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 = \(\frac{Volume\;of\;one\;particle}{Volume\;of\;cubic\;unit\;cell}\)x 100%

= \(\frac{\frac{4}{3}\pi r^3}{8r^3}\) x 100%

= \(\frac{1}{6}\pi\)x 100%

= \(\frac{1}{6}\pi \times \frac{22}{7}\)x 100%

= 52.4%

(ii) Body-centred cubic

 figure.png

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,

b² = a² + a²

⇒ b² = 2a²

⇒ b = √2a

Again, from ΔAFD, we have

c² = a² + b²

⇒ c² = a² + 2a²

⇒ c² = 3a²   (Since b² = 2a²)

⇒ c = √3a

Let the radius of the atom be r.

Length of the body diagonal, c = 4π

⇒ √3a = 4r

⇒ a = \(\frac{4r}{\sqrt 3}\)

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

Volume of the cube, a³ = \(\Big(\frac{4r}{\sqrt{3}}\Big)^3\)

A body-centred cubic lattice contains 2 atoms.

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

= \(\frac{8}{3}\pi r^3\)

∴ Packing efficiency = \(\frac{Volume\;occupied\;by\;two\;spheres\;in\;the\;unit\;cell}{Total\;volume\;of\;the\;unit\;cell}\)x 100%

= \(\frac{\frac{8}{3}\pi r^3}{(\frac{4}{\sqrt3})^3}\)x 100%

= \(\frac{\frac{8}{3}\pi r^3}{(\frac{64}{3\sqrt{3}}r)^3}\)x 100%

= 68%

(iii) Face-centred cubic

 figure.png

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

From ΔABC, we have:

AC² = BC² + AB²

⇒ b² = a² + a²

⇒ b² = 2a²

⇒ b = √2a