 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2a/section-b/21/a/ii/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section B / 21 / a / ii / Solution by

Codex  0 2026-10-20

The

K = 0

lines give two equations independent of

D_{J K}

1.50130 = 6 B - 108 D_{J}, 2.00171 = 8 B - 256 D_{J} .

(1)

Solving yields $B ≃ 0.25022 c m^{- 1}$ and $D_{J} ≃ 2.1 \times 1 0^{- 7} c m^{- 1}$ . The small distortion is obtained by subtracting close numbers, so its precision is considerably worse than that of

B

For components in a fixed group,

ν (J, 0) - ν (J, K) = 2 D_{J K} (J + 1) K^{2} .

(2)

For example the

J = 2, K = 2

displacement gives

D_{J K} = 0.00008/24

, while

J = 3, K = 3

gives

0.00024/72

. Both give $D_{J K} ≃ 3.3 \times 1 0^{- 6} c m^{- 1}$ ; the smaller gaps are consistent after rounding. The pure rotational transitions have

Δ K = 0

, so the

(A - B) K^{2}

and

D_{K} K^{4}

terms cancel. Neither $A$ nor $D_{K}$ can be determined from these zero-field lines.

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