# User:Paul V Klimov/Notebook/JuniorLab307L/2008/10/13

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SJK 03:35, 1 November 2008 (EDT)
03:35, 1 November 2008 (EDT)
Your lab notebook and analysis are fantastic! Outstanding analysis, thinking, and graphs. Very impressive.

## Electron Diffraction

In this experiment, we will be exploring the wave-particle duality of matter. We will do so by diffracting electrons through the hexagonal crystal lattice in graphite.

## Theory

In the early 1900s, deBroglie postulated that the symmetry in nature required that matter must have some properties of a wave, in certain limits. de Broglie was likely influenced by the then known fact that light had certain properties resembling those of a wave and others resembling those of a particle. de Broglie said that matter waves have the following wavelength:

$\lambda = \frac{h}{p}$

Not long after his proposition, experimental evidence for his postulate was provided, when electrons were observed to diffract (on accident).

It was found that electrons (and other matter waves) diffract in the same way that light diffracts. The condition for electron diffraction, often referred to as the Bragg condition, is as follows. 'd' is the lattice spacing, n is the diffraction order, and lambda is the wavelength of the particle. The angle theta here is not the angle of diffraction on the screen, but the angle that opposes the path length difference for a matter wave. It also turns out that the angle of diffraction from the incident particles is 2*theta.

2dsinθ = 2dθ = nλ

The angle subtended from the center of the screen to the first maxima ('R' is the radius of the diffraction ring) is found to be 2*theta:

2θ = R / 2L

For small angles, this relationship simplifies to:

$\frac{Rd}{L}=\lambda$

where D is the spacing between the maxima on the screen, a distance L away.

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_{k}}}=\frac{h}{\sqrt{2meV_{a}}} =\frac{2\pi\hbar}{\sqrt{2meV_{a}}}$

Setting the wavelength of the electron equal to the wavelength in diffraction, we get:

$\frac{Rd}{L}=\frac{2\pi\hbar}{\sqrt{2meV_{a}}}$

using the fact that R = D/2 we find the following relationship for the lattice spacing, 'd':

$d=\frac{4\pi\hbar L}{D\sqrt{2meV_{a}}}$

The above formula can be simplified, but is left like this to match the formula given in the lab manual.

## Equipment

• Tel 2501 Universal stand
• Electron Diffractor 2555 (5Kv .3mA)
• Teltron Limited London England 813 KV Power Unit
• HP 6216B Power Supply
• Wavetek Meterman 85XT multimeter
• Shock proof Calipers number 6020
• Electronic Digital Caliper: Carrera Percision (used on second day) *no model number found anywhere*.

## Cautions

Given that the current will be much lower than a milliampSJK 02:54, 1 November 2008 (EDT)
02:54, 1 November 2008 (EDT)
It's microamp current when the electrons are traveling through the vacuum bulb...if you made a direct connection with your body, much more current could flow! Thus you should assess the maximum current of the power supply, not the maximum current draw of the device
, electrical shock is probably the least of our worries. However, the diffraction device, and especially its contents appear to be quite fragile. We should avoid shaking it so as to not damage the graphite foil in any way, or offset the filament (if that is even possible). In addition, it is in our best interest to keep the heater current below the .25mA rating of the diffraction device to prevent damage to the graphite foil. As given in the lab manual, the foil will get red if the particle current is too great, and we will be sure to look out for this.

## Procedure

• We set up 2 power supplies in our circuit but only used one, as suggested. The setup matched the schematic provided in Dr.Golds manual.
• The heater was turned on and we waited for several minutes to wait for the filament to heat up. The filament was glowing shortly after it was turned on.
• We have an ammeter set up in series to measure the current at the heater because we cannot let it exceed .25mA. Attaining a current of comparable magnitude could result in the graphite foil being punctured.
• The diffraction maxima will be measured with the calipers. As suggested by Dr.Koch, we will be measuring the inner diameter of the outer ring and the outer diameter of the inner ring. (the second day we tried another measuring technique! I will discuss this when it comes up)SJK 02:56, 1 November 2008 (EDT)
02:56, 1 November 2008 (EDT)
Hmmm, I don't remember suggesting this...does it still make sense?
• Voltage was read directly off of the power supply.

