 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/iv/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 / iv / Solution by

Codex  0 2026-10-20

The largest applied field gives

\frac{μ E _{field}}{h c} = \frac{( 1.6 ) ( 3.33564 \times 1 0 ^{- 30} ) ( 1 0 ^{4} )}{( 6.62607015 \times 1 0 ^{- 34} ) ( 2.99792458 \times 1 0 ^{10} )} ≃ 2.69 \times 1 0^{- 3} c m^{- 1} .

For parallel polarization, adjacent allowed M values in a

J, K

component are separated by

Δ_{M} ν = \frac{2 ( 2.69 \times 1 0 ^{- 3} ) ∣ K ∣}{J ( J + 1 ) ( J + 2 )} .

Already at

K = 1

this is $2.24 \times 1 0^{- 4} c m^{- 1}$ for $J = 2$ and $8.96 \times 1 0^{- 5} c m^{- 1}$ for $J = 3$ . Both exceed the stated

2 \times 1 0^{- 5} c m^{- 1}

resolution, and larger

∣ K ∣

gives larger first-order spacings. The spectrometer should resolve suitable Stark components, provided their physical linewidths and signal-to-noise permit it. This predicts resolvable splittings rather than incorrectly comparing the maximum level shift directly with instrumental resolution.

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