Solutions 5

1.      X-ray diffraction for diatomic chain.

This problem is designed to give you some additional exercise in calculating structure factors, form factors and other properties useful in understanding x-ray data. It is also an entry-level problem in how to use x-ray diffraction data to determine the structure within the unit cell. By generalizing the approach here to multiple dimensions and atoms, you can see how to make a model for predicting the observed x-ray diffraction peak heights. Generally, this is one part of the typical approach to understanding x-ray data. First, you take the scatter diagram. Next, you model the structure and compare the predicted pattern to the observed one. Finally, you refine the model as much as the data justifies.

In the case we consider, we have a hypothetical 1-D crystal, which changes its unit cell with changing temperature. Here's the picture again:


You are asked to assume localized Gaussian charge distributions about each atom:

Also, assume that the atoms are each surrounded by Z1 electrons, that the atoms are each surrounded by Z2 electrons.

a.†† The direct lattice and reciprocal lattice vectors for both phases are identical; only the structure within the cell changes. Because this is a 1-D problem, there is only one direct lattice vector. It has length, b, and an arbitrary origin. In the rest of the problem, I refer to the direction to the right along the chain as the x‑direction. I choose as my origin a point midway between a filled atom to the left and an open atom to the right. The vector is shown above as a1. In terms of length and direction, we have:

Similarly, the reciprocal lattice basis vector is defined so that its dot product with the direct lattice basis is just 2. Then, we get:

b.†† Form factors.

From the discussion in class, the form factors are calculated by evaluating the integral:

In this case, the unit cell is 1-D and the charge densities to use are just the Gaussians for each atom type, offset within the unit cell by some temperature dependent distance. Using the atom symbol as a subscript to label the various quantities, the form factor is then:

The charge densities are just the Gaussian forms above. We are further allowed to assume that the Gaussians are very narrow in space, so that the integral above can be approximated by an integral over the entire line. Then,

These are just the Fourier Transforms of Gaussians, which are just k-space Gaussians. By making the simple transformation of variables so that integration is about the center of the charge density, we get:

This pair of results is valid for both the high and low temperature phases.


c)†† Structure factor and diffraction peak intensities.

In part a) we showed that the reciprocal lattice is also a 1-D chain just like the direct lattice. The Bragg scattering condition tells us that the difference in wave vector between the incident and scattered waves must be a vector of the reciprocal lattice. In other words,

Remember that the magnitudes of the incident and scattered wave vectors are equal, so we can immediately draw a picture of how an incident wave is scattered:

This is just the type of diffraction you see when a laser is incident on a screen with a diffraction grating in it.

The only issue which we need to address is how the shape of the openings in the diffraction lines affects the scattered pattern. In the language of x-ray scattering, the atomic form factors must be combined to give the structure factor for scattering into a particular G.

From the discussion in class, the total structure factor is defined as:

Here, the sum is over the atoms in the unit cell. In this particular case, the unit cell has two atoms. Given the definition I used in part a) for the direct lattice vector, the positions of the atoms within the cell are written as respectively:

Then, the structure factor is:

This result is general; now let's discuss specific cases.

i)††† High temperature limit.

In the high temperature case, the value of the atomic separation within the unit cell is fixed at a=b/2. Under this condition, the structure factor takes on a simple form. Plugging in this value for a leads to the result:

The structure factor then has four possible combinations:

where n is an integer (including 0 and negative numbers). The scattered wave intensity goes as the squared magnitude of the structure factor, so these cases have amplitudes which are the same for all the even numbered peaks, related to the sum of the charge densities and again equal for the odd numbered peaks, but this time related to the difference in the charge densities.

As an interesting special case to consider, imagine that the two atomic types happen to have equal charge densities. Then, the charge densities cancel out the odd-order peaks and only the even numbered peaks appear. Physically, for the high temperature phase with equal charge densities, this result is sensible because the periodicity is twice the lattice spacing, b, so that the diffraction peaks are spaced twice as far apart. This effect is a special example where the structure factor completely eliminates peaks due to basis effects.

i)††† Low temperature behavior.

In this case, the value of the atomic separation within the unit cell is temperature dependent so that the x-ray intensities become functions of temperature. Here, a=(b/2)(T/Tc) and under this condition, the structure factor takes on a simple form. Plugging in this value for a leads to the result:

Clearly, the structure factor has become temperature dependent. Of course, this temperature dependence will cause the diffraction peaks to become temperature dependent too.

