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
- $mathrm\{APF\}\; =\; frac\{N\_mathrm\{atoms\}\; V\_mathrm\{atom\}\}\{V\_mathrm\{crystal\}\}$
where
N_{atoms} is the
number of atoms in the
crystal V_{atom} is the
volume of an
atom and
V_{crystal} 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 4
r 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(
$(4/3)$pi r
^{3}) it becomes possible to calculate the APF as follows:
- $mathrm\{APF\}\; =\; frac\{N\_mathrm\{atoms\}\; V\_mathrm\{atom\}\}\{V\_mathrm\{crystal\}\}$
- $=\; frac\{2\; (4/3)pi\; r^3\}\{(4r/sqrt\{3\})^3\}$
- $=\; frac\{pisqrt\{3\}\}\{8\}$
- $approx\; 068!$
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:
- $a\; =\; 2\; times\; r$
- $c\; =\; (sqrt\{frac\{2\}\{3\}\})(4r)$
It is then possible to calculate the APF as follows:
- $mathrm\{APF\}\; =\; frac\{N\_mathrm\{atoms\}\; V\_mathrm\{atom\}\}\{V\_mathrm\{crystal\}\}$
- $=\; frac\{6\; (4/3)pi\; r^3\}\{[1](a^2)(c)\}$
- $=\; frac\{6\; (4/3)pi\; r^3\}\{[2](2r)^2(sqrt\{frac\{2\}\{3\}\})(4r)\}$
- $=\; frac\{6\; (4/3)pi\; r^3\}\{[3](sqrt\{frac\{2\}\{3\}\})(16r^3)\}$
- $=\; frac\{pi\}\{sqrt\{18\}\}$
- $approx\; 074!$
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
See also
References
External links