Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A9 / a / i

Table of contents
- Solution i

Solution

The stipulated conventional cell is a square of side

2 a

. Although introduced as a centered rectangular lattice, this equal-side case is a square Bravais lattice viewed with a rotated primitive basis. A corner-to-center separation is

d_{nn} = a^{2} + a^{2} = 2 a,

which is shorter than the

2 a

separation between adjacent conventional corners. Consequently the first contact of equal circles occurs at

2 r = d_{nn}, r = \frac{a}{2} .

The conventional square represents two lattice points: one from its four shared corners and one from its center. Its packing fraction is therefore

η = \frac{2 π r ^{2}}{( 2 a ) ^{2}} = \frac{π}{4} ≃ 0.7854 .

Equivalently, one circle occupies each primitive area

2 a^{2}

. This is square circle packing; the drawing does not define a triangular close-packed lattice.

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