Diffraction Principles

In the usual geometric “derivation” of Bragg’s Law one assumes that crystalline

arrays of atoms diffract X-rays just as the regularly etched lines of a grating diffract light.

While this quickly yields the desired Bragg expression, it presents an analogy that is

inadequate to the task of explaining the results of an X-ray diffraction measurement. A

somewhat better physical analogy of X-ray diffraction by the atoms of crystals might go

as follows:

Suppose we had a large, perfectly still, lake on which water waves were able to travel

without the attenuation that we normally expect. Further suppose that at one edge of the

lake we had a large number of posts that emerge from the water to form periodic array

(like a very large dock). The posts might be of various sizes, some as large as the pilings

of a pier, some no bigger than broomsticks. But whatever assortment of posts, they are

arranged in an extended periodic array as they extend out into the lake. If a series of

waves impinge on this array, there will be pattern of reflected waves that emerge and

propagate out into the lake. We can imagine that, by studying the pattern of the reflected

waves, we could work out the pattern of posts underneath the dock. The situation

described parallels the problem that faces a crystallographer when s/he hopes to

determine the structure of a crystal using diffraction techniques; the reader may want to

bear this analogy in mind when thinking about diffraction from crystals.

An even better physical analogy of the diffraction phenomenon is that offered by

optical transforms, which we shall see demonstrated in class.

In handling the elastic scattering of X-rays by matter, we will use a classical wave

description of the X-ray radiation. The incident electromagnetic radiation is

characterized by specifying its wavelength, ! , and its direction of propagation:

E = E

0exp 2! i k "r #$t( ){ }; H = H

0exp 2! i k "r #$t( ){ } ,

where E and H are the oscillating electric and

magnetic fields to which some point, r , is

subject as the wave travels in the k direction.

If the wave’s electric field oscillates with

frequency ν = c/λ in the x direction, the

magnetic field oscillates in the y direction,

then the wavevector, k, points in the z

direction (E0 = E0^x ; H0 = H0

^y; k = k z). In

this expression, the magnitude of the

wavevector, |k| = k, is related to the wavelength, ! , by the relation k = 1/λ .

When an X-ray is elastically scattered by an atom, it emerges with the same

wavelength. We say that an X-ray with wavevector k is scattered to wavevector k ′, but

k ′ = k = 1/λ . However, the scattered photon is emitted as a kind of spherical wave but

with an amplitude that decreases to some extent as the angle by which it is diffracted

increases. Representing the direction of propagation of the wavefronts by arrows, we

will refer the angle with respect to the direction of the incoming diffracted wave as 2θ :

2!

Every atom has a scattering power that scales with the number of electrons of the

atom (or ion) and which falls off with scattering angle (2θ) in a characteristic manner as a

function of wavelength, as illustrated in the figure below for the chlorine atom and ion:

E

H

z

x

y

In the absence of the angle dependence of the

scattering, the scattered X-ray would propagate

outward like a circular wave in water:

If we include the angle dependence, there is

stronger scattering in the forward direction (in the

direction of the incoming X-ray, 2θ = 0), the 2-

dimensional analog would look something like the

figure at right.

X-ray diffraction for structure determination is

concerned with interference effects that result from scattering by periodic, crystalline

arrays of atoms. To see how this interference arises, first consider the interference effect

caused by two scatterers, separated by a vector d. In the immediately following figure,

2θ dependence of the atomic scat-tering represented in 2 dimensions. The green-tipped, red arrow represents the incident X-ray propagation direction.

we see a pictorial representation of the wavefronts of an incident X-ray as they pass two

scatterers at successively later times. In 1, the first wavefront has passed the first

scatterer and the scattered wave is shown propagating outward from that scatterer. In 2,

we see the situation just as the first wavefront reaches the second scatterer and scattered

waves continue to emanate from the first scatterer as successive wavefronts pass its

location.

k

! k

k

1 2

3 4

d d

d

d

In 3, time is moved forward by 1/ ν seconds (=λ/c) so that the second wavefront is just

reaching the second scatterer. Finally, in 4, we see the interference pattern that is being

built up by the wave being scattered off the two scatterers.

With the diagram below we analyze the interference phenomenon, where some vector

algebra is used to specify the terms that concern us. k points in the direction of the

incoming wave; k′ points in a direction for which we want to know the scattered

intensity:

k =k

k— a unit vector in the k direction

— a unit vector in the k direction!ˆ!k!k =!k

ˆ ˆpath length difference = d•(k – k )!

d

k

d•!k

ˆd•k

2"

2"

The amplitude of the scattered wave is f1(2θ) + f2(2θ) × (phase shift factor), where

f1(2θ) + f2(2θ) are just the scattering amplitudes of atoms 1 and 2, respectively, and the

