# User:Paul V Klimov/Notebook/JuniorLab307L/2008/09/15

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# Lab 2: The Ratio e/m for Electrons

SJK 23:01, 3 October 2008 (EDT)
23:01, 3 October 2008 (EDT)

Notes By: Paul Klimov and Garrett McMath

## 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. Using the Lorentz force law for charged particles in a magnetic field, the e/m ratio will be determined. The calculations for this experiment will be based on derivations made in the 'Theory' section that follows.

## Theory

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

$\vec{F}=e(\vec{v} \times \vec{B}) = m \frac{\vec{v}^{2}}{R}$

this implies:

$\frac{e}{m}=\frac{|\vec{v}|}{R|\vec{B}|}$

The electrons are accelerated through a potential V, implying:

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

$v=\sqrt{\frac{2eV}{m}}$

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

$\frac{e}{m}=\sqrt{\frac{e}{m}}\frac{\sqrt{2V}}{RB}$

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:

$\frac{e}{m}=\frac{2V}{(RB)^{2}}$

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:

$d\vec{B}=\frac{\mu_{0}i}{4\pi}\frac{ \vec{dl}\times\vec{r}}{r^{3}}$

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

dl = Rdθ

$B_{y}=\frac{N\mu_{o}i}{4\pi}\int_{0\leq\theta\leq2\pi}\frac{Rd\theta}{(a^{2}+R^{2})}\frac{R}{\sqrt{a^{2}+R^{2}}}$

$B_{y}=\frac{N\mu_{o}i}{4\pi}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}\int_{0\leq\theta\leq2\pi}d\theta$

$B_{y}=\frac{N\mu_{o}i}{2}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}$

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:

$B_{y}(i)=\frac{N\mu_{o}iR^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}$

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

## Experimental Setup

SJK 14:49, 3 October 2008 (EDT)
14:49, 3 October 2008 (EDT)
Excellent notes! A picture would help too, but this is very good.
• 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/m experimental apparatus to e/m measure (instead of Electrical deflect).

4.) Paul V Klimov 11:17, 24 September 2008 (EDT) The last thing that was turned on was the SOAR power supply running current through the Helmholtz coils. Nothing was connected to ground. The current was run throuh a multimeter to measure the current. The applied potential was measured on a seperate multimeter that was connected in parallel "around" the coils.

## 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. Mine and Garrett's 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. SJK 14:52, 3 October 2008 (EDT)
14:52, 3 October 2008 (EDT)
This is a reasonable way to estimate the uncertainty. In principle, you can be more certain than the width of the beam (you could pick an edge, or the middle, e.g., with better precision than the width)...but this is a good estimate.

NOTE: A COMPREHENSIVE LIST OF POSSIBLE SOURCES OF ERROR ARE LISTED LATER!

## Data

### 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

3.)

Paul V Klimov 11:23, 24 September 2008 (EDT) The set of data immediately above this was abandoned due to the time constraint. In addition, we had noticed that we could only attain measurable (large) radii for a small combination of currents and voltages. This range was very close to our first two sets of measurements, and so we ended up going through and checking some of our first measurements instead. During this checking process, we observed radii that were consistent with our first results. I will add more later...

