Of Electrons, Energy, and Excitons

Transcription

1 Of,, and Gert van der Zwan July 17, / 35

2 Bacterial 2 / 35

3 Bacterial Bacterial LH1, LH2: excitonic interaction and energy transfer. RC, cytochromes: electron transfer reactions. Q, UQ: proton transfer reactions. Source: M. Brederode, Thesis. 3 / 35

4 Bacterial II Bacterial 4 / 35

5 Bacterial III Bacterial Net result: for every 870 nm photon two protons are transported over the membrane against a ph difference of ph = 3. 5 / 35

6 Bacterial IV Bacterial Electron tranfers reactions, redox potentials and transfer times in the RC. 6 / 35

7 Reactions Marcus Theory Crossing Point rate constant 7 / 35

8 Transition State Theory+ Reactions Marcus Theory Crossing Point rate constant Equilibrium in the reactant well. Single reaction coordinate on adiabatic surface. Other (vibrational) modes irrelevant. No electronic switching. Classical motion over the barrier. Rate constant: k = κe G /k B T = κe S /k B e E a/k B T κ depends on dynamics in the barrier region (between S R and S P. For TST: κ = k BT h Trans. Faraday Soc., 1938, / 35

9 Marcus Theory Reactions Marcus Theory Crossing Point rate constant Electron transfer reactions do not involve bond breaking and bond making, so there is no barrier? A charge has a reaction potential, (just like a dipole): ψ R = q(ǫ r 1) 4πǫ 0 ǫ r a A qq The free energy of a charge in its own reaction field is G eq = 1 2 A qq 2 9 / 35

10 Marcus Theory II Reactions Marcus Theory Crossing Point rate constant If the amount of charge in the cavity is changed instantaneously, the order in the solvent remains the same (entropy), but the energy changes. Non equilibrium free energy: G neq = 1 2 ω2 s(q δq) 2 with ω 2 s = A q Reaction Coordinate=Solvent Polarization 10 / 35

11 Marcus Theory III Reactions Marcus Theory Crossing Point rate constant Solvent perspective: for a given polarization there is an equivalent charge. G = G eq,a G eq,d : driving force λ: reorganization energy. : transition state. X D,A : Solvent in equilibrium with Donor, Acceptor. At the crossing point X C it does not matter for the solvent where the electron is, and transfer can take place. 11 / 35

12 Crossing Point Reactions Marcus Theory Crossing Point rate constant Donor Potential: Acceptor potential: G D = G eq,d ω2 s(x X D ) 2 G A = G eq,a ω2 s(x X A ) 2 Crossing point: X C = 1 2 (X A +X D )+ G ω 2 s(x A X D ) Reorganization λ = 1 2 ω2 s(x A X D ) 2 Activation free energy: G = (λ+ G ) 2 4λ 12 / 35

13 Marcus Rate Constant Reactions Marcus Theory Crossing Point rate constant The reaction rate is the product of the Arrhenius factor containing the activation free energy, and a transmission coefficient κ: ) 2 k ET = κe (λ+ G 4λk B T A charge (or charge distribution) creates its own barrier by polarizing the environment. J. R. Miller, L. T. Calcaterra, G. L. Closs J. Am. Chem. Soc., (1984), 106, / 35

14 Transfer Förster Properties Examples 14 / 35

15 Transfer Transfer Förster Properties Examples 15 / 35

16 Transfer II Transfer Förster Properties Examples What can happen when a donor gets excited? Fluorescence or radiationless decay. Electron and Hole transfer: Marcus Theory. Excitation transfer Förster: dipole dipole interaction, distance between donor and acceptor large. Dexter: overlap of wavefunctions Exciton coupling: dipole dipole interaction, short distance between donor and acceptor. In photosynthesis the chromophores are organized in such a way that energy or electrons are transfered before fluorescence occurs. 16 / 35

17 Förster Mechanism Transfer Förster Properties Examples Fermi s golden rule: Γ nm = 4π 2 V nm 2 δ(ν n ν m ) δ(ν n ν m ): energy conservation V nm : dipole dipole interaction. V nm = f 2 4πǫ 0 ǫ r r 3 µ n ( 1 3 r r ) r 2 µ m 17 / 35

18 Förster Mechanism II Transfer Förster Properties Examples Total transfer rate (sum over all vibrational levels): Γ DA = f4 κ 2 4ǫ 2 0 n4 r 6 n,m µ 2 Dµ 2 AS 2 ms 2 nδ(ν n ν m ) with orientation factor (carets denote unit vectors) κ = ˆµ D (1 3ˆrˆr) ˆµ A 18 / 35

19 Förster Mechanism III Transfer Förster Properties Examples Normalized Fluorescence Spectrum: m F(ν) = µ2 D S2 mδ(ν ν m ) m µ2 D S2 m Absorption spectrum ǫ A (ν) = n 8π 3 N a hcln10 S 2 nδ(ν ν n ) µ 2 A f2 n in mol L 1 cm 1 19 / 35

