Electron
Diffraction
Read the notes on scattering of waves from crystals first. Then read the write-up provided by the manufacturer of the tube. Use this write-up to provide any detail left out of the following description of the experiment, but be aware that that write-up has several very misleading presentations and has not proved to be entirely adequate in the past.
In this experiment you will radiate two samples with a beam of monoenergetic electrons and will observe the diffraction pattern on a fluorescent screen. [BE CAREFUL TO MONOTOR THE TARGET CURRENT AND NEVER EXCEED 10 MICROAMPS.] The electrons scatter as “matter waves” whose wavelength is related to the momentum of the electrons by the deBroglie relationship.
1. Aluminum Target:
The first
target is A1, which is polycrystalline and produces a ring pattern similar to
the powder pattern observed in the MgO experiment
done with the x-ray apparatus. Figure 1
shows the geometry for scattering from a polycrystalline target for a
particular set of planes. The presence
within the size of the beam of a large number of crystals provides access to
ALL 
for a given crystal,
and each will produce one ring
on the screen.
***Q: Why do you get rings on the screen?
You should measure the radius of
each ring for a given voltage, and use the Bragg relationship (Eqs.(1) and (4) in the scattering
notes write-up) to calculate the wavelength of the radiation required to
produce this radius. In order to do
this, you will have to identify the associated with each
ring, which is a bit of a puzzle for you to solve. You will also have to know what the lattice
constant of Al is. This you can
calculate from the density of Al, if you accept for the moment that A1 is known
to be FCC. (You can also look up the
lattice constant, but this is less satisfactorily aesthetically.) If you make a spreadsheet with one column
being the observed radii,
one being an assumed value of [h,k,l]
for that ring, and one being something like sin(q)
divided by the square root of (h2+k2+l2) , you
should be able to assign h, k and l such that you get the same value in the
last column for all rings. If you now put in a known value for the lattice
constant, you can calculate a wavelength for each ring. All rings should give the same value, so you
can calculate an average and standard deviation. Compare this wavelength to that you calculate
from the known tube voltage and deBroglie’s
relationship. Perform this experiment
for at least two tube voltages, possibly more.
If you wish, you may also turn around this analysis, as the manufacturer’s
write-up would have you do, to calculate the lattice constant, given the deBroglie relationship, and compare the result with the
known lattice constant.
2. Pyrolytic Graphite:
This target is in single crystals of carbon large enough that it is possible, with some care, to focus the electron beam on a single crystal at a time. [NOTE: this crystal has been slowly destroyed over the years, so that now you will have to hunt long and hard to find a good one which produced the simple hexagonal picture on the screen.] The crystal structure is in sheets, with the carbon atoms lying in a hexagonal array in a plane perpendicular to the motion of the electrons. The sheets are very thin, perhaps as little as 10 Angstroms, so that the scattering pattern is best understood in terms of an elaboration of a two-slit interference pattern. Figure 2 shows the geometry of the carbon target. The hexagonal symmetry of the crystal structure gives rise to hexagonal symmetry in the pattern appearing on the screen. [Note: The manufacturer’s write-up tries to draw both the lattice constant a (which refers to the crystal dimensions) and the separation r of the dots (which refers to the dimensions on the screen) on the same figure, resulting in a very misleading picture.] From Fig. 2, one can see that constructive interference between scattering of electrons from cells 1 and 2 will occur if
***Q: Why is this
true? What is 
? Be sure you
know. It is not Bragg’s Law.
Note: The black dots in the left hand panel of figure 2 represent “scattering units”. You might want to think of these as “atoms”, but they really represent unit cells. If you will look at the graphite model in the laboratory, you will see that it is composed of interlocking hexagonal cells. The lattice constant is the distance between hexagonal cells, not the distance between atoms in a cell. You should not try to apply the concept of Miller Indices to graphite.
This condition for constructive interference FOR THESE TWO ATOMS ALONE will occur along solid lines shown in Fig. 2b, screen part of diagram.
***Q: Why lines, instead of points, on the screen?
These lines
will be separated by a distance 
, where D is the distance from crystal to screen, as shown on
the diagram. But atoms 1 and 3 will also
have a radiate in phase, which will give rise to a second set of lines for
constructive interference, as also shown as dashed lines in the right hand panel
of figure 2.b. One could
continue this line of reasoning using atoms 2 and 3, although this is
not done here nor is it necessary. It is
clear that constructive interference between atom 1 and ALL of the neighboring
atoms will occur only where the various lines intersect. From the right hand
panel of Fig. 2b one can see that the perpendicular spacing of the “lines” is
given by the young’s double slit expression, but that this is not the
separation of the dots on the screen. The separation of the dots is the spacing
divided by the cosine of thirty degrees.
***Q: Work out the details of this geometry. Note that the pattern of atoms is rotated by 30 degrees relative to the pattern of dots on the screen. Why? All calculations are done in the small angle approximation for the scattering angle.
Measure the distance r in the above discussion from the screen. It is probably more accurate to measure not just from the center to the first bright dot, but farther out on the pattern. You want a single scale parameter characterizing the size of the pattern on the screen. From this and the known voltage, calculate the constant a for pyrolytic graphite and compare with the “accepted” value. Do this for at least two tube voltages.
