Question
Calculate packing efficiency in face$-$centred cubic lattice.
✓
Answer
Step $1 :$ Radius of sphere :
In the unit cell of face$-$centred cubic lattice, there $8$ atoms at $8$ corners and $6$ atoms at $6$ face centres. Consider the face $\text{ABCD}$.
The atoms are in contact along the face diagonal $BD$. Let a be the edge length and $r$, the radius of an atom. Consider a triangle $B C D$.
$BD ^2= BC ^2+ CD ^2$
$ = a ^2+ a ^2=2 a$
$ \therefore BD =\sqrt{2} a$
From figure, $B D=4 r$
$\therefore 4 r =\sqrt{2} a$
$\therefore r =\frac{\sqrt{2}}{4} a=\frac{a}{2 \sqrt{2}}$
$r =\frac{a}{2 \sqrt{2}} \quad$
Step $2 :$ Volume of sphere :
$\text { Volume of one particle } =\frac{4}{3} \pi r^3$
$ =\frac{4 \pi}{3} \times\left(\frac{a}{2 \sqrt{2}}\right)^3$
$ =\frac{4}{3} \pi a^3\left(\frac{1}{2 \sqrt{2}}\right)^3$
$ =\frac{\pi a^3}{12 \sqrt{2}}$
Step $3 : $Total volume of particles :
The unit cell of fee crystal lattice contains 4 particles.
$\therefore \text { Volume occupied by } 4 \text { particles }=4 \times \frac{P i a^3}{12 \sqrt{2}}$
$=\frac{P i a^3}{3 \sqrt{2}}$
Step $4 :$ Packing efficiency :
$\text { Packing efficiency }$
$ =\frac{\text { Volume occupied by particles in unit cell }}{\text { Volume of unit cell }} \times 100$
$ =\frac{\pi a^3 / 3 \sqrt{2}}{a^3} \times 100$
$ =\frac{3.142}{3 \sqrt{2}} \times 100$
$ =74 \% \quad$
$\therefore$ Packing efficiency $=74 \%$
$\therefore$ Percentage of void space $=100-74$
$=26 \%$
Edge length and particle parameters in cubic system
Coordination number and packing efficiency in systems
In the unit cell of face$-$centred cubic lattice, there $8$ atoms at $8$ corners and $6$ atoms at $6$ face centres. Consider the face $\text{ABCD}$.
The atoms are in contact along the face diagonal $BD$. Let a be the edge length and $r$, the radius of an atom. Consider a triangle $B C D$.
$BD ^2= BC ^2+ CD ^2$
$ = a ^2+ a ^2=2 a$
$ \therefore BD =\sqrt{2} a$
From figure, $B D=4 r$
$\therefore 4 r =\sqrt{2} a$
$\therefore r =\frac{\sqrt{2}}{4} a=\frac{a}{2 \sqrt{2}}$
$r =\frac{a}{2 \sqrt{2}} \quad$
Step $2 :$ Volume of sphere :
$\text { Volume of one particle } =\frac{4}{3} \pi r^3$
$ =\frac{4 \pi}{3} \times\left(\frac{a}{2 \sqrt{2}}\right)^3$
$ =\frac{4}{3} \pi a^3\left(\frac{1}{2 \sqrt{2}}\right)^3$
$ =\frac{\pi a^3}{12 \sqrt{2}}$
Step $3 : $Total volume of particles :
The unit cell of fee crystal lattice contains 4 particles.
$\therefore \text { Volume occupied by } 4 \text { particles }=4 \times \frac{P i a^3}{12 \sqrt{2}}$
$=\frac{P i a^3}{3 \sqrt{2}}$
Step $4 :$ Packing efficiency :
$\text { Packing efficiency }$
$ =\frac{\text { Volume occupied by particles in unit cell }}{\text { Volume of unit cell }} \times 100$
$ =\frac{\pi a^3 / 3 \sqrt{2}}{a^3} \times 100$
$ =\frac{3.142}{3 \sqrt{2}} \times 100$
$ =74 \% \quad$
$\therefore$ Packing efficiency $=74 \%$
$\therefore$ Percentage of void space $=100-74$
$=26 \%$
Edge length and particle parameters in cubic system
| Unit cell | Relation between a and r | Volume of one particle | Total volume occupied by particles in unit cell |
| $1.$ scc | $ r=a / 2=$ $ 0.5000 a$ |
$ \pi a^3 / 6=$ $ 0.5237 a^3$ |
$ \pi a^3 / 6=$ $ 0.5237 a^3$ |
| $2.$ bec | $r=\sqrt{3} a / 4$ $=0.4330 a$ |
$ \sqrt{3} \pi a^3 / 16$ $ =0.34 a^3$ |
$ \sqrt{3} \pi a^3 / 8$ $ =0.68 a^3$ |
| $3.$ fcc/ccp | $ r=\sqrt{2} a / 4=$ $ 0.3535 a$ |
$ \pi a^3 / 12 \sqrt{2}$ $ =0.185 a^3$ |
$ \pi a^3 / 3 \sqrt{2}$ $ =0.74 a^3$ |
| Lattice | Coordination number of atoms | Packing efficiency |
| $1.$ scc | $6 :$ four in the same layer, one directly above and one directly below | $52.4\%$ |
| $2.$ bec | $8 :$ four in the layer below and four in the layer above | $68 \%$ |
| $3.$ fcc/ccp | $12 :$ six in its own layer, three above and three below | $74 \%$ |
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