Question

Consider a body centered cubic (BCC) arrangement, let $d_e,d_{fd},d_{bd}$ 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. $d_{bd}>d_e>d_{fd}$
• B. $d_e<d_{fd}<d_{bd}$
• C. $d_{fd}>d_{bd}>d_e$
• D. $d_{bd}>d_{fd}>d_e$

Answer 1

The correct option is D $d_{bd}>d_{fd}>d_e$

The edge length of the unit cell is $d_e$. The structure of cubic unit cell is shown below.

The length of face diagonal is $d_{fd}$
Thus, in $△ABC$
$A{C}^2=A{B}^2+B{C}^2{d}_{fd}^2=d_e^2+d_e^2{d}_{fd}^2=2d_e^2{d}_{fd}=d_e\sqrt{2}$

The length of body diagonal is $d_{bd}$
In $△AEF$
$A{E}^2=A{F}^2+E{F}^2{d}_{bd}^2=d_e^2+2d_e^2{d}_{bd}^2=3d_e^2{d}_{bd}=d_e\sqrt{3}$

Thus,
$d_{bd}>d_{fd}>d_e$