Users: Alexandra S. Andrego and Anastasia A. Ierides/Notebook/Physics 307L/2009/10/26

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e/m Ratio for Electrons <html><img src="/images/9/94/Report.png" border="0" /></html> Alex's Main project page
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Please note that this lab notebook page is the combined efforts of Alex Andrego and Anastasia A. Ierides

Purpose

SJK 15:41, 14 November 2009 (EST)
15:41, 14 November 2009 (EST)
As usual, excellent primary lab notebook! Really great notes, very easy to follow. My main criticism in this lab would be that you jumped to conclusions on explanation of systematic error, and didn't explore other possibilities at all.
The purpose of this lab is to measure the charge to mass ratio (e/m) for electrons whilst studying the effects that electric and magnetic fields have on a charged particle. For N coils carrying current I with radius R, the magnetic field along the symmetry axis x is given by:
[math]\displaystyle{ B=\frac{\mu R^2NI}{(R^2+x^2)^{3/2}}\,\! }[/math]
You can see a more detailed purpose in Professor Gold's e/m Ratio for Electrons.

Brief Description of e/m

According to J.J. Thompson if the electron has finite charge e and has mass m it will obey the laws of motion for a charged particle moving through a magnetic/electric field. The e/m ratio is the charge to mass ratio of an electron which can be determined if the energy of the electrons and the magnetic field strength are known. "In this experiment [we will be] observing the behavior of electrons in a magnetic field and determine a value for the electron charge-to-mass ratio e/m. The apparatus consists of a large vacuum tube supported at the center of a pair of Helmholtz coils,... The vacuum tube contains an electron gun which produces a collimated beam of electrons that is deflected by a magnetic field. An electron gun has two main parts: a filament that produces electrons through thermionic emission, and an anode that is placed at high positive potential so as to accelerate thermal electrons from the filament to the main region of the vacuum tube,... The magnetic field produced by the Helmholtz coils deflects the electrons into circular trajectories and these paths are made visible through collisions by the electrons with a trace amount of mercury vapor present in the vacuum tube." The Electron Charge-to-Mass Ratio e/m

Equipment

  • Hewlett-Packard DC Power Supply (Model 6384A, 4-5.5V, 0-8A)
  • SOAR corporation DC Power Supply (Model 7403, 0-36V, 3A)
  • Gelman Instrument Company Deluxe Regulated Power Supply (500 V, 100 mA)
  • 2 BK PRECISION Digital Multimeter (Model 2831B, (1)SER. NO. 000-03-0618 & (2)SER. NO. 099-10-0357 , 5 WATTS, 8 VA,50~60 Hz)
  • e/m Experimental Apparatus (Model TG-13)

Safety

Before we begin, some points of safety must be noted:
  1. First and foremost your safety comes first and then the equipments'
  2. Check the cords, cables, and machinery in use for any damage or possible electrocution points on fuses of machinery by making sure the power cords' protective grounding conductor must be connected to ground
  3. Be careful to ground all power supplies properly before use
  4. Be careful while handling and working with the mercury tube
  5. Make sure the areas containing and around the experiment are clear of obstacles

Set Up

Total Apparatus
(bottom) Hewlett-Packard DC Power Supply (Model 6384A, 4-5.5V, 0-8A) and (top) BK PRECISION Digital Multimeter (Model 2831B, SER. NO. 000-03-0618 , 5 WATTS, 8 VA,50~60 Hz)
e/m Experimental Apparatus (Model TG-13)
(left of e/m E. A.) SOAR corporation DC Power Supply (Model 7403, 0-36V, 3A), (top right) BK PRECISION Digital Multimeter (Model 2831B, SER. NO. 099-10-0357 , 5 WATTS, 8 VA,50~60 Hz), (bottom right) Gelman Instrument Company Deluxe Regulated Power Supply (500 V, 100 mA)
measuring scale
image of electron beam


The procedure we followed was based on the descriptions given in Professor Gold's manual


  • Connect a regulated 6-9 Vdc supply rated at 2 A to the Helmholtz coil jacks using BCN cables
  • Connect the ammeter in series between the supply and the coil jacks
  • Connect the 6.3 V power supply (rated at 1.5 A0 to the heater jacks of the electron gun
  • Connect a high voltage source of 150-300 V dc rated at 40 mA to the Electrode jacks of the electron gun
    • The value of this voltage determines the average velocity of the electrons in the beam
  • Connect the dc voltmeter at the jacks labeled Voltmeter on the base panel
  • Turn the Current Adjust control to zero and set the switch on the panel to the e/m position
    • Nothing should be connected to the jacks labeled Deflection Plates at this time
  • Turn on the Heater supply and allow the electron gun filament to heat up for two minutes
  • Apply a 200 Vdc potential from the high voltage supply to the Electrodes
  • Turn off the light when ready to begin experimentation
  • Use a black cloth hood to mask the tube and to backdrop the beam while witnessing the beam of electrons
  • Turn on the coil current and increase the Current Adjust control until the beam forms a complete circle
  • Rotate the tube socket until the end of the curving beam strikes between the two wire leads
  • Use the scale located behind the bulb to measure the radius of the loop of the beam

