Latest revision as of 18:58, 6 August 2010

Electron's e/m Ratio

Experimentalists: Nik and me I (actually, isn't Nik and "I" correct, as your ":" implies "are" or "were"?)

Goal

If a beam of electrons moving at a certain speed is bent by a magnetic field, the entire beam will curve into a circle of fixed radius. This result means that there is a charge to mass ratio for the electron, and it can be calculated! The accepted value for the charge to mass ratio of an electron is

[math]\displaystyle{ \frac{e}{m}=1.759\times10^{11} \frac{C}{kg} }[/math].

Theory

^{SJK 01:52, 10 October 2007 (CDT)}

01:52, 10 October 2007 (CDT)
Excellent derivation in this section!

I will find two expressions for velocity of an electron beam moving in a circle and then set them equal to each other. I do this in hopes of finding a useful expression for e/m.

First expression for velocity

If the beam is moving in a circle, the velocity, [math]\displaystyle{ v }[/math], can be discovered in terms of magnetic field strength (B), charge (e), radius of circle (r), and mass of electron (m). The force (radial) required to keep anything in circular motion is

[math]\displaystyle{ F_r=-\frac{m v^2}{r} }[/math].

The force (radial) from the perpendicular magnetic field is

[math]\displaystyle{ F_r=-e v B \, }[/math].

Solving for [math]\displaystyle{ v }[/math] in these two equations by first setting [math]\displaystyle{ F_r=F_r }[/math] gives

[math]\displaystyle{ v=\frac{reB}{m} }[/math].

Second expression for velocity

To create the electron beam, the electrons will be accelerated between two plates. The voltage between these plates, [math]\displaystyle{ V }[/math], is the only significant means for the electron to speed up since the electric field practically goes to zero on the other side of the plate and since gravity is negligible with the speeds we are dealing with. Therefor, kinetic energy (K) equals the negative change in potential energy ([math]\displaystyle{ \Delta }[/math]U) between the two plates, which equals charge (e) times the voltage between the plates (V):

[math]\displaystyle{ K=\frac{1}{2}m v^2=-\Delta U=Ve }[/math].

Solving for [math]\displaystyle{ v }[/math] gives

[math]\displaystyle{ v=\sqrt{\frac{2Ve}{m}} }[/math].

Why find these expressions?

Setting both expression for velocity equal to each other and solving for e/m gives

[math]\displaystyle{ \frac{e}{m}=\frac{2V}{B^2 r^2} }[/math].

This equation is nice since one must only measure [math]\displaystyle{ r }[/math], [math]\displaystyle{ V }[/math], and [math]\displaystyle{ B }[/math].

An alteration to our result

Instead of measuring [math]\displaystyle{ B }[/math], which can be difficult, a relationship between current (I) and magnetic field would allow one to measure the current instead. Since I will be using a Helmholtz coil to generate [math]\displaystyle{ B }[/math], the following derivations are appropriate.

Whenever there are [math]\displaystyle{ n }[/math] circular loops of current of radius [math]\displaystyle{ R }[/math] that are in basically the same spot, the magnetic field along the axis perpendicular to and in the center of the loops is given by

[math]\displaystyle{ B=\frac{\mu I R^2 n}{2\left(x^2+R^2\right)^{\frac{3}{2}}} }[/math],

where [math]\displaystyle{ \mu }[/math] is magnetic permeability and [math]\displaystyle{ x }[/math] is the distance along the axis from the plane containing the loop.

For any well-made Helmholtz coil, the distance between both coils is R, so, since I am doing the experiment half way between both coils, [math]\displaystyle{ x=R/2 }[/math]. Also, since there are two coils, [math]\displaystyle{ n=2N }[/math], where N is the number of loops in one of the two coils, we can manipulate the above equation into

[math]\displaystyle{ B=\left(\frac{4}{5}\right)^{\frac{3}{2}}\frac{\mu N I}{R} }[/math].

Since N=130, R=0.15m, and [math]\displaystyle{ \mu=\mu_0=4\pi }[/math]x10^-7 Wb/(A*m), the Helmholtz coil I am using abides (approximately) by the following equation at the point on the axis exactly between both coils:

[math]\displaystyle{ B=(7.8\times10^{-4} \frac{Wb}{A\cdot m^2})\cdot I }[/math].

^{SJK 01:51, 10 October 2007 (CDT)}

01:51, 10 October 2007 (CDT)
But since the B field is squared, do you think it may actually be more important that the e-beam is off-axis? Can you think of a method for testing this (computationally)?

