Single Choice

Packing fraction in BCC lattice is

A$$\cfrac { 1 }{ 6 } \pi $$
B$$\cfrac { \sqrt { 2 } }{ 6 } \pi $$
C$$\cfrac { \sqrt { 3 } }{ 8 } \pi $$
Correct Answer
D$$\cfrac { \sqrt { 3 } }{ 2 } \pi $$

Solution

The primitive unit cell for the body-centered cubic crystal structure contains several fractions taken from nine atoms (if the particles in the crystal are atoms): one on each corner of the cube and one atom in the center. Because the volume of each of the eight corner atoms is shared between eight adjacent cells, each BCC cell contains the equivalent volume of two atoms (one central and one on the corner).

Each corner atom touches the center atom. A line that is drawn from one corner of the cube through the center and to the other corner passes through 4r, where r is the radius of an atom. By geometry, the length of the diagonal is $$a\sqrt{3} $$. Therefore, the length of each side of the BCC structure can be related to the radius of the atom by

$$ a = \dfrac{4r}{\sqrt{3}} $$

Packing fraction = $$ \dfrac{number\ of\ atoms \times volume\ of\ atom}{volume-of-cell} $$

$$ \Rightarrow \dfrac{2 \times \frac{4}{3} \times \pi r^{3}}{(\frac{4r}{\sqrt{3}})^{3}} $$

$$ Packing fraction = \dfrac{\pi \sqrt{3}}{8} $$

SIMILAR QUESTIONS

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