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 creates apertures that cause diffraction of electrons. I want to calculate certain dimensions of these hexagons by measuring the diffraction of electrons.

I have no clue what exactly I am measuring since 3D crystal latices form very complex diffraction patterns, but 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...).

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 as it 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. 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.

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. 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.

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 [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].

[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 the exact formulas using chord lengths and other bullshit, but these formulas were WAY too messy to simply fit a line to). 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