To derive this I have assumed that the electron beam will be on the axis, when, in fact, it will make a loop around the axis. However, this expression for [math]\displaystyle{ B }[/math] can still be used in the expression for [math]\displaystyle{ e/m }[/math] since the magnetic field weakens only slightly as one deviates from the axis as long as the deviation is small (not larger than R/2).

And now, here's The Final Equation!

[math]\displaystyle{ \frac{e}{m}= (3.29\times10^6\frac{A^2\cdot m^4}{Wb^2}) \frac{V}{B^2 r^2} }[/math].

Equipment

^{SJK 01:54, 10 October 2007 (CDT)}

01:54, 10 October 2007 (CDT)
Just to be nitpicky: "strong" is ambiguous in terms of power supplies. "High Voltage" "High Current" would be more descriptive.

2 weak direct current power supplies (10V and 2A are the max outputs needed)
1 strong direct current power supply (500V needed)
2 multimeters
Banana plugs
The Device! (Uchida Model TG-13)
- Helmholtz coil
- Vacuum tube with helium gas and electron gun inside
- Mirror with ticks for measuring length
- Black cloth hood to block out light so electron and helium collisions can be seen

Setup

To plug everything in we

acquired The Device!
positioned The Device! so that its axis lie in the East/West direction to minimize interference from the Earth's magnetic field
plugged in the three power supplies and the two multimeters into the wall
connected one of the weak power supplies into the heater for the electron gun with banana plugs and turned it on not exceeding 6.3V
plugged in the other weak power supply into the Helmholtz coil using banana plugs, but intercepted the circuit to connect a multimeter (in series) to measure current, and then turned on the power to the Helmholtz coil and multimeter so there were about 2 amps flowing
plugged in the large power supply to power the electrodes of the electron gun using banana plugs, and turned it on so that about 300V are applied
connected a multimeter to The Device! to measure voltage across electrodes

To adjust the electron beam we

turned off lights
waited for heater to heat up (after about two minutes it is at equilibrium)
adjusted voltage to electrodes and current to Helmholtz coils so that the beam made a helix
rotated the vacuum tube so the helix became a circle
moved the power supplies until their magnetic fields no longer affected the size of the circle
adjusted the focus on The Device! until the circle size was maximized (the focus aims the electron beam from the anode to the hole in the cathode)

My Procedure

Once everything was set up, the procedure was very easy. We simply took many radius measurements of the electron beam loop at various voltages at a constant current, and then we took more radius measurements at various currents at a constant voltage. The beam appeared as a faint blue glow due to the electrons colliding with the helium atoms. Oh, and we had the voltage for the heating element at 6.3V.

Measuring the radius was slightly tricky. One thing we kept in mind was that the helium would make the circle smaller due to drag, so we measured the outermost radius of the circle (the beam width is large enough to give a substantial distribution of circle radii).

Another difficulty was our inability to put a ruler into the vacuum tube. The mirror in the back of the device had a ruler on it. So, to measure the radius, we simply had to line up the real beam with its image on the mirror and take the reading off of the mirror where they line up (it's hard to explain how we did this unless The Device! is in front of me to show you, but it's really not that difficult). Since the zero mark on the mirror is between the two edges of the circle, we took measurements from the left and right edges and then averaged them together. Our measurements from the right edge were longer because the mirror was not centered and because by the time the electrons had gone to the left side, they had encountered significant drag and the radius of the circle had become smaller.

Data

In the following tables, I refer to the "Radius," but I am technically incorrect to do this. The true radius is the average of the left and right "radii."

Current=1.5A and variable voltage

I am throwing out trial 9 since drag from helium is too much at this large of a radius.

Trial #	1	2	3	4	5	6	7	8	9
Voltage (V)	230	250	270	290	310	330	350	370	390
Left Radius (cm)	2.6	2.8	2.9	3.0	3.0	3.1	3.2	3.2	3.2
Right Radius (cm)	3.8	4.0	4.2	4.3	4.5	4.6	4.7	4.8	4.8

Voltage=275V and variable current

I am throwing out trial 18 since drag from helium is obviously too much at this large of a radius (note how the left radius goes down from trial 17 since the beam has traveled the farthest at the left).