## Possible Sources of Error

• Magnetic Field of Earth: At first we thought this could be an issue, because we hadn't aligned the coils with the magnetic field of the earth. However, after looking up the B field here in Albuquerque, we found that it had negligible effect on the e/m ratio. Paul V Klimov 01:39, 28 September 2008 (EDT) I have had second thoughts and have decided to include the east component of the earths magnetic field in the calculations for the em ratio mean and the em ratio calculated from the slope. Eastern B-Field in Albuquerque: 3853.8 nT. This will be added to the magnetic field produced by the coils, as this is about the same direction that the magnetic field in the coils was pointed in.
• Width of the Beam: This was definitely a concern of ours. The beam is roughly 1mm in width when the radius was small, and appeared to grow as the radius got larger. This uncertainty was noted in each measurement. SJK 14:55, 3 October 2008 (EDT)
14:55, 3 October 2008 (EDT)
What does the width of the beam represent? And based on your hypothesis for that, which would be a better point to measure, the inside or the outside of the beam?
• Not Measuring B Field Directly: The magnetic field due to the Helmholtz coils was derived above. However, we never actually measured the field directly. Not only could our result from theory be slightly off, but we could have significant fringing, which would have changed the magnitude of the magnetic field.
• Optical distortions: Refraction and reflection due to the glass envelope undoubtedly affected our measurement of the radius.
• Ruler is not movable : As the radius of the beam decreases, the diameter falls out of the horizontal plane of the ruler. This could definitely cause errors in our measurement of the radius of the beam.
• Beam is somewhat elliptical: A possible cause for this is that the beam electrons are constantly colliding with He atoms. Therefore, as the beam goes further along its circumference, the beam loses more and more energy, which would affect the radius of the beam.
• Beam Intensity: The beam intensity is related to the temperature of the filament. If the filament is hotter, more electrons will be ejected than if the filament had been colder. With more electrons going around in the beam, it could be possible to get a more uniform radius. This remark relies on the assumption that the electrons at the front of the beam, at any instantaneous moment, would collide with a bulk of the helium atoms in the way, while the electrons behind them pass through relatively unobstructed. This could be a false assumption, however. To see if this will have an effect on the experiment, we will adjust the heater voltage and current and see if the radius of the beam changes:::
• When changing the current, the radius of the ring changed significantly. When lowering the current the ring got smaller, and when raising the current, the ring got larger, as one would expect. Therefore, there is absolutely no doubt that this had an effect on our measured e/m ratio.
SJK 15:20, 3 October 2008 (EDT)
15:20, 3 October 2008 (EDT)
This is a very cool and thoughtful analysis! I agree with your assessment about the "UV" (or higher energy) tails being the electrons that get sufficient energy to surpass the work function. Maxwell velocity distribution is probably the relevant distribution. I haven't thought about this carefully, but it would seem that since the peak of the distribution (apparently chi-squared distribution for energy) is below the work function, that the vast majority of electrons that actually are above the work function will not be much higher than the work function. That is, the probability distribution function will be dropping off, so the most likely kinetic energy (after losing work function) will be zero. Also, it may be an even steeper fall-off if we consider it's only one dimension we care about? Hmmm...don't trust any of my analysis. I agree with your conclusion, though that I wouldn't expect the electrons to have much more than a few KT's of excess energy. Another piece of evidence would be the fact that cathode ray tubes (like TVs) can produce pretty sharp electron beams?

Paul V Klimov 17:16, 23 September 2008 (EDT) Above I mention that the energy of electrons ejected increases as the filament gets hotter. I thought that the energy given to the electrons just by the filament could be significant enough to change the radius of the beam. After thinking about this, however, I have changed my mind (with the help of my discussion with Dr.Koch.) I made some pretty crude order of magnitude calculations to justify this. First, I computed the temperature of the filament, based on the fact that it was radiating at about 580nm (orangish). Thus, for an ideal blackbody, by Weins law:

$\lambda_{m}T = 2.9\cdot10^{-3} m\cdot K$

Now, because the filament was glowing orange, I will assume its wavelength was roughly 580nm. Wiens law returns:

$T \approx 5000K$

Then, approximating (once again very crudely!) the energy of an electron as a gas molecule with 3 degrees of freedom, we can see that its energy is:

$E_{e} \approx \frac{3}{2}k_{B}T$

This, then, returns an energy of:

$E \approx .7eV$

Clearly, this calculation is not perfect, because even Cesium doesn't have that low of a work function. I think that there is probably a significant amount of radiation at lower wavelengths, including even UV, which I didn't account for in the above calculations, which would raise the energy above typical work-functions. Nonetheless, I think that a reasonable upper bound on the kinetic energy of the electron as soon as it leaves the filament is under 10eV. Clearly this energy is quite a bit lower than the 200+eV the electrons gain after being accelerated through the 200+ Volt difference in the 'gun'. Therefore, I conclude that the initial energy of the electrons, immediately after being ejected from the filament, is not significant enough to change our results considerably.

