INTRODUCTION: Does matter have a wavelike nature? A calculation using deBroglie's equation (λ = h/p) shows that electrons accelerated through a potential difference of 4 kV would have a wavelength of about 0.02 nm. Although it is impossible to produce a diffraction grating with rulings of approximately d = 0.2 nm (to produce measureable interference), we have available natural "gratings" in the form of crystals. The tube you will use has a graphite (carbon) polycrystal with an appropriate inter-atomic distance. You will be determining that distance, d.
  1. Look up both "electron diffraction" and "double-slit interference" in either your Modern Physics text or an Introductory Physics text. Take good notes. Read these instructions before class as well.
  2. Find library resources that tell you more about the role of Louis deBroglie, Clinton Joseph Davisson and Lester Germer, and George Paget Thomson in the history of this experiment. Be sure to include this in your Introduction, as well as some discussion of “particle-wave duality”.

THE EXPERIMENT: If the material we were using as a target were a single "one-dimensional" crystal, we would see an electron diffraction pattern consisting of vertical lines on a screen. If it were a perfect two- or three-dimensional pattern, we would see a two dimensional dot pattern on our screen. Instead, it is not a single crystal, and so you will see a diffraction pattern with two rings, as seen in Fig. 1:
Fig. 1: Side-view of the electron diffraction tube. Note that the diameter of the rings, D, is only approximately proportional to the angle, θ, because the center of curvature of the tube is not located at the carbon foil. (Users Guide: Electron Diffraction Tube 555, Tel-Atomics)

(There are other rings, but they will be too dim and at angles too large for us to observe.) Measuring the relation between accelerating voltage and diffraction ring diameter will allow us to determine the interatomic distances in the carbon foil inside.

To pull those distances out of our raw data, we will need to start with the deBroglie wavelength of a particle, λ = h/p. For non-relativistic electrons (KE << mc2), we also have p = mv. The individual electrons' kinetic energy is equal to KE = ˝ mv2, This kinetic energy is the result of an accelerating voltage, Va, and is equal to eVa.
Question 1: Derive an expression for electron wavelength as a function of accelerating voltage.

Question 2: In this experiment, the accelerating voltage is limited by our power supply and multimeter to no more than 5kV. Can we ignore relativity? Why?

The diffraction rings follow the same physics as the interference maxima in a two-slit or multiple-slit interference pattern. Find the equation -- in a physics textbook -- in terms of the slit distance, d, the angle, θ, of the "nth" interference maximum, and the wavelength, λ. Note that for small angles, sinθ = D/2L.

Question 3: Derive an expression for D as a function of Va. Rearrange this expression so that you can plot a linear graph. Verify with the instructor that your result is correct before continuing with the experiment.
Fig. 2: Wiring diagram of the electron diffraction tube. Do not exceed the following: Filament voltage 8.0V, Anode current 0.20mA. (Users Guide: Electron Diffraction Tube 555, Tel-Atomics)

Connect the tube into the circuit shown in Figure 2, paying special attention to the precautionary notes. Do not turn on the power supply until your circuit has been checked by an instructor, and when you do turn it on, let the filament warm up for 1 minute before increasing Va . Turn up Va until you have a diffraction pattern, adjusting the bias voltage so that the current never exceeds 0.175mA.

Measure the inner diameters of the two most visible rings, Dinner and Douter for several values of Va , and plot your results in such a way that you get a linear graph. Plot as you go, so as to determine the best strategy for amassing good data points. When you are done, use appropriate software to find the slope -- and uncertainty -- of your straight line fit. From this you can determine the ratio of the two Values -- including uncertainty -- and their actual magnitudes.

ANALYSIS: Here comes the hard part. How do you interpret two different values for the interatomic spacing? Both of these numbers correspond to the distance between planes of atoms in the carbon (graphite) lattice. If this lattice were a simple cubic lattice -- as in Fig. 3 -- this would be simpler, but not trivial. If you look at it straight on at a square lattice, you will observe that the "atoms" line up behind each other in simple rows. But rotate it a little and you will see different atoms start to line up behind the atoms in front. For this point of view, the distance between the rows of atoms has changed. It turns out that the ratio of the distance between planes for the rows marked (100) and (110) in Fig. 3 is 1.414 : 1.

Graphite consists of two-dimensional sheets loosely bound to other parallel two-dimensional sheets. Within each sheet, the atoms are arranged in a hexagonal lattice. (Fig. 4 & 5.) Your first task is to locate the planes. Take the attached sheet with the hexagonal lattice on it and find two different sets of planes. Notice that the distance between planes has to repeat itself exactly. Most likely you have found the "(10)" and the "(11)" planes, which differ by a factor of 1.732 in size. Measure these two interplanar distances , d10 and d11 carefully on your sheet, and compare them. What is the the square of the ratio of these two distances? Does the uncertainty allow for the square to equal 3?

Finally, calculate the two "d"s from your data, using the slopes you calculated from your raw data. Don't forget to calculate the uncertainties. From these, calculate the closest distance between two atoms in the graphite. You can estimate the approximate spacing of carbon atoms in the crystal knowing that 12 grams of carbon contain 6 x 1023 atoms and the density of carbon is about 2 g/cc. Does this result agree with your earlier measurements? Bear in mind that the graphite crystal is not cubic, so this is only a crude estimate.
Figure 3: The geometry of a cubic lattice. Note that the first two interplanar distances, (100) and (110) differ by a factor of 1.414. (Instruction Manual and Experiment Guide for the PASCO scientific WA-9314B. (PASCO Scientific: 1991) Fig. 12.2, p. 33.)

Figure 4: The geometry of the hexagonal lattice of carbon. Note that the two interplanar distances, d10 and d11 correspond to the two blue rings observed in the electron diffraction tube of Fig. 1. (Users Guide: Electron Diffraction Tube 555, Tel-Atomics)

Figure 5: A hexagonal lattice. Directly measure the distances d10 and d11 on this page by measuring the distances between say ten such planes and dividing by the appropriate multiple. Compare this ratio to the ratio of the distances, d, that you derive from your electron diffraction data.