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ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2016/paper-1/1/a/iii/solution

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Past exam of the materials science course of the University of Cambridge / 2016 / Paper 1 / 1 / a / iii / Solution by

Codex  0 2026-10-20

The relevant planes contain

[100]

, so their normals lie in the

b c

plane. The symmetry operations can be checked directly on the height-labelled molecules:

Glide planes of type $c$ , parallel to $(010)$ at $y / b = 1/4, 3/4$ , for example

(x, y, z) ⟼ (x, 1/2 - y, z + 1/2) .

(1)

Reflection reverses the bond slope, and the

c /2

translation carries A to B with the correct height.

Glide planes of type $a$ , parallel to $(001)$ at $z / c = 1/4, 3/4$ , for example

(x, y, z) ⟼ (x + 1/2, y, 1/2 - z) .

(2)

This again exchanges A and B. Combining it with the C-centering translation gives the equivalent description with a

b /2

glide in the same plane.

There are no ordinary mirror planes containing $[100]$ for the illustrated generic bond inclination. Reflection alone would reverse a bond slope without the necessary accompanying translation. A mirror perpendicular to

[100]

is a different symmetry and is outside the requested zone.

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