BRAVAIS NETWORKS: CONCEPT, CHARACTERISTICS, EXAMPLES, EXERCISES - PHYSICAL - 2025

The Bravais lattices are all the fourteen dimensional unit cells that can be placed in the atoms of a crystal. These cells consist of a three-dimensional arrangement of points that form a basic structure that is repeated periodically in the three spatial directions.

The origin of this name for basic crystal structures dates back to 1850, when Auguste Bravais demonstrated that there are only 14 possible three-dimensional basic unit cells.

Figure 1. Bravais lattices are the set of 14 unit cells necessary and sufficient to describe any crystalline structure. (wikimedia commons)

The set of 14 Bravais networks are subdivided into seven groups or structures according to the geometry of the cells, these seven groups are:

1- Cubic

2- Tetragonal

3- Orthorhombic

4- Trigonal-Hexagonal

5- Monoclinic

6- Triclinic

7- Trigonal

Each of these structures defines a unit cell, this being the smallest portion that preserves the geometric arrangement of the atoms in the crystal.

Characteristics of Bravais networks

The fourteen Bravais networks, as mentioned above, are subdivided into seven groups. But each of these groups has its unit cells with its characteristic parameters which are:

1- The network parameter (a, b, c)

2- Number of atoms per cell

3- Relationship between network parameter and atomic radius

4- Coordination number

5- Packing factor

6- interstitial spaces

7- By translations along the vectors a, b, c the crystal structure is repeated.

Cubic networks

It consists of the simple or cubic lattice P, face-centered lattice or cubic lattice F, and body-centered lattice or cubic lattice I.

All cubic networks have the three network parameters corresponding to the x, y, z directions of the same value:

a = b = c

Cubic network P

It is convenient to note that atoms are represented by spheres whose centers are at the vertices of the cubic unit cell P.

In the case of the cubic lattice P the number of atoms per cell is 1, because at each vertex only one eighth of the atom is inside the unit cell, so 8 * ⅛ = 1.

The coordination number indicates the number of atoms that are close neighbors in the crystal lattice. In the case of the cubic lattice P the coordination number is 6.

Cubic network I

In this type of network, in addition to the atoms at the vertices of the cube, there is an atom in the center of the cube. So the number of atom per unit cell in the cubic lattice P is 2 atoms.

Figure 2. Body-centered cubic lattice.

Cubic network F

It is the cubic lattice that in addition to the atoms in the vertices has an atom in the center of the face of each cube. The number of atoms per cell is 4, since each of the six face atoms has half inside the cell, that is, 6 * ½ = 3 plus 8 * ⅛ = 1 at the vertices.

Figure 3. Face-centered cubic lattice.

Hexagonal net

In this case the unit cell is a straight prism with a hexagonal base. Hexagonal networks have the three corresponding network parameters fulfilling the following relationship:

a = b ≠ c

The angle between vector a and b being 120º, as shown in the figure. While between vectors a and c, as well as between b and c, right angles are formed.

Figure 4. Hexagonal network.

The number of atoms per cell will be calculated as follows:

- In each of the 2 bases of the hexagonal prism there are 6 atoms in the six vertices. Each of these atoms occupies ⅙ of the unit cell.

- In the center of each of the 2 hexagonal bases there is 1 atom that occupies 1/2 unit cell.

- On the 6 lateral faces of the hexagonal prism there are 3 atoms each occupying ⅔ of the unit cell, and 3 atoms each occupying ⅓ of the volume of the unit cell.

(6 x ⅙) x 2 + ½ x 2 + ⅔ x 3 + ⅓ x 3 = 6

The relationship between the lattice parameters a and b with the atomic radius R under the assumption that all the atoms are of equal radius and are in contact is:

a / R = b / R = 2

Examples

Metals are the main examples of crystalline structures and also the simplest because they generally consist of only one type of atom. But there are other non-metallic compounds that also form crystalline structures, such as diamond, quartz, and many others.

- The iron

Iron has a simple cubic unit cell with lattice or edge parameter a = 0.297 nm. In 1 mm there are 3.48 x 10 ^ 6 unit cells.

- Copper

It has a face-centered cubic crystal structure, made up of only copper atoms.

- Precious gems

Precious gems are crystalline structures of basically the same compound, but with small portions of impurities that are often responsible for their color.

Diamond

It is composed solely of carbon and contains no impurities, which is why it is colorless. Diamond has a cubic (isometric-hexoctahedral) crystal structure and is the hardest known material.

Quartz

It is composed of silica oxide, it is generally colorless or white. Its crystalline structure is trigonal-trapezohedral.

Ruby

Gemstone generally green in color, has a monoclinic structure and is composed of iron-magnesium-calcium silicate.

Topaz

Exercise 1

Find the relationship between the lattice parameter and the atomic radius for a cubic lattice F.

Solution: First, it is assumed that the atoms are represented as spheres all of radius R in "contact" with each other, as shown in the figure. A right triangle is formed in which it is true that:

(4 R) ^ 2 = a ^ 2 + a ^ 2 = 2 a ^ 2

Therefore, the edge-radius relationship is:

a / R = 4 / √2

Exercise 2

Find the relationship between the lattice parameter and the atomic radius for a cubic lattice I (body-centered).

Solution: Atoms are assumed to be represented as spheres all of radius R in "contact" with each other, as shown in the figure.

Two right triangles are formed, one of hypotenuse √2a and the other of hypotenuse √3a as can be proved by using the Pythagorean theorem. From there we have that the relationship between the lattice parameter and the atomic radius for a cubic lattice I (centered in the body) is:

a / R = 4 / √3

Exercise 3

Find the packing factor F for a unit cell of a cubic structure F (face-centered cubic) in which the atoms have radius R and are in "contact".

Solution: The packing factor F is defined as the quotient between the volume occupied by the atoms in the unit cell and the volume of the cell:

F = V atoms / V cell

As demonstrated above, the number of atoms per unit cell in a face-centered cubic lattice is 4, so the packing factor will be:

F = 4 / =…

… 4 / ^ 3 = (√2) π / 6 = 0.74

References

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