Crystallography and Crystal Defects. Anthony Kelly. Читать онлайн. Newlib. NEWLIB.NET

Автор: Anthony Kelly
Издательство: John Wiley & Sons Limited
Серия:
Жанр произведения: Физика
Год издания: 0
isbn: 9781119420163
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determine [uvw]. The result is [uvw] = (k1l2k2l1) a + (l1h2l2h1) b + (h1k2h2k1) c.

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

equation

      and so [uvw] = [−5, −5, 5] ≡ images. Likewise, given two directions [u1v1w1] and [u2v2w2], we can obtain the plane (hkl) containing these two directions by solving the simultaneous equations:

      (1.16)equation

      (1.17)equation

      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)equation

      to find that (hkl) = (v1w2v2w1, w1u2w2u1, u1v2u2v1). The method equivalent to Eq. (1.15) is to evaluate the determinant:

      (1.19)equation

      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)equation

      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.

      1.4.1 Translational Symmetry

      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 r joining any two lattice points (or the operation of translational symmetry bringing one lattice point into coincidence with another) can always be written as:

Translation symmetry in a crystal.

      1.4.2 Rotational Symmetry

      If one stood at the point marked H in Figure 1.1a and regarded the surroundings in a particular direction, say that indicated by one of the arrows, then on turning through an angle of 60° = 360°/6 the outlook would be identical. We say an axis of sixfold rotational symmetry passes normal to the paper through the point H. Similarly, at O′ an axis of threefold symmetry passes normal to the paper, since an identical outlook is found after a rotation of 360°/3 = 120°. A crystal possesses an n‐fold axis of rotational symmetry if it coincides with itself upon rotation about the axis of 360°/n = 2π/n radians. In crystals, axes of rotational symmetry with values of n equal to one, two, three, four and six are the only ones found which are compatible with translational symmetry. These correspond to repetition every 360°, 180°, 120°, 90° and 60° and are called monad, diad, triad, tetrad and hexad axes, respectively. The reasons for these limitations on the value of n are explained in Section 1.5.

      1.4.3 Centre of Symmetry

      A