## Questions

1 In vector notation, adiabatic and diabatic electronic states are related to one another by a 2 x 2 rotation, according to Equation 4 from the lecture: Equation 4

In the case of a symmetric molecule at the avoided crossing, γ = π/4.

What are the ground (ψ1) and excited (ψ2) states of the molecule in terms of the diabatic states (φi, φf) at the avoided crossing?

2 Electron transfer within the Marcus-Hush framework can be described by the intersection of two parabolas, as shown below. Figure 1

where the rate of electron transfer is given by Equation 14 from the lecture: and Equation 12 Prove the second relationship geometrically.

Hint: Set up the parabolas with equations (origin of plane is at the minimum of the left parabola)

yi = ax2

and

yf = a( x – b )2 - c

and solve for the intersection point – call it (x’, y’).

3 The following systems undergo electron transfer from a 1,4-dimethoxynaphthalene donor (left chromophore) to a dicyanomethylene acceptor (right chromophore). The rate of electron transfer for systems 1-4 is too fast to be measured. However, the charge-transfer bands, observed spectroscopically, give a good estimation of the electronic couplings. Estimate the attenuation coefficient (β) in nm-1.

System 1 has 4 (σ) bonds separating the chromophores, 2 has 6, 3 has 8 and 4 has 10. The given couplings are in the usual spectroscopic units, cm-1. Assume that each σ-bond represents a distance of 0.15 nm. Figure 2

1  2 The parabolas can be set up as here, where the origin is at the vertex of parabola i. Figure 1

The first step is to write out the equation for the two parabolas. We assume that the origin of the (x,y) plane is at the vertex of the left parabola (the initial state).

yi = ax2

yf = a( x – b )2 - c

Now set the equations equal at the intersection point (where the avoided crossing would appear).

y’ = a (x’)2 = a(x’ – b’)2 – c

a (x’)2 = a (x’)22abx’ + (ab2 – c)

2abx’ = ab2 – c  Now, one can see that if 3 The two equations that are required are Equations 7 and 16 from the lecture:

kET = ∝|Vif|2

and

kET = k0 exp[-βR]

Noting that the rate of electron transfer is proportional to the square of the coupling, we can estimate the relative rates of electron transfer.

Also, we can transform Equation 17 into an equation for a straight line, with β being the slope, i.e.

ln(kET) = ln(k0) - βR

 System Vif kET ln(kET) R/nm 1 370 1.37 x 105 11.8 0.60 2 112 1.25 x 104 9.43 0.90 3 40 1.60 x 103 7.38 1.20 4 18 3.24 x 102 5.78 1.50

Plotting the data gives an attenuation coefficient of β~ 7 nm-1. Note that this is slightly lower than normal. Open the possibility for discussion of why. The assumption that was we can just add the length of the σ-bonds, and this introduces a slight error. This is the maximum distance (triangle inequality for adding the σ-bonds will ensure that the distances will actually be less).