Question

The packing efficiency of the face centred cubic (fcc), body-centred cubic (bcc) and simple primitive cubic (pc) lattices follows the order:
A. fcc > bcc > pc
B. bcc > fcc > pc
C. pc > bcc > fcc
D. bcc > pc > fcc

Answer 1

Hint: We know that each cube has 8 corners, 12 edges, 6 faces, 12 face diagonals and 8 body diagonals. Keeping this information in mind we need to proceed for the comparison.

Step by step answer:
The percentage efficiency of a simple cubic unit cell is:
Suppose,
Length of the unit cell $$\text{= a}$$
Radius of the sphere (atom) $$\text{= r}$$
Total volume of the unit cell $$\text{= }{\text{a}}^{\text{3}}\text{= }{\left(\text{2r}\right)}^{\text{3}}\text{= 8}{\text{r}}^{\text{3}}$$
Number of atoms per unit cell $$\text{= 8 }\times {\displaystyle \frac{\text{1}}{\text{8}}}\text{ = 1}$$
Volume of the atom $$\text{r = }{\displaystyle \frac{\text{1}}{\text{2}\sqrt{\text{2}}\text{a}}}$$$$\text{= }{\displaystyle \frac{\text{4}}{\text{3}}}\pi {\text{r}}^{\text{3}}$$
$$\therefore$$packing fraction $$\text{= }{\displaystyle \frac{\text{Occupied volume }}{\text{Total volume}}}$$
$$\text{= }{\displaystyle \frac{\left(\text{4/3}\right)\pi {\text{r}}^{\text{3}}}{\text{8}{\text{r}}^{\text{3}}}}\text{= 0}\text{.5233}$$
Thus, the percentage of occupied volume or packing efficiency $$\text{= 0}\text{.5233 }\times \text{ 100 = 52}\text{.33 }$$
The percentage efficiency of a body-centred unit cell is:
Suppose,
Length of the unit cell $$\text{= a}$$
Radius of the sphere (atom) $$\text{= r}$$
In this unit cell,
$$\text{a = }{\displaystyle \frac{\text{4}}{\sqrt{\text{3}}}}\times \text{ r}$$
Total volume of the unit cell $$\text{= }{\text{a}}^{\text{3}}\text{= }{\left({\displaystyle \frac{\text{4}}{\sqrt{\text{3}}}}\right)}^{\text{3}}{\text{r}}^{\text{3}}\text{= }{\displaystyle \frac{\text{64}}{\text{3}\sqrt{\text{3}}}}{\text{r}}^{\text{3}}$$
Number of atoms per unit cell $$\text{= 2}$$
Volume of two atoms $$\text{= 2 }\times {\displaystyle \frac{\text{4}}{\text{3}}}\pi {\text{r}}^{\text{3}}$$
Therefore, packing fraction (3D) $$\text{= }{\displaystyle \frac{\text{Occupied volume }}{\text{Total volume}}}$$
$$\text{=}{\displaystyle \frac{\text{2 }\times {\displaystyle \frac{\text{4}}{\text{3}}}\pi {\text{r}}^{\text{3}}}{{\displaystyle \frac{\text{64}}{\text{3}\sqrt{\text{3}}}}{\text{r}}^{\text{3}}}}\text{=0}\text{.68}$$

Thus, the percentage of occupied volume or packing efficiency $$\text{= 68 }$$
The percentage efficiency of a face-centred cubic unit cell:
Suppose,
Length of the unit cell $$\text{= a}$$
Radius of the sphere (atom) $$\text{= r}$$
In this unit cell,
$$\text{a=2}\sqrt{\text{2}}\times \text{r}$$
Total volume of the unit cell $$\text{=}{\text{a}}^{\text{3}}\text{=(2}\sqrt{\text{2}}{\text{)}}^{\text{3}}{\text{r}}^{\text{3}}\text{=16}\sqrt{\text{2}}{\text{r}}^{\text{3}}$$
Number of atoms per unit cell $$\text{= 4}$$
Volume of four atoms $$\text{= 4 }\times {\displaystyle \frac{\text{4}}{\text{3}}}\pi {\text{r}}^{\text{3}}$$
                (This is the occupied volume)
Therefore, packing fraction (3D) $$\text{=}{\displaystyle \frac{\text{Occupied volume }}{\text{ Total volume}}}$$
$$\text{=}{\displaystyle \frac{\text{4 }\times {\displaystyle \frac{\text{4}}{\text{3}}}\pi {\text{r}}^{\text{3}}}{\text{16}\sqrt{\text{2}}{\text{r}}^{\text{3}}}}\text{=0}\text{.7401}$$
Thus, the percentage of occupied volume or packing efficiency $$\text{= 74}\text{.01 }$$
Hence we can see that the packing efficiency goes in to order:
fcc > bcc > pc
So Option A is the correct answer.

Note:

Atomic radius of simple cubic unit cell is: $$\text{r = }{\displaystyle \frac{\text{a}}{\text{2}}}$$

Atomic radius of body-centred unit cell is: $$\text{r = }{\displaystyle \frac{\sqrt{\text{3}}}{\text{4 a}}}$$

Atomic radius of face-centred unit cell is: $$\text{r = }{\displaystyle \frac{\text{1}}{\text{2}\sqrt{\text{2}}\text{a}}}$$