Copper FCC: atoms, packing, concentration and density

Copper crystallises in the face centred cubic (FCC) structure: atoms at the 8 corners and at the centres of the 6 faces, touching 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

Given: $R = 0.128$ nm, $M = 63.55$ g/mol, $N_A = 6.022\times10^{23}$ mol⁻¹.

(i) Number of atoms per unit cell

• Corner atoms: each shared by 8 cells, $8\times\frac{1}{8} = 1$
• Face atoms: each shared by 2 cells, $6\times\frac{1}{2} = 3$

$$N = 1 + 3 = 4$$

Lattice constant

Along the face diagonal $\sqrt{2}\,a = 4R$:

$$\begin{aligned} a &= 2\sqrt{2}\,R = 2\sqrt{2}\times0.128 = 0.3620\ \text{nm} = 3.620\times10^{-10}\ \text{m} \\ a^3 &= 4.745\times10^{-29}\ \text{m}^3 \end{aligned}$$

(ii) Packing density

$$\text{APF} = \frac{4\times\frac{4}{3}\pi R^3}{(2\sqrt{2}R)^3} = \frac{\pi}{3\sqrt{2}} = 0.74$$

(iii) Atomic concentration

$$n_{at} = \frac{N}{a^3} = \frac{4}{4.745\times10^{-29}} = 8.43\times10^{28}\ \text{atoms m}^{-3}$$

(iv) Density

$$\begin{aligned} \rho &= \frac{N M}{N_A a^3} = \frac{4\times63.55\times10^{-3}}{6.022\times10^{23}\times4.745\times10^{-29}} \\ &= 8.90\times10^{3}\ \text{kg m}^{-3} = 8.90\ \text{g cm}^{-3} \end{aligned}$$

Answer: (i) 4 atoms per cell; (ii) packing density 0.74; (iii) $8.43\times10^{28}$ m⁻³; (iv) $\rho = 8.90\times10^3$ kg m⁻³ (close to the measured 8.96 g cm⁻³).