# Diamond Structure

In this article we will have a look at the crystal structure which is formed by many elements of the 4th main group of the periodic table.^{[1]} ^{[2]} Besides carbon these are germanium and silicon which are both very important for semiconductor physics. Remember that the common feature of these elements is the electron configuration of the outer shell:

One could in principle expect that these atoms have a filled s orbital and two half-filled p orbitals. However, it is possible that these orbitals merge and form four new equivalent so-called sp3 hybrid orbitals all being only half-filled. This rearrangement entails initially some energy expense but afterwards the atoms are able to form four very strong covalent bond which compensate this expense by far.

Therefore it is evident that such atoms try to form a three-dimensional structure in which every atom has four uniformly distributed nearest neighbours as binding partners.^{[3]}

## Conventional Unit Cell

How can this structure be classified in our previous classification (14 Bravais lattices)? The structure is not a Bravais lattice by itself because there are two types of lattice points with different environments. But when we choose a proper perspective, we can see that the underlying structure is actually a fcc structure with a two-atomic basis. Thus there are two atoms attached to each fcc lattice point: One located just at the position of the lattice point and one being shifted by the vector ${\left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right) }$. Thereby the number of atoms per conventional unit cell is doubled from 4 to 8.^{[4]} ^{[5]} ^{[6]}

## Packing Density

To calculate the packing density of a crystal structure one thinks of the atoms as inflated spheres (of volume $V_\text{sph}$) which just touch each other, i.e. they cannot be increased any further without overlapping. The packing density $\varrho$ is then defined as the ratio of the volume filled by the spheres to the total volume.

The easiest way to calculate $\varrho$ is to consider the conventional unit cell: There are $n=4$ lattice points per unit cell with $N=2$ atoms sitting on each such lattice point. Neighboured atoms are shifted by a vector of length $d = \sqrt{3} \cdot \frac{a}{4}$. Thus the atoms are assigned a radius of $r = \frac{d}{2}$. Then the packing density reads \begin{align} \varrho &= \frac{n \cdot N \cdot V_\text{sph}}{V_\text{uc}} \nonumber \\ &= \frac{ 2 \cdot 4 \cdot \frac{4}{3} \pi \left( \frac{\sqrt{3}}{8} a \right)^3 }{ a^3} = \frac{\sqrt{3}}{16}\pi \nonumber \\ &\approx 34\% \end{align} with $V_\text{uc} = a^3$ being the volume of the unit cell.

This value is really small compared to the close-packed structures (74%). But even though there are not many neighbours to form bonds with, the diamond structure is very resistant because the few existing bonds are extremely tight.^{[7]} ^{[8]}

## Coordination number

The atoms in the diamond structure have $c_1 = 4$ nearest neighbours (coordination number) at a distance of $d_{c_1} = 2r = \frac{\sqrt{3}}{4}a$ as discussed above and $c_2 = 12$ next-nearest neighbours at the neighboured faces of the cube with a distance of $d_{c_2} = \frac{1}{\sqrt{2}}$.^{[9]}

^{[10]}

## References

[1] | Festkörperphysik Oldenbourg 2001 (pp. 96-97) |

[2] | Einführung in die Festkörperphysik Oldenbourg 2006 (p. 20) |

[3] | Festkörperphysik De Gruyter 2014 (ch. 1.2.8) |

[4] | Festkörperphysik De Gruyter 2014 (ch. 1.2.8) |

[5] | Festkörperphysik Oldenbourg 2001 (pp. 96-97) |

[6] | Einführung in die Festkörperphysik Oldenbourg 2006 (p. 20) |

[7] | Festkörperphysik De Gruyter 2014 (ch. 1.2.8.) |

[8] | Einführung in die Festkörperphysik Oldenbourg 2006 (p. 20) |

[9] | Festkörperphysik Oldenbourg 2001 (pp. 96-97) |

[10] | Einführung in die Festkörperphysik Oldenbourg 2006 (p. 20) |