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# Atomic packing factor

In crystallography '''atomic packing factor''' (APF) 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 For one-component crystals (those that contain only one type of atom) the APF is represented mathematically by
$mathrm\left\{APF\right\} = frac\left\{N_mathrm\left\{atoms\right\} V_mathrm\left\{atom\right\}\right\}\left\{V_mathrm\left\{crystal\right\}\right\}$

where Natoms is the number of atoms in the crystal Vatom is the volume of an atom and Vcrystal is the volume occupied by the crystal It can be proven mathematically that for one-component structures the most dense arrangement of atoms has an APF of about 074 In reality this number can be higher due to specific intermolecular factors For multiple-component structures the APF can exceed 074

## Worked out example

The primitive unit cell for the body-centered cubic (BCC) crystal structure contains nine atoms: one on each corner of the cube and one atom in the center Because the volume of each corner atom is shared between adjacent cells each BCC cell contains two 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
$a = frac\left\{4r\right\}\left\{sqrt\left\{3\right\}\right\}$

Knowing this and the formula for the volume of a sphere($\left(4/3\right)$pi r3) it becomes possible to calculate the APF as follows:
$mathrm\left\{APF\right\} = frac\left\{N_mathrm\left\{atoms\right\} V_mathrm\left\{atom\right\}\right\}\left\{V_mathrm\left\{crystal\right\}\right\}$

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

$= frac\left\{pisqrt\left\{3\right\}\right\}\left\{8\right\}$

$approx 068!$

For the Hexagonal system|hexagonal close-packed] (HCP) structure the derivation is similar The side length of the hexagon will be denoted as a while the height of the hexagon will be denoted as c Then:
$a = 2 times r$

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

It is then possible to calculate the APF as follows:
$mathrm\left\{APF\right\} = frac\left\{N_mathrm\left\{atoms\right\} V_mathrm\left\{atom\right\}\right\}\left\{V_mathrm\left\{crystal\right\}\right\}$

$= frac\left\{6 \left(4/3\right)pi r^3\right\}\left\{\left[1\right]\left(a^2\right)\left(c\right)\right\}$

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

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

$= frac\left\{pi\right\}\left\{sqrt\left\{18\right\}\right\}$

$approx 074!$

## APF of common structures

By similar procedures the ideal atomic packing factors of all crystal structures can be found The common ones are collected here as reference rounded to the nearest hundredth