Question
Calculate the packing efficiency for bcc lattice.
✓
Answer
Packing efficiency of metal crystal in body-centred cubic (bcc) lattice:
- Step 1: Radius of sphere (particle):
In bcc unit cell, particles occupy the corners and in addition one particle is at the centre of the cube. The particle at the centre of the cube touches two corner particles along the diagonal of the cube. To obtain radius of the particle (sphere) Pythagoras theorem is applied. For triangle FED, $\angle FED =90^{\circ}$.
$\left.\therefore F D^2=F E^2+E D^2=a^2+a^2=2 a^2 \text { (because } F E=E D=a\right)$
For triangle $A F D, \angle A D F=90^{\circ}$
$\therefore A F^2=A D^2+F D^2$
Substitution of equation (1) into (2) yields
$\left.A F^2=a^2+2 a^2=3 a^2 \text { (because } A D=a\right)$
or $AF =\sqrt{3} a$
From the figure, $A F=4 r$.
Substitution for AF from equation (3) gives
$\sqrt{3} a =4 r \text { and hence, } r =\frac{\sqrt{3}}{4} a$
- Step 2: Volume of sphere:
Volume of sphere particle $=\frac{4}{3} \pi r^3$
Substitution for $r$ from equation (4) gives
Volume of one particle $=\frac{4}{3} \pi \times\left(\frac{\sqrt{3}}{4} a \right)^3=\frac{4}{3} \pi \times \frac{(\sqrt{3})^3}{64} a ^3=\frac{\sqrt{3} \pi a ^3}{16}$
- Step 3: Total volume of particles:
Unit cell bcc contains 2 particles.
Hence, volume occupied by particles in bcc unit cell
$=\frac{2 \sqrt{3} \pi a ^3}{16}=\frac{\sqrt{3} \pi a ^3}{8} \cdots(5)$
- Step 4: Packing efficiency:
$ \text { Packing efficiency }=\frac{\text { Volume occupied by particles in unit cell }}{\text { total volume of unit cell }} \times 100$
$=\frac{\sqrt{3} \pi a ^3}{8 a ^3} \times 100=68 \% $
Thus, $68 \%$ of the total volume in bcc unit lattice is occupied by atoms and $32 \%$ is empty space or void volume.
- Step 1: Radius of sphere (particle):
In bcc unit cell, particles occupy the corners and in addition one particle is at the centre of the cube. The particle at the centre of the cube touches two corner particles along the diagonal of the cube. To obtain radius of the particle (sphere) Pythagoras theorem is applied. For triangle FED, $\angle FED =90^{\circ}$.
$\left.\therefore F D^2=F E^2+E D^2=a^2+a^2=2 a^2 \text { (because } F E=E D=a\right)$
For triangle $A F D, \angle A D F=90^{\circ}$
$\therefore A F^2=A D^2+F D^2$
Substitution of equation (1) into (2) yields
$\left.A F^2=a^2+2 a^2=3 a^2 \text { (because } A D=a\right)$
or $AF =\sqrt{3} a$
From the figure, $A F=4 r$.
Substitution for AF from equation (3) gives
$\sqrt{3} a =4 r \text { and hence, } r =\frac{\sqrt{3}}{4} a$
- Step 2: Volume of sphere:
Volume of sphere particle $=\frac{4}{3} \pi r^3$
Substitution for $r$ from equation (4) gives
Volume of one particle $=\frac{4}{3} \pi \times\left(\frac{\sqrt{3}}{4} a \right)^3=\frac{4}{3} \pi \times \frac{(\sqrt{3})^3}{64} a ^3=\frac{\sqrt{3} \pi a ^3}{16}$
- Step 3: Total volume of particles:
Unit cell bcc contains 2 particles.
Hence, volume occupied by particles in bcc unit cell
$=\frac{2 \sqrt{3} \pi a ^3}{16}=\frac{\sqrt{3} \pi a ^3}{8} \cdots(5)$
- Step 4: Packing efficiency:
$ \text { Packing efficiency }=\frac{\text { Volume occupied by particles in unit cell }}{\text { total volume of unit cell }} \times 100$
$=\frac{\sqrt{3} \pi a ^3}{8 a ^3} \times 100=68 \% $
Thus, $68 \%$ of the total volume in bcc unit lattice is occupied by atoms and $32 \%$ is empty space or void volume.
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