Physics in a nutshell

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# Close Packed Structures: fcc and hcp

## Close Packing of Spheres

### Two Dimensions

One can easily see that the closest packing of spheres in two dimensions is realised by a hexagonal structure: Each sphere is in contact with six neighboured spheres.

### Three Dimensions

In three dimensions one can now go ahead and add another equivalent layer. However, for ideal packing it is necessary to shift this layer with respect to first one such that it just fits into the first layer's gaps.

Now the third layer can be either exactly above the first one or shifted with respect to both the first and the second one. How close-packed structures of spheres can be constructed: In a first layer the spheres are arranged in a hexagonal pattern, each sphere being surrounded by six others (A). Then a second layer with the same structure is added. But this layer is slightly shifted and hence just filling the gaps of the first layer (B). In a third step another equivalent layer is added filling the gaps just as before but now there are two opportunities: Either this layer lies exactly above the first one (A) or it is shifted with respect to both A and B and thus has its own position C.

So there are three relative positions of these layers possible (denoted by A, B and C). These could in principle be combined arbitrarily but in nature the sequences A-B-A-B-A-B and A-B-C-A-B-C are most common. In the following we will see that the lattice that forms the latter one is just the fcc lattice which is one of the 14 Bravais lattices we encountered before. The other one is called hcp (hexagonal close packing) but not a Bravais lattice because the single lattice sites (lattice points) are not completely equivalent! There are two different types of lattice sites which have different environments. Therefore the hcp structure can only be represented as a Bravais lattice if a two-atomic basis is added to each lattice site.

## The fcc Structure

### Conventional Unit Cell

The conventional unit cell is a cube with edge length $a$ and 8 lattice sites at the corners and 6 additional ones at the faces of the cube.

What is the number of lattice points per unit cell>? The 8 vertices are shared by 8 cells and contribute therefore only with $\frac{1}{8}$ each (think of them as extended spheres). The 6 additional points at the faces are shared by two cells and therefore contribute with $\frac{1}{2}$ respectively. So this adds up to a total of \begin{align} n = 8 \cdot \frac{1}{8} + 6 \cdot \frac{1}{2} = 4 \end{align} points per unit cell. (a) The conventional unit cell of the fcc structure (b) The primitive unit cell can be used as well but does not reveal the symmetry of the lattice. (c) The hexagonal structure of the fcc lattice: Three hexagonal layers are stacked while being shifted against each other (type A-B-C-A-B-C).

### Packing Density When the lattice points are inflated gradually, at some point they start to touch each other along the diagonals of the faces of the cube. One can now interpret them as close packed spheres with a radius defined geometrically by ${4r = \sqrt{2}a}$ ${\Leftrightarrow \; r = \frac{\sqrt{2}}{4}a}$.

The packing density $\varrho$ is the ratio of the volume filled by the spherical atoms within a unit cell to the total volume $V_\text{uc}$ of the unit cell. Overall there are $n=4$ atoms per unit cell with a volume of $V_\text{sph} = \frac{4}{3} \pi r^3$ each. Thus for the packing density one obtains \begin{align} \varrho &= \frac{n \cdot V_\text{sph}}{V_\text{uc}} = \frac{ 4 \cdot \frac{4}{3} \pi \cdot \left( \frac{\sqrt{2}}{4} \right)^3 a^3}{a^3} \nonumber \\[1ex] &= \frac{\sqrt{2} \pi}{6} \approx 74\% \end{align} which is the highest possible for spherical objects.

### Coordination Number Each atom has 12 nearest neighbours (the neighboured face atoms) and 6 next-nearest neighbours (located along the vertices of the lattice).

In the fcc structure each atom has $c_1 = 12$ nearest neighbours (coordination number) at a distance of \begin{align} d_{c_1} = 2r = \frac{a}{\sqrt{2}} \approx 0.707a \end{align} and $c_2 = 6$ next-nearest neighbours at a distance of \begin{align} d_{c_2} = a \approx 2.83r \approx 1.415 \, d_{c_1} . \end{align} 

## The hcp Structure (a) Position of the additional atom within the conventional unit cell: $\left( \frac{2}{3}, \frac{1}{3}, \frac{1}{2} \right)$. (b) The hexagonal layers form a close-packed structure.

The hcp structure is characterised by two nested hexagonal lattice that are shifted by the vector $\left( \frac{2}{3},\frac{1}{3},\frac{1}{2} \right)$ (in the conventional unit cell basis) against each other. The undelying lattice is not a Bravais lattice since the individual lattice points are not equivalent with respect to their environments. But it can be looked at as a hexagonal Bravais lattice with a two-atomic basis with the atoms sitting at the positions $\left( 0, 0, 0, \right)$ and $\left( \frac{2}{3},\frac{1}{3},\frac{1}{2} \right)$.

Even though the term hcp is used for all those lattices, a close-packing of equal atoms is only obtained for a certain ratio $\frac{c}{a} = \sqrt{\frac{8}{3}}$ of the lattice constants.

## References

  S. Hunklinger Festkörperphysik De Gruyter 2014 (pp. 69-70)  S. H. Simon Oxford Solid State Basics Oxford 2013 (ch. 12.2.2)  R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch. 1.2.2)  S. Hunklinger Festkörperphysik De Gruyter 2014 (pp. 69-70)  R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch. 1.2.2)  R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch. 1.2.2)  H. Ibach, H. Lüth Solid-State Physics Springer 2009 (p. 31)  R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch 1.2.4)  N. W. Ashcroft, N. D. Mermin Festkörperphysik Oldenbourg 2001 (pp. 97-98)

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