## Data

### Week1

V (V) outer diameter (mm) inner diameter (mm)
4900 38.62324±1.5 24.1555±1.5
4800 39.76624±1.5 24.38400±1.5
4700 40.3098±1.5 24.4792±1.5
4600 40.89400±1.5 24.58212±1.5
4500 41.66794±1.5 25.1206±1.5
4400 42.0015±1.5 25.89022±1.5
4300 42.38752±1.5 28.70200±1.5
4200 43.7388±1.5 28.88234±1.5
4100 43.36288±1.5 29.45384±1.5
4000 45.31106±1.5 29.7307±1.5
3900 45.62094±1.5 30.1498±1.7
3800 45.6438±1.7 30.23616±1.7
3700 47.7393±1.7 31.0515±1.7
3600 48.38700±1.7 31.87700±1.7
3400 50.01006±1.7 32.2072±1.7
3200 52.83454±1.7 32.4739±1.7
SECOND TRIAL

V (V) outer diameter (mm) inner diameter (mm)
4900 39.20236±1.5 21.84400±1.5
4800
4700
4600
4500 41.82618±1.5 25.1206±1.5

We started taking more data, but the measurements were so similar that we thought it would be pointless to continue. We will fit it before the next period and see if we have any major outliers. If this is the case, we will go back and make some adjustments.

### Week2

After looking at our data from the first week, we decided that it would be a good idea to retake some measurements, adjusting our measuring technique. An important distinction is that this time we will be measuring the inner diameter of the inner ring, not the outer diameter as we did the first time. The outer ring will be measured as before, on the inner diameter. We also used a new digital caliper that was mentioned in the equipment section:

V (V) outer diameter (mm) inner diameter (mm)
4900 36.99±1.5 N/A
4800 37.11±1.5 20.35±1.5
4700 37.31±1.5 21.74±1.5
4600 37.63±1.5 22.04±1.5
4500 38.02±1.5 22.50±1.5
4400 38.28±1.5 22.86±1.5
4300 38.64±1.5 22.24±1.5
4200 39.06±1.5 22.97±1.5
4100 39.45±1.5 23.03±1.5
4000 39.86±1.5 23.23±1.5
3900 40.15±1.7 23.55±1.7
3800 40.55±1.7 23.74±1.7
3700 40.92±1.7 24.17±1.7
3600 41.19±1.7 24.41±1.7
3400 41.87±1.7 24.78±1.7
3200 42.23±1.7 25.04±1.7

## Possible Sources of Error

• Width of Rings: Perhaps the most prominent source of error in this experiment arises from the fact that the maxima rings are very blurry and quite wide. And in fact, when we think about the orientation of things in the device, it becomes apparent that for larger diffraction angles we should expect wider maxima. This is a result of the fact that the graphite crystal is not at the center of the glass sphere; if it was, all diffraction maxima would have the same width, given symmetry. Of course this was not done by whoever built the machine because this would shrink the diameter of the maxima, unless the diameter of the spherical tube was increased. This was probably avoided due to the fact that the device would have to be larger and it would cost more to produce. However, If I were performing a diffraction experiment for real research this is something that would be addressed without question! To get good and consistent results our diffraction maxima must have the same widths!
• Lattice spacing: A graphite crystal has hexagonal cells. At first this seemed like a fairly simple lattice to me, however, after some thought, I realized the complexity of the system. Looking at a hexagonal crystal, it is easy to identify many diffraction centers, depending on the angle at which the crystal is oriented. After all, the electrons in the 'beam' will not be seeing the same lattice after the graphite is rotated even through small angles. How can we be sure that the crystal is perfectly aligned so that diffraction patterns that we see are actually due to the 'characteristic' spacings suggested in the manual? In the process of researching this, I stumbled upon a diagram of the various interplanar spacings in graphite.SJK 03:01, 1 November 2008 (EDT)
03:01, 1 November 2008 (EDT)
This is a really cool link!
Clearly there are many diffraction centers so we cannot be sure that the patterns we are seeing are due to the 2 'characteristic spacings' noted in the lab manual.
I did some quick calculations to find the diameters of the diffraction maxima corresponding to some of the spacings in the above diagram (which had the spacings listed on the | website that the image came from). Those calculations show that we would not be seeing most of the spacings other than the ones given in the lab manual. However, what still gets me is what happens if the crystal is turned. Wouldn't the spacings look different to the electrons? Wouldn't the same spacing give us a different diffraction diameter after a small rotation? I'm not sure if my argument is relevant, so please let me know if I am missing some crucial concept. SJK 03:06, 1 November 2008 (EDT)
03:06, 1 November 2008 (EDT)
this is a very good question to ask, and yes, I think you are missing a key point, but once you discover it, I think you'll have a really great understanding. I must not have bugged you guys with a question I usually ask students during this lab: why do you see circular diffraction rings? Why don't you see diffraction spots? The answer to this is related to your question. I can give you more clues if you need them.
• Geometry Errors: One thing that was mentioned in the lab manual was that corrections could be made based on the geometry of the diffraction device. I could think of two possible errors due the shape of the device, both of which were incorporated into my calculations:

