ME588: Quiz Three
Spring 1999

This page last updated May 24, 1999

1) (40 points) The diffraction pattern below was obtained using Cu Ka radiation on a powder of silicon (Si). Note that only the high angle peaks are shown. Silicon crystallizes as diamond-cubic. Using the data given, answer the following questions:

a) (20 points) Why does each of the peaks appear to have two maxima?

 

b) (20 points) What are the Miller Indices for the five peaks shown?

 

DATA FOR Si DIFFRACTION

Peak #

2 THETA (degrees)

Peak #

2 THETA (degrees)

1

106.723

5

127.556

2

107.127

6

128.141

3

114.101

7

136.933

4

114.551

8

137.669

9

158.684

2) (30 points) POLE FIGURES: We are going to perform a measurement of a thin film of a simple cubic metal. We know the following things about the sample:

a) it has fairly good WIRE texture along a [110] fiber axis,

 

b) it is oriented so that the (110) plane is parallel to the sample surface.

 

If we set the CHI angle so that we will only observe diffraction from (210) planes, sketch the expected pole figure. (Indicate the high intensity points on the PHI-CHI plot).

(HINT: Table 2-3 and Figure 2-36 in Cullity may be helpful.)

3) (30 points) Using the standard (001) stereographic projection for a CUBIC crystal shown below as a starting point, show the positions of the

(-1 0 0), (-1 0 1) (1 0 1) and (1 0 0) poles for a TETRAGONAL crystal, with c = 2 a.

(HINT: Think about how the projected poles move as the crystal distorts from cubic to tetragonal.)

SOLUTIONS:QUIZ 3

1a) The peaks each appear as with two maxima because of the K-a1 and K-a2 Cu x-rays, which have slightly diffferent wavelengths, and therefore satisfy Braggs Law for each hkl at a slightly different angle.

1b) We know that, from the Bragg's Law equation and the plane spacing equation for cubic materials, we can write the following relation:

Making a table of values, we can find the s-values for each diffraction angle. Using Appendix 10 of Cullity, we can index the peaks as follows:

Peak #

2 THETA (degrees)

s

(hkl)

1+2

106.5

31.94

440

3+4

114.3

35.11

531

5+6

127.8

40.12

620

7+8

137.3

43.16

533

9

158.7

48.05

444

2) Because the sample has wire texture around the [110] axis, we know that the grains will have all their (110) planes parallel to each other, as well as to the sample surface.

Around the [110] axis, though, we expect random orientation. The pole figure should therefore provide uniform rings, located at CHI angles determined from the angle between the (110) planes and the (210) family of planes.

We can either calculate this angle, using the interplanar angle equation from Appendix 3, or we can use the Table 2-3, which lists the interplanar angles for the cubic planes.

There are three different orientations of the (210) family of planes relative to the (110) planes. These are at interplanar angles of 18.4, 50.8, and 71.6 degrees.

The pole figure will therefore shown three diffraction rings at CHI values of 18.4, 50.8, and 71.6 degrees. The width of the rings is determined by how well textured the sample is along the [110] direction (we don't have any information about that, so we can't display anything about that).

3) Because of symmetry, we really only have to think about two of these, say the (-100) and (-101) poles. In the figure below, I have sketched a cubic unit cell transforming into a tetragonal cell. You can see that the effect of increasing the c-lattice parameter is to increase the angle between the (-101) plane and the (001) plane. This appears on the stereographic projection as a movement of the (-101) pole further to the outside of the projection.

From simple geometry, we can determine the angle OMEGA between the (-101) plane and the (001) plane (see figure below).

We can also see, from this figure, that the change in the c-lattice parameter has no effect on the angle between the (-100), (100) planes and the (001) plane-- it is still 90 degrees. The final answer then is shown below:

End of file.