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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 253
isotopomer ν̃/cm−1 ν̃/cm−1
if C binds if O binds
12C16O 1421 1640
13C16O 1421 1576
12C18O 1339 1640
13C18O 1339 1576
�e force constant is found using ν̃ = (1/2πc) (kf/µ)1/2, rearranged to kf =
µ(2πν̃c)2.
P7E.10 �e ground-state wavefunction is found from [7E.7–275] as ψ0(y) = N0e−y
2/2.
�e normalization constant is found by evaluating
N20 ∫
+∞
−∞
e−y
2
dy = 2N20 ∫
+∞
0
e−y
2
dy = 2N20 × ( 12π1/2) = 1
where required integral is of the formof IntegralG.1with k = 1.�e normalized
wavefunction is therefore ψ0(y) = π−1/4e−y
2/2.
(a) For the expectation value of y, the integral to evaluate is
⟨y⟩ = ∫
∞
−∞
ψ∗0 yψ0 dy = π−1/2 ∫
∞
−∞
y e−y
2
dy
�e integrand is an odd function, meaning that its value at −y′ is the
negative of its value at y′, and so when integrated over a symmetrical
range about y = 0 the result is necessarily zero. Hence ⟨y⟩ = 0 .
(b) To calculate ⟨y2⟩ the required integral is of the form of Integral G.3 with
k = 1
⟨y2⟩ = ∫
∞
−∞
ψ∗0 y
2ψ0 dy = π−1/2 ∫
∞
−∞
y2e−y
2
dy
= 2π−1/2 ∫
∞
0
y2e−y
2
dy = 2π−1/2 × 1
4π1/2 = 1
2
(c) �e �rst excited state is ψ1(y) = N1 ye−y
2/2. �e normalization constant
is found by evaluating
N21 ∫
+∞
−∞
y2e−y
2
dy = 2N21 ∫
+∞
0
y2e−y
2
dy = 2N21 × ( 14π1/2) = 1
where required integral is of the form of Integral G.3 with k = 1. �e
normalized wavefunction is therefore ψ1(y) = (4/π)1/4 y e−y
2/2.
�e expectation value of y is zero for the same reason as in (a). To calcu-
late ⟨y2⟩ the required integral is of the form of Integral G.5 with k = 1
⟨y2⟩ = ∫
∞
−∞
ψ∗1 y
2ψ1 dy = (4/π)1/2 ∫
∞
−∞
y4e−y
2
dy
= 2(4/π)1/2 ∫
∞
0
y4e−y
2
dy = 2(4/π)1/2 × 3
8π1/2 = 3
2