“phase shift factor” depends on the path length difference, d•(^k –

^k´). For completely

constructive interference in the scattered wave, the difference in the length of the path

that it traverses as it is scattered by the two scatterers is an integral multiple of the

wavelength:

d ! k " #k( ) = n$ where n = 1,2,3,…

or d ! k " #k( ) = n since k = 1 $

More generally, the amplitude of the scattered wave, A(2θ), is

A(2!) " f1(2!) + f

2(2!) # exp(2$ i(k % &k ) 'd) =

f1(2!) + f

2(2!) #

1 when (k % &k ) 'd = n

%1 when (k % &k ) 'd = n +1 2

()*+

,-.+

The intensity of the scattered waves is given by the square of the amplitude, I(2θ) =

|A(2θ)|2 = A*A. Note that if λ is large compared to d there can be no completely

constructive interference; in order to get significant interference effect the wavelength of

the incident radiation should be comparable to or smaller than the distances between the

scattering atoms.

The translational symmetry of a crystal allows us to quickly generalize these results.

In a crystal, the position of every atom is related to the position of an equivalent atom

( f1(2!) = f 2(2! ) = f 3(2!) =!) in another unit cell via translation by a lattice vector,

Ri . Thus, if we wish to specify the conditions on k and k′ so that all of a set of a

crystal’s equivalent atoms will scatter X-rays with constructive interference, we have:

Ri ! k " # k ( ) = n for all the lattice vectors Ri

or e2$i k" # k ( )!Ri = 1

We have found that the vectors k – k′ are just the reciprocal lattice vectors. A simple

way to state the condition for constructive interference (diffraction) is to say that the X-

ray will be diffracted by a reciprocal lattice vector:

! k = k + Khkl .

The importance of the reciprocal lattice in

interpreting diffraction patterns is clear; the

direction of every diffracted beam (the so-called

reflections) can be determined from the above

equation. We will have more to say about the

geometrical relationships that relate the

reflections to the lattice below.

k

k

Khkl

k´

Of course, most interesting materials possess more than one atom per unit cell. To

see how the effect of having more than one kind of scatterer, let us consider the case of

crystal with two atoms, A and B, in each unit cell. For convenience we will choose the

origin of the cell such that it is at the position of atom A, and we let d be the vector that

joins A and B within the cell (see the figure below). In general, the atoms A and B will

differ in their ability to scatter X-rays and the atoms’ scattering amplitudes will depend

on the angle, 2! , through which the beam is scattered (2! is just the angle between k

and k′ + Khkl), the number of electrons on atoms A and B, and the distribution of electron

density. These atomic scattering strengths are labeled as

�

fA (2!) and fB (2!) - these are

called the atomic form factors.

d

A

B

d

A

B

d

A

B

d

A

B

d

A

B

d

A

B

A

d

A

B

d

A

B

A A A

B BB

The amplitude of the scattered wave, Fhkl , associated with a given reciprocal lattice

vector, Khkl, is

Fhkl ! f A(2") + f B(2") # exp(2$iKhkl %d)

If there are N atoms in the unit cell, this expression, which is known as the geometric

structure factor, is generalized to

�

Fhkl ! f j (2") # exp(2$iK hkl %d j )j

N

& ;

where fj (2θ) is the form factor for the jth atom and dj specifies the position of jth atom.

While this expression may seem intimidating, its use is straightforward in practice.

There exists a useful geometrical construction aid in understanding how one can

predict when a crystal’s orientation is such that the diffraction condition, Khkl = k – k′, is

satisfied. The wavevector of the incident X-ray is placed in the proper rotational

orientation with respect to the reciprocal lattice (and with the proper length, determined

by constructing it such that 1/λ is properly scaled relative to the reciprocal lattice

dimensions). The origin of k is placed on the origin of the reciprocal lattice and a sphere

of radius k (=1/λ) is constructed with its center at the tip of k, guaranteeing that the origin

lies on the sphere’s surface. When the

crystal’s orientation is changed relative

to k such that some other reciprocal

lattice vector, Khkl, also lies on the

sphere, the diffraction condition is

satisfied (as illustrated). The tips of k

and k′ touch in the center of what is

called the Ewald sphere.

A rotation of a crystal in the incident X-ray beam can, of course, be viewed as

rotation of the angle of incidence of the beam in the opposite direction, viewed from the

frame of reference of the crystal. The illustration below shows the ac-plane of a

monoclinic direct lattice and a*c*-plane of the associated reciprocal lattice. A rotation of

the direct lattice in the sense indicated will bring about a rotation of the reciprocal lattice

through an equal angle. In the frame of reference of the crystal then, the Ewald sphere

rotates in the opposite sense and as it rotates, successive reflections are brought “in and

out of the sphere of reflection.”

The Ewald sphere concept helps us visualize several important aspects of diffraction.

First, the radius of the sphere is inversely proportional to the wavelength of the radiation

used in the diffraction experiment, while the spacing of the reciprocal lattice is inversely

proportional to the unit cell dimensions. Thus, more reflections are swept through the

Ewald sphere when shorter wavelength radiation is used or when the unit cell dimensions

are larger.

