Question

For an FCC crystal with lattice parameter "a", the atomic packing factor is :-

• A. 0.74
• B. 0.84
• C. 0.34
• D. 1.04

Hint:

FCC and HCP (Hexagonal Close-Packed) structures both have the maximum possible packing efficiency for spheres of equal size, which is 0.74. If you see FCC or HCP, 0.74 is your go-to answer!

Answer 1

The Correct Option is A

Concept: The Atomic Packing Factor (APF) is a dimensionless quantity that represents the fraction of volume in a crystal structure that is actually occupied by constituent particles (atoms), modeled as hard spheres. \[ APF = \frac{N_{atoms} \times V_{atom}}{V_{unit cell}} \]

Step 1: Determine the number of atoms in an FCC unit cell.
In a Face-Centered Cubic (FCC) lattice:
• Atoms at 8 corners: \( 8 \times \frac{1}{8} = 1 \) atom.
• Atoms at 6 faces: \( 6 \times \frac{1}{2} = 3 \) atoms.
• Total atoms (\(N\)) = 4 atoms per unit cell.

Step 2: Establish the relationship between radius (r) and lattice parameter (a).
In FCC, atoms touch along the face diagonal. \[ \text{Face Diagonal} = \sqrt{a^2 + a^2} = a\sqrt{2} \] Since the diagonal consists of 4 radii: \[ 4r = a\sqrt{2} \quad \Rightarrow \quad r = \frac{a\sqrt{2}}{4} \]

Step 3: Calculate the APF.
\[ APF = \frac{4 \times (\frac{4}{3}\pi r^3)}{a^3} \] Substituting \( r \): \[ APF = \frac{\frac{16}{3}\pi (\frac{a\sqrt{2}}{4})^3}{a^3} = \frac{\frac{16}{3}\pi \frac{2\sqrt{2}a^3}{64}}{a^3} = \frac{\pi\sqrt{2}}{6} \approx 0.74 \]