 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1/a3/a/ii/solution

For a transition starting at

J

and ending at

J + 2

, the molecule gains

Δ ε = B [(J + 2) (J + 3) - J (J + 1)] = B (4 J + 6) .

Energy conservation therefore gives the scattered-light wavenumbers

ν_{S} (J) = ν_{L} - (4 J + 6) B, J = 0, 1, 2, \dots .

The first shifts below the laser are $6 B, 10 B, 14 B, 18 B$ . Adjacent Stokes lines are separated by $4 B$ , while the first line is

6 B

from the Rayleigh line. The figure uses positive Raman shift

ν_{L} - ν_{S}

; increasing shift means decreasing scattered-light wavenumber.

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