Question

Calculate the packing efficiency of a cubic lattice when an atom located at the centre remains in touch with the other two atoms located on the diagonal.

Hint:

In a BCC structure, atoms touch each other along the body diagonal, not along the edge. Always use \(\sqrt{3}a = 4r\) to relate radius and edge length.

Answer 1

Step 1: Understanding the lattice structure.
The atom at the centre touches the corner atoms along the body diagonal. This indicates that the structure is a
Body-Centred Cubic (BCC) lattice. Step 2: Relation between atomic radius and cube edge length.
In a BCC lattice, atoms touch each other along the body diagonal. The length of the body diagonal is: \[ \text{Body diagonal} = \sqrt{3}a \] where \(a\) is the edge length of the cube. The body diagonal passes through two radii from opposite corner atoms and two radii from the centre atom, i.e., a total of \(4r\). \[ \sqrt{3}a = 4r \quad \Rightarrow \quad a = \frac{4r}{\sqrt{3}} \] Step 3: Number of atoms per unit cell.
In BCC: \[ N = 2 \quad \text{atoms per unit cell.} \] Step 4: Volume of atoms in the unit cell.
\[ V_{\text{atoms}} = N \times \frac{4}{3}\pi r^3 = 2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3 \] Step 5: Volume of the unit cell.
\[ V_{\text{cell}} = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}} \] Step 6: Packing efficiency formula.
\[ \text{Packing Efficiency} = \frac{V_{\text{atoms}}}{V_{\text{cell}}} \times 100 \] \[ = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100 \] \[ = \frac{8\pi r^3 \sqrt{3}}{64r^3} \times 100 \] \[ = \frac{\pi \sqrt{3}}{8} \times 100 \approx 68% \] Conclusion:
The packing efficiency of the cubic lattice in this case is approximately: \[ \boxed{68%} \]