Question:medium

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

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For quick questions on cubic unit cells:
• fcc packing efficiency = \(74%\)
• bcc packing efficiency = \(68%\)
• simple cubic packing efficiency = \(52.4%\)
Updated On: May 14, 2026
  • \(4.088 \times 10^{-23} \text{ cm}^3\)
  • \(2.156 \times 10^{-23} \text{ cm}^3\)
  • \(1.184 \times 10^{-23} \text{ cm}^3\)
  • \(3.226 \times 10^{-23} \text{ cm}^3\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The packing efficiency of a crystal lattice is the fraction of the total volume of the unit cell that is actually occupied by the constituent particles (atoms, ions, or molecules).
For a Face-Centered Cubic (FCC) lattice, the atoms are packed very efficiently.
Step 2: Key Formula or Approach:
The formula for the volume occupied by particles is \(\text{Occupied Volume} = \text{Packing Fraction} \times \text{Total Volume}\).
For an fcc unit cell, the packing fraction is \(0.74\).
This means that \(74%\) of the total volume of the unit cell is occupied by the spheres, and the remaining \(26%\) is empty space.
Step 3: Detailed Explanation:
We are given the total volume of the unit cell:
\(\text{Volume}_{\text{unit cell}} = 1.6 \times 10^{-23} \text{ cm}^3\).
Using the packing fraction for FCC (\(0.74\)), we calculate the occupied volume:
\[ \text{Volume occupied} = 0.74 \times (1.6 \times 10^{-23} \text{ cm}^3) \] \[ \text{Volume occupied} = 1.184 \times 10^{-23} \text{ cm}^3 \] Step 4: Final Answer:
The volume occupied by all particles in the given FCC unit cell is \(1.184 \times 10^{-23} \text{ cm}^3\).
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