Physical Chemistry , 1st ed.

many possible planes, which suggests that X-ray crystallography is rather com-

plicated. In some cases it is; but Miller indices give us a way to model how dif-

ferent crystal planes can interact with radiation.*

Example 21.7

Assuming that there are atoms at the proper positions in the crystal, deter-

mine the Miller indices of the plane illustrated in Figure 21.21. The fact that

the crystal is not cubic is irrelevant! Notice also that only a single unit cell is

shown.

Solution

We will use the lower left-hand corner as our origin. The indicated plane in-

tercepts the a,b, and ccrystal axes at 1 unit ofa, 1 unit ofb, and ^12 unit ofc.

Taking the reciprocal of the number of unit vectors, we get ^11 ,^11 , and  11 /2as the

Miller indices, or 1, 1, and 2 (respectively). We therefore express the Miller

indices of this crystal plane as (112).

Example 21.8

Cesium chloride, CsCl, has a simple cubic lattice that has a lattice parameter

of 4.11 Å. What wavelength of radiation is primarily diffracted at an angle of

20.0° by the (111) plane of CsCl?

Solution

This is a simple question, but it brings together several ideas. We will apply

Bragg’s law in a straightforward fashion, but first we must determine the dis-

tance, the dspacing, between the parallel (111) planes of CsCl. In order to do

that, we need to use geometry to determine dspacings from the lattice para-

meter. Figure 21.22 shows the sequential steps. From Figure 21.22, we find

that the dspacing between parallel (111) planes is 2.91 Å. Using Bragg’s law:

746 CHAPTER 21 The Solid State: Crystals

Origin

Origin

(100)

(100)

Figure 21.20 This diagram shows how the
(100) plane can be defined using negative Miller
indices.

*For fcc crystals, the (100) planes experience destructive interference of diffracted X rays,

so no net radiation is diffracted. This phenomenon is called systematic extinction.The (110)

and (210) planes in fcc crystals also experience extinction. See Table 21.3 later in this sec-

tion for additional planes that exhibit extinction.

a

c

b

length

Length

1

2

a

d

c

b

(length)

Step 1. length 4.11 Å 2 5.81 Å

Step 2.

Step 3. d one leg of right triangle

(the triangle indicated by bold lines)

1

2 (5.81 Å)^ 2.91 Å^

1

2

Other leg 2.91 Å,

hypotenuse 4.11 Å

Using Pythagorean theorem:

(4.11 Å)^2 (2.91 Å)^2  d^2

8.424 Å^2 d^2

d 2.91 Å

Figure 21.22 Determining the diffraction of X rays using Miller indices. Application of a

little geometry lets us calculate the dspacing from the unit cell parameters. Then we can use the

Bragg equation. See Example 21.8.

Figure 21.21 Atoms in the indicated plane
have what Miller indices? See Example 21.7.