Calculate the efficiency of packing in case of a metal crystal for

(i) Simple cubic

In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

Let the edge length of the cube be ‘a’ and the radius of each particle be r.

So, we can write:

a = 2r

Now, volume of the cubic unit cell = a3

= (2r)3

= 8r3

We know that the number of particles per unit cell is 1.

Therefore, volume of the occupied unit cell $=\frac{4}{3} \pi r^{3}$

Hence, packing efficiency $=\frac{\text { Volume of one particle }}{\text { Volume of cubic unit cell }} \times 100 \%$

$=\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100 \%$

$=\frac{1}{6} \pi \times 100 \%$

$=\frac{1}{6} \times \frac{22}{7} \times 100 \%$

$=52.4 \%$

(ii) Body-centred cubic

It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.

From ΔFED, we have:

$b^{2}=a^{2}+a^{2}$

$\Rightarrow b^{2}=2 a^{2}$

$\Rightarrow b=\sqrt{2} a$

Again, from ΔAFD, we have:

$c^{2}=a^{2}+b^{2}$

$\Rightarrow c^{2}=a^{2}+2 a^{2} \quad\left(\right.$ Since $\left.b^{2}=2 a^{2}\right)$

$\Rightarrow c^{2}=3 a^{2}$

$\Rightarrow c=\sqrt{3} a$

Let the radius of the atom be r.

Length of the body diagonal, c = 4π

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

$\Rightarrow a=\frac{4 r}{\sqrt{3}}$

or, $r=\frac{\sqrt{3} a}{4}$

Volume of the cube, $a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3}$

A body-centred cubic lattice contains 2 atoms.

So, volume of the occupied cubic lattice $=2 \pi \frac{4}{3} r^{3}$

$=\frac{8}{3} \pi r^{3}$

$\therefore$ Packing efficiency $=\frac{\text { Volume occupied by two spheres in the unit cell }}{\text { Total volume of the unit cell }} \times 100 \%$

$=\frac{\frac{8}{3} \pi r^{3}}{\left(\frac{4}{\sqrt{3}} r\right)^{3}} \times 100 \%$

$=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100 \%$

= 68%

(iii) Face-centred cubic

Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.

From ΔABC, we have:

$\mathrm{AC}^{2}=\mathrm{BC}^{2}+\mathrm{AB}^{2}$

$\Rightarrow b^{2}=a^{2}+a^{2}$

$\Rightarrow b^{2}=2 a^{2}$

$\Rightarrow b=\sqrt{2 a}$

Let r be the radius of the atom.

Now, from the figure, it can be observed that:

$b=4 r$

$\Rightarrow \sqrt{2} a=4 r$

$\Rightarrow a=2 \sqrt{2} r$

Now, volume of the cube, $a^{3}=(2 \sqrt{2} r)^{3}$

We know that the number of atoms per unit cell is 4.

So, volume of the occupied unit cell $=4 \pi \frac{4}{3} r^{3}$

$\therefore$ Packing efficiency $=\frac{\text { Volume occupied by four spheres in the unit cell }}{\text { Total volume of the unit cell }} \times 100 \%$

$=\frac{4 \pi \frac{4}{3} r^{3}}{(2 \sqrt{2} r)^{3}} \times 100 \%$

$=\frac{\frac{16}{3} \pi r^{3}}{16 \sqrt{2} r^{3}} \times 100 \%$

= 74 %