Calculate packing efficiency of FCC and BCC

Question

Calculate the packing efficiency of (a) Face-Centred Cubic (FCC) and (b) Body-Centred Cubic (BCC) unit cells.

Solution — Step by Step

Packing efficiency is the fraction of the total volume of a unit cell occupied by the atoms (assuming atoms are spheres touching each other):

\text{Packing efficiency} = \frac{\text{Volume occupied by atoms in unit cell}}{\text{Total volume of unit cell}} \times 100\%

Atoms per FCC unit cell:

8 corner atoms × (1/8) = 1 atom
6 face-centred atoms × (1/2) = 3 atoms
Total: 4 atoms per unit cell

Radius-edge length relationship: In FCC, atoms touch along the face diagonal. Face diagonal = $a\sqrt{2}$ = 4r, where $a$ = edge length, $r$ = atomic radius.

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

Volume occupied by 4 atoms:

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

Substitute $r = \frac{a}{2\sqrt{2}}$ :

r^3 = \frac{a^3}{16\sqrt{2}}

V_{atoms} = \frac{16}{3}\pi \times \frac{a^3}{16\sqrt{2}} = \frac{\pi a^3}{3\sqrt{2}}

Volume of unit cell: $V_{cell} = a^3$

\text{Packing efficiency} = \frac{\pi a^3}{3\sqrt{2} \times a^3} = \frac{\pi}{3\sqrt{2}} = \frac{3.14159}{4.2426} \approx 0.7405

\boxed{\text{FCC packing efficiency} = 74.05\%}

Atoms per BCC unit cell:

8 corner atoms × (1/8) = 1 atom
1 body-centred atom × 1 = 1 atom
Total: 2 atoms per unit cell

Radius-edge length relationship: In BCC, atoms touch along the body diagonal. Body diagonal = $a\sqrt{3}$ = 4r.

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

Volume occupied by 2 atoms:

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

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

r^3 = \frac{a^3 \times 3\sqrt{3}}{64}

V_{atoms} = \frac{8}{3}\pi \times \frac{3\sqrt{3}a^3}{64} = \frac{8 \times 3\sqrt{3}\pi a^3}{192} = \frac{\sqrt{3}\pi a^3}{8}

\text{Packing efficiency} = \frac{\sqrt{3}\pi}{8} = \frac{1.7321 \times 3.14159}{8} = \frac{5.441}{8} \approx 0.6802

\boxed{\text{BCC packing efficiency} = 68.02\%}

Summary Table

Structure	Atoms/cell	r vs a	Packing Efficiency
Simple Cubic (SC)	1	$r = a/2$	52.4%
BCC	2	$r = a\sqrt{3}/4$	68.0%
FCC	4	$r = a/2\sqrt{2}$	74.0%
HCP	6	$r = a/2$	74.0%

FCC and HCP both achieve the same maximum packing efficiency of 74% — this is the theoretical maximum for equal sphere packing (Kepler’s conjecture, proved by Hales in 1998).

Why This Works

Higher packing efficiency means atoms fit more closely together → denser material. FCC metals (like copper, aluminium, gold, silver) are denser and more ductile than BCC metals (like iron, chromium, tungsten) partly because of packing differences. BCC metals tend to be harder due to the less close-packed structure, which impedes dislocation movement.

The two numbers to memorise for exams: FCC = 74% and BCC = 68%. The derivation is a 5-mark question in JEE Main and CBSE Class 12. For the derivation, focus on (1) correctly identifying the touching direction, (2) correctly finding atoms per unit cell, (3) the correct radius-edge formula for each structure.

Common Mistake

Students often use the WRONG touching direction for FCC — they use the body diagonal instead of the face diagonal. In FCC, the body-centre position is EMPTY — atoms only touch along the face diagonal. In BCC, the face-centre is empty — atoms only touch along the body diagonal. Using the wrong diagonal gives wrong $r$ vs $a$ relationship and wrong packing efficiency. Draw the unit cell and mark which atoms are touching before starting the calculation.

Question

Solution — Step by Step

Summary Table

Why This Works

Common Mistake

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