## Qualitative Experiments and Other Interesting Sightings

• When the glass envelope is turned sideways a spiral forms. This spiral is because the electrons come out with a velocity that is parallel to the magnetic field. Therefore the field doesn't put any force on them. The velocity components that are perpendicular to the field are acted on by the B field, which causes the electrons to spin. This combination of spin and unaffected velocity results in a spiral.
• We changed the polarity of the coils by running current through them in the opposite direction as before. This change in polarity causes the magnetic field to reverse direction. By the Lorentz force law, we expect the electrons to bend in the opposite direction as before, hitting the glass envelope. This is what is observed.
• Deflection experiment: Here we ran current through the deflection plates, in parallel with the electron gun. The power supply powering the coils was turned off, to get rid of the magnetic field. When we follow the given procedure, the electrons get deflected vertically immediately. This is because of the newly present electric field.
• Paul V Klimov 11:12, 24 September 2008 (EDT) Another interesting thing that was noticed, but not written down during lab (because of time) was the color of the beam atlow radii. Specifically, the beam seemed to disappear when it reached the top of the circle, then reappearing again as it finished its trip. This is very interesting because clearly the electrons have more energy at some points than at others. What could be causing this is not very clear, however. One possibility could be that the field produced by the coils fringes considerably (I will have to think about this).SJK 15:22, 3 October 2008 (EDT)
15:22, 3 October 2008 (EDT)
You mean the electric field from the plates, right? I agree that this is a very cool effect, and my best guess, too, is that the field fringing is significant

## Post Experimental Data Analysis

NOTE: Earths magnetic field not represented in the Figures. But this field is taken into account in some of the results, where mentioned! I used the eastward magnetic field in Albuquerque (3853.8 nT), because this is about the direction that the magnetic field in the coils was pointed in (I had to use Google maps for this one.). This probably didn't affect our results significantly, but it is worth mentioning. Ideally, we would have measured this during the lab, and aligned the coils exactly with the field in some known direction.

SJK 15:25, 3 October 2008 (EDT)
15:25, 3 October 2008 (EDT)
I haven't read through everything yet, but I don't understand why the red line (presumably a fit) does not go through the data points. An explanation of what the red line is in the figure captions would head off confusion.

### Inputing Data

After the data was gathered, I wrote a program in MATLAB that took the data, calculated least squares fits, and calculated the e/m ratios. The e/m ratios were calculated using both the least squares fits, and also using the general equation that was developed in the theory section. The program was also coded to produce means and standard deviations, when I thought necessary. I also coded the program to produce several figures. These figures and calculation methods will be discussed below.

### Uncertainty

When obtaining measurements, it was clear that the minimum uncertainty for each measurement would be 1mm, which was the roughly the width of the beam (which we actually realized got larger as the radius increased). Then, after deciding the radius of the beam, my partner and I decided on an uncertainty based on other factors that could contribute to an erroneous observation of the radius. For example, sometimes the beam seemed to be thicker than other times, which was reflected in our decision for a reasonable uncertainty. Likewise, as the beam radius got smaller, the beam moved below the horizontal plane of the ruler, which would have also added to our uncertainty, if we though it was necessary.

No uncertainty was noted for the currents or voltages of any of the devices. The currents were monitored at all times for all connections with multimeters that produced several significant figures. Likewise, the potential across the heater and the electron gun were also monitored constantly. Before taking our data, we waited for all values to stabilize. When the data was taken all available digits were noted, without rounding.

Error bars were produced differently for each figure, based on our uncertainties and reasoning. The way in which I calculated the error bars will be discussed below as I analyze each experiment and plot individually:

### Constant Current Discussion

Figure 1a Constant Current
Figure 1b Constant Current with modified slope

The first set of data was taken at constant current (of the Helmholtz coils), when only the voltage of the electron gun was changed. After producing a plot of the radius^2 vs the accelerating voltage (see Figure1), I had MATLAB calculate the least squares fit of the line, the corresponding em ratio, mean, and standard deviation. The e/m ratio was extrapolated from the slope by writing r^2 as a function of the accelerating voltage.

$r^{2}(V)=(\frac{m}{e})\frac{2}{B^{2}}V$

$slope=(\frac{m}{e})\frac{2}{B^{2}} \rightarrow \frac{e}{m}=\frac{2}{B^{2} slope}$

Error bars for the plots were made in the following way: First, I calculated the radius without uncertainty and squared it. In a separate calculation, I added the radius to its uncertainty and squared that. I then subtracted these quantities, taking the resulting value to be the spread of the error bars above and blow the measured point.

