 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/section-b/20/d/i/solution

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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section B / 20 / d / i / Solution by

Codex  0 2026-10-20

For an axial spin Hamiltonian

H_{D} = D S_{z}^{2}

with

D / k_{B} = - 25

K, the minima are

m_{S} = \pm 5/2

. The highest levels along the reversal ladder are the

m_{S} = \pm 1/2

doublet. Hence

U / k_{B} = ∣ D ∣ {(5/2)^{2} - (1/2)^{2}} = 25 (6) = 150 K,

about 1.25 kJ mol

^{- 1}

or 12.9 meV per molecule. Using

∣ D ∣ S^{2}

would incorrectly put a half-integer spin at a nonexistent

m_{S} = 0

level. Transverse anisotropy and quantum tunnelling of magnetization can create faster relaxation pathways, so this axial ladder barrier need not equal an experimentally fitted effective barrier.

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