Question

The ratio of the nearest neighbor atomic distances in body-centered cubic (bcc) and face-centered cubic (fcc) crystals with the same unit cell edge length is

• A. \( \frac{\sqrt{3}}{2} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{2} \)

Hint:

In bcc crystals, the nearest neighbor distance is \( \frac{\sqrt{3}}{2} \) of the edge length, while in fcc crystals, it is \( \frac{1}{\sqrt{2}} \) of the edge length.

Answer 1

The Correct Option is A

Step 1: Analyzing the crystal structures.
The nearest neighbor distance in a bcc crystal is \( \frac{\sqrt{3}}{2} \times \) edge length, and in an fcc crystal, it is \( \frac{1}{\sqrt{2}} \times \) edge length.
Step 2: Conclusion.
Thus, the ratio of the nearest neighbor distances in bcc and fcc is \( \frac{\sqrt{3}}{2} \), corresponding to option (A).