Question

The fraction of the FCC unit cell volume filled with hard sphere is :

• A. $\frac{\pi\sqrt{2}}{5}$
• B. $\frac{\pi\sqrt{2}}{6}$
• C. $\frac{\pi\sqrt{2}}{7}$
• D. $\frac{\pi\sqrt{2}}{8}$

Hint:

The numerical value for FCC packing is $0.74$. It is one of the most efficient packing arrangements, also known as Cubic Close Packing (CCP).

Answer 1

The Correct Option is B

Concept: Packing fraction (P.F.) is the ratio of the volume occupied by atoms to the total volume of the unit cell: \[ \text{P.F.} = \frac{Z \times \frac{4}{3}\pi r^3}{a^3} \] For a Face-Centered Cubic (FCC) lattice:
• Number of atoms per unit cell ($Z$) = 4.
• Relationship between radius ($r$) and edge length ($a$): $4r = a\sqrt{2} \implies r = \frac{a\sqrt{2}}{4}$.

Step 1: Substitute values into the packing fraction formula.
\[ \text{P.F.} = \frac{4 \times \frac{4}{3}\pi \left( \frac{a\sqrt{2}}{4} \right)^3}{a^3} = \frac{\frac{16}{3}\pi \cdot \frac{2\sqrt{2}a^3}{64}}{a^3} \] \[ \text{P.F.} = \frac{16 \cdot 2\sqrt{2} \cdot \pi}{3 \cdot 64} = \frac{32\sqrt{2}\pi}{192} \] \[ \text{P.F.} = \frac{\sqrt{2}\pi}{6} \]