ME588: Homework Three
Fall 2000

Assignment Due Monday, October 23

• the interplanar spacing (d),
• the angle between the plane and the [100] direction, and
• the angle between two planes.

The calculation should be done for cubic lattices, but should be easily adaptable for use in other crystal systems as well.

Use the spreadsheet to determine the following in a cubic lattice with a=0.405 nm:

1. the d-spacing of the (2 3 1) planes;
2. the angle between the [100] direction and the (-4 2 7) plane,
3. the angle between the (3 0 2) plane and the (1 1 0) plane.

NOTE: Think about organizing your spreadsheet in a way that will allow you to improve it for future homeworks, where we will be calculating diffraction patterns, indexing diffraction patterns, and more.

1. simple cubic (a = 0.4 nm)
2. simple tetragonal (a = 0.4 nm, c = 0.3 nm)
3. simple tetragonal (a = 0.3 nm, c = 0.4 nm)
4. simple rhombohedral (a = 0.4 nm, ALPHA = 75 degrees).

a) Cell A, base-centered tetragonal containing two atoms per unit cell at locations of 0, 0, 0 and 1/2, 1/2, 0, for which a = 0.30 nm and c = 0.4 nm;

b) Cell B, simple tetragonal with one atom per unit cell at coordinates 0, 0, 0.

Determine the simplified structure factor equations for each cell and the positions (2THETA values) of the first four lines that would be observed (F^2 not equal to zero) on a powder pattern made with Cu Ka x-rays. Make a table of the 2THETA values, and the (hkl) values referenced to each unit cell, similar to that below:

2THETA [degrees]	Cell A	Cell B
Lowest angle
----
----
Highest angle

Use a drawing of the two cells to show that the planes associated with any line refer to the same set of planes, even though they have different Miller Indices.

SOLUTIONS

1) I wrote my spreadsheet using EXCEL, which looks something like this:

The answers are:

• d-spacing of (231): 0.1082 nm
• angle from [100] to (-4 2 7): -28.79 deg
• angle between (302) and (110): 53.96 deg.

1. simple cubic (a = 0.4 nm)
2. simple tetragonal (a = 0.4 nm, c = 0.3 nm)
3. simple tetragonal (a = 0.3 nm, c = 0.4 nm)
4. simple rhombohedral (a = 0.4 nm, ALPHA = 75 degrees).

First, wehave to put together the plane spacing equations and Braggs Law for each of the crystal systems:

I put these into my crystal calculator from problem 1 and fiddled with the h, k, l values to find the three planes with the largest d spacing. My spreadsheet then finds the diffraction angle using the d-spacing and the Cu Ka x-ray wavelength.

CELL A) Part of this was done in class as the derivation of the structure factor for the base centered cell with atoms at 000 and 1/2 1/2 0:

.

This gives F = 2f for h, k unmixed, and F = 0 otherwise.

The lattice parameters for cell A are a = 0.30 nm and c = 0.40 nm.

Look at the first several planes and make a table using the lattice parameters given (it might be helpful to use a spreadsheet program like the one you constructed for problem 1). I get the following list:

PLANE	d [nm]	2THETA [degrees]	F
(001)	0.4	22.2	2f
(100)	0.3	29.7	0
(101)	0.24	37.4	0
(110)	0.21	42.6	2f
(002)	0.2	45.3	2f
(111)	0.187	48.5	2f
(102)	0.166	55.1	0
(200)	0.15	61.8	2f

CELL B) For the primitive cell, we have only one atom at 000, in which case (see example in class) the structure factor is independent of h, k, and l. Remember: the lattice parameters (a and c) are DIFFERENT in this cell from those in the A-type unit cell!

The lattice parameters for cell B are a = 0.212 nm and c = 0.40 nm. Again, a list of the first several planes shows:

PLANE	d [nm]	2THETA [degrees]	F
(001)	0.4	22.2	f
(100)	0.21	42.6	f
(002)	0.2	45.3	f
(101)	0.187	48.5	f

The table from the homework assignment looks like this:

2THETA [degrees]	Cell A	Cell B
22.2	(001)	(001)
42.6	(110)	(100)
45.3	(002)	(002)
48.5	(111)	(101)

Comparison of the first four lines for both CELLS A and B shows identical values of 2THETA. This has to be the case since they are different mathematical descriptions of the same crystal. The figure below shows the correspondence between the different planes in the two crystal structures.