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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 383
Finally, note that the harmonic frequency can be expressed as a wavenumber:
ν̃ = ωHO/2πc, hence ω2HO = 4π2c2 ν̃2, and that the rotational constant can also
be expressed as a wavenumber B = cB̃
Ẽ2 = J2(J + 1)2
4(hcB̃)3
ch3(4π2c2 ν̃2)/4π2
= J2(J + 1)2 4B̃
3
ν̃2
If follows that D̃J = 4B̃3/ν̃2, as required.
For completeness it is necessary to consider the term E1 in eqn 11.3 in some
more detail.�is term contains the moment of inertia which, as a result of the
stretching of the bond, is given by I = meffR2c .
E1 =
J 2
2I
= J 2
2meffR2c
From eqn 11.2 Rc = Re/(1 −meffω2/kf). It therefore follows that
= J
2
2meff
(1 −meffω2/kf)2
R2e
If it is assumed that meffω2/kf ≪ 1 (which is the case for typical molecules),
then this term may be ignored and hence
E1 =
J 2
2meffR2e
= J
2
2Ieq
where Ieq is themoment of inertia corresponding to the equilibrium separation
– that is, in the absence of any centrifugal distortion.
P11C.18 According to [11C.15b–448] ν̃R(J) − ν̃P(J) = 4B̃1(J + 1
2 ), and according to
[11C.15a–448] ν̃R(J − 1) − ν̃P(J + 1) = 4B̃0(J + 1
2 ). Hence
ν̃R(1) − ν̃P(1) = 6B̃1 and ν̃R(0) − ν̃P(2) = 6B̃0
�erefore
B̃1 = [(2150.858 cm−1) − (2139.427 cm−1)]/6 = 1.905 cm−1
B̃0 = [(2147.084 cm−1) − (2135.548 cm−1)]/6 = 1.923 cm−1
Because only data from a few lines have been used, high precision is not war-
ranted.