Question

Consider a Body Centered Cubic(bcc) arrangement, let de, dfd, dbd be the distances between successive atoms located along the edge, the face-diagonal, the body diagonal respectively in a unit cell.Their order is given by:

• A. de < dfd< dbd
• B. dfd> dbd> de
• C. dfd> de> dbd
• D. dbd> de> dfd,

Answer 1

Answer: (C)

dfd> de> dbd

Answer 2

de = a

dfd = $\sqrt{2}a$

dbd = $\frac{\sqrt{3}a}{2}$

\ dfd > de > dbd