 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-1/a3/a/iv/solution

A hot band starts from a thermally populated vibrationally excited state, for example

v = 1 \to 2

. It can satisfy

Δ v = 1

and need not have a small transition moment, but its lower-state population is reduced by the Boltzmann distribution:

\frac{N _{1}}{N _{0}} ≃ exp [- (E_{1} - E_{0}) / (k_{B} T)] .

It is therefore usually weak at low temperature and grows relative to the fundamental on heating. Anharmonicity shifts its frequency away from the

0 \to 1

transition; in a polyatomic molecule, excitation in another mode can also produce a hot-band sequence.

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