Miller effect

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. By convention, the APF of a crystal structure is determined by assuming that atoms are rigid spheres. The radius of the spheres is taken to be the maximal value such that the atoms do not overlap. For one-component crystals (those that contain only one type of atom), the APF is represented mathematically by

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

where N_atoms is the number of atoms in the unit cell, V_atom is the volume of an atom, and V_{unit cell} is the volume occupied by the unit cell. It can be proven mathematically that for one-component structures, the most dense arrangement of atoms has an APF of about 0.74. For multiple-component structures, the APF can exceed 0.74.

Worked out example

Body-centered cubic crystal structure

BCC structure

The primitive unit cell for the body-centered cubic (BCC) crystal structure contains several fractions taken from 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

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

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

$\mathrm {APF} ={\frac {N_{\mathrm {atoms} }V_{\mathrm {atom} }}{V_{\mathrm {crystal} }}}={\frac {2(4/3)\pi r^{3}}{(4r/{\sqrt {3}})^{3}}}$

$={\frac {\pi {\sqrt {3}}}{8}}\approx 0.68.\,\!$

Hexagonal close-packed crystal structure

HCP structure

For the hexagonal close-packed (HCP) structure the derivation is similar. Here the unit cell is a hexagonal prism containing six atoms. Let a be the side length of its base and c be its height. Then:

$a=2r$

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

It is then possible to calculate the APF as follows:

$\mathrm {APF} ={\frac {N_{\mathrm {atoms} }\cdot V_{\mathrm {atom} }}{V_{\mathrm {crystal} }}}={\frac {6\cdot (4/3)\pi r^{3}}{[(3{\sqrt {3}})/2](a^{2})(c)}}$

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

$={\frac {\pi }{\sqrt {18}}}\approx 0.74.\,\!$

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.

g৳

• Hexagonal close-packed: 0.74
• Face-centered cubic: 0.74
• Diamond cubic: 0.34

See also

• Crystal

References

1. 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
   
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2. 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
   
   My blog: http://www.primaboinca.com/view_profile.php?userid=5889534