Question

The atomic packing factor of a diamond cube structure is:

• A. 78%
• B. 68%
• C. 34%
• D. 52%

Hint:

Diamond cubic has 8 atoms per cell with \(r = \tfrac{\sqrt3}{8}a\). It is an open, directionally bonded structure, so its APF is much lower than FCC.

Answer 1

The Correct Option is C

Step 1: The diamond cubic structure has 8 atoms per conventional cubic cell of edge \(a\).

Step 2: Atoms touch along the body diagonal direction between a corner/face atom and a tetrahedral atom. The nearest-neighbour distance is one quarter of the body diagonal: \[2r = \frac{\sqrt{3}}{4}a \;\Rightarrow\; r = \frac{\sqrt{3}}{8}a.\]

Step 3: The packing factor is total atomic volume divided by cell volume: \[\text{APF} = \frac{8\cdot\tfrac{4}{3}\pi r^3}{a^3}.\]

Step 4: Substitute \(r = \tfrac{\sqrt{3}}{8}a\): \[\text{APF} = \frac{32}{3}\pi\left(\frac{\sqrt{3}}{8}\right)^3 = \frac{32}{3}\pi\cdot\frac{3\sqrt{3}}{512} = \frac{\pi\sqrt{3}}{16}.\]

Step 5: Evaluate numerically: \[\text{APF} = \frac{3.1416\times 1.732}{16} \approx 0.34 = 34\%.\]\[\boxed{\text{APF} \approx 34\%}\]