b + A a D a + A b Vibration relaxation rapidly converts resultant acceptor singlet (A b ) and donor singlet (D a ) to their respective ground vibrational levels. Therefore, k T >> k -T . k T k -T
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Donor fluorescence becomes quenched as a result of the energy transfer. Acceptor becomes excited and subsequently it can fluoresce. This process is called sensitized emission. Donor absorption Donor emission Sensitized emission (if no resonance from acceptor energy transfer) λ Overlap here key to process = = resonance energy transfer D b D a A b A a rapid vibrational relaxation Vibrational relaxation Donor Acceptor sensitized acceptor emission Energy transfer rate k T = ? Consider the following D-A pair. μ D μ A R R > 10 Å Donor Acceptor Electronic Hamiltonian for D-A pair : H D-A = H D + H A + V(D, A) where H D , H A denote the electronic hamiltonian for the isolated donor and acceptor , respectively, and V(D, A) is the interaction between them. For R > 10 Å, V(D, A) may be approximated by electric dipole-dipole interaction. Then, V(D, A) = ( μ D μ A / R³ - 3 ( μ D R )( R μ A )/ R 5 . If we begin the analysis by assuming that the Donor is excited into its first excited electronic singlet state by absorbing a photon at t = 0, with the Acceptor in the ground vibrational level of its ground electronic singlet, then the problem we need to solve is a non-stationary state one.
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Solution to the non-stationary state problem After donor excitation at t = 0, Ψ ( D , A, t = 0 ) = Φ b ( D ) Φ a ( A ) exp -i ( ω b D t )exp -i ( ω a A t ) = Φ b ( D ) Φ a ( A ) where ε b D = ħ ω b D , ε a A = ħ ω a A At time t after excitation, Ψ ( D , A, t ) = C( D b A a , t ) Φ b ( D ) Φ a ( A ) exp -i ( ω b D t )exp -i ( ω a A t ) + C( D a A b , t ) Φ a ( D ) Φ b ( A ) exp -i ( ω a D t )exp -i ( ω b A t ) k T = d [C*(D a A b , t)•C(D a A b , t)] /dt k T = (½ ħ ²) ( |< Φ a (D) Φ b (A) | V| Φ b (D) Φ a (A) >| ² ) proportional to ( ( κ /R³)² ) ( |< Φ a (D) Φ b (A) | | μ D || μ A | | Φ b (D) Φ a (A) >| ² ) = ( κ ²/R 6 ) ( |< Φ a (D) | μ D | Φ b (D) >< Φ b (A)
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This note was uploaded on 10/27/2010 for the course BI 110 taught by Professor Richards,j during the Winter '08 term at Caltech.

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L9 - Biophysical Chemistry Chemistry 24a Winter Term...

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