Experience based on Millikan's experiment

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Experiment based on Millikan's experiment of calculating the value of the electron's charge
Hello! As I said in the summary, this is inspired by Millikan's experiment.

Suppose you have 19 (closed) packages, all filled with different quantities of the same unknown object (which has a mass of m). You then weigh these packages one by one. How could you estimate the mass "m" of that unknown object?

You can't open the packages, and naturally, you don't know how many of these objects are in every package. Also, the MCD is 1, and I'm not sure if m is an integer.

I have asked this on other math forums, and nobody has been able to help. The condition of the answers they gave me is that the m has to be an integer, but I'm not sure if this is the case. Does anybody know what to do? I haven't been able to found what Millikan did with the gathered data after doing the experiment.
 

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  • #2
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I am guessing mcd = ‘maximum common divisor’ (the more common term to me would be gcd = ‘greatest common divisor’)? In that case, it does not even make sense to talk of a mcd/gcd if you are not dealing with integers in the first place. If the weights of the individual packages are integers but you dont want to assume that m is also an integer, how will you ever exclude the possibility that m is half as much as your guess but all packages contain double the quantity?
 
  • #3
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The largest mass such that all 19 masses are multiples of it, within the expected deviations from your experimental uncertainties. If you plot the likelihood as function of mass it should have many peaks. Take the largest peak that has a reasonable likelihood. It doesn't have to be the right answer, but there is a good chance it is.

If your packages have many particles and/or uncertainties are large this might be difficult to determine. Otherwise the right peak should be pretty obvious.
 
  • #4
jtbell
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If your packages have many particles and/or uncertainties are large this might be difficult to determine.
For many years, I taught a Millikan "oil drop" experiment lab using an earlier version of this apparatus. Almost always, students were unable to get good results on the first go because of these factors. Most of the "oil drops" (actually, microscopic latex spheres) had large numbers of extra electrons on them. With the uncertainties/limitations introduced by the apparatus, it wasn't possible to distinguish between e.g. 20e and 21e.

On a second go, I told students to predict how drops with < 5 electrons should behave, and look for those. With a lot of patience and time, some students were able to collect enough data to show groups of trials clustering around 1e, 2e, 3e, etc. Above about 7e (IIRC) the clusters merged because of the experimental uncertainties.

We didn't use sophisticated mathematical techniques to identify clusters of trials. We simply plotted the charge measurements as points along a line, and "eyeballed" the clusters.
 
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Vanadium 50
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Some comments:

(1) Picking the right sphere is essential. A half a dozen good measurements beats 100 lousy ones. This usually means you want one with very little horizontal motion so you can watch it for a good long time.

(2) There are two styles of this experiment - one in which you stop the sphere/drop, and one where you measure its speed when its velocity is constant. The second is easier to do, but the first gets much better results.

Sometimes while you are watching a hanging sphere, it will take off. This is a good thing. It means it gained or lost an electron. If you can stop it and remeasure it, the differences are likely to be 1e of charge.

(3) As said, there's no way you're going to distinguish between 20e and 21e. That's a 5% difference. Did you make a measurement good to 5%? Probably not - so this can only confuse you.

OK, now to the question at hand.

You have two estimators: the values and their differences. The smallest of the values or their differences is a good guess as to what 1e is. But select the differences from relatively low valued values: if you measure 2, 3, 4 , 19 and 19.1, I'd pick 1 as the difference, not 0.1.
 
  • #6
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When I did this experiment I found the elementary charge "by eye", but I was curious about a more mathematical approach.

Quick and dirty:
* Pick a number for the smallest mass unit
* Uniformly randomly multiply this by 1 to 10, then vary it by up to 5% with a uniform distribution. Add a uniform noise of -0.3 to 0.3 of that smallest unit. Repeat this 40 times.
* For each given mass unit hypothesis, find the relative difference between closest multiple and simulated measurement, square it. Sum over all 40 values. Smaller means a better fit.

This isn't given higher weight to smaller values, or actually calculating the likelihood - these things would improve the quality.

Here is the resulting graph. Guess which value I chose for the smallest unit.
2.7246, right in that prominent deep dip

milliscan.png
 
  • #7
Vanadium 50
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boy, that's pretty noisy data. Not much difference between right and almost right. I'd be interested in how it looks if you just looked at the 20 smallest values. Maybe even 10.

If you are going down this path, I would look to do a transform of the data. For the correct value, you will see a peak at e, but also a peak at e/2, e/3, etc. Your search should incorporate this.

That said, while there is value in doing this experiment as if it were 1909, it's not 1909 any more. What forces are on the sphere? Gravity, buoyancy, drag and electrostatics. Why do we try and do this with spheres that are moving with constant velocity or better still, stopped? Ease of measurement and ease of calculation - in particular, drag has F as a function of v, which is not so nice.

Instead of using eyeballs, take movies. Camera can't cost more than $75. Stick a grid on the view, project time and voltage, and now you can use any sphere in the field of view. Position vs. time lets you infer F(t), calculate the drag force, and ultimately q, and better still, the uncertainty on q.
 
  • #8
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Well, quick and dirty.

But it's quite robust, the strongest negative peak is always there. Keep in mind that this is not a log likelihood function - this is something hacked together in a spreadsheet, so the y axis is arbitrary.
 
  • #9
Vanadium 50
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There's more information here. Your dip is at y=.7. Since you have 40 points, that means the average difference between the points in the dip and the nearest integer is 0.2. The other candidate peaks are at 2.5, which means their average is 0.25, which is the average distance to the nearest integer for a uniformly distributed random number.

The range -0.3 to 0.3 has a standard deviation of 0.6/sqrt(12) = 0.17. That's pretty close to 0.2, which suggests that the whole measurement is driven by a small number of well-measured points.

That's fore sure. It's speculation that these well measured points are also the points with a small number of electrons, but I bet it is right.
 

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