Trial #	10	11	12	13	14	15	16	17	18
Current (A)	1.8	1.7	1.6	1.5	1.4	1.3	1.2	1.1	1.0
Left Radius (cm)	2.2	2.4	2.6	2.8	3.1	3.2	3.4	3.5	3.4
Right Radius (cm)	3.4	3.6	3.9	4.1	4.4	4.7	4.9	5.1	5.3

Koch comment

When I look at your apparent ruler offset for the above data, I get the following:

Supposed offset = 0.6, 0.6, 0.65, 0.65, 0.75, 0.75, 0.75, 0.8, 0.8

Supposed offset = 0.6, 0.6, 0.65, 0.65, 0.65, 0.75, 0.75, 0.8, 0.95

Because the apparent offset increases with radius, I think averaging them is not at all correct, though one still has to wonder about the offset. If their ruler is that far off (0.6 cm?) then why the hell do we trust the numbers written on the front of the apparatus?

A dose of wisdom

^{SJK 01:58, 10 October 2007 (CDT)}

01:58, 10 October 2007 (CDT)
Some day I want to publish a paper from a single unrepeated measurement, and then use "datum" throughout the paper.

"A datum is good but data are better."
-a perceptive truth worth heeding

Calculations

^{SJK 02:05, 10 October 2007 (CDT)}

02:05, 10 October 2007 (CDT)
See above: I don't agree with averaging the two radii, for any of the datum left and datum right pairs.

Using the equation for e/m I derived and

[math]\displaystyle{ Radius=\frac{Radius_{left}+Radius_{right}}{2} }[/math],

I get the following two tables if I round many of the radius values up when applicable (e.g. 0.0375 [math]\displaystyle{ \rightarrow }[/math] 0.038) when displaying them, but I do not use the rounded value for further calculations...

This table is for constant current and increasing voltage
Trial #	1	2	3	4	5	6	7	8
Radius (m)	0.032	0.034	0.036	0.037	0.038	0.039	0.040	0.040
e/m (x10¹¹ C/kg)	3.28	3.16	3.13	3.18	3.22	3.26	3.28	3.38

This table is for constant voltage and decreasing current
Trial #	10	11	12	13	14	15	16	17
Radius (m)	0.028	0.030	0.033	0.035	0.038	0.040	0.042	0.043
e/m (x10¹¹ C/kg)	3.56	3.48	3.34	3.38	3.28	3.43	3.65	4.04

Rather than providing my best guess of the e/m ratio now, I want to first discuss sources of error to justify my guess. I say "guess" because it turns out that my data is so bad that there really is not any data analysis to be done.

Error Analysis

For an enormous number of reasons, there is much systematic error in this experiment. There is human error due to recording the radius, but this does not account for the decrease and then increase of the e/m ratio as the radius gets larger (human error would create a Gaussian that was not a function of the radius).

Before I discuss the disgustingly huge sources of error, I will build up to them by talking about some minor sources of systematic error.

If the voltage is at 300V, the electrons have 300eV of kinetic energy. I don't know the material of the electrodes, but I assume the work function for removing at electron is about 10eV. If I was very ambitious (which I am not since the other causes of error will make this correction imperceivable), I would learn about the work function of the electrode and take it into account.
External magnetic fields, such as the Earth's magnetic field and the fields of nearby equipment caused error, even though I did my best to prevent it.
I calculated the magnetic field in the center of the Helmholtz coil even though the electron beam is not centered but instead travels around the center. However, this error is very small. The Helmholtz coil is designed to generate an almost uniform magnetic field.
Only between infinitely large plates is the electric field constant and there is no electric field leaking out around the plates. This problem caused a noticeable "tail" at the end of the electron beam as the electrons, having finished their trek around the circle, approach the anode. The electrons were repulsed by the anode and were bent away. Ideally, there should not be any field behind the anode plate, but experimental physics is never ideal. I feel reasonably certain that this error is minor, which is good because I don't want to try to account for it.
An electron with 300eV is going about 3.5% the speed of light, not nearly fast enough to concern me, but still creates minuscule error.
The vacuum tube was made from curved glass. The error due to refraction probably creates a noticeably different value for the radius measurement.
The electrons should repel each other making the circle slightly bigger, but this is hardly a problem since the strongest forces are from the nearest electrons which are in line with each other and not creating any force perpendicular to motion.

^{SJK 02:14, 10 October 2007 (CDT)}

02:14, 10 October 2007 (CDT)
Interesting! I didn't think about any magnetic field from the beam itself. Did you do any calculations to estimate if it is a big deal or not? I suppose it is possible, but on the other hand, I would guess that the current in the beam is pretty tiny. After all, the resistance of a big ass loop of air is quite a bit higher than the resistance of the Helmholtz coils...

