Lattice, basis and FCC packing density

Lattice and basis

• Lattice: a regular, periodic array of imaginary points in space such that every point has exactly the same surroundings. It shows only the geometry of repetition.
• Basis: the atom or group of atoms attached to each lattice point. Repeating the basis at every lattice point builds the real crystal.

$$\text{Crystal structure} = \text{Lattice} + \text{Basis}$$

Example: copper = FCC lattice + basis of one Cu atom; NaCl = FCC lattice + basis of one Na⁺ and one Cl⁻ ion.

FCC unit cell

An atom sits at each of the 8 corners and at the centre of each of the 6 faces. The atoms touch along the face diagonal.

         o------------o
        /|    x      /|
       / |          / |     o = corner atom
      o------------o  |     x = face-centre atom
      |  |  x      |x |
      |x |    x    |  |
      |  o---------|--o
      | /    x     | /
      |/           |/
      o------------o

  One face:   o-------o
              | \   / |    face diagonal
              |   x   |    = R + 2R + R = 4R
              | /   \ |
              o-------o

Let $R$ = atomic radius, $a$ = lattice constant.

Along a face diagonal: $\sqrt{2}\,a = 4R \Rightarrow a = 2\sqrt{2}\,R = 2.828R$.

Body diagonal

$$\text{Body diagonal} = \sqrt{3}\,a = \sqrt{3}\times2\sqrt{2}\,R = 2\sqrt{6}\,R = 4.899R$$

Packing density

Atoms per cell: $N = 8\times\frac{1}{8} + 6\times\frac{1}{2} = 4$.

$$\begin{aligned} \text{APF} &= \frac{N\cdot\frac{4}{3}\pi R^3}{a^3} = \frac{4\times\frac{4}{3}\pi R^3}{16\sqrt{2}\,R^3} \\ &= \frac{\pi}{3\sqrt{2}} = 0.74 \end{aligned}$$

Answer: body diagonal $= 2\sqrt{6}\,R \approx 4.90R$; packing density $= 0.74$ (74 % of the cell volume is filled). FCC is close-packed with coordination number 12; examples are Cu, Al, Ag, Au.