As a special case, let's assume that the charge densities are equal for simplicity. Then, the structure factor can be rewritten by factoring out the temperature dependent terms:

where the plus or minus refers to n even or odd respectively.

For T=0, this gives the expected result that the peak intensity is that due to a charge distribution with twice the charge of one atom in isolation. Also, for T=Tc we get the result above for the high temperature phase, namely that all the odd peaks are zero in amplitude and the even peaks alternate in sign. In terms of the squared magnitude, the peaks will be of equal height.

For intermediate temperatures, the peaks have an amplitude modulation due to the Cos term. Interesting intermediate cases occur each time the ratio of T to Tc is a whole number over n. For even whole numbers, we get unity and for odd numbers we get zero.

To plot the result, we need to select a particular temperature and also a particular value for the b parameter in the Gaussian. I'll consider the case where the temperature is half way to the transition and where the Gaussian is highly peaked in space so that the Gaussian prefactor is unity. Then, the structure factor is:

Here's the Mathematica lines to produce a plot:

First make a function for evaluating the structure factor at various points. In this case, I will plot the square of the structure factor to simulate the intensity envelope we would observe:


Now make a list of the points. Note that my list has some bogus points to make it look more like an x-ray diffraction peak plot:

SG={†††††††† {-7,0},{-7,sg[-7]},{-7,0},†††††††††

††††† {-6,0},{-6,sg[-6]},{-6,0},

††††† {-5,0},{-5,sg[-5]},{-5,0},

††††† {-4,0},{-4,sg[-4]},{-4,0},

††††† {-3,0},{-3,sg[-3]},{-3,0},

††††† {-2,0},{-2,sg[-2]},{-2,0},

††††† {-1,0},{-1,sg[-1]},{-1,0},

††††† {0,0},{0,sg[0]},{0,0},

††††† {1,0},{1,sg[1]},{1,0},

††††† {2,0},{2,sg[2]},{2,0},

††††† {3,0},{3,sg[3]},{3,0},

††††† {4,0},{4,sg[4]},{4,0},

††††† {5,0},{5,sg[5]},{5,0},

††††† {6,0},{6,sg[6]},{6,0},

††††† {7,0},{7,sg[7]},{7,0}

††††† };

Show[ Graphics[Line[SG]],

††††† PlotRange->{0,1.25},

††††† Frame -> True,

††††† FrameLabel -> {"Peak Number","(Sg/2Zo)^2"}]


  1. Marder 2.1

Here, we are asked to consider the bcc and fcc lattices as simple cubic lattices with appropriate basis atoms. It's clear that the choice of lattice and basis is not unique. Here are my choices:

a)      Number of basis vectors and their value.

For the body centered cubic, the standard cubic unit cell is shown in Marder Figure 2.5. If you look at this standard cubic cell, you can see that it contains one atom at the center of the cube, and 1/8 of each atom at the eight corners, for a total of two atoms in the basis. I choose the atom in the front, lower, left corner as the origin. Therefore, the basis vectors are to this corner atom at (0, 0, 0) and the centered atom at (a/2, a/2, a/2) as the second atom in the basis.

Similarly, for the fcc case, there are four atoms in the unit cell (Marder Figure 2.2) and they have basis vectors of (0,0,0), (a/2,0,a/2), (a/2,a/2,0), and (0,a/2,a/2).

  1. Marder 3.1

a)      Marder wants you to show that the reciprocal of the bcc and fcc lattices are respectively fcc and bcc lattices. Here's how it works for the bcc case:

What are the bcc direct lattice vectors?

In the figure, I show an example choice of the direct lattice vectors.

Once you have these vectors, you are in good shape to work out the reciprocal lattice vectors by using Marder's equation 3.24.

Here's how this particular choice of direct lattice vectors actually work out:


Show that the reciprocal lattice of the bcc direct lattice is an fcc lattice.

Recall that the definition of the Reciprocal Lattice Vectors is:

I set up three direct lattice vectors:

At this point, generating the RLVs is just a matter of grinding out the vector equations. The easiest is for the third of the RLVs. It looks like this:

The other two vectors come out to be:

This particular set of RLVs is not particularly symmetric. However, we know that the RLVs are non-unique (just like the direct lattice vectors). Even if they are a bit messy, they should allow us to construct a lattice. In the following figure, Iíve drawn the three RLVs and included several points for a face-centered-cubic lattice so that you can convince yourself that the three RLVs really could be used to reach any of these points (and of course, all the rest of the lattice). Here is what they look like:


I emphasize that the particular set of direct lattice vectors that you start with will determine a different set of RLVs, but the point is to determine what the lattice actually looks like. The vectors (both direct and reciprocal) are non-unique, but the lattice is not.

