Bragg’s law
An early approach to the analysis of diffraction patterns produced by crystals was to regard a lattice plane as a semi-transparent mirror, and to model a crystal as stacks of reflecting lattice planes of separation d (Fig. 20.19). The model makes it easy to calculate the angle the crystal must make to the incoming beam of X-rays for constructive interference to occur. It has also given rise to the name reflection to denote an intense beam arising from constructive interference. Consider the reflection of two parallel rays of the same wavelength by two adjacent planes of a lattice, as shown in Fig. 20.19. One ray strikes point D on the upper plane but the other ray must travel an additional distance AB before striking the plane immediately below. Similarly, the reflected rays will differ in path length by a distance BC. The net path length difference of the two rays is then , AB+BC=2dsinθ
where θ is the glancing angle. For many glancing angles the path-length difference is not an integer number of wavelengths, and the waves interfere largely destructively. However, when the path-length difference is an integer number of wavelengths (AB+BC=nλ), the reflected waves are in phase and interfere constructively. It follows that a reflection should be observed when the glancing angle satisfies Bragg’s law:
nλ=2dsinθ
Reflections with n = 2, 3,...are called second-order, third-order, and so on; they correspond to path-length differences of 2, 3,...wavelengths. In modern work it is normal to absorb the n into d, to write the Bragg law as , λ=2dsinθ
and to regard the nth-order reflection as arising from the {nh,nk,nl} planes (see Example 20.1). The primary use of Bragg’s law is in the determination of the spacing between the layers in the lattice for, once the angle θ corresponding to a reflection has been determined, d may readily be calculated.
Example 20.2 Using Bragg’s law
A first-order reflection from the {111} planes of a cubic crystal was observed at a glancing angle of 11.2° when Cu(Kα) X-rays of wavelength 154 pm were used. What is the length of the side of the unit cell? Method The separation of the planes can be determined from Bragg’s law. Because the crystal is cubic, the separation is related to the length of the side of the unit cell, a, by eqn 20.2, which may therefore be solved for a.
Answer According to eqn 20.5, the {111} planes responsible for the diffraction have separation , d111 =
The separation of the {111} planes of a cubic lattice of side a is given by eqn 20.2 as , d111 =
Therefore,
Some types of unit cell give characteristic and easily recognizable patterns of lines. For example, in a cubic lattice of unit cell dimension a the spacing is given by eqn 20.2, so the angles at which the {hkl} planes give first-order reflections are given by
sin θ = (h2 +k2+l2)1/2
The reflections are then predicted by substituting the values of h, k, and l: {hkl} {100} {110} {111} {200} {210} {211} {220} {300} {221} {310} ...
h2+k2+l2 1 2 3 4 5 6 8 9 9 10...
Notice that 7 (and 15, . . .) is missing because the sum of the squares of three integers cannot equal 7 (or 15, . . .). Therefore the pattern has absences that are characteristic of the cubic P lattice.
