Deduce the Packing Fraction of BCC and FCC Crystal Structures (with Neat Sketch)

In a crystalline solid, the packing fraction (also called volume fraction) quantifies how efficiently space is filled by atoms modeled as hard spheres. For a chosen unit cell,

• compute the total volume of atoms contained (according to the unit-cell sharing of atoms), then
• divide by the unit-cell volume.

We will deduce the packing fractions for BCC and FCC by relating the lattice parameter $a$ to the atomic (sphere) radius $r$, using the geometry of contacting spheres in the unit cell.

Key concepts:

• packing fraction
• unit cell
• [lattice parameter]{def="Edge length of a cubic unit cell, usually denoted by $a$"}
• [hard-sphere model]{def="Treat atoms as non-overlapping spheres of radius $r$"}
• coordination number

Neat sketch: Unit-cell contact geometry (conceptual)

BCC (body-centered cubic)

In BCC, spheres touch along the body diagonal: from one corner atom to the body-center atom.

A sketch of the body diagonal contact idea:

• Corner atoms touch the body-center atom.
• The body diagonal length equals $4r$.

FCC (face-centered cubic)

In FCC, spheres touch along the face diagonal: from one corner atom to a face-centered atom (within the same face).

• The face diagonal length equals $4r$.

These are the two key geometric relations we will use to deduce $a$ in terms of $r$.

Step 1: Packing fraction definition for cubic crystals

For a cubic unit cell:

$$\phi = \\frac{V_{\\text{atoms in unit cell}}}{V_{\\text{unit cell}}}$$

Unit-cell volume for cubic:

$$V_{\\text{unit cell}} = a^3$$

Total atomic volume depends on how many atoms’ worth of material are inside the unit cell:

$$V_{\\text{atoms}} = N\\cdot \\frac{4}{3}\\pi r^3$$

where $N$ is the effective number of atoms per unit cell.

Key packing-fraction workflow:

1. Determine $N$ for the structure (BCC vs FCC).
2. Determine $a$ in terms of $r$ from the touching-sphere geometry.
3. Substitute into $\phi = \dfrac{N\\frac{4}{3}\pi r^3}{a^3}$.

Step 2: Derive BCC packing fraction

2.1 Atoms per unit cell in BCC

BCC has atoms at:

• 8 corners (each corner atom contributes $1/8$)
• 1 body-centered atom (fully inside)

So,

$$N_{\\text{BCC}} = 8\\left(\\frac{1}{8}\\right) + 1 = 2$$

This is the effective number of atoms per BCC unit cell.

2.2 Geometry: relation between $a$ and $r$ in BCC

In BCC, the body diagonal contains two radii at each end plus two more? More precisely: the body diagonal length spans four radii:

• corner sphere radius $r$
• to body center (distance between centers along diagonal)
• to opposite corner sphere

Thus,

$$\\sqrt{3}a = 4r$$

so

$$a = \\frac{4r}{\\sqrt{3}}$$

2.3 Compute packing fraction

Atomic volume in unit cell:

$$V_{\\text{atoms}} = 2\\cdot \\frac{4}{3}\\pi r^3 = \\frac{8}{3}\\pi r^3$$

Unit cell volume:

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

Therefore,

$$\\phi_{\\text{BCC}} = \\frac{\\frac{8}{3}\\pi r^3}{\\frac{64r^3}{3\\sqrt{3}}} = \\frac{8\\pi}{64}\\cdot \\sqrt{3} = \\frac{\\pi\\sqrt{3}}{8}$$

Final result:

$$\\boxed{\\phi_{\\text{BCC}} = \\frac{\\pi\\sqrt{3}}{8} \\approx 0.680}$$

Step 3: Derive FCC packing fraction

3.1 Atoms per unit cell in FCC

FCC has atoms at:

• 8 corners (each contributes $1/8$)
• 6 faces, with 1 atom per face center (each face atom contributes $1/2$ because it’s shared by 2 unit cells)

Thus,

$$N_{\\text{FCC}} = 8\\left(\\frac{1}{8}\\right) + 6\\left(\\frac{1}{2}\\right) = 1 + 3 = 4$$

3.2 Geometry: relation between $a$ and $r$ in FCC

In FCC, the face diagonal consists of two corner-to-face-centered segments and equals four radii:

$$\\sqrt{2}a = 4r$$

so

$$a = \\frac{4r}{\\sqrt{2}} = 2\\sqrt{2}\\,r$$

3.3 Compute packing fraction

Atomic volume in unit cell:

$$V_{\\text{atoms}} = 4\\cdot \\frac{4}{3}\\pi r^3 = \\frac{16}{3}\\pi r^3$$

Unit cell volume:

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

Therefore,

$$\\phi_{\\text{FCC}} = \\frac{\\frac{16}{3}\\pi r^3}{16\\sqrt{2}\\,r^3} = \\frac{\\pi}{3\\sqrt{2}}$$

Final result:

$$\\boxed{\\phi_{\\text{FCC}} = \\frac{\\pi}{3\\sqrt{2}} \\approx 0.740}$$