Question

If the body centered unit cell is assumed to be a cube of edge length 'a' containing spherical particles of radius 'r' then the relation between the diameter d of particle and surface area 'S' of the BCC is:

• A. $S=32d^4/3$
• B. $S=2d^2$
• C. $S=4d^2$
• D. $S=8d^2$

Answer 1

The correct option is D $S=8d^2$

For BCC unit cell the relation between radius of a particle 'r' and edge length of unit cell, a is
$r=\frac{\sqrt{3}}{4}a$
Since,
$\text{Diameter}\left(d\right)=2r=2\times \frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{2}a$

Implying
$d^2=\frac{3}{4}{a}^2$
Therefore,
$4d^2/3=a^2$
Multiplying by 6 on both sides gives
$S=6a^2=8d^2,$
where S is the surface area of the cube
$S=6a^2$.