Measurements and Data

{{#widget:Google Spreadsheet

key=tE2JkJPxeYd_iIjcg2Tr-8Q width=530 height=445

}}

Calculations and Analysis

SJK 15:37, 14 November 2009 (EST)
15:37, 14 November 2009 (EST)
Excellent raw data notebook & nice graphs. Was easy for me to follow your data and check things out. Also, looks like you took nice data, based on looking at your best fit lines.
In the Helmholtz configuration, from Professor Gold's Manual we are given:
[math]\displaystyle{ x=R/2\,\! }[/math], [math]\displaystyle{ N=130\,\! }[/math], and [math]\displaystyle{ R=0.15 m\,\! }[/math]
The permeability of free space is given as
[math]\displaystyle{ \mu=4\pi\times10^{-7}\frac{weber}{amp-meter}\,\! }[/math]
From these values we can calculate:
[math]\displaystyle{ B=\frac{\mu R^2NI}{(R^2+x^2)^{3/2}}\,\! }[/math]
[math]\displaystyle{ B=\frac{\mu R^2NI}{(\frac{5}{4}R^2)^{3/2}}\,\! }[/math]
[math]\displaystyle{ B=\frac{\mu NI}{(\frac{5}{4})^{3/2}R}\,\! }[/math]
[math]\displaystyle{ B=(7.8\times10^{-4}\frac{weber}{amp-meter})\times I\,\! }[/math]
We know that:
[math]\displaystyle{ {e}{V}=\frac{1}{2}{m}{v}^{2}\,\! }[/math]
[math]\displaystyle{ {F}_{B}={q}{v}{B}\,\! }[/math]
So,
[math]\displaystyle{ \frac{e}{m}=\frac{{2}{V}}{{r}^{2}{B}^{2}}=\frac{{2}{V}}{{r}^{2}{({{7.8}\times10}^{-4}{I})}^{2}}\,\! }[/math]
In order to find [math]\displaystyle{ \frac{e}{m}\,\! }[/math] we need to plot
  1. [math]\displaystyle{ r^2\,\! }[/math] vs. [math]\displaystyle{ V\,\! }[/math], where [math]\displaystyle{ I\,\! }[/math] is constant
  2. [math]\displaystyle{ \frac{1}{r}\,\! }[/math] vs. [math]\displaystyle{ I\,\! }[/math], where [math]\displaystyle{ V\,\! }[/math] is constant
The current accepted value of [math]\displaystyle{ \frac{e}{m}\,\! }[/math] is:
[math]\displaystyle{ \frac{e}{m}=1.76\times10^{11}\frac{C}{kg}\,\! }[/math]
  1. We can plot [math]\displaystyle{ r^2\,\! }[/math] vs. [math]\displaystyle{ V\,\! }[/math] with this equation:
[math]\displaystyle{ r^2=\frac{2V}{({7.8\times10^{-4}{I})}^{2}}\times\frac{m}{e}\,\! }[/math]
treating [math]\displaystyle{ \frac{2}{({7.8\times10^{-4}{I})}^{2}}\times\frac{m}{e}\,\! }[/math] as a constant which gives us a linear relationship with this as our slope.
In order to determine the slope we need to plot our data:

