1 Consider a notional photosynthetic antenna complex with eightfold rotational symmetry, in which the transition dipole moments of the chromophores lie in the plane, perpendicular to the rotational symmetry axis, and disposed at an angle at 60o to the radial direction.

Calculate the orientational factor κ, for transfer between any two adjacent chromophores.

Equation 1
Equation 1

(Note that the caret over each vector in the defining formula denotes a unit vector.)

Compare the result with the case of a simple dimer having parallel transition moments aligned along their mutual displacement vector, and comment on the relative magnitude of κ.

2 Equation 2 from the lecture states that

Show that, if applied to values of transition dipoles in the traditional debye units (D), and distances Rmn in nanometres, it gives the following result:

Equation 2
Equation 2

1 D = 3.338 x 10-30 C m

vacuum permittivity ε0 = 8.854 x 10-19 F m-1 .

3 The Förster distance R0 is the distance of separation from an acceptor at which a donor has the same probability of releasing its energy by spontaneous decay as by transfer to that acceptor (see Slide 11). What are the relative efficiencies of spontaneous fluorescence by the donor, compared toRET, at distances of (a) 0.5 R0 and (b) 1.5R0?

4 At what distance of donor-acceptor separation will the R-4 term in the full RET rate equation (Equation 17 and Slide 21) become sufficiently significant to represent 10% the magnitude of the leading, more familiar R-6 term?

Assume for the value of K, which is equal to 2 π/λwhere λ is the wavelength corresponding to the conveyed energy, K = 7.6 x 106 m-1.


1 The angles are found by simple geometry:

Figure 1
Figure 1

κ = cos (52.5o – 7.5o = 45o) – 3 cos(52.5o)cos(7.5o)

= 0.707 – 3(0.609 x 0.991)

= 0.707 – 1.811

≈ -1.1.

The sign is not really important since it’s the square of κ that appears in the energy transfer rate equation (Equations 8 and 9). Often, when the data to evaluate its exact value is not available, κ is assumed to have a value of unity.

The numerically largest value of kappa, |κ| = 2, arises when the two transition moments and the displacement vector are collinear, as shown in the figure on the left at the foot of Slide 9. This case gives the maximum possible value of kappa (giving optimal transfer efficiency). However, a simple sketch suffices to show that this is not achievable for an array with more than twofold rotational symmetry.


Equation 1
Equation 1

(to 3 significant figures.)

3 (a) (0.5)6 = 0.015

(i.e. 0.015/1.015 = 1.5%, i.e. 98.5% RET);

(b) (1.5)6 = 11.39

(i.e. 11.39/12.39 = 91.9%, i.e. 8.1% RET).

4 (K2/R4) ÷ (3/R6) = K2R2/3 = 0.1.

So R2 = 0.3/K2 = 51.9 x 10-16 m. Hence R = 7.2 x 10-8 m = 72 nM