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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 447
given data βε is evaluated at the above temperature as
βε = h2
8mkTX2
= h2N2A
8MRTX2
= (6.0221 × 1023mol−1)2 × (6.6261 × 10−34 J s)2
8 × (1.008 × 10−3 kgmol−1) × (8.3145 JK−1mol−1)
× 1
(0.0302... K) × (100 × 10−9 m)2
= 0.00785...
With this value the partition function is then evaluated term by term to give
qTX =
∞
∑
n=1
e−(n
2−1)βε = 1 + e−3βε + e−8βε + e−15βε + ...
= 1 + 0.977... + 0.939... + 0.889... + ... = 9.57
Even under these conditions, the integral approximation deviates by less than
5% from the explicit sum.
P13B.6 �e partition function is given by [13B.1b–538], q = ∑i g ie−βε i , where g i is
degeneracy and the corresponding energy is given as ε i = hcν̃ i , and β = 1/kT .
Here g i = 2J + 1, where J is the right subscript in the term symbol. At 3287 ○C,
which is 3556 K
hcβ = (6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1)
(1.3806 × 10−23 JK−1) × (3556 K)
= 4.04... × 10−4 cm
�erefore the electronic partition function is
qE = 5 + 7 × e−(4.04...×10
−4 cm)×(170 cm−1)
+ 9 × e−(4.04...×10
−4 cm)×(387 cm−1) + 3 × e−(4.04...×10
−4 cm)×(6557 cm−1)
= 5 + 6.53... + 7.69... + 0.211... = 19.4....
�e population of level i is N i/N = (g i × e−hcβν̃ i )/qE.�us
N(3F2)
N
= 5
19.4...
= 0.257 N(3F3)
N
= 6.53...
19.4...
= 0.336
N(3F4)
N
= 7.69...
19.4...
= 0.396 N(5F1)
N
= 0.211...
19.4...
= 0.0109
P13B.8 �e wavenumbers of the lines in the pure rotational spectrum of a diatomic
are given by [11B.20a–436], ν̃(J + 1 ← J) = 2B̃(J + 1). �erefore a plot of
ν̃(J + 1← J) against 2(J + 1) is expected to be a straight line with slope B̃.�e
separation of successive lines is 2B̃, and from the �rst few lines a consistent
value of B̃ ≈ 10.6 cm−1 is found. Using this, the assignment of the lines is
con�rmed as 1 ← 0 for that at 21.19 cm−1, and 2 ← 1 for that at 42.37 cm−1.
�e given lines are therefore successive members of ν̃(J + 1← J) starting from