Question:medium

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

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

Solution and Explanation

Step 1: Understanding the Concept:
In a crystalline solid, particles (atoms, ions, or molecules) occupy a specific fraction of the total unit cell volume. This fraction is known as the packing efficiency.
For a Face-Centered Cubic (fcc) unit cell, the packing efficiency is 74% or 0.74.
Step 2: Key Formula or Approach:
Packing Efficiency = \(\frac{\text{Volume occupied by particles in a unit cell}}{\text{Total volume of unit cell}}\)
For fcc, Packing Efficiency = 0.74.
Number of particles per unit cell (\(Z\)) for fcc = 4.
Volume occupied by one particle = \(\frac{\text{Total volume occupied by all particles in the unit cell}}{Z}\).
Step 3: Detailed Explanation:
Total volume of unit cell = 1.6 \(\times\) 10\(^{-23}\) cm\(^3\).
Total volume occupied by all 4 particles = Packing Efficiency \(\times\) Total unit cell volume.
Total volume occupied = 0.74 \(\times\) 1.6 \(\times\) 10\(^{-23}\) cm\(^3\) = 1.184 \(\times\) 10\(^{-23}\) cm\(^3\).
Since there are 4 particles in an fcc unit cell, the volume of a single particle is:
Volume of one particle = \(\frac{1.184 \times 10^{-23}}{4}\) cm\(^3\).
Volume of one particle = 0.296 \(\times\) 10\(^{-23}\) cm\(^3\) = 2.96 \(\times\) 10\(^{-24}\) cm\(^3\).
Step 4: Final Answer:
The volume occupied by a single particle in the given fcc unit cell is 2.96 \(\times\) 10\(^{-24}\) cm\(^3\).
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