Question:medium

Calculate the volume occupied by a particle in fcc unit cell if volume of unit cell is $1.6 \times 10^{-23} \text{ cm}^3$.

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For FCC: Volume of 1 atom $= \frac{0.74 \times \text{Volume of Unit Cell}}{4}$.
Updated On: May 7, 2026
  • $5.44 \times 10^{-24} \text{ cm}^3$
  • $2.96 \times 10^{-24} \text{ cm}^3$
  • $8.37 \times 10^{-24} \text{ cm}^3$
  • $6.15 \times 10^{-24} \text{ cm}^3$
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
In a face-centered cubic (fcc) unit cell, the particles (atoms) occupy exactly 74% of the total volume of the unit cell. This is known as its packing efficiency.
Step 2: Key Formula or Approach:
Packing efficiency of fcc = 0.74.
Number of particles in fcc unit cell (\(Z\)) = 4.
Volume occupied by all particles = \(0.74 \times \text{Volume of unit cell}\).
Volume of one particle = \(\frac{\text{Total volume occupied by particles}}{\text{Number of particles (Z)}}\).
Step 3: Detailed Explanation:
Given volume of unit cell (\(V_{\text{cell}}\)) = \(1.6 \times 10^{-23} \text{ cm}^3\).
Total volume occupied by all 4 particles = \(0.74 \times 1.6 \times 10^{-23} \text{ cm}^3 = 1.184 \times 10^{-23} \text{ cm}^3\).
Volume occupied by one particle = \(\frac{1.184 \times 10^{-23}}{4} \text{ cm}^3 = 0.296 \times 10^{-23} \text{ cm}^3 = 2.96 \times 10^{-24} \text{ cm}^3\).
Step 4: Final Answer:
The volume occupied by a particle is \(2.96 \times 10^{-24} \text{ cm}^3\).
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