Calculate packing efficiency of BCC and FCC unit cells

Question

Calculate the packing efficiency of: (a) Body-Centred Cubic (BCC) unit cell (b) Face-Centred Cubic (FCC) unit cell

Solution — Step by Step

Packing efficiency = (Volume occupied by atoms in unit cell ÷ Total volume of unit cell) × 100

Part A: BCC Unit Cell

BCC has 1 atom at the body centre (fully inside) + 8 corner atoms each shared by 8 unit cells.

Total atoms = 1 + (8 × 1/8) = 2 atoms

In BCC, atoms touch along the body diagonal. The body diagonal = $a\sqrt{3}$ .

Along this diagonal: 4r = $a\sqrt{3}$ , so $r = \dfrac{a\sqrt{3}}{4}$

Volume of 2 atoms = $2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi \left(\dfrac{a\sqrt{3}}{4}\right)^3$

= \dfrac{8}{3}\pi \times \dfrac{3\sqrt{3}\,a^3}{64} = \dfrac{\sqrt{3}\,\pi\, a^3}{8}

Volume of unit cell = $a^3$

\text{Packing efficiency} = \dfrac{\sqrt{3}\,\pi}{8} \times 100 = \dfrac{1.732 \times 3.14159}{8} \times 100 \approx \mathbf{68\%}

Part B: FCC Unit Cell

FCC has 8 corner atoms (each 1/8 inside) + 6 face-centred atoms (each 1/2 inside).

Total atoms = (8 × 1/8) + (6 × 1/2) = 1 + 3 = 4 atoms

In FCC, atoms touch along the face diagonal. Face diagonal = $a\sqrt{2}$ .

Along this diagonal: 4r = $a\sqrt{2}$ , so $r = \dfrac{a\sqrt{2}}{4} = \dfrac{a}{2\sqrt{2}}$

Volume of 4 atoms = $4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi \left(\dfrac{a}{2\sqrt{2}}\right)^3$

= \dfrac{16}{3}\pi \times \dfrac{a^3}{16\sqrt{2}} = \dfrac{\pi\, a^3}{3\sqrt{2}}

\text{Packing efficiency} = \dfrac{\pi}{3\sqrt{2}} \times 100 = \dfrac{3.14159}{4.2426} \times 100 \approx \mathbf{74\%}

Structure	Atoms/Cell	r in terms of a	Packing Efficiency
BCC	2	$r = \dfrac{a\sqrt{3}}{4}$	68%
FCC (CCP)	4	$r = \dfrac{a}{2\sqrt{2}}$	74%
Simple Cubic	1	$r = \dfrac{a}{2}$	52.4%

Why This Works

The key insight is which direction atoms touch. In BCC, the body centre atom is in contact with all 8 corners — so the body diagonal is the “contact direction.” In FCC, atoms on adjacent face centres touch each other along the face diagonal.

Once we know the contact direction, we can express the radius $r$ in terms of edge length $a$ . After that, it’s straightforward geometry: pack the volume of all atoms in the cell divided by $a^3$ .

FCC wins at 74% because the face-centred arrangement allows atoms to sit in the “gaps” more efficiently — this is why metals like Cu, Al, and Au prefer FCC. It’s the most efficient packing without going to hexagonal close-packing (which is also 74%).

Alternative Method — Using the Formula Directly

If you remember the $r$ relations, you can skip the derivation entirely in an exam.

For BCC: $\text{PE} = \dfrac{2 \times \frac{4}{3}\pi r^3}{a^3}$ . Substitute $a = \dfrac{4r}{\sqrt{3}}$ :

\text{PE} = \dfrac{\frac{8\pi r^3}{3}}{\left(\frac{4r}{\sqrt{3}}\right)^3} = \dfrac{8\pi/3}{64/(3\sqrt{3})} = \dfrac{8\pi \cdot 3\sqrt{3}}{3 \cdot 64} = \dfrac{\pi\sqrt{3}}{8} \approx 68\%

In JEE, memorise the final values — 52% (SC), 68% (BCC), 74% (FCC/HCP). The derivation appeared as a 4-mark question in JEE Main 2021 February Shift 2, but for Single Correct options, direct recall saves 2 minutes.

Common Mistake

Confusing which diagonal to use for each structure.

Students often write 4r = $a\sqrt{2}$ for BCC (the face diagonal) instead of 4r = $a\sqrt{3}$ (the body diagonal). In BCC, the body centre atom touches corner atoms along the body diagonal — there’s no contact along the face. If you use the wrong diagonal, your radius is wrong, and your packing efficiency comes out as ~60%, which matches no standard structure and will cost you the full marks.

Quick check: if BCC → body diagonal ( $\sqrt{3}$ ), FCC → face diagonal ( $\sqrt{2}$ ).