Question

The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbours. Which of these structures represents the most efficient packing? That is, which one of them packs with the least amount of unused space?

Short answer

The percentage of free space in each of the three cubic lattices are 47.6% in case of simple cubic lattice, 32% in case of FCC, and 26% in case of BCC.

BCC lattice has the most efficient packing.

Step-by-step solution

Definition of cubic lattice

Cubic lattices are of three types: simple cubic, face centered cubic (FCC), and body centered cubic (BCC).

The dfraction of total space occupied by the atoms in a unit cell is called the packing dfraction.

Packing efficiency \( = \dfrac{{{\rm{ Volume occupied by atoms in the unit cell }}}}{{{\rm{ Volume of the unit cell }}}} \times 100\)-.-(1)

\(\% \) free space \( = 100\) - Packing efficiency - - (2)

Determining which of these structures represents the most efficient packing 

1. For a simple cubic lattice:

Number of atoms per unit cell \( = 1\)

Volume occupied by the atoms \( = \dfrac{4}{3}\pi {r^3}\)

Volume of the unit cell \( = 8{r^3}\)

Packing efficiency \( = \dfrac{{\dfrac{4}{3}\pi {r^3}}}{{8{r^3}}} \times 100 = 52.4\% \)

\(\% \)free space \( = 100 - 52.4 = 47.6\% \)

2) For a FCC lattice:

Number of atoms per unit cell \( = 2\)

Volume occupied by the atoms \( = \dfrac{4}{3}\pi {{\rm{r}}^3}\)

Volume of the unit cell \( = \dfrac{{64}}{{3\sqrt 3 }}{r^3}\)

Packing efficiency \( = \dfrac{{2\left( {\dfrac{4}{3}\pi {{\rm{r}}^3}} \right)}}{{(64/3\sqrt 3 ){{\rm{r}}^3}}} \times 100 = 68\% \)

\(\% \)free space \( = 100 - 68 = 32\% \)

3) For a BCC lattice:

Number of atoms per unit cell \( = 4\)

Volume occupied by the atoms \( = \dfrac{4}{3}\pi {r^3}\)

Volume of the unit cell \( = 8\sqrt 8 {r^3}\)

Packing efficiency \( = \dfrac{{4\left( {\dfrac{4}{3}\pi {r^3}} \right)}}{{(8\sqrt 8 ){{\rm{r}}^3}}} \times 100 = 74\% \)

\(\% \)free space \( = 100 - 74 = 26\% \)

Therefore, the BCC lattice has the least free space and thus the most efficient packing.