FCC unit cell: body diagonal and packing density

In a face centred cubic (FCC) unit cell there is an atom at each of the 8 corners and one 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    |  |     atoms touch along
      |  o---------|--o     each face diagonal
      | /    x     | /
      |/           |/
      o------------o

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

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

Lattice constant

Along a face diagonal: corner radius + face atom diameter + corner radius.

$$\sqrt{2}\,a = 4R \quad\Rightarrow\quad a = \frac{4R}{\sqrt{2}} = 2\sqrt{2}\,R = 2.828\,R$$

Body diagonal

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

Number of atoms per cell

$$N = 8\times\frac{1}{8} + 6\times\frac{1}{2} = 1 + 3 = 4$$

Packing density

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

Answer: body diagonal $= 2\sqrt{6}\,R = 4.90R$ (with $a = 2\sqrt{2}R$), packing density $= \pi/(3\sqrt{2}) = 0.74$ (74%). FCC is a close-packed structure with coordination number 12. Examples: Cu, Al, Ag, Au, Ni.