Copper FCC: packing, density and Fermi energy

Copper has an FCC unit cell: 4 atoms per cell, atoms touching along the face diagonal, so $\sqrt{2}a = 4R$.

Given: $R = 0.13$ nm, $M = 63.55$ g/mol $= 63.55\times10^{-3}$ kg/mol, $N_A = 6.022\times10^{23}$ mol⁻¹, $h = 6.626\times10^{-34}$ J s, $m_e = 9.1\times10^{-31}$ kg. Copper is monovalent (one free electron per atom).

Lattice constant:

$$a = 2\sqrt{2}\,R = 2\sqrt{2}\times0.13 = 0.3677\ \text{nm} = 3.677\times10^{-10}\ \text{m}$$ $$a^3 = 4.971\times10^{-29}\ \text{m}^3$$

(i) Packing density

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

(ii) Density of copper

$$\begin{aligned} \rho &= \frac{N M}{N_A a^3} = \frac{4\times63.55\times10^{-3}}{6.022\times10^{23}\times4.971\times10^{-29}} \\ &= 8.49\times10^{3}\ \text{kg m}^{-3} = 8.49\ \text{g cm}^{-3} \end{aligned}$$

(iii) Atomic concentration

$$n_{at} = \frac{N}{a^3} = \frac{4}{4.971\times10^{-29}} = 8.05\times10^{28}\ \text{m}^{-3}$$

(iv) Fermi energy

With one conduction electron per atom, electron concentration $n = n_{at} = 8.05\times10^{28}$ m⁻³. At 0 K,

$$\begin{aligned} E_F &= \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} \\ &= \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}}\left(\frac{3\times8.05\times10^{28}}{\pi}\right)^{2/3} \\ &= 6.03\times10^{-38}\times1.807\times10^{19} = 1.09\times10^{-18}\ \text{J} \\ &= \frac{1.09\times10^{-18}}{1.6\times10^{-19}} = 6.81\ \text{eV} \end{aligned}$$

Answer: (i) APF $= 0.74$; (ii) $\rho = 8.49\times10^3$ kg m⁻³ (8.49 g cm⁻³); (iii) $n_{at} = 8.05\times10^{28}$ m⁻³; (iv) $E_F = 1.09\times10^{-18}$ J $= 6.81$ eV.

(With the more accurate $R = 0.128$ nm these become 8.90 g cm⁻³, $8.43\times10^{28}$ m⁻³ and 7.0 eV, close to measured values.)