Principles of Fluorescence Spectroscopy

Answers to Problems
CHAPTER 1
A1.1. A. The natural lifetimes and radiative decay rates
can be calculated from the quantum yields and
experimental lifetimes:
τN(eosin) τ/Q 3.1
4.77 ns
0.65
τN(EB) τQ 0.61
5.08 ns
0.12
(1.18)
(1.19)
Hence eosin and erythrosin B have similar natural lifetimes
and radiative decay rates (eq. 1.3). This is
because both molecules have similar absorption and
emission wavelengths and extinction coefficients (eq.
1.4).
The non-radiative decay rates can be calculated
from eq. 1.2, which can be rearranged to
1 1
knr τ τN (1.20)
For eosin and erythrosin B the non-radiative decay
rates are 1.1 x 10 8 s –1 and 1.44 x 10 9 s –1 , respectively.
The larger non-radiative decay rate of erythrosin B is
the reason for its shorter lifetime and lower quantum
yield than eosin.
B. The phosphorescence quantum yield (Q p) can be
estimated from an expression analogous to eq
1.1:
Γ p
Qp
Γp knr (1.21)
Using the assumed natural lifetime of 10 ms, and k nr =
1 x 10 8 s –1 , Q p = 10 –6 . If k nr is larger, Q p is still smaller,
so that Q p ; 10 –7 for ErB. This explains why it is difficult
to observe phosphorescence at room temperature:
most of the molecules that undergo intersystem
crossing return to the ground state by non-radiative
paths prior to emission.
A1.2. The quantum yield (Q 2 ) of S 2 can be estimated from
Q2
Γ
Γ knr (1.22)
The value of k nr is given by the rate of internal conversion
to S 1 ,10 13 s –1 . Using Γ = 2.1 x 10 8 , one can estimate
Q 2 = 2 x 10 –5 . Observation of emission from S 2
is unlikely because the molecules relax to S 1 prior to
emission from S 2 .
A1.3. The energy spacing between the various vibrational
energy levels is revealed by the emission spectrum
of perylene (Figure 1.3). The individual emission
maxima (and hence vibrational energy levels) are
about 1500 cm –1 apart. The Boltzmann distribution
describes the relative number of perylene molecules
in the 0 and 1 vibrational states. The ratio (R)
of molecules in each state is given by
R e ∆E/kT
(1.23)
where ∆E is the energy difference, k is the Boltzmann
constant, and T is the temperature in degrees kelvin
(K). Assuming a room temperature of 300 K, this ratio
is about 0.01. Hence most molecules will be present in
the lowest vibrational state, and light absorption
results mainly from molecules in this energy level.
Because of the larger energy difference between S0 and S1, essentially no fluorophores can populate S1 as
a result of thermal energy.
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