In this particular case, you can see that the reciprocal lattice is just a face centered cubic lattice.

The fcc to bcc case is quite similar and is also implied already by the calculation above as the reciprocal lattice of the reciprocal lattice is just the original bravis lattice.


b)      Next, you are asked for the reciprocal lattice vectors of smallest magnitude for aluminum (fcc lattice), berylium (hcp lattice) and bcc iron. These three elements represent the three most common crystal structure, so you get exercise with all three.

Aluminum: The fcc lattice with conventional cubic cell of side, a, has a bcc reciprocal lattice of conventional cubic cell (in k-space) of side 4p/a . Therefore, the shortest reciprocal lattice vector would be from the corner to the center position, (4p/a)(1/2,1/2,1/2) with a total length of side (see Marder example on page 51) . For aluminum, the atomic spacing has a=4.05 angstroms (Marder Table 2.1). Thus, the shortest RLV is roughly 27 inverse nanometers. This shortest RLV points along the body diagonal of the cube, and therefore has equal x, y, and z-components. The Miller indices are (111).

Beryllium: The hcp lattice is composed of a simple hexagonal lattice, with a two-atom basis. Never the less, the reciprocal lattice is just another simple hexagonal lattice, rotated 30 degrees to the direct lattice. The shortest RLVs are associated with the triangular plane directions. Assuming a spacing between points in the triangles of a, and spacing between sheets of c, then the triangular array in k-space has spacing of (see below) and respectively. Therefore, the shortest reciprocal lattice vector could be either in the triangular plane, or perpendicular to the triangular plane, depending upon the actual values of a and c for the direct lattice. In the case of Be, we have a=2.29 angstroms and c=3.58 angstroms. Therefore, the shortest RLV is in the c-direction: 17.6 inverse nanometers. Miller indices are (001).

Iron: The bcc lattice with conventional cubic cell of side, a, has an fcc reciprocal lattice of conventional cubic cell (in k-space) of side 4p/a . Therefore, the shortest reciprocal lattice vector would be from the corner to the face-center position (any face will do) e.g., (4p/a)(1/2,1/2,0) with a total length of side . For iron, the atomic spacing has a=3.59 angstroms. Thus, the shortest RLV is roughly 24.8 inverse nanometers. Miller indices are (110).


  1. Marder 3.2

OK, for the simple hexagonal, let's just grunt out the RLVs. I will assume a sheet of triangularly coordinated points in the x-y plane, with one point at the origin, and points located along the x-axis. Then, a set of direct lattice vectors would be:

The reciprocal lattice vectors are then given by plowing through Marder's equation 3.24. Let's press on with the first reciprocal lattice vector:

Ugly... Next, do the second vector:

Ah! This result looks simpler. It shows that we do have a triangular lattice in the x-y plane of k-space, that the characteristic spacing is as above, and that one of the directions is along the y-axis. In other words, the direction of the second reciprocal lattice vector is 90 degrees to the x-axis and 30 degrees beyond the second direct lattice vector.

Finally, the third RLV is:

OK, now that we know the RLVs for the hexagonal lattice, we are ready to consider the hexagonal close packed (hcp) case. For hcp, we need a lattice with a basis. The lattice is hexagonal. The basis is a two-atom basis, with atoms at (see Marder's equations 2.5a and b) positions and . From the x-ray discussion, we know that the overall structure factor is a sum of atomic form factors, each weighted by an exponential phase factor appropriate for the location in the unit cell. In this case, the atomic form factors are identical, so we factor them out, leaving just the sum of two exponentials. Therefore, the structure factor is proportional to:

Each RLV is an integral multiple of the b-vectors we found above, so that we have:

Plug this result for the general RLV into the structure factor to find:

Or, putting this result into the form that Marder is looking for, we find:

The x-ray scattered electric field would be proportional to this structure factor. The intensity is then proportional to the squared magnitude.

Cancellation of scattering due to the basis will occur each time the structure factor is zero. This condition will in turn occur each time the overall phase is an odd multiple of p radians:

As an obvious example, if the RLV is only along the b3 direction, then the structure factor will cause cancellation for odd values of n3.