User:Garrett E. McMath/Notebook/Junior Lab/2008/09/15

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Lab 2: The Ratio e/m for Electrons

Notes By: Garrett McMath and Paul Klimov

SJK 00:22, 4 October 2008 (EDT)
00:22, 4 October 2008 (EDT)
Your raw data notebook is very good, reflecting the very good work you did in class. As I mention below, make sure to take a look at Paul's notebook for more feedback.

Purpose and Objectives

The purpose of this laboratory is to measure the e/m ratio of electrons. This will be done by accelerating electrons into a strong magnetic field, which will cause the electrons to rotate in a circle on a plane perpendicular to the field.

Theory

The Lorentz force, in the absence of an Electric field gives us:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \vec{F}=e(\vec{v} \times \vec{B}) = m \frac{\vec{v}^{2}}{R}}

this implies:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{e}{m}=\frac{|\vec{v}|}{R|\vec{B}|}}

The electrons are accelerated through a potential V, implying:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{1}{2}mv^{2}=eV}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle v=\sqrt{\frac{2eV}{m}}}

Now, we use this velocity in the above equation for the ratio e/m:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{e}{m}=\sqrt{\frac{e}{m}}\frac{\sqrt{2V}}{RB}}

This, then, boils down to give e/m ratio in terms of V, r, and B, all of which are variables that we can find:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{e}{m}=\frac{2V}{(RB)^{2}} }

From here, we must find the strength of the magnetic field which is produced by the Helmholtz coils: We use the Biot-Savart Law, to find the field due to one coil, and then add the field due to two coils. A single coil, with N loops, of radius R, stands with its opening aligned with the y-axis. The point of interest will be at y=a. Due to symmetry, we know that there will only be a field in the positive y direction (where current flows counter-clockwise, as seen looking from positive y). Theta will rotate in the xz plane, which will allow us to define a small current element. Biot-Savart Law:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle d\vec{B}=\frac{\mu_{0}i}{4\pi}\frac{ \vec{dl}\times\vec{r}}{r^{3}}}

we know that dl is perpendicular to r always, and so we can turn this into a scalar equation for the y coordinate.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle dl = R d\theta }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}=\frac{N\mu_{o}i}{4\pi}\int_{0\leq\theta\leq2\pi}\frac{Rd\theta}{(a^{2}+R^{2})}\frac{R}{\sqrt{a^{2}+R^{2}}}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}=\frac{N\mu_{o}i}{4\pi}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}\int_{0\leq\theta\leq2\pi}d\theta }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}=\frac{N\mu_{o}i}{2}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}} }

Now, because we have 2 coils, which make up the Helmholtz coils, we simply multiply the above expression to find the field as a function of the current:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}(i)=\frac{N\mu_{o}iR^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}} }

with this information, we should be able to complete the lab.

Experimental Setup

  • UCHIDA YOKO e/m experimental apparatus. Model TG-13.

Heater connected to SOAR PS-3630. Accelerating Electrodes connected in series with AMPROBE multimeter and then to the Gilman power supply. Voltmeter jacks connected to BK Precision multimeter Helmholtz coils connected in series with FLUKE multimeter to SOAR 7403 DC Power Supply.

  • SOAR DC Power supply rated 36V, 4A. Model Number PS-3630.

Connected in series with an ammeter/voltmeter and connected to the heater.

  • SOAR 7403 DC Power supply rated 36V, 4A. Serial Number 303018.

Connected in series with an ammeter/voltmeter and connected to the Helmholtz coils.

  • BK Precision Digital Multimeter Model Number 2831B.

Connected directly to the voltmeter jacks on the e/m experimental apparatus.

  • (Gelman Instrument Company. Deluxe Regulated Power Supply rated 500V, 100mA.

Connected in series with an ammeter/voltmeter and connected to the electron gun.

  • FLUKE 111 True RMS multimeter 10A fused. CATIII 600V.

Connected in series to the SOAR 7403 and e/m experimental apparatus.

  • AMPROBE 37XR-A. TRUE RMS CATII 1000V, CATIII 600V. 10A Max fused

Connected in series between e/m apparatus and Gelman power supply.

  • Cloth placed over the coils so that the electron beam can be seen later on in the experiment.


NOTE:

  • During the second week, we set up everything the same as above, except for one exception. The BK percision multimeters appeared to be working incorrectly. Because of this, we set up an extra multimeter to measure the accelerating potential.
  • AMPROBE 37XR-A. TRUE RMS CATII 1000V, CATIII 600V. 10A Max fused.

