Solid StateHard
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:
Options
A.de < dfd < dbd
B.dfd > dbd > de
C.dfd > de > dbd
D.dbd > de > dfd,
Solution
de = a
dfd = √2a
dbd =
∴ dfd > de > dbd
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