# Spacing of Planes

Spacing of Planes

The distance between successive lattice planes of the same type is called the spacing of planes or inter-planar distance between the planes. On the basis of this aspect, the lattices may be divided into following classes:

1. Simple cubic lattice: “When one only unit is situated at each corner of the cubic cell, it is known as simple cubic lattice”. In other words, it contains eight points situated at eight corners of a cube. There is no lattice point inside the cube.

2. The body centered cubic (bcc) lattice; “When one unit is situated at each corner and one at the center of a cube, it is known as body centered cubic lattice”. The lattice will consist of nine points.

3. Face centered cubic (fcc) lattice:When one unit is situated at each corner and one at the centre of each of the faces it is known as face centered cubic lattice”. The lattice will consist of  fourteen points.

It must be clearly understood, from the above discussion that it is the points and not the lines which constitute a space lattice. The lines have no actual existence but they are put merely for convenience of representation of shape or dimensions of the unit cell.

On the basis of the above classification of symmetry, the crystals have been divided into seven systems. These can be grouped into 32 classes which, in turn, can be regrouped into 7 crystal systems. These seven systems with the characteristics of their axes (angles and intercepts) along with some examples of each are given in the following table.

 S.No Name of System Axes Angles Bravais Lattices 1. Cubic a =b=c $\alpha = \beta = \gamma =90^0$ Primitive face-centered, Body centered =3 2. Tetragonal A = b $\neq$ c $\alpha = \beta = \gamma =90^0$ Primitive, Body centered = 2 3. Rhombohedral or tigonal a = b = c $\alpha = \beta \neq 90^0$ Primitive =1 4. Orthorhombic or Rhombic $a \neq b \neq c$ $\alpha = \beta = \gamma = 90 ^0$ Primitive, Face-Centered, Body centered, End sentered = 4 5. Monoclinic $a \neq b \neq c$ $\alpha = \gamma = 90^0 \\ \beta \neq 90^0$ Primitive, End-centered = 2 6. Triclinic $a \neq b \neq c$ $\alpha \neq \gamma \neq 90^0$ Primitive =1 7. Hexagonal $a = b \neq c$ $\alpha = \beta = 90^0 \\ \gamma = 120^0$ Primitive =1 Total = 14

These crystal systems differ in length of unit cell edges (a, b and c) and the angles between the unit cell edges.

In cubic and trigonal (rhombohedral) systems, the three unit edges are of equal length but for the rest five systems it is not so. The interfacial angles are all $90^0$ in the cubic, tetragonal and orthorhombic systems but it not so for the rest four systems.

Co-ordination Number: To give a better idea of the structure of crystal and to facilitate the determination of the space occupied by particles the term co-ordination number has been introduced.

“The co-ordination number is defined as the number of equidistant neighbours that an atom has in the given structure”.

For the three cubic lattices the co-ordination number is obtained as follows:

1. Simple cubic lattices: In this case there is one atom at each of the eight corners of the unit cell considering an atom at one corner as the center, it will be found that this atom is surrounded by six equidistant neighbours (atoms) and thus the co-ordination number will be six. If ‘a’ is the side of the unit cell, then the distance between the nearest neighbours shall be equal to ‘a’.

2. Body centered cubic (b.c.c.) lattice: Here the central atom is surrounded by eight equidistant atoms and hence the co-ordination number is eight. The nearest distance between two atoms will be $\sqrt{(3/2)}$.

3. Face centered cubic (f.c.c.) lattice: In this case there are eight atoms at the eight corners of the unit cell and six atoms at the center of six faces. Considering an atom at the face center as origin, it will be found that this face is common to two cubes and there are twelve points surrounding it situated at a distance which is equal to half the face diagonal of the unit cell. Thus the co-ordination number will be twelve and the distance between the two nearest atoms will be $\sqrt{(2)}$.

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