Principles of Fluorescence Spectroscopy

PRINCIPLES OF FLUORESCENCE SPECTROSCOPY 193
but the derivation is rarely given. These expressions have
been derived by several routes. 41,138–139 The simplest
approach uses the kinetic equations and algebraic manipulation.
41,138 The excitation is assumed to be sinusoidally
modulated light
(5.40)
so that b/a = m L is the modulation of the incident light. The
fluorescence emission is forced to respond with the same
frequency, but the phase shift and modulation will be different.
One can assume the excited-state population is given as
follows:
(5.41)
and determine the relationship between fluorescence lifetime
and the phase shift (φ) and the demodulation (m). The
intensity I(t) at any time is proportional to the number of
molecules in the excited state N(t).
Suppose the intensity decay following δ-function excitation
is a single exponential:
(5.42)
For a single-exponential decay the differential equation
describing the time-dependent excited-state population is
dI(t)
dt
Substitution of 5.41 into eq. 5.43 yields
ωB cos (ωt φ)
L(t) a b sin ωt
N(t) A B sin (ωt φ)
I(t) I 0 exp (t/τ)
1
I(t) L(t)
τ
(5.43)
(5.44)
This equation must be valid for all times. The relationship
between the values of a, b, A, and B and the fluorescence
lifetime τ can be obtained by expansion of the sine and
cosine functions, followed by equating of the constant
terms and terms in sin ωt and cos ωt. This yields
1
A B sin (ωt φ) a b sin ωt
τ
a (1/τ)A 0
ω cos φ (1/τ) sin φ 0
ω sin φ (1/τ) cos φ b/B
(5.45)
(5.46)
(5.47)
From eq. 5.46 one obtains the familiar relationship
(5.48)
Squaring eqs. 5.46 and 5.47, followed by addition, yields
Recalling that A = aτ [eq. 5.45], one obtains
(5.49)
(5.50)
which is the usual relationship between the lifetime and the
demodulation factor.
An alternative derivation is by the convolution integral.
134 The time-dependent intensity I(t) is given by the
convolution of excitation function (eq. 5.40) with the
impulse response function (eq. 5.42):
Substitution of eqs. 5.40 and 5.42 yields
∞
I(t) I0 exp(t’/τ) a b cos(ωt ωt’)dt’
0
(5.51)
(5.52)
These integrals can be calculated by recalling the identities
cos (x y) cos x cos y sin x sin y
∞
0
∞
0
sin φ
cos φ tan φ ωτ φ
ω 2 (1/τ) 2 (b/B) 2
m B/A
b/a 1 ω2 τ 2 m 1/2
∞
I(t) L(t’)I(t t’)dt’
0
exp (kx) sin mx dx
m
k2 m2 a
exp (kx) cos mx dx
k2 m2 (5.53)
(5.54)
(5.55)