1.) Width of the tube: The width of the tube is 1.5mm, meaning that each measurement could return an overestimate of the diameter by: $2\cdot(1.5mm) sin(\theta) = 3mm \frac{D}{2L}$

2.) Correction for the distance from diffraction to the maxima, 'L' : Because of the circular cross-section of the tube, the actual distance of travel for the electrons would be less than the distance from the graphite to the center of the tube, 'L'. The correction for this can be computed using the radius of curvature of the spherical tube, and the angle from the center of curvature to the maxima, which I will denote by phi: The effective contraction of the length is:

$\Delta L = R - R\cdot cos(\phi) = R(1-cos(\phi))$

but,

$\phi = arcsin(\frac{D}{2}\frac{1}{R})$

Where D/2 is the radius of the diffraction maxima. Plugging this in, you get:

$\Delta L = R(1-cos(arcsin(\frac{D}{2R}))$

Not only did these corrections have a considerable effect on my calculations, but they also brought us closer to the actual value (which is of course something that would be unknown in most real experiments). Clearly such corrections are important and must be carried out when they are identified.

## Post Experimental Data Analysis

Figure 1: Calculated lattice spacings from the outer maxima ring. The blue error bars represent a >95% confidence interval. The green error bars represent the SEM about the mean (red line), and thus a ≈68% confidence interval. The dark green line represents the actual lattice spacing.
Figure 2: Calculated lattice spacings from the inner maxima ring. The blue error bars represent a >95% confidence interval. The green error bars represent the SEM about the mean (red line), and thus a ≈68% confidence interval. The dark green line represents the actual lattice spacing.
Figure 3: Outer maxima diameter versus inverse of the root of the accelerating voltage. The blue error bars represent a ≥95% confidence interval. The line is a least squares line that was constrained to go through the origin (R-Value: .8746). No SEM provided because only 1 measurement at each potential was taken
Figure 4: Inner maxima diameter versus inverse of the root of the accelerating voltage. The blue error bars represent a ≥95% confidence interval. The line is a least squares line that was constrained to go through the origin (R-Value: .9396). No SEM provided because only 1 measurement at each potential was taken

All calculations done on MATLAB, except for the lsr slope, which was found on EXCEL. All relevant algorithms will be mentioned as they come up.

### Data Selection

We took data on both days and I have decided to work with only data gathered on the second day for several reasons: On the first day, we were getting used to the device, and I believe our measuring techniques had not been well established. This quite visible given some large jumps in our data. In addition to these jumps, some quick data analysis showed that we had trends in our data that were greater than those seen in our measurements on the second day. In addition, on the second day, we took measurements of the inner diameter for both maxima, which I believe is more consistent than our approach on the first day, where we measured the inner diameter of the outer maxima and the outer diameter of the inner maxima.

Therefore, for all of these reasons, I believe that the data gathered on the second day better represents the physical system.

### Error Propagation

Given that we had uncertainty in our measurement of the diameter of the diffraction maxima, error propagation was necessary to find our uncertainty in the calculated lattice spacing. Error propagation was performed on only one variable, the diameter of the maxima, because this was the only variable for which we thought uncertainties were necessary. The uncertainty for the lattice spacing was computed individually for each point, and is represented by the error bars in Figure 1 and Figure 2. Error propagation was done as follows, with 'd' the lattice spacing, 'D' the calculated diameter, and a new variable, 'eta', which was introduced to represent the constants in the equation to clean up the math.

$d=\frac{\eta}{D}$

$\Delta d = |\frac{-\eta}{D^{2}}|\Delta D = \frac{\eta}{D}\frac{\Delta D}{D}=d\frac{\Delta D}{D}$

So, my results in the figures are reported as follows:

$d \pm ErrBars = d \pm d\frac{\Delta D}{D}$

Given that our uncertainty was based on the absolute largest error we could have had in a measurement of the diameter, I think it is safe to say that this uncertainty gives us a ≥95% confidence interval. (as suggested by Dr. Koch)

### Using a Least Squares Line

The measured diameter versus the inverse root of the accelerating voltage was plotted in Figure 3 and Figure 4. This was done to produce a straight line whose slope could be related to the lattice spacings. Least squares lines were constrained to go through the origin, as required by theory.

