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354 11MOLECULAR SPECTROSCOPY
For this case the Beer–Lambert law is lg(I/I0) = −ε[J]0x0(1 − e−L/x0) , where
ε = κ/ ln 10.
In the limit L ≫ x0, e−L/x0 → 0 and hence lg(I/I0) = −ε[J]0x0 . In this limit
the absorption is independent of the length of the sample. If x0 ≫ L then
the exponential may be expanded up to the linear term to give (1 − e−L/x0) ≈
(1 − 1 + L/x0) ≈ L/x0. �en lg(I/I0) = −ε[J]0L , which is the form of the
Beer–Lambert law for a constant concentration.
P11A.4 Suppose that species A and B associate to give a complex AB according to the
equilibrium: A + B ÐÐ⇀↽ÐÐ AB. Suppose that species A absorbs at a particular
wavelength with absorption coe�cient εA, and that species AB absorbs at the
same wavelength but with absorption coe�cient εAB; B does not absorb at this
wavelength.�e absorbance at this wavelength is
Aλ = εA[A]L + εAB[AB]L
If the initial amount of A is [A]0 and at equilibrium the amount of AB is [AB]e,
mass balance requires [A] = [A]0 − [AB]e. It follows that
Aλ = εA ([A]0 − [AB]e) L + εAB[AB]eL = εA[A]0L
´¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¶
term 1
+ (εAB − εA)[AB]eL
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
term 2
As the amount of added B is changed, term 1 remains the same, but in gen-
eral term 2 will change because of the resulting change in [AB]e. However, if
the wavelength is chosen such that εAB = εA, term 2 will go to zero and the
absorbance will be invariant to the amount of B added. �is is the isosbestic
point.
P11A.6 Because only the area under the absorption band is of interest it can be as-
sumed, without loss of generality, that the band is centred on ν̃ = 0.�e band-
shape is then given by ε(ν̃) = εmaxe−a ν̃2 , where the parameter a determines the
width.�is parameter is found by �nding the wavenumbers ν̃1/2 at which ε has
fallen to half its maximum value
1
2 εmax = εmaxe−a(ν̃1/2)2
hence ln 12 = −a(ν̃1/2)2
hence (ν̃1/2)2 = ln 2/a
where to go to the last line ln 12 = − ln 2 is used.�e wavenumbers of the points
at which ε has fallen to half its maximum value are therefore ±(ln 2/a)1/2, and
hence the separation between these is 2(ln 2/a)1/2.�is separation is the width
at half-height ∆ν̃1/2
∆ν̃1/2 = 2(ln 2/a)1/2 hence a = 4 ln 2
(∆ν̃1/2)2