Procedure

1.) Turn on SOAR PS-3630 PSU for heater to:

  • 1.500A Rated. (.7A)
  • 6.302V

Nothing connected to ground. This starts the heating of the filament. It starts radiating as it heats up.

2.) Turn on Gilman power supply for electron gun

  • 200V (this will be changed as necessary when we carry out our experiments)
  • .012mA (this will be changed as necessary when we carry out our experiments)

3.) Settings on e/m experimental apparatus:

  • Current Adjust turned to zero at start. We will turn this up later on to get current flowing through the coils to bend the beam.
  • Focus kept in place for now -- this will be changed to get a good beam later.
  • Set to e/ SJK 23:23, 3 October 2008 (EDT)
    23:23, 3 October 2008 (EDT)
    I notice this sentence is not completed here...I'm guessing you guys used Paul's notebook to take the original notes? I think linking to the original notes (instead of copying them) would be fine. In either case, it's a good idea to look them over to correct mistakes like these (if you can remember what you meant to write).

    You and Paul took great notes, and I put some comments in his book, potentially regarding stuff you both did. So make sure to look at those comments on his page. (Lots of positive comments and some negative)

Experimental Procedure

  • To measure the radius of the beam, we lined up the actual beam with the reflection on the mirror on the back of the apparatus. Our measurements were compared and we decided on a radius, and decided on a reasonable uncertainty.
  • Possible Source of error: Our voltmeters accuracy doesn't go into the decimals.
  • Each data point has an uncertainty of at least ±.1 cm because this is roughly the width of the beam. We will have to see later what kind of error this will produce on our measured e/m ratio. m experimental apparatus to e/m measure (instead of Electrical deflect).

Data

Source of error: Our voltmeters accuracy doesn't go into the decimals. Each data point has an uncertainty of at least ±.1 cm because this is roughly the width of the beam

Constant Current

  • i=1.069A

1.)200V, 4.1±.1cm

2.)210V, 4.35±.1cm

3.)220V, 4.45±.1cm

4.)230V, 4.5±.1cm

5.)240V, 4.55±.1cm

6.)250V, 4.60±.1cm

7.)260V, 4.65±.1cm

8.)270V, 4.70±.1cm

9.)280V, 4.75±.2cm

10.) 290V, 4.80±.3cm

11.) 300V, 4.90±.3cm

At the end, the beam seemed to get a bit thicker. This introduced more uncertainty into our measurements.

Constant Voltage

One source of experimental error: The beam seems like it is somewhat elliptical. The radius on the right is slightly smaller than on the left. We have decided to use only the left radius. This will definitely introduce uncertainty into our calculations. However, after rotating the bulb, we have a spiral forming. With the spiral, we cannot take any measurements because there is simply too much going on.

V=280V

1.)i=1.000A, 4.75±.2cm

2.)i=1.050A, 4.75±.2cm

3.)i=1.100A, 4.60±.1cm

4.)i=1.151A, 4.55±.2cm

5.)i=1.202A, 4.47±.1cm

6.)i=1.253A, 4.35±.1cm

7.)i=1.297A, 4.25±.2cm

8.)i=1.351A, 4.15±.2cm

9.)i=1.399A, 4.05±.2cm

10.)i=1.449A, 4.00±.2cm

ANOTHER ROUND OF DATA FOR CONSTANT VOLTAGES:

  • We repeated this in attempt to get better measurements. However, the circle keeps moving laterally. We will re center the ring each time. Here we must assume that the beam is a perfect circle each time. The voltage across the coils was untouched.
  • Another Source of error:The ruler is aligned with the center of the glass globe. Therefore, once the radius of the beam becomes too small, the beam is no longer aligned with the ruler. This could definitely cause error in our calculations. The diameter of the circle must be on the same horizontal plane for the measurements to be consistent.

V=299.8

1.)i=1.245A, R=3.90±.1cm

2.)i=0.989A, R=4.25±.2cm


  • The above data was abandoned. At that high of a voltage, the radii was only reasonable for a small range of currents.

V=249.6V

1.)i=.997A, R=3.95±.1cm

2.)i=1.252A, R=3.70±.2cm

LAB SUMMARY

SJK 00:34, 4 October 2008 (EDT)
00:34, 4 October 2008 (EDT)
I do want your lab summary to be on a separate page. And also, actually a bit more concise. You will want to have this much analysis written down in your lab notebook, but then you want to sumamarize it on a separate page. Check out PK's summary for an example.

