Anthony Kelly - Crystallography and Crystal Defects

for a constant of proportionality, ξ . The vectors a*, b* and c* in Eq. (1.8)are termed reciprocal lattice vectors , defined through the equations:

(1.9)

If the normal to the ( hkl ) set of planes is simply taken to be the vector

(1.10)

the magnitude of nis inversely proportional to the spacing of the hkl planes; that is, it is inversely proportional to the distance OP in Figure 1.10(Section A1.2), irrespective of the orientations of the x ‐, y ‐ and z ‐axes with respect to one another.

Furthermore, it is evident that the scalar product of a normal to a set of planes, n, with a vector r= [ uvw ] lying on one of these planes must be zero. That is, r· n= 0. Writing out this dot product explicitly, we obtain the result:

which is the Weiss zone law. This demonstrates that the Weiss zone law is a scalar product between two vectors, one of which lies in one of a set of planes and the other of which is normal to the set of planes.

Given the indices of any two planes, say ( h 1 k 1 l 1) and ( h 2 k 2 l 2), the indices of the zone [ uvw ] in which they lie are found by solving the simultaneous equations:

(1.11)

(1.12)

Since it is only the ratios u : v : w which are of interest, these equations can be solved to give:

(1.13)

There are other methods of producing the same result. For example, we could write down the planes in the form:

We then cross out the first and the last columns and evaluate the 2 × 2 determinants from (i) the second and third columns, (ii) the third and fourth columns, and (iii) the fourth and fifth columns:

(1.14)

Therefore, we find [ uvw ] = [ k 1 l 2− k 2 l 1, l 1 h 2− l 2 h 1, h 1 k 2− h 2 k 1]. A third method is to evaluate the determinant

(1.15)

to determine [ uvw ]. The result is [ uvw ] = ( k 1 l 2− k 2 l 1) a+ ( l 1 h 2− l 2 h 1) b+ ( h 1 k 2− h 2 k 1) c.

Thus, for example, supposing ( h 1 k 1 l 1) = (112) and ( h 2 k 2 l 2) = ( ), we would have:

and so [ uvw ] = [−5, −5, 5] ≡ . Likewise, given two directions [ u 1 v 1 w 1] and [ u 2 v 2 w 2], we can obtain the plane ( hkl ) containing these two directions by solving the simultaneous equations:

(1.16)

(1.17)

Using a similar method to the one used to produce Eq. (1.14), we draw up the three 2 × 2 determinants as follows:

(1.18)

to find that ( hkl ) = ( v 1 w 2− v 2 w 1, w 1 u 2− w 2 u 1, u 1 v 2− u 2 v 1). The method equivalent to Eq. (1.15)is to evaluate the determinant:

(1.19)

It is also evident from Eqs. (1.11)and (1.12), the conditions for two planes ( h 1 k 1 l 1) and ( h 2 k 2 l 2) to lie in the same zone [ uvw ], that by multiplying Eq. (1.11) by a number m and Eq. (1.12) by a number n and adding them, we have:

(1.20)

Therefore the plane ( mh 1+ nh 2, mk 1+ nk 2, ml 1+ nl 2) also lies in [ uvw ]. In other words, the indices formed by taking linear combinations of the indices of two planes in a given zone provide the indices of a further plane in that same zone. In general m and n can be positive or negative. If, however, m and n are both positive, then the normal to the plane under consideration must lie between the normals of ( h 1 k 1 l 1) and ( h 2 k 2 l 2): we will revisit this result in Section 2.2.

The symmetrical arrangement of atoms in crystals is described formally in terms of operations of symmetry. The symmetry arises because an atom or group of atoms is repeated in a regular way to form a pattern. Any operation of repetition can be described in terms of one of the following four different types of pure symmetry element or symmetry operator.

This describes the fact that similar atoms in identical surroundings are repeated at different points within the crystal. Any one of these points can be brought into coincidence with another by an operation of translational symmetry. For instance, in Figure 1.1a the carbon atoms at O, Y, N and Q occupy completely similar positions. We use the idea of the lattice to describe this symmetry. The lattice is a set of points each with an identical environment which can be found by inspection of the crystal structure, as in the example in Figure 1.1b. We can define the arrangement of lattice points in a three‐dimensional crystal by observing that the vector rjoining any two lattice points (or the operation of translational symmetry bringing one lattice point into coincidence with another) can always be written as: