Question:medium

What is the volume occupied by particles in $\mathrm{BCC}$ structure if 'a' is edge length of unit cell?

Show Hint

To verify your answers quickly during exams, remember that the Packing Efficiency of a lattice is defined as $\frac{\text{Occupied Volume}}{\text{Total Cell Volume } (a^3)}$. Since the packing efficiency of a $\mathrm{BCC}$ lattice is roughly $68\%$, checking which option yields $\approx 0.68a^3$ leads you directly to the correct option: $\frac{\sqrt{3}\pi}{8} \approx \frac{1.732 \times 3.1416}{8} \approx 0.68$.
Updated On: Jun 11, 2026
  • $\frac{\sqrt{3}\pi a^3}{8}$
  • $\frac{\pi a^3}{3\sqrt{2}}$
  • $\frac{\pi a^3}{12\sqrt{2}}$
  • $\frac{\sqrt{3}\pi a^3}{16}$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: State the goal cleanly.
We want the total volume of the spheres that actually live inside one body centred cubic unit cell, written using the edge $a$.
Step 2: Recall the count of atoms in BCC.
A BCC cell owns $8$ corner atoms shared as $\tfrac{1}{8}$ each plus one whole body atom, so $Z = 8\times\tfrac18 + 1 = 2$.
Step 3: Relate radius to edge using the body diagonal.
In BCC the atoms touch along the body diagonal of length $\sqrt{3}\,a$, which equals four radii, so $\sqrt{3}\,a = 4r$ giving $r = \dfrac{\sqrt{3}\,a}{4}$.
Step 4: Write the occupied volume as a product.
Occupied volume $= Z \times \dfrac{4}{3}\pi r^3 = 2 \times \dfrac{4}{3}\pi r^3$.
Step 5: Substitute the radius.
\[ 2 \times \frac{4}{3}\pi \left(\frac{\sqrt{3}\,a}{4}\right)^3 = 2 \times \frac{4}{3}\pi \cdot \frac{3\sqrt{3}\,a^3}{64} \]
Step 6: Simplify to the final form.
\[ = 2 \times \frac{\sqrt{3}\,\pi a^3}{16} = \frac{\sqrt{3}\,\pi a^3}{8} \]
So the spheres fill $\dfrac{\sqrt{3}\,\pi a^3}{8}$ of the cell, option (A).
\[ \boxed{\dfrac{\sqrt{3}\,\pi a^3}{8}\ \text{(option A)}} \]
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