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ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2022/paper-1/1/d/solution

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Past exam of the materials science course of the University of Cambridge / 2022 / Paper 1 / 1 / d / Solution by

Codex  0 2026-10-20

In the tetragonal crystal system, physical squared distances use

a^{2} (Δ x^{2} + Δ y^{2}) + c^{2} Δ z^{2}

. Two Ti–O distances from the corner Ti have length

2 u a

; the other four have length

2 a^{2} (1/2 - u)^{2} + c^{2} /4

. Equating both to

r

gives

a = \frac{r}{2 u}, c = 2 r 1 - \frac{( 1/2 - u ) ^{2}}{u ^{2}} .

Thus, within the equal-bond model,

a = 4.596 \overset{˚}{A}, c = 2.907 \overset{˚}{A} .

The rutile structure remains tetragonal and its octahedral angles remain distorted; these numbers are not a claim about measured ambient-pressure lattice parameters.

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