# Lattice:Diamond

The diamond lattice, which consists of tetrahedrally-arranged atoms/particles, has simple cubic symmetry.

Along an appropriate axis, the lattice has six-fold symmetry (hexagonal).

## Canonical Diamond

A canonical diamond lattice (single atom/particle type arranged as shown above) has symmetry Fd3m. The atoms occupy the positions (0,0,0;1/4,1/4,1/4)+face-centering.

### Symmetry

• Crystal Family: Cubic
• Crystal System: Cubic
• Bravais Lattice: F (fcc)
• Crystal class: Hexoctahedral
• Point Group: d3m
• Space Group: Fd3m
• Particles per unit cell: $n=8$ • Volume of unit cell: $V_{d}=a^{3}$ • Dimensionality: $d=3$ • Projected d-dimensional volume: $v_{d}=a^{3}$ • Solid angle: $\Omega _{d}=4\pi$ • Nearest-neighbor distance: $d_{nn}={\sqrt {3}}a/4\approx 0.433a$ • Assuming spherical particles of radius R:
• Particle volume fraction: $\phi =32\pi R^{3}/\left(3a^{3}\right)$ • Maximum volume fraction: $\phi _{max}=\pi {\sqrt {3}}/16\approx 0.340$ when $R=a{\sqrt {3}}/8$ ### Structure

The lattice may be thought of as two interpenetrating FCC lattices.

### Particle Positions

There are 18 positions. In total there are 8 particles in the unit cell:

• $8\,\mathrm {corners} \,\times \,{\frac {1}{8}}=1$ • $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6\,\mathrm {faces} \,\times \,{\frac {1}{2}}=3$ • $\left(0,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},0,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},0\right),\left(1,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},1,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},1\right)$ • $4\,\mathrm {internal\,\,tetrahedral\,\,sites} \,\times \,1=4$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+(0,0,0)=\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)=\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},0,{\frac {1}{2}}\right)=\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},{\frac {1}{2}},0\right)=\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)$ ### Examples

#### Elemental

6. Carbon (C) (a = 3.567 Å)
14. Silicon (Si) (a = 5.431 Å)
32. Germanium (Ge) (a = 5.657 Å)
50. Gray Tin (Sn) (a = 6.491 Å)

• TBD

## Diamond-like Two-particle

Also known as zincblende. This is effectively

## Double-filled Diamond-like Two-particle

The diamond lattice includes 8 "tetrahedral holes", with only 4 occupied in a 'normal' diamond structure. A two-particle lattice can be formed by filling all 8 internal holes with the 2nd particle-type, in which case the particles exist in a 1:2 ratio. This is often called a CaF2 lattice.

### Particle Positions

There are 22 positions. In total there are 12 particles in the unit cell:

• $8\,\mathrm {corners} \,\times \,{\frac {1}{8}}=1$ • $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6\,\mathrm {faces} \,\times \,{\frac {1}{2}}=3$ • $\left(0,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},0,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},0\right),\left(1,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},1,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},1\right)$ • $4\,\mathrm {principal\,\,tetrahedral\,\,sites} \,\times \,1=4$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+(0,0,0)=\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)=\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},0,{\frac {1}{2}}\right)=\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},{\frac {1}{2}},0\right)=\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)$ • $4\,\mathrm {tetrahedral\,\,holes} \,\times \,1=4$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-(0,0,0)=\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)=\left({\frac {3}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-\left({\frac {1}{2}},0,{\frac {1}{2}}\right)=\left({\frac {1}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-\left({\frac {1}{2}},{\frac {1}{2}},0\right)=\left({\frac {1}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)$ ### Neighbors

There are 12 particles in the unit cell: 4 of type A, and 8 of type B. The type A particles each connect to 8 particles of type B. The type B particles each connect (tetrahedrally) to 4 particles of type A.

The lattice may also be thought of as having cubes of type B with a connected type A particle at the center. However the extended lattice is not BCC-like, since only every second "BCC cube" has a central particle of type A.

## Double Inter-penetrating Diamond Two-Particle

This lattice is formed from two diamond lattices that inter-penetrate. It is often called the NaTl lattice.

### Particle Positions

There are 36 positions. In total there are 16 particles in the unit cell, 8 of each type.

#### Particle Type A

• $8\,\mathrm {corners} \,\times \,{\frac {1}{8}}=1$ • $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6\,\mathrm {faces} \,\times \,{\frac {1}{2}}=3$ • $\left(0,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},0,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},0\right),\left(1,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},1,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},1\right)$ • $4\,\mathrm {internal\,\,tetrahedral\,\,sites} \,\times \,1=4$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+(0,0,0)=\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)=\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},0,{\frac {1}{2}}\right)=\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},{\frac {1}{2}},0\right)=\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)$ #### Particle Type B

• $12\,\mathrm {edges} \,\times \,{\frac {1}{4}}=3$ • $\left({\frac {1}{2}},0,0\right),\left(0,{\frac {1}{2}},0\right),\left(0,0,{\frac {1}{2}}\right)$ • $\left(1,0,{\frac {1}{2}}\right),\left(1,1,{\frac {1}{2}}\right),\left(0,1,{\frac {1}{2}}\right),\left({\frac {1}{2}},1,0\right),\left(1,{\frac {1}{2}},0\right),\left({\frac {1}{2}},0,1\right),\left(0,{\frac {1}{2}},1\right),\left(1,{\frac {1}{2}},1\right),\left({\frac {1}{2}},1,1\right)$ • $1\,\mathrm {central} \,\times \,1=1$ • $\left({\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}}\right)$ • $4\,\mathrm {internal\,\,tetrahedral\,\,sites} \,\times \,1=4$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-(0,0,0)=\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)=\left({\frac {3}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-\left({\frac {1}{2}},0,{\frac {1}{2}}\right)=\left({\frac {1}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {3}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)-\left({\frac {1}{2}},{\frac {1}{2}},0\right)=\left({\frac {1}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)$ ### Examples

#### Atomics

• Sodium Thallium crystal (NaTl)

## Cristobalite

Cristobalite is a diamond-like lattice formed in silicon dioxide (SiO2). The Si atoms bond tetradrally, each to four O atoms. Each O atom is bonded to 2 Si atoms, effectively acting as a bridge.

