HCP & CCP Unit Cells: Hexagonal vs Cubic Close Packing

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⚛️ The Geometry of Matter:
Atomic Packing in 2‑D & 3‑D

APF · coordination number · FCC, HCP, BCC · Kepler bound
packing efficiency interstitial voids hard‑sphere model senior physics

🔷 Rigid‑sphere model & why geometry matters

In condensed matter physics we treat atoms as rigid, hard spheres. This classical picture ignores electron‑cloud overlap, yet it provides a powerful geometric foundation for density, symmetry, and mechanical properties. For a senior physics student the goal is to move beyond visualisation into quantitative analysis of \(APF\) and coordination environments. The atomic packing factor (APF) is the fraction of volume in a crystal structure occupied by constituent atoms. It is dimensionless and always less than unity.

📐 1. Packing in two dimensions

square packing

Spheres aligned in a grid: each sphere touches four neighbours.

  • coordination number \(CN = 4\)
  • relatively “loose” arrangement – large interstitial space.
  • Area packing fraction = \(\pi/4 \approx 0.785\)

hexagonal close‑packing (2‑D)

Second row nestled into valleys of first row.

  • \(CN = 6\)
  • most efficient packing of circles on a plane: area coverage ≈ 90.7%
\( \text{area packing fraction} = \frac{\pi}{2\sqrt{3}} \approx 0.907\)

📦 2. From layers to 3‑D crystals

Three‑dimensional structures are built by stacking 2‑D layers. The atomic packing factor is:

\[ APF = \frac{N_{\text{atoms}} \times V_{\text{atom}}}{V_{\text{unit cell}}}, \quad V_{\text{atom}} = \frac43 \pi r^3 \]

simple cubic (SC)

Stack square layers directly on top.

  • \(CN = 6\)
  • \(APF \approx 0.52\) (rare – polonium only)

Atoms touch along cube edge: \(a = 2r\). \(N=1\).

body‑centered cubic (BCC)

Corner atoms + one atom at cube centre.

  • \(CN = 8\)
  • \(APF \approx 0.68\) (Fe, W, Cr…)
\(a = \frac{4r}{\sqrt{3}}\)

🌟 3. The close‑packed structures: FCC & HCP

Stacking hexagonal 2‑D layers yields the maximum possible packing density 0.74 (the Kepler conjecture). The two types differ only in stacking sequence.

🔹 HCP (ABAB…)

Third layer directly above first layer. Hexagonal symmetry.
Examples: Mg, Ti, Zn

🔸 FCC (ABCABC…)

Third layer in “C” voids, fourth repeats. Cubic symmetry.
Examples: Au, Cu, Al

📊 4. Quantitative summary (3‑D lattices)

PropertySimple cubicBCCFCCHCP
Coordination number681212
Atoms per cell1246 (effective)
Atomic radius \(r\)\(a/2\)\(\frac{\sqrt{3}a}{4}\)\(\frac{\sqrt{2}a}{4}\)\(a/2\) (ideal \(c/a\))
Packing factor (APF)0.520.680.740.74
ExamplesPoFe, W, CrCu, Al, AuMg, Ti, Zn

🕳️ 5. Interstitial sites: the gaps

Even close‑packed structures have 26% empty space. Smaller “guest” atoms (carbon in steel) occupy interstitial voids.

  • tetrahedral voids (4 atoms) – smaller, more numerous
  • octahedral voids (6 atoms) – larger, accommodate solute atoms

In FCC: 4 octahedral sites & 8 tetrahedral sites per unit cell. Void geometry controls alloy solubility and diffusion.

🔬 6. FCC packing factor – full derivation

Step 1: number of atoms in FCC

8 corners × ⅛ + 6 faces × ½ = 1 + 3 = 4 atoms

Step 2: relation between \(a\) and \(r\)

Atoms touch along face diagonal: \(a\sqrt{2} = 4r\)

\[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}\,r \]

Step 3: volumes

\(V_{\text{atoms}} = 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3\)

\(V_{\text{cell}} = a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}\,r^3\)

Step 4: APF ratio

\[ APF = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2}r^3} = \frac{\pi}{3\sqrt{2}} \approx \frac{3.1416}{4.2426} = 0.74 \]

✅ FCC efficiency 74% – maximum for cubic system.

⚙️ 7. BCC packing factor – full derivation

Step 1: number of atoms in BCC

8 corners × ⅛ + 1 centre = 1 + 1 = 2 atoms

Step 2: relation between \(a\) and \(r\)

Atoms touch along body diagonal: \(a\sqrt{3} = 4r\)

\[ a = \frac{4r}{\sqrt{3}} \]

Step 3: volumes

\(V_{\text{atoms}} = 2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3\)

\(V_{\text{cell}} = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}}\)

Step 4: APF ratio

\[ APF = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} = \frac{8\pi}{64/\sqrt{3}} = \frac{\pi\sqrt{3}}{8} \approx \frac{3.1416\times1.732}{8} = 0.68 \]

✅ BCC packing factor ≈ 0.68

🧊 8. HCP – the “boss fight”

HCP derivation involves the height of the hexagonal prism (\(c/a\) ratio). For ideal close‑packing of spheres, \(c/a = \sqrt{8/3} \approx 1.633\). Then \(APF = \pi/(3\sqrt{2})\) again (same 0.74).

Magnesium and zinc have slight deviations from ideal ratio, affecting ductility.

Number of atoms in HCP unit cell

6 atoms: 12 corner × 1/6 + 2 face centres × 1/2 + 3 interior = 2 + 1 + 3 = 6.

Volume of hexagonal prism = \( \frac{3\sqrt{3}}{2}a^2 c\). With ideal c/a, APF = 0.74.

📌 Kepler conjecture & final word

Atomic packing is the bridge between microscopic symmetry and macroscopic physics. Coordination number dictates bonding energy; packing factor influences density, ductility, and void distribution. These real‑space geometries are the physical constraints behind every solid‑state system. The Kepler conjecture (proved by Hales 1998) confirms that no arrangement of identical spheres exceeds \(\pi/(3\sqrt{2}) \approx 0.74\) packing density – FCC and HCP achieve this fundamental bound.

❓ advanced FAQs

🔺 Why is FCC called “cubic” if layers are hexagonal?

The stacking of hexagonal layers follows an ABCABC sequence that repeats every three layers. The overall symmetry of the 3‑D lattice is face‑centered cubic – the unit cell is a cube with atoms on each face.

🔺 How many tetrahedral voids per atom in FCC?

There are 8 tetrahedral voids per FCC unit cell. Since the cell contains 4 atoms, there are 2 tetrahedral voids per atom. Octahedral voids: 4 per cell → 1 per atom.

🔺 What is the Kepler conjecture?

It states that no arrangement of identical spheres can have a packing density greater than \(\pi/(3\sqrt{2}) \approx 0.74\). Proven by Hales 1998 (formally 2014). FCC and HCP achieve this bound.

🔺 Why do BCC metals (Fe) have higher ductility than HCP (Mg)?

BCC has more slip systems (48) due to lower symmetry and non-close-packed planes, while HCP has limited slip systems (basal planes). Packing affects mechanical behaviour.

✏️ problem example: BCC packing factor (alternative)

Given atomic radius r = 0.124 nm for Fe (BCC), compute lattice constant and APF.

\(a = 4r/\sqrt{3} = 4(0.124)/1.732 = 0.2864 \text{ nm}\). APF = 0.68 as derived.

✍️ Advanced Physics Series – where matter meets geometry

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