Question

The packing efficiency (in %) of spheres for a body-centered cubic (bcc) lattice is approximately

• A. 74
• B. 68
• C. 60
• D. 52

Hint:

Packing efficiency for SC (simple cubic) = 52%.
Packing efficiency for BCC = 68%.
Packing efficiency for FCC/HCP = 74% (maximum possible for identical spheres).

Answer 1

The Correct Option is B

Step 1: Relation between atomic radius and unit cell edge in BCC
In a body-centered cubic lattice, atoms touch each other along the body diagonal. The body diagonal has length \( \sqrt{3}a \), where \(a\) is the edge length.
Along the body diagonal, we have 2 atomic radii from corner atoms and 2 radii from the center atom, i.e. total \(4r\). Hence, \[ \sqrt{3}a = 4r ⇒ a = \frac{4r}{\sqrt{3}} \] Step 2: Number of atoms per unit cell in BCC
A BCC unit cell contains 2 atoms (8 corners \(\times \frac{1}{8} = 1\), plus 1 body-centered atom).
Step 3: Volume occupied by atoms in the unit cell
The volume of 1 atom = \(\frac{4}{3}\pi r^3\).
For 2 atoms, total volume of spheres = \[ V_{\text{atoms}} = 2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3 \] Step 4: Volume of the unit cell
\[ V_{\text{cell}} = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}} \] Step 5: Packing efficiency
\[ \eta = \frac{V_{\text{atoms}}}{V_{\text{cell}}} \times 100 = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100 = \frac{\pi \sqrt{3}}{8} \times 100 \] \[ \eta \approx 0.680 \times 100 = 68% \] Hence, the packing efficiency of a BCC lattice is approximately 68%.