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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 251
through zero. �e wavefunction asymptotically approaches zero at y = ±∞,
but as the function does not pass through zero these limits do not count as
nodes.�e nodes in the wavefunction therefore correspond to the solutions of
Hυ(y) = 0. Hυ(y) is a polynomial of order υ, meaning that the highest power
of y that occurs is yυ ; such polynomials in general have υ solutions and hence
there are υ nodes.�erefore (a) the wavefunction with υ = 5 has 5 nodes; (b)
the wavefunction with υ = 35 has 35 nodes.
E7E.8(b) �e wavefunction with υ = 3 is ψ3(y) = N3(y3 −3y)e−y
2/2. Nodes occur when
the wavefunction passes through zero; the wavefunction approaches zero at
y = ±∞, but these do not count as nodes as the wavefunction does not pass
through zero. It is evident that the nodes occur when y3 − 3y = y(y2 − 3) = 0
which solves to give nodes at y = 0,±
√
3 .
E7E.9(b) �e wavefunction with υ = 3 is ψ3(y) = N3(8y3 − 12y)e−y
2/2, which gives a
probability density of P(y) = ∣ψ3(y)∣2= N23(8y3 − 12y)2e−y
2
.�e extrema are
located by di�erentiating the wavefunction, setting the result to 0 and solving
for y.�e di�erential is evaluated using the product rule
dψ3(y)
dy
= N23
⎛
⎝
d(8y3 − 12y)2
dy
e−y
2
+ (8y3 − 12y)2 de
−y2
dy
⎞
⎠
= N23 [2(8y3 − 12y)(24y2 − 12)e−y
2
+ (8y3 − 12y)2 × (−2ye−y
2
)]
It is necessary to usemathematical so�ware to �nd the solutions of dψ3(y)/dy =
0; these solutions are
0.000 ± 0.6021 ± 1.225 ± 2.0341 ±∞
It is evident from inspection of the wavefunction that y = 0 is a mimimum in
the probability density, and that there is also a minimum when (8y2 − 12) = 0,
which corresponds to y = ±1.225. It follows thatmaxima occur at y = ±0.6021
and y = ±2.034 .
Solutions to problems
P7E.2 (a) In the case that mB ≫ mA, mA + mB ≈ mB, and so µAB ≈ mAmB/mB =
mA.�e ratio of the vibrational frequencies is then
ωA′B = ωAB(µAB/µA′B)1/2 ≈ ωAB(mA/mA′)1/2
If the frequencies are expressed in Hz the analogous result is
νA′B ≈ νAB(mA/mA′)1/2
(b) Using the previous result gives ν 2H35Cl ≈ ν 1H35Cl(m 1H/m 2H)1/2 = (5.63 ×
1014 Hz) × (1 mu/2 mu)1/2 = 3.98 × 1014 Hz .