{{#widget:Google Spreadsheet

key=tpnytH1JPN416hUnQ12M-1w width=775 height=445

}}

From this graph we have that:
[math]\displaystyle{ slope=0.0000049942\frac{m^2}{V}\,\! }[/math]
We also have that the equation of slope is:
[math]\displaystyle{ slope=\frac{2}{({7.8\times10^{-4}{I})}^{2}}\times\frac{m}{e}\,\! }[/math]
Therefore we can calculate the ratio of [math]\displaystyle{ \frac{e}{m}\,\! }[/math] by:
[math]\displaystyle{ slope=0.0000049942\frac{m^2}{V}=\frac{2}{({7.8\times10^{-4}{I})}^{2}}\times\frac{m}{e}\,\! }[/math]
[math]\displaystyle{ \frac{e}{m}=\frac{2}{0.0000049942\times({7.8\times10^{-4}{I})}^{2}}\,\! }[/math]
Where,
[math]\displaystyle{ I=1.34 A\,\! }[/math]
So we have:
[math]\displaystyle{ \frac{e}{m}=\frac{2}{0.0000049942\times({7.8\times10^{-4}\times{1.34})}^{2}}\,\! }[/math]
[math]\displaystyle{ \simeq3.666\times10^{11}\frac{C}{kg}\,\! }[/math]
Taking into account our slope uncertainty value of [math]\displaystyle{ \pm 0.00000072021\,\! }[/math] the range of our measured ratio [math]\displaystyle{ \frac{e}{m}\,\! }[/math] is...
[math]\displaystyle{ 3.204\times10^{11}\frac{C}{kg}\leq \frac{e}{m}\leq 4.283\times10^{11}\frac{C}{kg}\,\! }[/math]
SJK 12:37, 14 November 2009 (EST)
12:37, 14 November 2009 (EST)
As opposed to below, here I get the same numbers as you do. So, I guess you just made a mistake below? In any case, even here it's important to provide some more information as to how you're calculating the error propagation. You can explicitly say that you're treating any other uncertainties (besides slope) as insignificant, and thus you can use this exact method
2. We can plot [math]\displaystyle{ \frac{1}{r}\,\! }[/math] vs. [math]\displaystyle{ I\,\! }[/math] with this equation:
[math]\displaystyle{ \frac{1}{r}=\sqrt{\frac{(7.8\times10^{-4})^{2}}{2V}\times\frac{e}{m}}\times I\,\! }[/math]
treating [math]\displaystyle{ \sqrt{\frac{(7.8\times10^{-4})^{2}}{2V}\times\frac{e}{m}}\,\! }[/math] as a constant which gives us a linear relationship with this as our slope.
In order to determine the slope we need to plot our data:

{{#widget:Google Spreadsheet

key=tAqH3es5Nr1JgnqrFSEEA-w width=775 height=445

}}

  • Note: Due to the fact that all values of current are negative, we took the absolute value of the current for graphing purposes because the negative sign was purely directional.
From the above graph we have that:
[math]\displaystyle{ slope=16.93945521\frac{1}{A m}\,\! }[/math]
We also have that the equation of slope is:
[math]\displaystyle{ slope=\sqrt{\frac{(7.8\times10^{-4})^{2}}{2V}\times\frac{e}{m}}\,\! }[/math]
Therefore we can calculate the ratio of [math]\displaystyle{ \frac{e}{m}\,\! }[/math] by:
[math]\displaystyle{ slope=16.93945521\frac{1}{A m}=\sqrt{\frac{(7.8\times10^{-4})^{2}}{2V}\times\frac{e}{m}}\,\! }[/math]
[math]\displaystyle{ \frac{e}{m}=\frac{(16.93945521)^2\times2V}{(7.8\times10^{-4})^{2}}\,\! }[/math]
Where,
[math]\displaystyle{ V=350 V\,\! }[/math]
So we have:
[math]\displaystyle{ \frac{e}{m}=\frac{(16.93945521)^2\times2\times350}{(7.8\times10^{-4})^{2}}\,\! }[/math]
[math]\displaystyle{ \simeq3.301\times10^{11}\frac{C}{kg}\,\! }[/math]
Taking into account our slope uncertainty value of [math]\displaystyle{ \pm 1.730792\,\! }[/math] the range of our measured ratio [math]\displaystyle{ \frac{e}{m}\,\! }[/math] is...
[math]\displaystyle{ 2.408\times10^{11}\frac{C}{kg}\leq \frac{e}{m}\leq 3.699\times10^{11}\frac{C}{kg}\,\! }[/math]
SJK 12:29, 14 November 2009 (EST)
12:29, 14 November 2009 (EST)
I get the same mean value you do (3.3E11), but cannot figure out how you get that range. More explanation is needed, especially since many people may think you're using error propagation, whereas you may also be plugging in (slope-uncertainty) and calculating low e/m etc.


Notes about Our Uncertainty

Systematic Error:

  • The only way to achieve measurements for the radius of our electron beam was to measure by eye using a fixed ruler in the back of our apparatus. This could cause a huge source of error because we were basically estimating for each measurement. This also caused a lot of rounding error because we were estimating we were not able to give very specific measurements.
Steve Koch 15:39, 14 November 2009 (EST): As I mentioned on your summary pages, I don't think this is the source of your systematic error. Unfortunately we didn't get to talk about this at all during the lab class, but the experiment is basically forced to have a lot of systematic error, and you can hypothesize on the reasons if you do this lab for your formal report.

Summary

If you wish to see Alex Andrego's informal summary of this lab follow this link
If you wish to see Anastasia Ierides's informal summary of this lab follow this link

Acknowledgments

Prof. Gold's Lab Manual served as a loose guideline for our lab procedure
The following link was used in our "Brief Description of the e/m ratio" above The Electron Charge-to-Mass Ratio e/m
Professor Koch and Pranav for always being of great help to us!


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