### Structure

Can be thought of as a diamond lattice, where the tetrahedral species sit on the positions of the canonical diamond sites, and the two-bonded species sites at the midway point of each bond.

The tetrahedral species sit on the 14 "FCC" sites, plus 4 "internal" diamond sites. The two-bonded species sit on the 16 internal bonds in the unit cell. The tetrahedral species sit at a distance from one another of:

• Next-nearest-neighbor distance: $d_{nnn}={\sqrt {3}}a/4\approx 0.433a$ The different species thus have a distance of half that:

• Nearest-neighbor distance: $d_{nn}={\sqrt {3}}a/8\approx 0.217a$ ### Particle Positions

There are 34 positions (24 in unit cell)

#### Particle Type A (bond tetrahedrally)

There are 18 positions. In total there are 8 particles in the unit cell:

• $8\,\mathrm {corners} \,\times \,{\frac {1}{8}}=1$ • $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$ • $6\,\mathrm {faces} \,\times \,{\frac {1}{2}}=3$ • $\left(0,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},0,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},0\right),\left(1,{\frac {1}{2}},{\frac {1}{2}}\right),\left({\frac {1}{2}},1,{\frac {1}{2}}\right),\left({\frac {1}{2}},{\frac {1}{2}},1\right)$ • $4\,\mathrm {internal\,\,tetrahedral\,\,sites} \,\times \,1=4$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+(0,0,0)=\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)=\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},0,{\frac {1}{2}}\right)=\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)$ • $\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},{\frac {1}{2}},0\right)=\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)$ #### Particle Type B (two-fold bonded)

There are 16 positions (all 16 in the unit cell):

• $4\,\mathrm {level\,\,one} \,\times \,1=4$ • ${\frac {(0,0,0)+\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)}{2}}=\left({\frac {1}{8}},{\frac {1}{8}},{\frac {1}{8}}\right)$ • ${\frac {\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},{\frac {1}{2}},0\right)}{2}}=\left({\frac {3}{8}},{\frac {3}{8}},{\frac {1}{8}}\right)$ • ${\frac {\left({\frac {1}{2}},{\frac {1}{2}},0\right)+\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)}{2}}=\left({\frac {5}{8}},{\frac {5}{8}},{\frac {1}{8}}\right)$ • ${\frac {\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)+(1,1,0)}{2}}=\left({\frac {7}{8}},{\frac {7}{8}},{\frac {1}{8}}\right)$ • $4\,\mathrm {level\,\,two} \,\times \,1=4$ • ${\frac {\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},0,{\frac {1}{2}}\right)}{2}}=\left({\frac {3}{8}},{\frac {1}{8}},{\frac {3}{8}}\right)$ • ${\frac {\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)+\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)}{2}}=\left({\frac {1}{8}},{\frac {3}{8}},{\frac {3}{8}}\right)$ • ${\frac {\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)+\left(1,{\frac {1}{2}},{\frac {1}{2}}\right)}{2}}=\left({\frac {7}{8}},{\frac {5}{8}},{\frac {3}{8}}\right)$ • ${\frac {\left({\frac {3}{4}},{\frac {3}{4}},{\frac {1}{4}}\right)+\left({\frac {1}{2}},1,{\frac {1}{2}}\right)}{2}}=\left({\frac {5}{8}},{\frac {7}{8}},{\frac {3}{8}}\right)$ • $4\,\mathrm {level\,\,three} \,\times \,1=4$ • ${\frac {\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)+\left({\frac {1}{2}},0,{\frac {1}{2}}\right)}{2}}=\left({\frac {5}{8}},{\frac {1}{8}},{\frac {5}{8}}\right)$ • ${\frac {\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)+\left(1,{\frac {1}{2}},{\frac {1}{2}}\right)}{2}}=\left({\frac {7}{8}},{\frac {3}{8}},{\frac {5}{8}}\right)$ • ${\frac {\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)+\left(0,{\frac {1}{2}},{\frac {1}{2}}\right)}{2}}=\left({\frac {1}{8}},{\frac {5}{8}},{\frac {5}{8}}\right)$ • ${\frac {\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)+\left({\frac {1}{2}},1,{\frac {1}{2}}\right)}{2}}=\left({\frac {3}{8}},{\frac {7}{8}},{\frac {5}{8}}\right)$ • $4\,\mathrm {level\,\,four} \,\times \,1=4$ • ${\frac {(1,0,1)+\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)}{2}}=\left({\frac {7}{8}},{\frac {1}{8}},{\frac {7}{8}}\right)$ • ${\frac {\left({\frac {3}{4}},{\frac {1}{4}},{\frac {3}{4}}\right)+\left({\frac {1}{2}},{\frac {1}{2}},1\right)}{2}}=\left({\frac {5}{8}},{\frac {3}{8}},{\frac {7}{8}}\right)$ • ${\frac {\left({\frac {1}{2}},{\frac {1}{2}},1\right)+\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)}{2}}=\left({\frac {3}{8}},{\frac {5}{8}},{\frac {7}{8}}\right)$ • ${\frac {\left({\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}}\right)+(0,1,1)}{2}}=\left({\frac {1}{8}},{\frac {7}{8}},{\frac {7}{8}}\right)$ 