Lab Manual Questions

1. Why do we see the electron beam at all?

The visual beam that we see is merely the ionization of the helium gas caused by the electrons losing energy to the helium causing the helium to emit photons of such wavelength in the visible light spectrum.

2. We ignored the Earth’s magnetic field in our procedure. How much error does this introduce into this experiment?

Using the website http://www.ngdc.noaa.gov/geomagmodels/IGRFWMM.jsp, you can get the magnectic field and its components of your location. The total magnetic field in Albuquerque is 50,292.7 nT, with components of 23,027.7 nT North, 3858.3 nT East, 23,348.7 nT Horizontal, and 44,544.3 nT Upward. Our apperatus was more or less aligned with the east component of the earths B-field so plugging this value in using 300 V will give a reasonable approximation of the error the earths field could have caused.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle R = \sqrt{\frac{2Vm}{e}}\frac{1}{B} = 15.15m }

This value is actually pretty large using MatLab paul analyzed this result which you can see in the data analysis section.


3. Suppose that protons were emitted in the vacuum tube instead of electrons. How would this effect the experiment?

Using the same formula as in question 2:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{R_{p}}{R_{e}}=42.81 }

So other than the difference in polarity the major difference would be a much stronger magnetic field would be needed to bend the beam.

4. Show that if the magnetic field is held constant, the time t required for an electron to make a complete circle in your e/m tube and return to the anode is independent of the accelerating voltage by deriving an expression for this time.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle T=\frac{2\pi R}{v} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle R = \sqrt{\frac{2Vm}{e}}\frac{1}{B} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle v = \sqrt{\frac{2eV}{m}} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle T = \frac{2\pi m}{e B} }

Data Analysis

SJK 00:06, 4 October 2008 (EDT)
00:06, 4 October 2008 (EDT)
First of all, it is really great how you cite Paul for creating the MatLab codes and tell the reader where to find those codes. That is really great scientific practice. However, I'm going to clarify next week in lecture this point: For the lab summaries, your goal should be to do the data analysis and discussion "yourself." Meaning, it's OK to discuss and work with other people, but you want to learn how to do the calculations and analysis in the absence of the other person. For example, in this case, you'd want to know how to write your own matlab code. Or, if you don't have access or know how to use matlab, then you'd need to use another software application (like Excel). From what I read in your book, I don't think you learned as much as you could have, though I do know you were discussing these things with Paul. An example of what I'm saying is you put down a significant amount of results without explaining what they mean or where they came from. Like I said, we'll talk about this on Monday and please ask me for clarification as well. In your next lab, here's an example of what you could say, "During lab session, Paul and I analyzed our data using Matlab software that he wrote. We generated a bunch of figures and results that are here (link). I wrote my own Excel sheet to replicate these results, and the sheet is here (link)..." Then go on to discuss the methods of analysis and what you think it means.
  • Data analysis was done using programs written by Paul Klimov(for MatLab code of programs used see Paul Klimov's Lab notebook under MATLAB CODE)

File:ConstantI2.tif File:ConstantIDiffSlope.tif File:ConstantV.tif File:ConstantVDiffSlope.tif File:Empoints2.tif File:Emvsradius.tif

Programs

Using data analysis programs written in MATLAB by Paul klimov we calculated least squares regressions for both sets of data, constant voltage and constant current. The program then calculated an e/m ratio for both sets of data using the formula that is shown in theory section.

Margin of Error

As will be discussed later in sources of error we had uncertainty in all of our measurements. The most obvious reason being the width of the beam caused a minimum uncertainty of one millimeter, though we often had larger uncertainty either due to thicker beams caused by higher voltage, or simply a large discrepency between Paul's and my measurements.

Constant Current

Using the MatLab programs, the following information was obtained:

  • Slope: .0836 ± 0.00113e-4 m^2/V
  • Corresponding e/m Ratio: 3.440955866544257e+11 C/kg
  • Corresponding e/m Ratio with Earths Magnetic Field: 3.370766636622313e+11 C/kg
  • Mean e/m Ratio:3.415607994557289e+11 C/kg
  • Mean e/m Ratio with Earths Magnetic Field: 3.391161980958560e+11 C/kg
  • Standard Deviation: .1538626188222192e+11 C/kg


Slope Modification: Paul also had Matlab modify the linear regresion to fit within the error bars that data is included here along with the modified graph, figure 1b.