As for those giant sources of systematic error, they originate from the drag force from the helium and the magnetic field produced by the beam itself.

I speculate that at small radii, the magnetic field created by the loop itself becomes strong enough to make the total magnetic field small enough to significantly increase our calculated e/m ratio.

The drag force from the helium is HUGE! By simply adjusting the focus of the electron beam so that more electrons make it through the hole of the cathode drastically changed the radius. Changing the temperature of the heating filament to boil off more electrons also drastically changed the radius. The more electrons emitted, the larger the radius. I speculate that this is because an electron encountering drag will not be affected as much if there are many more electrons behind it to push it through and to keep the helium atoms out of the path. We were brave and turned the heating element above the recommended voltage, and the radius must have grown 50% until we got scared and turned it down. The larger the radius, the larger the affect of drag; therefor, one should expect that large radii produce larger e/m ratio. This is in fact the case except that very small radii also had larger e/m ratio's due to the magnetic field generated by the beam.

My actual data shows that small radii produce a larger e/m ratio (due to the beam's magnetic field), and large radii also produce a larger e/m ratio (due to drag). The e/m ratio is smallest at radii that are neither large nor small. The smallest value is the best since all error is trying to increase the e/m ratio. However, even the smallest value is far too large since momentarily increasing the voltage to the heating filament by a small amount drastically increased the radius. I hope I am justified in saying that the only conclusion I can reach is

[math]\displaystyle{ \frac{e}{m} \lt 3.13\times10^{11}\frac{C}{kg} }[/math].

Assuming the actual e/m ration to be 1.76x10¹¹ C/kg, relative error is less than %78 and only equals %78 percent when e/m = 3.13x10¹¹ C/kg.

^{SJK 02:16, 10 October 2007 (CDT)}

02:16, 10 October 2007 (CDT)
OK, so I agree that given the huge systematic error, there isn't a whole lot of practical benefit from analyzing the random error. However, since I know you can do it (and did do it), and since in real life it's important for people assessing the work, I want that kind of information to be reported!

The large problems with error could easily be fixed by using a much hotter filament and much fewer helium atoms (the beam would glow just as brightly but drag would be much less of a problem). Also, the magnetic field would have to be measured directly instead of indirectly through the current through the coils. The latter suggestion is more difficult to implement than the former, but it could be done.

Lab Critique

^{SJK Steven J. Koch 02:19, 10 October 2007 (CDT)}

Steven J. Koch 02:19, 10 October 2007 (CDT)
Wow! You sound pissed off! That's probably a good thing as a sign of a good physicist. I am still not convinced that the magnetic field from the beam is a big deal. However, drag from helium definitely seems to be, so your description of the "real" experiment does sound much better. Do we have that one lying around? I don't think so. Also, with the better experiment, you don't get to see the cool electron beam.

Overall, outstanding job on this lab. Sorry the numbers enraged you so much, but like you say, maybe it's a good and painful lesson. Your analysis is very thorough and I know the data you took were very careful.

The whole concept of this experiment is terrible. How the e/m ratio was initially measured was by not using a circular beam, but by a straight beam. Knowing that electrons exist and have a definite charge and mass, the velocity of accelerated electrons between plates can be written as

[math]\displaystyle{ v=\sqrt{\frac{2Ve}{m}} }[/math],

which is the same equation I used. The key to this experiment is to put the beam in both an electric field (E) and magnetic field such that they cancel each other and the beam is straight. When this cancellation occurs,

[math]\displaystyle{ F=eE-evB=0 \, }[/math].

We can solve for [math]\displaystyle{ v }[/math] and get

[math]\displaystyle{ v=\frac{E}{B} }[/math].

Setting both expressions for [math]\displaystyle{ v }[/math] equal to each other and solving for e/m gives

[math]\displaystyle{ \frac{e}{m}=\frac{E^2}{2 V B^2} }[/math].

Unlike my experiment, this method avoids much error that results from circular motion. Drag can be avoided if a detector to find the end of the beam is used instead of actually illuminating the beam with helium to see it and because there is no loop to create a magnetic field as a result of Lenz's law. Perhaps the goal of the experiment I did was to show that one must really think an experiment through for it to be worth anything.

Where to mail the Nobel Prize money

Bradley Knockel
Bahama Street
Bahamas

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