Читать книгу Crystallography and Crystal Defects - Anthony Kelly - Страница 15

This law expresses the mathematical condition for a vector [uvw] to lie in a plane (hkl). This condition can be determined through elementary vector considerations. Consider the plane (hkl) in Figure 1.10 with the normal to the plane .

Figure 1.10 The plane (hkl) in a crystal making intercepts of a/h, b/k and c/l along the x‐, y‐ and z‐axes, respectively, together with the vector normal to (hkl)

A general vector, r, lying in (hkl) can be expressed as a linear combination of any two vectors lying in this plane, such as and . That is:

(1.3)

for suitable λ and μ. Hence, expressing and in terms of a, b and c, it follows that:

(1.4)

If we re‐express this as r = ua + vb + wc – a general vector [uvw] lying in (hkl) – it follows that:

(1.5)

and so:

(1.6)

which is the condition for a vector [uvw] to lie in the plane (hkl): Weiss zone law.⁵ It is evident from this derivation that it is valid for arbitrary orientations of the x‐, y‐ and z‐axes with respect to one another.

Frequently, a number of important crystal lattice planes all lie in the same zone; that is, they intersect one another in parallel lines. For instance, in Figure 1.9 the planes (100), and (110) are all parallel to the direction [001]. They would be said to lie in the zone [001], since [001] is a common direction lying in all of them. The normals to all of these planes are perpendicular to [001]. This is not an accident – the normals are constrained to be perpendicular to [001] by the Weiss zone law.

To see why, we can make use of elementary vector algebra relationships discussed in Appendix 1, Section A1.1. Consider the plane (hkl) shown in Figure 1.10. The vector normal to this plane, n, must be parallel to the cross product . Hence:

(1.7)

and so, after some straightforward mathematical manipulation, making use of the identities

it is apparent that n is parallel to the vector ha* + kb* + lc*. That is:

(1.8)

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 n is 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 (h₁k₁l₁) and (h₂k₂l₂), 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] = [k₁l₂ − k₂l₁, l₁h₂ − l₂h₁, h₁k₂ − h₂k₁]. A third method is to evaluate the determinant

(1.15)

to determine [uvw]. The result is [uvw] = (k₁l₂ − k₂l₁) a + (l₁h₂ − l₂h₁) b + (h₁k₂ − h₂k₁) c.

Thus, for example, supposing (h₁k₁l₁) = (112) and (h₂k₂l₂) = (), we would have:

and so [uvw] = [−5, −5, 5] ≡ . Likewise, given two directions [u₁v₁w₁] and [u₂v₂w₂], 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) = (v₁w₂ − v₂w₁, w₁u₂ − w₂u₁, u₁v₂ − u₂v₁). 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 (h₁k₁l₁) and (h₂k₂l₂) 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 (mh₁ + nh₂, mk₁ + nk₂, ml₁ + nl₂) 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 (h₁k₁l₁) and (h₂k₂l₂): we will revisit this result in Section 2.2.