$D=\frac{2hL}{d\sqrt{2me}} \frac{1}{\sqrt{V}} = Slope \frac{1}{\sqrt{V}}$

$Slope = \frac{2hL}{d\sqrt{2me}} = \frac{\gamma}{d}, \gamma \equiv \frac{2hL}{\sqrt{2me}}$

$d = \frac{\gamma}{Slope}$

In addition, an R value was calculated to tell me the quality of the fit (although I did perform any other statistical analysis involving the number). Given that the least squares fits were not exceptionally great, I avoided adjusting the slope because it would require too much personal interpretation, which would undoubtedly give me biased results. The outcome of these methods are reported in the Results sections below.

### Calculating Means, SEMs, and Errors

Deciding whether or not a mean should be calculated for some of our data was a tough decision. After plotting the calculated lattice spacing, that was calculated from the outer maxima, versus the accelerating voltage (or just measurement number), a clear trend was present. However, no trend was seen when plotting the lattice spacing calculated from the inner maxima. I just wanted to mention this here, but I talk some more about it in the 'Some Discussion' section later. The algorithms that I used are listed at the very bottom of the page in the MATLAB section.

## Results

### Important Outer Maxima Results

• Actual Atomic Spacing:.123nm
• Slope of Outer Diameter vs. (Voltage)^-1/2 plot: 2.510nm/sqrt(V) (Steve Koch:I think you mean units of m/sqrt(V), not nm, right?)
• Atomic Spacing from slope: .1239nm (R-value: .8746) SJK 03:14, 1 November 2008 (EDT)
03:14, 1 November 2008 (EDT)
You can obtain an uncertainty of the slope using linear regression (assuming normally distributed residuals)...Excel does this, and if Matlab doesn't, you can program in the formulas yourself. See chapter 8 of Taylor
• Percent Error:.79%
• Atomic Spacing from Mean (SEM~68% C.I.): .1237(6)nm
• Atomic Spacing from Mean (Propagated Error~95% C.I.): .1237(52)nm ****Reported Value
• Percent Error:.55%

### Important Inner Maxima Results

• Actual Atomic Spacing:.213nm
• Slope of Inner Diameter vs. (Voltage)^-1/2 plot: 1.465 nm/sqrt(V)
• Atomic Spacing from slope: .2159nm (R-value: .9396)
• Percent Error:1.4%
• Atomic Spacing from Mean (SEM~68% C.I.): .2157(11)nm
• Atomic Spacing from Mean (Propagated Error~95% C.I.): .2157(165)nm ***Reported Value
• Percent Error:1.3%

### Reporting Values

I have decided to report the calculated spacing from the outer diameter with a 95% confidence interval, given by the largest propagated error. The reason I have chosen not to report the SEM as the best 68% confidence interval is because the data for that portion of the experiment had a clear trend. Although the SEM essentially includes the actual value of the lattice spacing, this is something that would not be known in an original experiment. Therefore, I chose to stick with the propagated error, which comfortably includes the actual value. I gave the propagated error a 95% confidence interval (after discussing the issue with Dr. Koch), because I believe me and my lab partner overestimated the error to stay on the safe side.

SJK 03:25, 1 November 2008 (EDT)
03:25, 1 November 2008 (EDT)
I am really impressed by your line of reasoning here! You have a very strong talent to analyze the problem from so many angles and to really push yourself to be objective and to question your assumptions at all the steps. This is going to be a huge asset for you as an experimentalist and it is really fun to see. All that said, however, at this point, I don't think I agree with your conclusion! (It's possible, though, that I haven't thought about deeply enough yet.) I was interested by your proposition of "error propagation in reverse" (and yes I think it's a legitimate thing to try). I think the key to the reason why I disagree with your conclusion is because you're seeing what the difference in radius would be for a given accelerating voltage. I agree that 100 microns is unreasonable uncertainty for a single measurement. However, you're able to measure the radius at many different accelerating voltages, and combined can reduce the SEM of the lattice spacing. This is related to the effect that the SEM decreases as the sqrt of number of measurements, for a fixed standard deviation of the parent distribution. Here's another way of looking at it: why is it any more ridiculous to claim that you can predict the diffraction radius to 100 micron precision than it is to claim you have sub-angstrom precision on lattice spacing?

