Principles of Fluorescence Spectroscopy

900 ANSWERS TO PROBLEMS
expected to decrease the excited state energy by 9554
cm –1 , to 19,017 cm –1 , which corresponds to 525.8 nm.
The units for ν A – ν F are as follows:
(esu cm) 2
(ergs) (cm/s(cm 3 )
(6.24)
Recalling that erg = g cm 2 /s 2 and esu = g 1/2 cm 3/2 /s,
one obtains ν A – ν F in cm –1 .
A6.2. The change in dipole moment can be estimated from
the Lippert plot (Figure 6.53). This plot shows biphasic
behavior. In low-polarity solvents the emission is
probably due to the LE state, and in higher-polarity
solvents the emission is due to the ICT state. The
slopes for each region of the Lippert plot are
slope (LE)= 7000 cm –1
slope (ICT) = 33,000 cm –1
The slope is equal to 2(µ E – µ G) 2 /hca 3 . Assuming a
radius of 4.2 Å used previously, 42
(µ E µ G) 2 7000
2 hca3
(6.25)
7000
2 (6.626x1027 )(3x10 10 )(4.2x10 8 ) 3 5.15x10 35
The units of (µ E – µ G ) 2 are (cm –1 ) (erg s)(cm/s)(cm 3 ).
Using erg = g cm 2 /s 2 , one obtains the units (g cm 3 /s 2 )
(cm 2 ). Taking the square root yields)
g1/2 cm3/2 cm
s
(6.26)
Since esu = g 1/2 cm 3/2 /s, the result (µ E – µ G ) is in esu
cm. This yields (µ E – µ G ) = 7.1 x 10 –18 esu cm = 7.1D.
The dipole moment of Prodan is estimated to change
by 7.1 Debye units upon excitation. An electron separated
from a unit positive charge by 1 Å has a dipole
moment of 4.8D. Hence there is only partial charge
separation in the LE state. It should be noted that this
value is smaller than initially reported 42 due to an trivial
error during the calculations. 58
For the ICT state a similar calculation yields (µ E –
µ G ) 2 = 2.42 x 10 –34 and ∆µ = 1.56 x 10 –17 esu cm =
Figure 6.53. Lippert plot of the Stokes shift of Pordan. Data from [42].
CHAPTER 7
15.6D. This change in dipole moment is equivalent to
separation of a unit change by 3.2 Å, which suggests
nearly complete charge separation in the ICT state of
Prodan.
A7.1. Assume the decay is a single exponential. Then the
time where the intensity has decayed to 37% of its
original intensity is the fluorescence lifetime. These
values are τ F = 1 ns at 390 nm and τ = 5 ns at 435 nm.
The decay time of 5 ns at 435 nm is the decay time the
F-state would display in the absence of relaxation. The
lifetime of the F-state at 390 nm is given by 1/τ F = 1/τ
+ 1/τ S . This is equivalent to stating the decay time of
the F-state (γ F ) is equal to the sum of the rates that
depopulate the F-state, γ F = 1/τ + k S . Hence τ S = 1.25
ns.
A7.2. In the fluid solvents ethanol or dioxane the apparent
lifetimes of TNS are independent of wavelength, indicating
spectral relaxation is complete in these solvents.
In glycerol or DOPC vesicles the apparent lifetimes
increase with wavelength, suggesting timedependent
spectral relaxation. The observation of τ φ >
τ m at long wavelength is equivalent to observing a
negative pre-exponential factor, and proves that relax