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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 373
E11C.3(b) �e wavenumber of the fundamental vibrational transition is simply equal to
the vibrational frequency expressed as a wavenumber.�is is given by [11C.4b–
443], ν̃ = (1/2πc)(kf/meff)1/2, wheremeff is the e�ectivemass, given bymeff =
m1m2/(m1 +m2). It follows that kf = meff(2πcν̃)2. With the data given
kf =
78.9183 × 80.9163
78.9183 + 80.9163
× (1.6605 × 10−27 kg)
× [2π × (2.9979 × 1010 cms−1) × (323.2 cm−1)]2
= 245.9 Nm−1
Note the conversion of the mass to kg.
E11C.4(b) �e wavenumber of the fundamental vibrational transition is simply equal to
the vibrational frequency expressed as a wavenumber.�is is given by [11C.4b–
443], ν̃ = (1/2πc)(kf/meff)1/2, wheremeff is the e�ectivemass, given bymeff =
m1m2/(m1 +m2). Assuming that the force constants of the two isotopologues
are the same, ν̃ simply scales as (meff)−1/2.�erefore
ν̃ 2HX
ν̃ 1HX
= (
meff , 1HX
meff , 2HX
)
1/2
hence ν̃ 2HX = ν̃ 1HX × (
meff , 1HX
meff , 2HX
)
1/2
Using these relationships, the following table is drawn up
nH19F nH35Cl nH81Br nH127I
ν̃ 1HX/cm−1 4141.3 2988.9 2649.7 2309.5
meff , 1HX/mu 0.9570 0.9796 0.9954 0.9999
meff , 2HX/mu 1.8210 1.9043 1.9651 1.9825
ν̃ 2HX/cm−1 3002.3 2143.7 1885.8 1640.1
E11C.5(b) �e terms (energies expressed as wavenumbers) of the harmonic oscillator are
given by [11C.4b–443], G̃(υ) = (υ + 1
2 )ν̃; these are wavenumbers and so can
be converted to energy by multiplying by hc to give E(υ) = (υ + 1
2 )hcν̃. �e
�rst excited state has υ = 1, and the second has υ = 2. �e relative popula-
tion of these levels is therefore given by the Boltzmann distribution, n2/n1 =
e−(E2−E1)/kT .�e energy di�erence E2−E1 = hcν̃, and hence n2/n1 = e−hc ν̃/kT .
It is convenient to compute the quantity hcν̃/k �rst to give
hcν̃/k = (6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1) × (321 cm−1)
1.3806 × 10−23 JK−1
= 461.8... K
It follows that n2/n1 = e−(461.8... K)/T
(i) At 298 K, n2/n1 = e−(461.8... K)/(298 K) = 0.212
(ii) At 800 K, n2/n1 = e−(461.8... K)/(800 K) = 0.561