Читать книгу Solid State Chemistry and its Applications - Anthony R. West - Страница 25

The concept of lattice planes causes considerable confusion because there are two separate ideas which can easily become mixed. Any simple structure, such as a metal or an ionic structure, may, in certain orientations, be regarded as built of layers or planes of atoms stacked to form a 3D structure. These layers are often related in a simple manner to the unit cell; for example, a unit cell face may coincide with a layer of atoms. The reverse is not necessarily true, however, especially in more complex structures and unit cell faces or simple sections through the unit cell often do not coincide with layers of atoms in the crystal. Lattice planes, a concept introduced with Bragg's law of diffraction (Chapter 5), are defined purely from the shape and dimensions of the unit cell. Lattice planes are entirely imaginary and simply provide a reference grid to which the atoms in the crystal structure may be referred. Sometimes, a given set of lattice planes coincides with layers of atoms, but not usually.

Consider the 2D array of lattice points shown in Fig. 1.13(a). This array may be divided into many different sets of rows and for each set there is a characteristic perpendicular distance, d, between pairs of adjacent rows. In three dimensions, these rows become lattice planes and adjacent planes are separated by the interplanar d‐spacing, d. (Bragg's law treats X‐rays as being diffracted from these various sets of lattice planes and the Bragg diffraction angle, θ, for each set is related to the d‐spacing by Bragg's law.)

Lattice planes are labelled by assigning three numbers known as Miller indices to each set. The derivation of Miller indices is illustrated in Fig. 1.13(b) (and those for a hexagonal lattice are shown in Fig. 1.14). The origin of the unit cell is at point 0. Two planes are shown which are parallel and pass obliquely through the unit cell. A third plane in this set must, by definition, pass through the origin. Each of these planes continues out to the surface of the crystal and in so doing cuts through many more unit cells; also, there are many more planes in this set parallel to the two shown, but which do not pass through this particular unit cell.

Figure 1.13 (a) Lattice planes (in projection); (b) derivation of Miller indices.

Figure 1.14 Miller indices for a hexagonal lattice.

In order to assign Miller indices to a set of planes, there are four stages:

1 From the array of lattice points, or the crystal structure, identify the unit cell, choose the origin and label the axes a, b, c and angles α (between b and c), β (between a and c) and γ (between a and b).

2 For a particular set of lattice planes, identify that plane which is adjacent to the one that passes through the origin.

3 Find the intersection of this plane on the three axes of the cell and write these intersections as fractions of the cell edges. The plane in question, Fig. 1.13(b), cuts the x axis at a/2, the y axis at b and the z axis at c/3; the fractional intersections are therefore ½, 1, ⅓.

4 Take reciprocals of these fractions and write the three numbers in parentheses; this gives (213). These three integers, (213), are the Miller indices of the plane and all other planes parallel to it and separated from adjacent planes by the same d‐spacing.

Some other examples are shown in Fig. 1.15. In (a), the shaded plane cuts x, y and z at 1a, ∞b and 1c, i.e. the plane is parallel to b. Taking reciprocals of 1, ∞ and 1 gives us (101) for the Miller indices. A Miller index of 0 means, therefore, that the plane is parallel to that axis. In Fig. 1.15(b), the planes of interest comprise opposite faces of the unit cell. We cannot determine directly the indices of plane 1 as it passes through the origin. Plane 2 has intercepts of 1a, ∞b and ∞ C and Miller indices of (100).

Figure 1.15(c) is similar to (b) but there are now twice as many planes as in (b). To find the Miller indices, consider plane 2, which is the one that is closest to the origin but without passing through it. Its intercepts are ½, ∞ and ∞ and the Miller indices are (200). A Miller index of 2, therefore indicates that the plane cuts the relevant axis at half the cell edge. This illustrates an important point. After taking reciprocals, do not divide through by the highest common factor. A common source of error is to regard, say, the (200) set of planes as those planes interleaved between the (100) planes, to give the sequence (100), (200), (100), (200), (100), …. The correct labelling is shown in Fig. 1.15(d). If extra planes are interleaved between adjacent (100) planes then all planes are labelled as (200).

The general symbol for Miller indices is (hkl). It is not necessary to use commas to separate the three letters or numbers and the indices are enclosed in parentheses, (). The brackets {} are used to indicate sets of planes that are equivalent; for example, the sets (100), (010) and (001) are equivalent in cubic crystals and may be represented collectively as {100}. In the examples chosen so far, all of the Miller indices are either 0 or positive but, as we shall see later, it is also important to define lattice planes that intersect axes in their negative directions. In these cases the relevant index has a bar above the number; thus planes are referred to as ‘bar h, bar k, bar l’.

Figure 1.15 Examples of Miller indices: (a) (101); (b) (100); (c) (200); (d) (h00); (e) indices of directions [210] and [].

To label lattice planes in hexagonal unit cells, a modification to the method is required. For hexagonal unit cells, with a = a ≠ c, there are three possible choices for the two a axes in the basal plane, a ₁ , a ₂ and a ₃, as shown in Fig. 1.14. Let us choose a ₁ and a ₂ to define the unit cell with the origin at O. Next, consider the set of planes shown, which are parallel to c; assign Miller indices in the usual way but in this case, find the intercepts of plane 1 on all three a axes as well as the c axis; since these planes in Fig. 1.14 are parallel to c, plane 1 does not intersect the c axis. This leads to four indices for this set, representing the intercepts on a ₁, a ₂, a ₃ and c, i.e. . The third index is redundant since, writing the indices as (hkil), the condition

holds (i.e. –1 + 3 – 2 = 0). Sometimes, all four indices are specified; sometimes, only three are specified (i.e. in the usual way); sometimes, the third index may be represented as a dot (i.e. ).