SJK 15:35, 3 October 2008 (EDT)
15:35, 3 October 2008 (EDT)
I think we talked about this a bit in class this week. And we'll definitely be talking about this a lot more this semester. I really like that you noticed this and thought about it! BTW: There's nothing wrong with asymmetric error bars on a plot, as long as you explain what they represent (__% confidence interval, e.g.)

NOTE: I realize this must not be the best way to make error bars in this situation. If I had subtracted the uncertainty in the radius in the equation, and compared the results, I would have gotten a slightly different magnitude for the error bars. The reason I avoided doing this is because I would have had error bars that spread in different magnitudes above and below the measured point. For these reasons, the given error bars are just estimates of uncertainties, and for this reason I will not include them as uncertainties in my final calculations of the e over m ratio. I will have to ask Dr. Koch how to deal with multiplying uncertainties!!***

Using the methods described above, the following information was obtained:

SJK 15:39, 3 October 2008 (EDT)
15:39, 3 October 2008 (EDT)
Some comments here. I think you may have some typos and / or used the wrong graph images? The graphs look the same to me (or very similar), and also your "slope" and "modified slope" have the same value of 0.0836. Also, I'm not a stickler for significant figures in a raw data notebook. However your numbers could be a lot more readable with less digits. we'll discuss this in lecture upcoming.

I'm surprised the Earth's field had such a big effect?
• 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: One thing that was suggested during lecture was to mess around with the slope to see how the final data changes. To do this, I had MATLAB tweak the slope of the linear fit within the error bars. This new slope is plotted in Figure 1b.

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

Because of the way the error bars were made, and because the slope is not great to begin with, I will not trust these results. However, it is interesting just to see how the ratio changed!

### Constant Voltage Discussion

SJK 15:41, 3 October 2008 (EDT)
15:41, 3 October 2008 (EDT)
In this case, I can see that the graphs are different, but the slopes don't look very different? If anything, I see a different intercept, not slope???
Figure 2a Constant Voltage
Figure 2b Constant Voltage with modified slope

Another set of data was taken at constant voltage across the electron gun, varying only the current of the coils. In the same way as above, I had MATLAB take in the data and produce a plot of radius vs 1/current (See Figure 2). This allowed me to relate the e/m ratio to the slope, as above.

$r^{2}(V)=(\frac{m}{e})\frac{2}{B^{2}}V$

B = ik

$r(i)=\sqrt{\frac{2mV}{e}}\frac{1}{k}\frac{1}{i}$

$slope = \sqrt{\frac{2mV}{e}}\frac{1}{k}$

$\frac{e}{m}=\frac{2V}{slope^{2}k^{2}}$

Error bars were produced in a different manner from the one discussed above. In this set of data, r was plotted vs. 1/current. Therefore, the spread of the error bars above and below the measured point was chosen to be simply the uncertainty in the measurement of the radius. I decided to do it this way because there are only terms linear in r in this calculation.

• 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: As above, I had MATLAB find a reasonable slope within the error bars. This slope is plotted in Figure 2b. After doing so, the e/m ratio was recalculated. Here are the results:

• Modified Slope: 5.369 e-2 m/A
• Modified Corresponding em ratio: 3.193096279082144e+11 C/kg
The fit doesn't look great to start with however, so I dont think it would be reasonable for me to use this e/m ratio.SJK 15:51, 3 October 2008 (EDT)
15:51, 3 October 2008 (EDT)
I agree. Actually, I don't understand that fit. It's clearly not the "best" fit, so I wonder what algorithm you were using in Matlab? By "best" fit, I mean it's clearly not a maximum likelihood or a minimization of the deviations from the line. My best guess is that it was constrained to go through the origin? Do you know what Matlab was doing? Unfortunately, I don't "read" matlab code very well, so I cannot tell you. If you know, it would be an essential thing to mention in your analysis (like you do below, saying what your algorithm for mean, std. dev. was)

### Composite Information

Figure 3 em ratios for all points
Figure 4 em ratios vs. radius

The data for e/m ratios was then computed using the general e/m ratio derived in the theory section. These are plotted in Figure 3. The mean from the constant voltage and constant current were averaged. The "best e/m ratio" was hand picked, after having MATLAB calculate the ratio for each data point. However, this best value is clearly not an accurate representation of our results, because it is clearly an outlier. Therefore, in the summary only the mean will be used!