  • Modified Slope: 0836 + 0.00113e-4 m^2/V
  • Modified Corresponding e/m Ratio:3.395065625434909e+11 C/kg

Constant Voltage

  • Slope:0=5.269 ± 0.1 e-2 m/A
  • Corresponding e/m Ratio:3.315449555220567e+11 C/kg
  • Corresponding e/m Ratio with Earths Magnetic Field: 2.879693734013608e+11 C/kg
  • Mean e/m Ratio:3.258205230221071e+11 C/kg
  • Mean e/m Ratio with Earths Magnetic Field: 3.237264954654076e+11 C/kg
  • Standard Deviation: .4224672028400931e+11 C/kg

Slope Modification: Figure 2b

  • Modified Slope: 5.369 e-2 m/A
  • Modified Corresponding em ratio: 3.193096279082144e+11 C/kg

Final Stats

SJK 23:46, 3 October 2008 (EDT)
23:46, 3 October 2008 (EDT)
There is missing any description of what "overall" and "best" mean. There are also the last two figures without any discussion of why or what they are.
  • overall mean e/m ratio: 3.336906612389180e+11 C/kg
  • overall mean e/m ratio with earths magnetic field:3.314213467806318e+11 C/kg
  • overall standard deviation: .3140717276887458e+11 C/kg
  • overall mean percent error: 89.99%
  • overall mean percent error with earths magnetic field:88.70%


  • best e/m ratio: 2.739946406166065 C/kg
  • best e/m ratio with earths magnetic field: 2.725458559266957 C/kg
  • best percent error: 56.01%
  • best percent error with earths magnetic field:55.18%


Sources of Error

SJK 00:01, 4 October 2008 (EDT)
00:01, 4 October 2008 (EDT)
This is a good discussion of error sources.

This lab we were told has one of the worst percentage errors of any of the labs we will be doing this semester. With that in mind we watched very closely for sources of this while doing the lab.

  • The first and most obvious source of error is the extremely difficult task of getting an accurate reading of the radius of the electron beam. The only way using the equipment we had is to line up one of the sides of the beam with its reflection on a ruler behind the beam. This presented a multitude of problems. First of all we had to do this in the dark and it was difficult to get a reading at all. Secondly the ruler was at a fixed height while the beam radius would change quite a bit often bringing the center of the circle below the ruler making a direct reading impossible. We had to visually infer where we thought the edge of the beam would hit the ruler and take reading that way. In addition to all those problems the beam was rarely an actual circle. More often than not the beam was elliptical probably due to the loss of energy to the Helium atoms. Also the beams thickness made a minimum margin of error on every reading given that it was at least a millimeter wide. This aspect of the lab produced a lot of error into our data though looking at the amount we would have to be off by to get the accepted value of the e/m ratio this source of error would not account for the massive discrepency between our value and that of the accepted.SJK 00:29, 4 October 2008 (EDT)
    00:29, 4 October 2008 (EDT)
    This is a really good point...your random errors, while noticeable, are much smaller than the obvious systematic error.
  • A second source of error we discussed was we did not align the coils with the magnetic field of the earth which could cause an unaccounted for B-field to affect the beam. Through calculation we determined that this source of error while real had negligible effects on the expiriment due to the high current we were using to create the magnetic field in the coils.
  • A source of error that discussed though never really proved was not taking into account was the energy loss of the electron in the collision with the gas. This is probably one of the more influencial sources of error in this expiriment. An interesting sidenote was the difference of color of of the beam at times which seems to indicate a difference in energy in the beam itself. Paul had a theory of the leading electrons losing more energy than the following but this would not explain the difference in color in certain pieces of the beam. Another explanation could be fluxuations or non-symetries in the B-field controlling the beam.
  • An interesting note to make is that the heating filament was actually adjustable too. We did this at the very end of the lab so we did not have much time to expiriment with this but it definatly had a huge effect on the beam radius. Turning up the voltage even slightly up the radius would almost exceed the limits of the bulb.SJK 00:31, 4 October 2008 (EDT)
    00:31, 4 October 2008 (EDT)
    I am still intrigued by this and the color effect. Did you try the filament in combination with the color effect? I think like you say there's some things to explore if there is more time.

Improvements

  • I would say most of the easiest fixs are for the visual readings that we determined are not an important factor in the error. The main fix is to somehow reduce the amount of energy loss to Helium, perphaps by changing the gas used so the energy loss is very low though we would probably need some optical device for seeing the emitted photons as they would hopefully not be in the visble light range anymore. SJK 00:23, 4 October 2008 (EDT)
    00:23, 4 October 2008 (EDT)
    Good thought. Perhaps a camera with open shutter would do the trick. Or a phosphors screen like PK mentioned.