Principles of Fluorescence Spectroscopy

908 ANSWERS TO PROBLEMS
emission. The transfer efficiency can be calculated
from the emission intensity at 450 nm for 290-nm
excitation, which reflects acceptor emission due to
excitation of the donor and direct excitation of the
acceptor. Dansyl-L-propyl-hydrazide does not contain
a donor, and hence this excitation spectrum defines
that expected for 0% transfer. For dansyl-L-propyl-αnaphthyl,
in which the donor and acceptor are closely
spaced, energy transfer is 100% efficient. For this
donor–acceptor pair the greatest sensitivity of the
excitation spectrum to energy transfer is seen near 290
nm, the absorption maximum of the naphthyl donor.
For the other derivatives the intensity is intermediate
and dependent upon the length of the spacer. For 290nm
excitation the transfer efficiency can be calculated
from the relative intensity between 0 and 100% transfer.
The efficiency of energy transfer decreases as the
length of the spacer is increased.
The object of these experiments was to determine
the distance dependence of radiationless energy transfer.
Hence we assume that the efficiency of energy
transfer depends on distance according to
E (R 0/r) j
(R 0/r) j 1
(13.38)
where R 0 and r have their usual meanings, and j is an
exponent to be determined from the observed dependence
of E on r. Rearrangement of eq. 13.38 yields
ln(E 1 1) j ln r j ln R 0
(13.39)
Hence a plot of ln(E –1 – 1) versus ln r has a slope of j.
These data are shown in Figure 13.41. The slope was
found to be 5.9 " 0.3. 18 From this agreement with the
Figure 13.41. Distance dependence of the energy transfer efficiencies
in dansyl-(L-propyl) n -α-naphthyl; n = 1–12. Revised from [18].
predicted value of j = 6 these workers concluded that
energy transfer followed the predictions of Förster.
See [18] for additional details.
The value of R 0 can be found from the distance at
which the transfer efficiency is 50%. From Figure
13.41 (right) R 0 is seen to be near 33 Å.
A13.4. The lifetime of compound I is τ DA and the lifetime of
compound II is τ D . Compound II serves as a control
for the effect of solvent on the lifetime of the indole
moiety, in the absence of energy transfer. The rate of
energy transfer is given by k T = τ DA –1 - τ D –1.
k T C J
(13.40)
where C is a constant. Hence a plot of k T versus J
should be linear. The plot of k T vs. J is shown in Figure
13.42. The slope is 1.10. These data confirm the
expected dependence of k T on the overlap integral. See
[20] for additional details.
A13.5. If the wavelength (λ) is expressed in nm, the overlap
integral for Figure 13.9 can be calculated using eq.
13.3 and is found to be 4.4 x 10 13 M –1 cm –1 (nm) 4 .
Using eq. 13.5, with n = 1.33 and Q D = 0.21, one finds
R 0 = 23.6 Å. If λ is expressed in cm then J(λ) = 4.4 x
10 –15 M –1 cm 3 , and using eq. 13.8 yields R 0 = 23.6 Å.
A13.6. The disassociation reaction of cAMP (A) from protein
kinase (PK) is described by
PK.cAMP ⇌ PK cAMP
B ⇌ F A
(13.41)
where B represents PK with bound cAMP, F the PK
without bound cAMP, and A the concentration of
Figure 13.42. Dependence of the rate of energy transfer on the magnitude
of the overlap integral. Revised from [20].