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374 11MOLECULAR SPECTROSCOPY
As expected, the relative population of the upper level increases with tempera-
ture.
E11C.6(b) Note: the data in the Exercise are in error, the transitions should be at 2329.91 cm−1,
4631.20 cm−1, and 6903.69 cm−1.
Taking ye = 0 is equivalent to using the terms for the Morse oscillator, which
are given in [11C.8–444], G̃(υ) = (υ + 1
2 )ν̃ − (υ + 1
2 )
2 ν̃xe.�e transition υ ← 0
has wavenumber
∆G̃(υ) = G̃(υ) − G̃(0)
= [(υ + 1
2 )ν̃ − (υ + 1
2 )
2 ν̃xe] − [(0 + 1
2 )ν̃ − (0 + 1
2 )
2 ν̃xe]
= υν̃ − υ(υ + 1)ν̃xe
Data on three transitions are provided, but only two are needed to obtain values
for ν̃ and xe.�e ∆G̃(υ) values for the �rst two transitions are
1← 0 ν̃ − 2ν̃xe = 2329.91 cm−1
2← 0 2ν̃ − 6ν̃xe = 4631.20 cm−1
Multiplying the �rst expression by 3 and subtracting the second gives
3(ν̃ − 2ν̃xe) − (2ν̃ − 6ν̃xe) = ν̃
hence ν̃ = 3 × (2329.91 cm−1) − (4631.20 cm−1) = 2358.5 cm−1
�is value for ν̃ is used in the �rst equation, which is then solved for xe to give
xe = 1
2 − (2329.91 cm−1)/[2 × (2358.53 cm−1)] = 6.07 × 10−3 .
E11C.7(b) �e strategy here is to see if the data can bemodelled by the energy levels of the
Morse oscillator.�is is tested by using the result from [11C.9b–445], ∆G̃υ+1/2 =
ν̃ − 2(υ + 1)xe ν̃, where ∆G̃υ+1/2 = G̃(υ + 1) − G̃(υ). �is implies that a plot
of ∆G̃υ+1/2 against (υ + 1) will have slope −2xe ν̃ and intercept ν̃.�e data are
shown in the table and the plot in Fig. 11.6.
υ G̃υ/cm−1 ∆G̃υ+1/2/cm−1 υ + 1
0 1 144.83 2 230.07 1
1 3 374.90 2 150.61 2
2 5 525.51 2 071.15 3
3 7 596.66 1 991.69 4
4 9 588.35
�e data are a good �t to the line
∆G̃υ+1/2/cm−1 = −79.46 × (υ + 1) + 2 309.5