Question:
The fraction of the total volume occupied by the atoms present in a face centred cubic unit cell is
The fraction of the total volume occupied by the atoms present in a face centred cubic unit cell is
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Packing efficiencies:
\[
\begin{array}{c|c}
\textbf{Structure} & \textbf{Packing Efficiency}
\hline \text{Simple Cubic} & 52.4\%
\text{BCC} & 68\%
\text{FCC/HCP} & 74\% \end{array} \]
\hline \text{Simple Cubic} & 52.4\%
\text{BCC} & 68\%
\text{FCC/HCP} & 74\% \end{array} \]
Updated On: Jul 9, 2026
- \(\dfrac{\pi}{6}\)
- \(\dfrac{\pi}{3\sqrt{2}}\)
- \(\dfrac{\pi\sqrt{3}}{8}\)
- \(2\sqrt{2}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
The fraction of volume occupied by atoms in a unit cell is called the packing efficiency.
For an FCC unit cell,
\[
\boxed{\text{Packing efficiency}=74\%}
\]
or
\[
\boxed{\frac{\pi}{3\sqrt{2}}}
\]
Step 1: Use the FCC relation. Number of atoms per FCC unit cell \[ Z=4 \] Edge length \[ a=2\sqrt{2}\,r \]
Step 2: Calculate the packing fraction. \[ \text{Packing fraction} =\frac{4\times\frac{4}{3}\pi r^3}{a^3} \] Substituting \(a=2\sqrt{2}\,r\), \[ =\frac{16\pi r^3/3}{(2\sqrt2\,r)^3} =\frac{16\pi/3}{16\sqrt2} =\frac{\pi}{3\sqrt2} \]
Step 3: Final conclusion. \[ \boxed{\frac{\pi}{3\sqrt2}} \] Hence, the correct option is \(\boxed{(B)}\).
Step 1: Use the FCC relation. Number of atoms per FCC unit cell \[ Z=4 \] Edge length \[ a=2\sqrt{2}\,r \]
Step 2: Calculate the packing fraction. \[ \text{Packing fraction} =\frac{4\times\frac{4}{3}\pi r^3}{a^3} \] Substituting \(a=2\sqrt{2}\,r\), \[ =\frac{16\pi r^3/3}{(2\sqrt2\,r)^3} =\frac{16\pi/3}{16\sqrt2} =\frac{\pi}{3\sqrt2} \]
Step 3: Final conclusion. \[ \boxed{\frac{\pi}{3\sqrt2}} \] Hence, the correct option is \(\boxed{(B)}\).
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