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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 375
1 2 3 4
2 000
2 100
2 200
υ + 1
∆G̃
υ+
1/
2/
cm
−1
Figure 11.6
�is con�rms that the data are well-described by the Morse levels. From the
slope
xe ν̃/cm−1 = − 12 × slope = −
1
2 (−79.46) hence xe ν̃ = 39.73 cm−1
and from the intercept
ν̃/cm−1 = intercept = 2 309.5 hence ν̃ = 2 309.5 cm−1
�e depth of the well, D̃e is then found using [11C.8–444], xe = ν̃/4D̃e rear-
ranged to D̃e = ν̃/4xe = ν̃2/4ν̃xe. �e dissociation energy is D̃0 = D̃e − G̃(0)
(Fig. 11C.3 on page 444), hence
D̃0 = D̃e − G̃(0) =
ν̃2
4ν̃xe
− G̃(0)
= (2309.5 cm−1)2
4 × (39.73 cm−1)
− (1144.83 cm−1) = 3.24 × 104 cm−1
To convert to eV, the conversion 1 eV = 8065.5 cm−1 from inside the front cover
is used to give D0 = 4.02 eV .
�e result is only quoted to modest precision because in e�ect the dissociation
limit is extrapolated from just a few levels near to the bottom of the potential
energy well.
E11C.8(b) �e wavenumber of the transition arising from the rotational state J in the P
branch (∆J = −1) of the fundamental transition (υ = 1 ← υ = 0) is given
by [11C.13a–447], ν̃P(J) = ν̃ − 2B̃J. In this case ν̃ = 2648.98 cm−1 and B̃ =
8.465 cm−1 hence
ν̃R(2) = (2648.98 cm−1) − 2 × (8.465 cm−1) × 2 = 2615.1 cm−1