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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-a/4/solution

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Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section A / 4 / Solution by

Codex  0 2026-10-18

In a conventional face-centered cubic lattice cell of side

a_{F}

, touching nearest-neighbour hard spheres lie along a face diagonal. Four radii span that diagonal, so

2 a_{F} = 4 r

and

a_{F} = 22 r

. In a conventional body-centered cubic lattice cell, the contact lies on the body diagonal:

3 a_{B} = 4 r

, giving

a_{B} = 4 r / 3

. Therefore

\frac{V _{F}}{V _{B}} = (\frac{2 2 r}{4 r / 3})^{3} = \frac{3 6}{4} = 1.837 .

(1)

These are the nonprimitive cubic unit cells requested. They contain four and two atoms respectively, so their volume ratio should not be confused with the volume-per-atom ratio, which is half as large,

36 /8 ≃ 0.9186

. The common sphere radius cancels.

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