Physics307L F09:People/Young/Formal

=e/m Ratio using Electron Diffraction=

Author: Daniel Young

Experimentalists: Arianna and Daniel Young

Location: UNM Department of Physics, Albuquerque, New Mexico, United States

Date of experiment: September 19, 2008

Abstract
We start with an electron gun consisting of a heater and some potential plates to create a constant stream of electrons that will appear as a purplish light when transmitted in a helium filled glass bulb. To find out the charge over mass ratio of the electron beam we will introduce a constant magnetic field perpendicular to the beam path using a set of Helmholtz coils that surround the apparatus. First we will measure the various radii of the electron beam by varying the accelerating Voltage in the electron gun for a fixed magnetic field. Second we will vary the B field and measure the resulting radii. The magnetic field will be changed by changing the current through the coils solely. Thirdly we will vary both magnetic field and accelerating Voltage and measure their effects on the radii of the electron beam. 

Introduction
In our experiment we hope to find an experimental value for the ratio of charge over mass for an electron. The electron charge over mass ratio is a well known value that we hope to confirm using the most practical methods available in out Lab. The electron is a fundamental building block of the atom. To understand more about the electron would mean a greater understanding of atoms and the interaction they undergo. Also the value of an electrons charge and mass is invaluable to other experimenters who wish to use electrons in any part of their experiment. 

Materials

 * 1 High Voltage power supply 5 kV max for the accelerating plates in the electron gun
 * 1 Low Voltage supply voltage 2.5V for the Helmholtz Coils
 * 1 Low Voltage supply not to exceed 6V for the Heater in the electron gun
 * banana plugs
 * multimeter's to monitor current and Voltage
 * air tight Helium gas filled bulb
 * electron gun
 * Helmholtz coils

Calibration

 * All of the power supplies are connected to Voltmeters and Ammeters to ensure that the values we view on the power supplies are correct.
 * if the bulb is turned too much one way or the other the beam can make a Helix rather than a loop so we made sure to line the beam up completely perpendicular to the appaeratus before taking measurements.

Derivation of e/m Relation to our input
Relation of charge over mass ratio using forces.

$$F_e = ma \ $$

$$B_Hq = \frac{mv}{r} \ $$

Relation of charge over mass ratio using Energy

$$KE = eV \ $$

$$\frac{1}{2}mv^2 = qV \ $$

Where,
 * $$V \ $$ is the accelerating Voltage in Volts
 * $$v \ $$ is the velocity of the electron in m/s
 * $$q \ $$ is the charge which in this case is the charge of an electron
 * $$m \ $$ is the mass of an electron which is
 * $$B_H \ $$ is the B field produced by the Helmholtz coils in (V s/m²)
 * $$r \ $$ is the radius of the electron beam path in meters

The general form of the equation for a Helmholtz coil is $$ B_H = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$$

We will estimate that the value for x is about R/2 within for our electron beam.

After simplifying all of our equations we have


 * $$\frac{q}{m} = (\frac{5}{4})^3 \frac{2V(R^2)}{{\mu_0}^2 n^2 I^2 r^2} \ $$

Where,
 * $$R \ $$ is the Radius of the Helmholtz coils which is .15 meters
 * $$\mu \ $$ permeability of free space 1.2566371 × 10-6 (N/A^2)
 * $$n \ $$ is the number of loops in the Helmholtz coils which is 130
 * $$I \ $$ is the current through the coils

Results and Discussion
I started by setting up the Power supplies to their respective slots with the banana plugs. All power supplies are connected to Multimeter's. The Helmholtz coils Multimeter is set up in series with the apparatus and the other power supplies are set up in parallel with the apparatus. The heater power supply Voltage, Helmholtz coil radius and number of loops are all held constant. The only possible variables are the accelerating Voltage and the Helmholtz coil current.

Vary Current
While keeping the Voltage constant we start with a large radius (High Current) and then lower the Current by small increments so we can measure the resulting radius of the electron Beam. Here the Magnetic field delegates the radius of the beam.

Since we are holding the Voltage constant.


 * $$r^2 \propto \frac{1}{I^2}$$

Constant Voltage: 449.3V

Vary Voltage
While holding the Current constant I start with a large radius (High Voltage) and lower the Voltage by small increments and measure the radius of the electron Beam. Here the tangential speed of the electron delegates the radius of the beam.

Since we are holding the Current constant.


 * $$r^2 \propto V$$

Constant Current of 1.6A

Vary Current and Voltage
Now we vary both Helmholtz Current and Voltage.


 * $$r^2 \propto \frac{V}{I^2}$$

The bold value in the table represents an outlier that was removed from the graph. It was the only value that was not within one standard deviation of the predicted slope.

Analysis
All calculations were made using matlab functions. A least squares fit was used to find the values of the final ratios.

The accepted value for the e/m ratio is
 * $$ \frac{e}{m}=1.76e11 \ $$

Constant Voltage Vary Current

 * $$ \frac{e}{m}=2.051e11 \ $$


 * σ=0.0526
 * sem=0.0066
 * %difference from accepted=15.53

Constant Current Vary Voltage

 * $$ \frac{e}{m}=1.0227e11 \ $$


 * σ=24.408
 * sem=3.051
 * %difference from accepted=41.89%

Vary Current & Vary Voltage

 * $$ \frac{e}{m}=1.2889e11 \ $$


 * σ=30.69
 * sem=4.38
 * %difference from accepted=26.77%

The value found in the first experiment leads me to believe that it is the most effective way of finding the e/m ratio. The accepted value falls within our 68% confidence interval of our data points. Also this method produced the best percent error and standerd error mean. Experiment 2 where only the current was varied produced the worst results.This leads me to believe that our the assumption of x=R/2 for the entire electron beam made a larger impact than originally predicted.

Conclusions
Since Experiment 2 produced large standard deviations and ratios that fall outside of the 68% confidence interval I chose Experiment 1 and 3 to be the most effective methods. Using weighted means I will now find the best value for the two values found in the two experiments.



\bar{x} = \frac{w_1 x_1 + w_2 x_2 + \cdots + w_n x_n}{w_1 + w_2 + \cdots + w_n}. $$



w_i = \frac{1}{\sigma_i^2}. $$

Since the standard deviation for Experiment three is so much larger than the standerd deviation for the first experiment the weighing the means shows that the third experiment is negligible in finding the best data point. Therefore my final value will remain.


 * $$ \frac{e}{m}=2.051e11 \ $$