20 Förster Mechanism IV Transfer Förster Properties Examples Barbatruc Γ DA = f4 κ 2 4ǫ 2 0 n4 r 6 = f4 κ 2 4ǫ 2 0 n4 r 6 n,m µ 2 Dµ 2 AS 2 ms 2 nδ(ν n ν m ) dν [ ][ ] µ 2 DSmδ(ν 2 n ν) µ 2 ASnδ(ν 2 n ν) m n After all the dust has settled: Γ DA = κ 2 n 4 τ D R 6 dν F D(ν)ǫ A (ν) ν 4 20 / 35

21 Properties of Förster energy transfer Transfer Förster Properties Examples Overlap integral: J DA = dν F D(ν)ǫ A (ν) ν 4 Overlap between the spectra is needed, otherwise energy can not be conserved. Overlap is usually small for the same molecule due to Stokes shift. Förster Length: R 6 0 = κ2 n 4 R 0 is usually of the order 10 nm. 21 / 35

22 Förster Transfer Examples Transfer Förster Properties Examples Two 6.3 D dipoles at 10 nm: V 0.2 cm 1. κ can range from 0 to 4 for fixed orientations. It is usually possible to find good labels for energy transfer. A. Sytnik et al., PNAS, 1996, / 35

23 Förster Transfer Examples II Transfer Förster Properties Examples B. Prevo and E.J.G. Peterman, Chem. Soc. Rev., (2014), 43, / 35

24 B820 Interaction LH2 Conclusions 24 / 35

25 B820 B820 Interaction LH2 Conclusions The LH1 antenna consists of 16 B820 dimers. A dimer binds two bacteriochlorophylla pigments. CD spectroscopy measures optical activity. Bacteriochlorophylla is not optically active. C. Koolhaas et al, Biophysical J., (1997), 72, / 35

26 Excitonic Interaction B820 Interaction LH2 Conclusions When pigments are close together, the interaction between transition dipoles starts to influence the electronic states. The pigments are no longer independent. In B820, the dipole dipole interaction is approximately 300 cm 1. As a consequence the absorption spectrum changes and the pigment pair becomes optically active, because rotational invariance is broken. This so called excitonic interaction plays an important role in energy transfer within the antenna systems, and in the optical properties of J aggregates. 26 / 35

27 Excitonically Coupled Pigment Pairs Hamiltonian B820 Interaction LH2 Conclusions H = H pigment1 +H pigment2 +V 12 In B820 the interaction strength is about 300 cm 1, and the orientation of the dipoles is almost parallel. An energy difference of 600 cm 1 at a wavelength of 800 nm (absorption of BChl), is about 40 nm. Therefore one transition is at 780 nm, the other at 820 nm. 27 / 35

28 Excitonically Coupled Pigment Pairs II B820 Interaction LH2 Conclusions 28 / 35

29 LH2 B820 Interaction LH2 Conclusions Hamiltonian: H = H pigmentn + nm n n,mv V nm : dipole dipole interactions between all the pigment. Excitonic states are delocalized states, where all the pigments are taking part in the excitation. C. Koolhaas et al, J. Phys. Chem. B, (1997), 101, / 35

30 LH2 B820 Interaction LH2 Conclusions Spectrum is shifted to the red. These bacteria live at the bottom of ponds where only red light penetrates. Lowest excitonic state is dark. This prevents fluorescence before energy is transfered to the RC. S. Georgakopoulou et al, Biophysical J., (2002), 82, / 35

31 Conclusions B820 Interaction LH2 Conclusions Molecules can be considered collections of state dipole moments and transition dipole moments. These are measurable quantities. Solvents can be considered collections of dipolar, polarizable molecules, described by their macroscopic dielectric properties. Much of the spectroscopy of molecules in solution or embedded in protein structures can be understood this way. Two main problems: dynamics of the interaction and protons. Want to know more: come to Finland. 31 / 35

32 Exercises Literature 32 / 35

33 Exercises Exercises Literature 1. How much work does it take to transport a proton over a membrane against a ph difference of 3? 2. Explain the behavior of the Marcus rate when the reorganization energy λ goes to zero. 3. Explain the behavior of the rate vs the driving force in the figure of p Explain the second figure on p. 22, What value does the paper use for κ 2? Can this be justified? 5. What causes the bleaching shown in the bottom right figure of p Explain how diagonal disorder (energy disorder) leads to the side peaks in the botton figure of p / 35

34 Literature Exercises Literature 1. J. Léonard, E. Portuondo Campa, A. Cannizzo, F. van Mourik, G. van der Zwan, J. Tittor, S. Haacke, and M. Chergui, Functional electric field changes in photoactivated proteins revealed by ultrafast Stark spectroscopy of the Trp residues, PNAS, (2009), 106, J. Joy R.T. Cheriya, K. Nagarajan, A. Shaji, and M. Hariharan, Breakdown of Exciton Splitting through Electron DonorAcceptor Interaction: A Caveat for the Application of Exciton Chirality Method in Macromolecules, J. Phys. Chem. C, (2013), 117, A. Sytnik and I. Litvinyuk, transfer to a proton transfer fluorescence probe: Tryptophan to a flavonol in human serum albumin, Proc. natl. Acad. Sci. USA, (1996) 93, B. Prevo and E.J.G. Peterman, Förster resonance energy transfer and kinesin motor proteins, Chem. Soc. Rev., (2014), 43, / 35