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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 393
As c approaches 1 the evaluation of the transitionmoment is made easier by the
following manipulations
1 − c2
(1 − c)1/2
= (1 − c2)(1 + c)1/2
(1 − c)1/2(1 + c)1/2
= (1 − c2)(1 + c)1/2
[(1 − c)(1 + c)]1/2
= (1 − c2)(1 + c)1/2
(1 − c2)1/2
= (1 − c2)1/2(1 + c)1/2 = (1 − c)1/2(1 + c)1/2(1 + c)1/2 = (1 − c)1/2(1 + c)
In the limit that c = 1 this term, and hence the transitionmoment, goes to zero.
�e value of c for which the moment is a maximum is found by di�erentiation
d
dc
(1 − c)1/2(1 + c) = − 12 (1 − c)
−1/2(1 + c) + (1 − c)1/2
setting the derivative to zero and multiplying by (1 − c)1/2 gives
0 = − 12 (1 + c) + (1 − c) hence c = 1
3
�emaximum transitionmoment occurs at c = 1
3 , and has the value (8/27)
1/2a
≈ 0.54 a.
E11F.12(b) �e Gaussian functions are written e−αx2/2, where the parameter α determines
the width. To evaluate the normalizing factor for the function e−αx2/2 requires
the integral
∫
+∞
−∞
e−αx2 dx
which is of the form of Integral G.1 with k = α and evaluates to (π/α)1/2.�e
normalizing factor is therefore N = (α/π)1/4. �e other Gaussian has half
the width of the �rst, so is of the form e−4αx2/2, and the normalizing factor is
N ′ = (4α/π)1/4.
�e transition moment is given by the integral
I = (2α/π)1/2 ∫
+∞
−∞
xe−αx2/2e−4αx2/2 dx
= (2α/π)1/2 ∫
+∞
−∞
xe−5αx2/2 dx
�e integrand is odd and the integral is over a symmetric interval, therefore the
integral, and hence the transition moment is zero .
E11F.13(b) �e absorption at 320 nm is likely to be due to an π∗ ← n transition. �e
wavelength is typical for such transitions, and they are weak because they are
symmetry forbidden.
�e absorption at 213 nm is likely to be due to a π∗ ← π transition. Of the two
transitions it is the stronger and it is at a somewhat longer wavelength than for
an isolated π bond as a result of the conjugation with the carbonyl π bond.