Crystal structure

A crystal is not just a neatly shaped solid. At the microscopic level it has long-range periodic order: a lattice supplies the repeating translations, while a basis (or motif) specifies what is attached to each lattice point.

Lattice, basis and unit cell

The lattice is an abstract periodic set of points. The basis is the atom, ion or group associated with each point. Their combination is the crystal structure. A unit cell is a repeating geometrical choice; it is not necessarily the smallest molecule-like object in the material.

The unit cell reproduces the whole pattern by translation. Different valid cells can describe the same crystal.

Counting particles shared by cells

Atoms drawn on cell boundaries are shared with neighbouring cells. A corner of a cubic cell contributes 1/8, a face-centred atom contributes 1/2, and an atom fully inside contributes 1. This bookkeeping gives 1 atom for simple cubic, 2 for body-centred cubic and 4 for face-centred cubic conventional cells.

Density of a cubic crystal: ρ = ZM/(N_Aa³)

Z is the number of atoms or formula units in the chosen cell.

Boundary atoms are fractions of an atom per cell. Z depends on how many cells share each pictured site.

Coordination and packing

Structure controls how many nearest neighbours surround each atom and how efficiently space is filled. BCC has coordination number 8; FCC has 12. Close packing does not by itself determine every property, but it strongly influences density, slip and mechanical behaviour.

Diffraction reveals periodic spacing

Crystal planes scatter X-rays. Constructive interference occurs when path differences satisfy Bragg’s condition nλ = 2d sin θ. The measured angles therefore encode interplanar spacings, while intensities carry information about where atoms sit within the cell.

A diffraction pattern is not a photograph of atoms: structure is inferred from a reciprocal-space measurement and a model.

Worked examples

1. Atoms in an FCC cell

Solution

8 corners contribute 8×1/8 = 1 and 6 faces contribute 6×1/2 = 3, for Z = 4.

2. Copper density

Solution

With Z=4, M=63.546 g mol⁻¹ and a=361.5 pm = 3.615×10⁻8 cm, ρ ≈ 8.9 g cm⁻³.

3. First-order Bragg angle

Solution

For λ=0.154 nm and d=0.200 nm, sinθ=λ/(2d)=0.385, giving θ ≈ 22.6°.