Error bars were made in the following way: First, the e/m ratio was calculated using the general formula (ie. not using any least squares slopes) with the measured radius, not including our uncertainty in the radius. Then, the charge-mass ratio was calculated again, this time using our radius plus our uncertainty as the radius in the equation. The two values were compared and the difference was set as the spread around the measured value.

NOTE: I realize this must not be the best way to make error bars in this situation. If I had subtracted the uncertainty in the radius in the equation, and compared the results, I would have gotten a slightly different magnitude for the error bars. The reason I avoided doing this is because I would have had error bars that spread in different magnitudes above and below the measured point. For these reasons, the given error bars are just estimates of uncertainties, and for this reason I will not include them as uncertainties in my final calculations of the e over m ratio. I will have to ask Dr. Koch how to deal with multiplying uncertainties!!***SJK 22:51, 3 October 2008 (EDT)
22:51, 3 October 2008 (EDT)
See above comment. I do think you are thinking along the correct lines. We'll learn the "correct" method soon, which involves taking partial derivatives, and what you're doing is close to a numerical estimation of the partial derivative.
• 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%

### Electron charge to mass ratio vs. beam radius

SJK 16:06, 3 October 2008 (EDT)
16:06, 3 October 2008 (EDT)
Plotting your apparent e/m ratio versus various parameters (such as beam radius) is an excellent way to study systematic error...good idea! Another good thing to do sometimes is to plot data versus time, do identify changes in your experimental technique or instrumentation drift.

During the experiment, it was noticed that as the beam got larger, its 'orbit' became more distorted. For this reason, the e/m ratio was plotted against the radius of the beam: see Figure 4. Clearly these distortions had a large impact on our resulting e/m ratio, as the results get progressively worse as the radius increased. As mentioned earlier, this is probably the result of:

• more collisions due to a larger orbit, resulting in a larger radius on the left side
• the beam itself getting wider, which was noted earlier

## Improvements for the lab in the future

SJK 22:52, 3 October 2008 (EDT)
22:52, 3 October 2008 (EDT)
Was this over at Regener? It does sound like a better way to do it, for several reasons.
• When I did the same experiment last year in one of the general physics labs, I remember using an apparatus which accelerated electrons upwards, and then curved them around onto a plate that was perpendicular to the beam. I think that was a much better way to do it for several reasons. First, we could get an accurate reading for the radius, since the beam landed on the ruler -- This way, even if the glass envelope was 2 inches thick you could still get an accurate reading since the beam comes into contact with the ruler. Second, the reading would be just as accurate if the beam was 1cm in radius or if it was 10cm in radius (ignoring collisions, which would cause more problems for a larger radius), unlike in this setup where the diameter of the beam leaves the plane of the ruler as it becomes small. And lastly, because the beam intercepts the ruler not long after it leaves the gun, it gets less distorted because the beam sustains fewer collisions -- Even though this could be accounted for, it is absurdly confusing when you're trying to center the elliptical beam in front of the ruler.
• Another good idea might be to take pictures of the beam, as was suggested by Dr. Koch.
• To avoid the beam from expanding due to kinetic energy losses, it might be a good idea to just evacuate the envelope. Of course we wouldn't see the beam, but it would make measurements more accurate. If guess if you were to try this you would have to use some sort of fluorescent screen which intercepted the beam.
• yet another thing that could make the experiment better would be to use a flat glass envelopeSJK 22:53, 3 October 2008 (EDT)
22:53, 3 October 2008 (EDT)
Ah, I see, like a pancake shape? Good idea...I wonder how hard that would be to make
, instead of a spherical one. This would make some qualitative experiments impossible, such as watching the beam spiral as it is rotated in the B field, but it would remove any uncertainty one might have due to the refraction and reflection caused by the spherical glass envelope.
• Perhaps a better way to do measure the e/m ratio would be to use the experiment that Thompson used. Here, he set up transverse electric and magnetic fields, and watched the deflection of electrons on a fluorescent screen. It can be proved (by the Lorentz force law) that with transverse fields, it is possible to have a 'drift velocity' for all values of the electric and magnetic fields that will allow the electrons pass through the fields without sustaining deflection. This 'velocity selector' would drop many variables and allow one to easily calculate the e/m ratio. This way we wouldn't have to worry about measuring elliptical radii, lining up rulers behind distorted glass envelopes, and other annoyances. A schematic of Thompson's apparatus can be seen here [|| Thompsons experiment].
• SJK 22:58, 3 October 2008 (EDT)
22:58, 3 October 2008 (EDT)
I think I see what you're saying here...sort of like calibrating the mercury line in the Balmer series...but I am guessing you couldn't turn the heater up high enough to get the "right answer" without frying it...that's my intuition.
An interesting thing that one could try in the future is to first 'calibrate' the filament heater (since we saw that it had a huge effect on the radius of the beam). To do so, one would first have to use some random setting for the accelerating voltage and coil current. From this, you could calculate what the radius of the beam should be, given the actual e/m ratio. Then, you would have to try to match this radius by adjusting only the settings on the heater. Clearly this wouldn't be possible if you were doing some original experiment where the "actual" value of something was unknown, which is likely the case outside of this lab. However, for our purposes it might be interesting to see how good of data the apparatus could return in this case. After all, if this method returned great data after repeating this experiment, we could throw out some of our possible systematic errors.

