QB_PH_SEC7_021: Crystal Structure & Lattice Dynamics

What is the atomic packing factor (APF) of a Face-Centered Cubic (FCC) crystal composed of identical hard spheres? (Give your answer to two decimal places)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **FCC Geometry & Touching Spheres**: - Face diagonal: $d = \sqrt{2} a = 4 R \implies a = \frac{4 R}{\sqrt{2}} = 2\sqrt{2} R$ - Unit cell volume: $V_{\text{cell}} = a^3 = (2\sqrt{2} R)^3 = 16\sqrt{2} R^3$ 2. **Number of Atoms per Unit Cell**: $N = 8 \times \frac{1}{8} \text{ (corners)} + 6 \times \frac{1}{2} \text{ (faces)} = 1 + 3 = 4$ 3. **Total Sphere Volume**: $V_{\text{spheres}} = 4 \times \left(\frac{4}{3}\pi R^3\right) = \frac{16}{3}\pi R^3$ 4. **Atomic Packing Factor (APF)**: $\text{APF} = \frac{V_{\text{spheres}}}{V_{\text{cell}}} = \frac{\frac{16}{3}\pi R^3}{16\sqrt{2} R^3} = \frac{\pi}{3\sqrt{2}} = \frac{3.14159}{4.24264} \approx 0.74048$ 5. **Official Value**: 0.74 (or 74%).