In neutron diffraction, if so-called thermal neutrons are used, a typical wavelength

would be 1.45 Å (= λn = h/pn = h•[3mnkBT]-1/2). In high-energy electron diffraction,

electrons with 100 keV kinetic energies are commonly used. Electrons of this energy

have deBroglie wavelengths much shorter than typical wavelengths used in X-ray

diffraction (λCu = 1.5418 Å, λMo = 0.71069 Å, λe,100keV = h/p = h•[2meE]-1/2 = 0.0039 Å)

and the Ewald sphere is very large in comparison with reciprocal lattice spacings. As a

result, by rotating a crystal by only a few degrees, large portions of a reciprocal lattice

plane nearly tangent to the Ewald sphere can be brought through it — making the

technique especially valuable for imaging.

k

!k

k

kk

!k!k

(h,0,0)

(0,0,l)

c

a

PROPERTIES OF DIRECT LATTICES AND RECIPROCAL LATTICES

In Chem 673, we introduced the reciprocal lattice via group theory. In the above, we

showed that the reciprocal lattice vectors are intimately related the condition for

constructive interference in diffraction, namely, the incoming x-ray is diffracted by a

reciprocal lattice vector: k′ = k + Khkl. Now we will further develop some geometrical

relationships between the direct lattice that underlies the translational periodicity of

crystals and the reciprocal lattice that serves as such a useful geometrical construct for

interpreting the diffraction patterns of crystals.

First we note two important mathematical relations; the first relates the direct and

reciprocal lattices and the second merely recalls a bit of vector algebra:

(1) The definition of the reciprocal lattice basis vectors, (a*, b*, and c*), guarantees

that the dot product between any direct lattice vector (DLV), Ri = ua + vb + wc, and any

reciprocal lattice vector (RLV), Khkl = ha* + kb* + lc*, will be an integer:

R

i• K

j= (ua + vb + wc) • (ha*+ kb*+ lc*) = uh + vk + wl = integer

!Ri• K

j= m ; m = integer

" e2#i(R

i•K

j)= 1

(2) The projection of any vector B on to an axis parallel to another vector A is given

by the quantity (B • A) A = B • A = Bcos! , where A^ is a unit vector parallel to A and θ

is the angle between A and B. This relation can be understood pictorially below:

A !

A

B

Bcos!

A•B = ABcos!

A•B = Bcos!ˆ

The direct lattice can be partitioned such that all the lattice points lie on planes.

Crystallographers classify these planes using intercepts on the unit cell axes cut by the

plane adjacent to the plane through the origin (see other handouts for examples). The

intercepts must be of the form (1 h ,0,0), (0,1 k ,0), and (0,0,1 l) . These planes are

normal to the RLV Khkl = ha* + kb* + lc*.

b

(1/h,0,0)

(hkl) plane

(0,0,1/l)(0,1/k,0)

-(1/h)a+(1/k)ba

c

-(1/h)a+(1/l)c

Proof:

The two vectors !(1 / h)a + (1 / k)b and !(1 / h)a + (1 / l)c respectively connect the

pairs of points (1 h ,0,0) & (0,1 k ,0) and (1 h ,0,0) & (0,0,1 l) - see the drawing. As

such, these two vectors must lie in the plane in question. Now,

K

hkl• [!(1 h)a + (1 k)b] = [!(h h)a • a

*+ (k k)b • b

*] = !1+1= 0

and

K

hkl• [!(1 h)a + (1 l)c] = [!(h h)a • a

*+ (l l)c • c

*] = !1+1= 0

So Khkl is normal to both vectors and therefore is normal to the plane in which they

lie and therefore is normal to the entire family of mutually parallel planes.

Since Khkl is perpendicular to the “hkl family” of planes, we can relate the distance

between planes to the magnitude of Khkl as follows:

Assume that Khkl is the shortest RLV that points in the exact direction that it happens

to point (i.e., Khkl = ha* + kb* + lc* where h, k, and l have no common factor).

a

b

dhkl = (1/h) a•Khkl = (1/h) a K cos!ˆ ˆ

Khkl

Khkl =ˆ Khkl

Khkl

- a unit vector!

(1/h,0,0)

lattice planes

The distance between planes, dhkl, is given by (see the 2D example above):

d

hkl= (a h) • K

hkl= (b k) • K

hkl= (c l) • K

hkl= (1 3)[(a h) + (b k) + (c l)] • (K

hklK

hkl)

where K

^ hkl is a unit vector parallel to Khkl. Substituting in the expression for Khkl we

obtain:

d

hkl= (1 3)[(h h)a • a

*+ (k k)b • b

*+ (k k)c • c

*] • (1 Khkl

) = (1 Khkl

)

or

Khkl

= 1

dhkl

This is an important result that we will use later. Simply stated, it says that

reciprocal lattice vectors are inversely proportional to the spacing between direct lattice

planes with which they are associated.