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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 249
As is shown in Fig. 7E.7 on page 277, as the quantumnumber becomes large the
probability density clusters more and more around the classical turning points
of the classical harmonic oscillator (that is, the points at which the kinetic
energy is zero). Because the classical oscillator is moving most slowly near
these points, they are the displacements at which it is most probable that the
oscillator will be found. Again, the quantum and classical results converge at
high quantum numbers
Solutions to exercises
E7E.1(b) �e zero-point energy of a harmonic oscillator is given by [7E.5–274], E0 =
1
2 ħω, where the frequency ω is given by [7E.3–274], ω = (kf/m)1/2. For this
system,
E0 = 1
2 × (1.0546 × 10−34 J s) × [(285 Nm−1)/(5.16 × 10−26 kg)]1/2
= 3.92 × 10−21 J
E7E.2(b) �e separation between adjacent energy levels of a harmonic oscillator is [7E.4–
274], ∆E = ħω, where the frequency, ω is given by [7E.3–274], ω = (kf/m)1/2.
�is is rearranged for the force constant as kf = m(∆E/ħ)2. Evaluating this
gives
kf = (2.88×10−25 kg)× [(3.17 × 10−21 J)/(1.0546 × 10−34 J s)]2 = 260 Nm−1
E7E.3(b) �e separation between adjacent energy levels of a harmonic oscillator is [7E.4–
274], ∆E = ħω, where the frequency, ω is given by [7E.3–274], ω = (kf/m)1/2.
�e Bohr frequency condition [7A.9–241], ∆E = hν, can be rewritten in terms
of the wavelength as ∆E = hc/λ.�e wavelength of the photon corresponding
to a transition between adjacent energy levels is therefore given by ħω = hc/λ,
or ħ(kf/m)1/2 = hc/λ. Solving for λ gives λ = 2πc/(kf/m)1/2; with the data
given
λ = 2π × (2.9979 × 108ms−1)
[(544 Nm−1)/(15.9949 × 1.6605 × 10−27 kg)]1/2
= 1.32 × 10−5 m
E7E.4(b) �e wavefunctions are depicted in Fig. 7E.6 on page 276; they are real. Two
wavefunctions are orthogonal if ∫ ψ∗i ψ j dτ = 0. In this case the wavefunctions
are ψ1(y) = N1 ye−y
2/2 and ψ2(y) = N2(y2 − 1)e−y
2/2, and the integration is
from y = −∞ to +∞. �e integrand is ψ2ψ1 = N2N1(y3 − y)e−y
2
, which is
an odd function, meaning that its value at −y is the negative of its value at +y.
�e integral of an odd function over a symmetric range is zero, hence these
wavefunctions are orthogonal.