This is a really interesting question, worth talking about I think.
At first I wanted to report the SEM, representing a 68% confidence interval, for the spacing calculated from the inner diameter. The reason I wanted to do this was because there was no clear trend in the data, and so I believed that the SEM was the best 68% confidence interval. However, after pondering this some more, I realized that this would be a fairly ridiculous claim. I found this out after doing error propagation in reverse (hopefully this is legitimate). I wanted to see how the diameter of the diffraction maxima would change if I could vary the lattice spacing by the value of the SEM. I used the averages of the relevant quantities, to get a feeling for the order of magnitude that this would result in:

$\Delta D = \frac{D \Delta d}{d} \sim .1 mm$

Reporting the SEM in this case would be too confident of a claim I believe. I would be saying that if the diffraction diameter would change by .1mm we could comfortably measure a difference, which is clearly not the case. Even though some of our uncertainties vary at this order of magnitude, they were rough estimates. I think that I would be more comfortable using the SEM as the best 68% c.i. if the result of this calculation was on the order of 1mm. Therefore, I have decided to report the largest value of my propagated error as the uncertainty, representing a ≈95% confidence interval.

## Some Discussion

As mentioned earlier, systematic error likely dominated in this lab. This was very evident when the calculated spacings from the outer maxima were plotted against the measurement numbers. Very clearly there was a trend which could not allow for a great least squares fit. In attempt to target the source of error, I had MATLAB vary several parameters in attempt to fix the fit. After some more thought and some discussion with Dr. Koch, we decided that the most likely source of this error is inaccuracies in measuring the accelerating voltage (Apparently it had been working poorly for some other groups). Unfortunately we didn't measure the accelerating voltage on a secondary device, to check the accuracy of the readings given on the power supply.

SJK 03:30, 1 November 2008 (EDT)
03:30, 1 November 2008 (EDT)
I think it's good that you're identifying this bias of knowing the right answer. However, even if you didn't know the answer, in a "real" experiment, you'd probably be able to make a good guess, given what you know about covalent bonds, etc. So, I think it's a good bet that you weren't off by anywhere close to 1 kV, and I think you would verify this if you checked with a voltmeter.

It is interesting that this straightens out your systematic problem...I wonder if it's a clue as to the true source of the systematic error.
I found that a better fit, for both the outer and inner maxima could be attained at a significant voltage increase of +1kV. However, while the least squares fits improved, the calculated atomic spacings became much worse, bumping our errors beyond 10% (compared to ≈1% errors). At this point I admit that my analysis comes with some bias, given that the actual spacings are known. Therefore, what I should really do is get back into the lab and retake some measurements, checking the accelerating voltages on a secondary device. Only then will I know if the source of the trend is bad voltage measurements. (I think I will be doing my final report on this so I will definitely do this when I come back to this experiment!)

## Improvements for the Future

• Better Screen: I am guessing that there is some fluorescing compound on the interior of the tube that gets activated when electrons collide into contact with the tube. Perhaps there is a more sensitive compound out there that would make the diffraction maxima easier to see. This is not necessarily a problem at higher voltages, but below 3500V it is insanely hard to see the diffraction rings.
• Flat screen: if the screen onto which the electrons are projected was flat, it would remove our need to make angle-dependent corrections for lengths. However, the outer rings would be wider than they are now if the screen was flat. The trade off might or might not be worth it.
• Photographs: It might be worth it to try to take photographs, perhaps without flash and with long exposure to try to get a better view of the diffraction rings. This would be especially helpful for lower accelerating voltages since the pattern quickly fades away after ≈3.5kV. If we did this, we would have to prepare some sort of measuring device so that we could calibrate measurements in the picture.
• Foil Manipulations: Like i mentioned earlier, I would like to see how severely the diffraction pattern would change after the foil is rotated through small angles. I would think this would have a huge impact, but maybe not since we got fairly good results.
• Simpler latices: If my earlier rants about the complexity of the hexagonal cells are at all justified, I would like to know why we don't use a foil of simpler structure. I would think that using something with a simple cubic bravais lattice would have fewer possible diffraction centers.SJK 03:33, 1 November 2008 (EDT)
03:33, 1 November 2008 (EDT)
well, see the puzzle above about why you even see rings. I'm not sure cubic would be any simpler.

## MATLAB

SJK 16:42, 1 November 2008 (EDT)
16:42, 1 November 2008 (EDT)
where is the matlab code?

All calculations done on MATLAB, not including the least squares fit which was found in EXCEL. Here are some algorithms that I used:

Algorhithm used for mean:

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

Algorithm used for standard deviation:

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

Algorithm used for standard error of the mean (SEM):

$SEM=\frac{\sigma_{x}}{\sqrt{N}}$

Algorithm used for Percent Error:

$Percent Error= 100\frac{|Actual-Measured|}{Actual}$