 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2a/section-b/23/c/iii/solution

$k ≃ 5.10 \times 1 0^{2} N m^{- 1}$ and $γ ≃ 0.638 N m^{- 1}$ . With

m = m_{u}

and

ω_{0} = 2 π c ν_{0}

, the dilute-gas stretch gives

k = \frac{m}{2} (2 π c 4161.2 cm^{- 1})^{2} = 510.10 N m^{- 1} .

Use

c

cm s^{- 1}

here. The shifted line is the higher totally symmetric mode; the other molecular stretch remains at

ω_{0}

. To first order,

γ ≃ m (2 π c)^{2} (ν_{+}^{2} - ν_{0}^{2}) = 0.63754 N m^{- 1} .

γ / k = 1.25 \times 1 0^{- 3}

validates the weak-coupling expansion. Inverting the exact eigenvalue instead, with

L = m ω_{+}^{2}

, gives

γ = \frac{L ( L - 2 k )}{2 ( L - k )} = 0.63715 N m^{- 1},

a negligible change at the precision of the model.

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