Physics307L:People/Knockel/Notebook/071121

Electron diffraction

experimentalists: pinky and the brain

Goal

Electrons are waves that can diffract when going through crystals. Graphite is an allotrope of carbon where the carbon forms sheets of hexagonal patterns. The sheets of hexagonal carbon stack in a specific way that forms a crystal lattice that causes diffraction of electrons. I want to calculate certain dimensions of this lattice by measuring the diffraction of electrons through it when the electrons are aimed perpendicular to the sheets.

I have no clue what exactly I am measuring since 3D crystal latices form very complex diffraction patterns, but I know I am measuring some dimension of the crystal and calling it d. Maybe the physically significant measurement is 2*d or maybe 1/2*d, but this minor scaling can be done quickly (you won't have to redo the experiment and calculations...).

Equipment

The CRT and two power sources I used (the one on the right produces the high-voltage and the voltage for the heater.

• Cathode ray tube (CRT) that fires electrons through a few graphite sheets and has a fluorescent coated as a target. (Model TEL 2507)
• Multimeter for measuring current on the order of 0.25 mA = 250 [math]\displaystyle{ \mu }[/math]A
• Caliper for measuring diffraction rings
• The following power sources:
  • high voltage up to 5 kV (for the accelerating voltage in the CRT)
  • Up to 50 V (for something that makes everything work somehow)
  • About 6.3 V AC (for the heating element in the CRT)

Setup

This is not important since figuring out a very specific technical device is not the purpose of the lab. Basically, we following a very elaborate circuit diagram from Dr. Gold to hook up the power supplies in a way that gives power to everything in the CRT. We also hooked up the multimeter so that it monitors the current to make sure that too many electrons are not sent through the graphite to destroy it (no more than 0.25 mA are allowed).

Procedure

Graphite is polycrystalline, which means that there are many subdivisions that each have a different orientation. These many subdivisions create circles of diffraction because each subdivision diffracts the light in a different direction. There happens to be 2 circles that are formed around the central beam (the central beam is made up of the electrons that went straight through the crystal without being diffracted).

All we are going to do is alter the voltage (V) that accelerates the electrons and measure the size of the circles (r). This [math]\displaystyle{ r }[/math] is the linear distance between where the central beam of electrons hits the screen and one of the circles, as measured by the caliper. This is not exactly the radius of the circle or any highly meaningful measurement because the fluorescent target is curved, but I will force it to be useful since it's the best measurement I can get. These measurements results can be used to find the slope of the line formed by graphing r(V^-1/2), which can be used to find a dimension of graphite (d).

The weaker (about 50 V) power source would only complicate voltage measurements, so we kept it in the circuit but did not use it to create a voltage.

Seeing the rings was difficult, especially at a low accelerating voltage. This is most likely the source of almost all random error. We used a less dense spacing of voltages when the voltages were low since this data is not too great. We stopped increasing the voltage when the power supply could go no farther.

Equations

As we all should but do not remember, the formula for the m=1 diffraction order is

[math]\displaystyle{ sin(\theta)d=\lambda\, }[/math],

where [math]\displaystyle{ \theta }[/math] is the angle of diffraction and [math]\displaystyle{ \lambda }[/math] is the wavelength of the diffracted particles. We eventually want to turn this into an equation containing only 3 variables: [math]\displaystyle{ d }[/math], [math]\displaystyle{ V }[/math], and [math]\displaystyle{ r }[/math].

[math]\displaystyle{ \lambda }[/math] in terms of [math]\displaystyle{ V }[/math]

The accelerated electrons convert all of their potential energy (PE=eV, where e is the charge of the electron) into kinetic energy (KE=p²/(2m), where m is the mass of the electron and p the momentum). Setting KE=PE and solving for [math]\displaystyle{ p }[/math] gives

[math]\displaystyle{ p=\sqrt{2meV} }[/math].

De Broglie's cute equation [math]\displaystyle{ \lambda }[/math]=[math]\displaystyle{ h }[/math]/[math]\displaystyle{ p }[/math] (h is Planck's constant) gives

[math]\displaystyle{ \lambda=\frac{h}{\sqrt{2meV}} }[/math].

The wavelength of the electrons depends on the accelerating voltage in the above way.

[math]\displaystyle{ sin(\theta) }[/math] in terms of [math]\displaystyle{ r }[/math]

I will make two approximations to do this:

1. [math]\displaystyle{ sin(\theta)=tan(\theta) }[/math]
2. The center of the fluorescent target is the point that is farthest away from the graphite on the opposite side of the CRT. I will assume that the distance (L) between the graphite and this point is the distance of a flat target instead of a curved target. I make this assumption so I can say that [math]\displaystyle{ tan(\theta)=r/L }[/math].

Both of these assumptions are reasonable for small [math]\displaystyle{ \theta }[/math].

As a result,

[math]\displaystyle{ sin(\theta)=\frac{r}{L} }[/math].

[math]\displaystyle{ r }[/math] as a function of 1/√[math]\displaystyle{ V }[/math]

Plugging in our [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ sin(\theta) }[/math] expressions into the topmost equation and solving for [math]\displaystyle{ r }[/math] gives

[math]\displaystyle{ r=slope\cdot\frac{1}{\sqrt{V}} }[/math],

where

[math]\displaystyle{ slope=\frac{Lh}{\sqrt{2me}\cdot d} }[/math].

Solving for [math]\displaystyle{ d }[/math] gives

[math]\displaystyle{ d=\frac{Lh}{\sqrt{2me}\cdot slope} }[/math].

Using these equations to find the slope from a linear fit and then [math]\displaystyle{ d }[/math] from the slope is not perfect because of the two approximations I made. The first approximation makes my [math]\displaystyle{ d }[/math] too small, and my second makes it too large (I know this because I also worked out much more exact formulas using chord lengths and other bullshit, but these formulas were WAY too messy for their slight benefit in accuracy). My second approximation makes [math]\displaystyle{ d }[/math] larger than the first approximation decreases it, so I should expect answers that are too large. I should also expect that, for the circle with larger [math]\displaystyle{ r }[/math] and therefor smaller [math]\displaystyle{ d }[/math], the systematic error should overshoot the actual value more than it overshoots the [math]\displaystyle{ d }[/math] for the smaller circle.

Values

The following are some fundamental constants needed to do the calculations:

• [math]\displaystyle{ h=6.626\times10^{-34} }[/math] J*s
• [math]\displaystyle{ m=9.1094\times10^{-31} }[/math] kg
• [math]\displaystyle{ e=1.6022\times10^{-19} }[/math] C

And the CRT length is

• [math]\displaystyle{ L=0.130(2)\, }[/math] m

Data

The "radii" were taken in inches up to three decimal places (eg 0.837 in), but I am converting these to meters (r1 is the smaller radius and r2 is the larger radius).

V (V)	r1 (m)	r2 (m)
2500	0.0167	0.0293
3000	0.0137	0.0262
3200	0.0135	0.0245
3400	0.0133	0.0229
3600	0.0127	0.0220
3800	0.0125	0.0212
4000	0.0121	0.0203
4200	0.0121	0.0208
4400	0.0110	0.0201
4600	0.0104	0.0198
4800	0.0103	0.0192

Calculations

In the following calculations, d1 corresponds to r1 and d2 to r2.

Finding d1

Finding d2