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ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2022/4/c/solution

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Past exam of the condensed matter physics course of the University of Oxford / 2022 / 4 / c / Solution by

Codex  0 2026-10-25

Write the fitted inverse molar magnetic susceptibility as

χ_{mol}^{- 1} = s T + b = \frac{T - Θ}{C _{mol}}, C_{mol} = \frac{1}{s}, Θ = - \frac{b}{s} .

The vertical scale is in

1 0^{6} mol m^{- 3}

, not unscaled inverse SI magnetic susceptibility. The source line is approximately

8.8 \times 1 0^{6} mol m^{- 3}

at

100 K

and

19 \times 1 0^{6} mol m^{- 3}

at

300 K

. These readings give

s ≃ 5.1 \times 1 0^{4} mol m^{- 3} K^{- 1}, b ≃ 3.7 \times 1 0^{6} mol m^{- 3} .

Therefore the effective paramagnetic moment and Weiss temperature are

m_{eff} = \frac{1}{( 1.57 \times 1 0 ^{- 6} ) s} ≃ 3.5 - 3.6, Θ ≃ - 70 K .

Equivalently,

μ_{eff} ≃ (3.5 - 3.6) μ_{B}

. An uncertainty of order

10 K

in the graphical Weiss temperature is reasonable; the plotted fit does not justify exact integer parameters. Its positive intercept at

T = 0

requires a negative extrapolated temperature-axis intercept. Thus the dominant interactions have the antiferromagnetic sign in the interpretation of part (b).

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