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# Atomic packing factor

In crystallography, atomic packing factor or packing fraction is the fraction of volume in a crystal structure that is occupied by atoms. It is dimensionless and always less than unity. For practical purposes, the APF of a crystal structure is determined by assuming that atoms are rigid spheres. It is represented mathematically by

$\mathrm{APF} = \frac{N_\mathrm{atoms} V_\mathrm{atom}}{V_\mathrm{crystal}}$

where N is the number of atoms in the crystal and V is the volume. It can be proven mathematically that one-component (one type of atom) close-packed structures, those that have the most dense arrangement of atoms, has an APF of 0.74. In reality, this number can be higher given specific intermolecular factors. For multiple-component structures, the APF can exceed 0.74.

## Worked example

The body-centered cubic crystal structure contains eight atoms on each corner of the cube and one atom in the center. Because the volume of the corner atoms are shared between adjacent cells, each BCC crystal only contains two whole atoms.

Each corner atom touches the center atom. A line that is drawn from one corner of the cube through the center and to the other corner passes through 4r, where r is the radius of an atom. By geometry, the length of the diagonal is a*√3. Therefore, the length of each side of the BCC structure can be related to the radius of the atom by

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

Knowing this and the formula for the volume of a sphere, it becomes possible to calculate the APF.

$\mathrm{APF} = \frac{N_\mathrm{atoms} V_\mathrm{atom}}{V_\mathrm{crystal}}$

$= \frac{2 (4/3)\pi r^3}{(4r/\sqrt{3})^3}$

$= \frac{\pi\sqrt{3}}{8}$

$= 0.68.\,\!$

For the Hexagonal structure (HCP) the derivation is different, since the volume and the area of the faces change. The side length of the hexagon will be denoted as "a" while the height of the hexagon would be denoted as "c"

$a = 2 \times r$

$c = (\sqrt{\frac{2}{3}})(4r).$

Knowing this and the formula for the volume of a sphere, it becomes possible to calculate the APF.

$\mathrm{APF} = \frac{N_\mathrm{atoms} V_\mathrm{atom}}{V_\mathrm{crystal}}$

$= \frac{6 (4/3)\pi r^3}{[(3\sqrt{3})/2](a^2)(c)}$

$= \frac{6 (4/3)\pi r^3}{[(3\sqrt{3})/2](2r)^2(\sqrt{\frac{2}{3}})(4r)}$

$= \frac{6 (4/3)\pi r^3}{[(3\sqrt{3})/2](\sqrt{\frac{2}{3}})(16r^3)}$
$= \frac{\pi}{\sqrt{18}}$

$= 0.74.\,\!$

## APF of common structures

By identical procedures, the ideal atomic packing factors of all crystal structures can be found. The common ones are collected here as reference.

## References

1. Schaffer, Saxena, Antolovich, Sanders, and Warner (1999). The Science and Design of Engineering Materials, Second Edition, New York: WCB/McGraw-Hill, 81-88.
2. Callister, W. (2002). Materials Science and Engineering, Sixth Edition, San Francisco: John Wiley and Sons, 105-114.