## Lab Manual Questions

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

We actually don't see the electrons. Instead, we see the helium that becomes ionized from collisions with electrons. The ionization energy of helium's outermost electron is roughly 25eV. Its innermost electron's binding energy is a bit higher at roughly 55eV [1]. Given the fact that our electrons were being accelerated through potentials of over 200V, it is safe to conclude that the helium is being ionized.

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

According to the NOAA, the total magnetic field in Albuquerque is roughly 50,292.7 nT, with 23,027.7 nT North, 3858.3 nT East, 23,348.7 nT Horizontal, and 44,544.3 nT Upward. A reasonable way to go about this is to calculate the radius of the electron beam, if it were accelerated into the earth's magnetic field, in the absence of any other magnetic sources. In my calculation, I will use an accelerating voltage of 300V (our maximum) and the east component of the earths magnetic field, which is roughly the direction that the field in the coils were pointed in.

$R = \sqrt{\frac{2Vm}{e}}\frac{1}{B} = 15.15m$

This is actually more significant than I imagined. After getting this calculation, I went back and had my program calculate the e/m ratio with this field added to the field produced by the coils. This was noted above in my "Post Experimental Data Analysis" Section.

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

Using the formula above, we see that the radius is proportional to the root of the mass of the particle. Therefore:

$\frac{R_{p}}{R_{e}}=42.81$

Therefore, if we used protons we would have had to use a much stronger magnetic field, or a weaker accelerating potential in order to acheive the same radii as if we were using electrons. However, it is possible that if we used a weaker accelerating potential, we could run into problems ionizing the helium in the envelope. In addition, the protons would have looped around the other way, since they are positively charged, which would have required us to change the polarity of the helmholtz coils.

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.

The period is just the circumference divided by the electrons velocity:

$T=\frac{2\pi R}{v}$

Now, we find the radius and velocity:

$R = \sqrt{\frac{2Vm}{e}}\frac{1}{B}$

$v = \sqrt{\frac{2eV}{m}}$

Plugging this into the formula for the period, we get:

$T = \frac{2\pi m}{e B}$

which is independent of the voltage, when B is held constant, as desired.

5. Would a relativistic correction for the electron’s momentum be appreciable for the present experiment?

Relativistic momentum is given by:

$\vec{p}=\gamma m\vec{v}$

where gamma is the relativistic factor:

$\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$.

Our most energetic electrons were on the order of 300eV. (will check, cant remember if this was max). Therefore, by the work energy theorem:

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

$v=\sqrt{\frac{2eV}{m}} \Leftrightarrow \frac{v^{2}}{c^{2}}=\frac{2eV}{mc^{2}} \Rightarrow .0011708$

γ = 1.0005859

The relativistic factor is, for our purposes, essentially 1. Therefore, relativistic calculations are unnecessary here, as they would provide little correction.

## MATLAB Code

Algorhithm used for mean:

$\bar{x}_{ij} = \frac{1}{N} \sum_{i=1}^N{x_{ij}}$

Algorithm used for standard deviation:

$\sigma_{x} =\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\bar{x})^{2}}$