NOTES ON SCATTERING
OF WAVES BY
When waves,
such as x rays or electron waves, fall onto a crystal, they will scatter with
high intensity at those angles for which the phase differences between the
scattered waves from all the atoms in the crystal differ by integral multiples
of 2
. That’s all there is
to the physics; the rest is geometry.
Unfortunately, the geometry can be complicated. There are two ways to go about formulating
the problem, one using nice pictures of planes (A) and one just writing down
the math and solving for the answers (B).
Both are presented here. It is
suggested that you go through A first, then try to
follow the development B as well.
A. In terms of plane and pictures:
Refer to Fig. 1.
A parallel beam enters from above onto the crystal. If one lines up the crystal so that a set of planes (there are many possible sets) lies perpendicular to the page and the incident beam is in the plane of the page, then two conditions must be met so that scattering from all of the atoms in the crystal will be in phase:
a) The angle of reflection must equal the angle of incidence. This ensures that the scattering from all atoms in a given plane will be in phase with each other.
***Q: Why is this true?
This is why a mirror reflects with the angle of incidence equal to the angle of reflection.
b) The scattering from an atom in one plane must be in the phase with that from an atom directly above it in another plane. This condition will hold if Bragg’s law holds:
(1)
Here d is the perpendicular distance between planes and 
is the wavelength.
***Q: Be sure that you can derive this from simple geometry. You only have to use the criterion that the path difference traveled by the two scattered beams differ by an integral number of wavelengths.
If both of the above conditions are met, then scattering from any atom anywhere in any plane will be in phase with that from any other atom anywhere, and constructive interference will occur.
Miller Indices: What is d for a real crystal? There are an infinite number of sets of planes, each set with its own d, and it is a matter of bookkeeping to identify them. It is useful to identify and described them using three integer numbers h, k and l, known as Miller indices. It can be shown that for a cubic lattice all planes can be identified by a unique h, k and l, for which
where a is the lattice
constant. A sort of picture of what
these indices mean in real space can be given.
Consider the two dimensional picture of a cubic lattice
(Fig. 2). 
Place one atom
at the origin of an x-y Cartesian coordinate system through which all planes
run. Now imagine another plane drawn at some angle and located so that it is
the closest plane of this angle to the origin which doesn’t pass the origin
itself. It will cross the x-axis
somewhere (say a/h) and the y-axis somewhere (say a/k). We will let you fool around with examples to
convince yourself that integers for h and k are enough to cover all cases; you
don’t need irrational numbers. Now from
geometry, the distance from the origin to the plane is 
. In the figure the
h=2, k=3 planes are shown. Note that if
h and k can be divided by a common factor n, this corresponds to nth order
reflection off the planes of the smallest non-divisible h, k. This is the two dimensional result; in three
dimensions, there is an additional index in the z direction and an additional 
in the
denominator. Figure 3 shows how the [1,0,0] and [1,1,0] planes look, for example. 
B. The more elegant but formal analysis:
I. A more complete
analysis starts from the observation that two plane waves, when reflected from
scattering centers separated by 
, will have travelled distances
different by (see figure)
(2)
and thus will be out of phase by
***Q: See if you can show from geometry that (2) gives the path difference. See figure below.
Then waves are in phase again if
, N some
integer.
Note that if you measure an angle 
symmetrically
then
(3)
II.
If you have an array of atoms in cubic arrangement, atoms are
located at 
where a is lattice
constant and N’s are any integers.
A beam of
x-rays will constructively scatter off all centers if 
or if
for all 
. This works only if
, 
, or 
, where h, k, 
are “Miller
indices.” The size of q is given by
,
so that (use (2))
or
(4)
C: Structure Factors
Equation (4) is exactly the same arrived at by more agricultural methods in section A. However, this formalism allows us to do more. Up to here we have assumed that the structure of the crystal is pure simple cubic, with one atom at the corner of each cell. This is seldom the case. For example, for a face centered cubic lattice, FCC, each atom is replaced by 4 atoms.
Under these conditions, the cell of 4 atoms all by itself
has a non-isotropic radiation pattern, and the result is that some h, k, planes don’t reflect
at all. For FCC, it is shown below that
all indices must be even or odd. Furthermore,
it the basic symmetry is FCC but each atom is replaced by two atoms, as is the
case for NaCl, it is shown below that the scattering will be strong if
the indices are even, weak if they are odd.
For example, consider FCC structure where atoms are located
at (x, y, z) = (0, 0, 0), (0, ˝, ˝), (˝, 0, ˝) and (˝, ˝, 0) (in units of
a). Each atom will reflect with same
strength but phase given by 
so the strength of the
scattering (F) is
where i
labels the atom. For FCC, using , etc., this gives 
.
Unless h, k, are either all even or
all odd, F=0.
Note further that if each atom is really 2, as in NaCl, then, if Na and Cl scattering
strengths are 
and 
respectively, the
scattering strength will be
So if 
, we get weaker lines for odd indices.
***Q: Can you figure out what the corresponding condition for BCC is?