Physics307L:People/Knockel/Notebook/071121

Electron diffraction

experimentalists: pinky and the brain

Goal

Electrons are waves that diffract when going through small aperture. 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 and the nature of how the electrons diffract through them by measuring the diffraction of electrons through them.

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.
• Multimeter for measuring current on the order of 0.25 mA
• 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)
  • ___ (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 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 have the same orientation of the hexagons within it but differ between each other. These many divisions create circles of diffraction because each subdivision diffracts the light in a different direction. Also, the stacking of carbon creates slits of two different sizes creating two circles. Another final thing is that only the m=0 ([math]\displaystyle{ \theta }[/math]=0) and m=1 (the first diffraction order) are bright enough to be visible, and only the m=1 is important.

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 the m=0 and m=1 diffractions 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 the widths that are diffracting the electrons (d).

I am not too sure how to get a very accurate reading for [math]\displaystyle{ V }[/math] because of the complex setup, so I plan on taking the value of whatever the meter on the high-voltage power supply reads.

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 electron 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].