# How do they ensure there is one electron on an oil drop in Millikan's oil drop experiment?

Since air is ionized using X-rays, in the 'observation chamber', there must be several electrons in the chamber, so the oil drops are exposed to several electrons so it seems intuitive that a single oil drop should catch more than one electron.

Then why is it that there's just one electron on a droplet or do they selectively choose those droplets which have a single electron residing on them?

• physics.stackexchange.com/questions/311522/… – andselisk Dec 24 '17 at 7:23
• And, more to the point, they can only catch an integer number of electrons. – Jon Custer Dec 24 '17 at 8:19
• Perhaps even more to the point, did Millikan ignore droplets which apparently had fractional charges? – matt_black Dec 24 '17 at 17:04
• @DavidRicherby the point is that, unless you know the answer, you can’t use this experiment to prove it. And even when you do know the answer, there are experimental errors. Questions have been raised about whether Millikan ignored some actual observations to make his results look better. – matt_black Dec 24 '17 at 18:38
• If only it were somehow possible to reproduce the experiment and verify the results... Oh wait. – barbecue Dec 25 '17 at 17:44

## 3 Answers

There does not have to be just one electron per drop. Say you have a drop which, in reality, picked up four electrons, another with five and a third drop with seven. None has one electronic charge, but when you measure charges you find they have a common factor; the first drop has four times that factor, the second has five times that factor and the third drop shows a multiplier of seven. It was this common factor, not necessarily the charge on any specific drop, that was recognized as one electronic charge.

The experiment showed all the drops had discrete amounts of charge. That means, the charges weren't all over the place (any random value). They only had specific values. Some had $2$ or $3$ or $4$ times the charge of others, but it was always some specific value that they had multiples of.

The conclusion was that oil drops didn't seem to pick up "any random amount" of charge, and the reason seemed likely to be because electric charge couldn't be just "any value". There seemed to be some basic unit of a "single electric charge", the smallest value that was found. Some oil drops had $1\times$ or $2\times$ or $5\times$ that charge, but no oil drops had (say) $3.77\times$ or $1.628\times$ that value.

• Actually, the initial formula $(1)$ that was employed in the experiment $$e'= 9\unicode[Times]{x3c0}\frac{v_0+v_E}{E}\sqrt{\frac{2\eta^3v_0}{g(\rho-\rho_0)}}\tag1$$ failed to give a constant common divisor, and Stokes law was consequently corrected for radii near the mean free path of a gas particle. $|$doi.org/10.1103/PhysRev.2.109 (Slightly different notation, here I used subscript '$0$' to denote having to do with air, and dropped subscript $n$ in the paper for an apostrophe.) – Linear Christmas Dec 27 '17 at 13:51

This is my absolute favorite physics experiment; I actually wrote a miniature essay on my lab report containing, in essence, the following:

Millikan's experiment is amazingly indirect. You don't directly measure the charge of the electron, nor do you compute it from an equation (like, say, the one relating electrostatic force to charge, though you do use this equation to deduce the non-unit charge on each oil drop). You don't even deduce it by fitting a curve to a set of data points. You do deduce it from a frequency graph, but it is, again, not any kind of fitting of the actual frequency data; rather, you identify spikes in the frequency counts and mark off the corresponding charges. Then you find their greatest common divisor and argue that your data was random enough that this must be the charge of one electron (since it is vanishingly unlikely after enough droplets that each one contained only, say, even numbers of electrons). It is actually a simple form of image recognition.

You don't need to luck into one-electron droplets. You just need to keep measuring until you have an unambiguous set of frequency spikes at sufficiently many different charges and apply (what?!) a number-theoretic computation.

## Addendum

The business of taking the greatest common divisor is a bit tricky in the presence of measurement errors: after all, the charges you find are not only subject to error from your lab equipment but also from your identification of the center of each spike. You can assume the charges are all integers by converting your limited-precision floating point numbers to fixed point, but those integers are almost certain to have a GCD of 1. For example, if you measure charges of 201 and 302, you'll find that the fundamental charge is not 100 (which is the obviously correct answer) but rather 1.

You can, of course, eyeball it: say, you can take various ratios and fit them to nearby rational numbers with a small common denominator (in the above example, the ratio is approximately 1.5025, so you easily find 1.5 = 3/2 as a likely "correct" ratio). A better way is to use an "error-tolerant" version of Euclid's algorthm. In short, proceed as usual by dividing (with remainder) the smallest number into all the others and repeating, except that rather than waiting for all the remainders to be 0 (indicating that the last remainder was the GCD), you wait for them all to be "small" in some sense. Say, an order of magnitude smaller than the previous one.

Take the above example: Euclid's algorithm gives you the following sequence of remainders: 302, 201, 101, 100, 1 (each one is the remainder of the division of the previous two). This suggests that 100 is the correct GCD, as indeed it is. Amazingly, the algorithm actually wiped out the measurement errors and got the exact correct GCD; I don't know if this kind of "focusing" effect is typical or if I just happened to use the right numbers.

This only increases my love for this experiment.

• I actually think 101 is a better fit for your initial values. Euclid's algorithm gives progressively "better" numbers on average (not necessarily on every step), so you will find a "good" but not exact fit before you complete the sequence, unless the sequence has a low true GCD. Try it with Fibonacci numbers: 21, 13, 8, 5, 3, 2, 1. 21 and 13 are both roughly divisible by 5, for example, and 3 divides them even better because it is smaller. Note: the result of each division is very close to a Lucas number: 1, 3, 4, 7, 11... – CJ Dennis Dec 27 '17 at 1:28
• I agree, and thanks for the discussion. One can make precise one's concept of "correct values" by putting error bars on the measurements, which may help you find a range of approximate gcds, and is better science anyway. I'm on my phone, so I don't want to work out how that would affect this example. I'm actually not entirely sure you could deduce a range just knowing the error bars, without trying every value in between. – Ryan Reich Dec 27 '17 at 1:54
• I was taught in high school that the experiment had to be reran for hours and hours to get enough ones to be reasonably sure there wasn't a smaller number, and the GCD was used only to cross-check afterwords. – Joshua Dec 27 '17 at 16:38
• You definitely have to go for hours :). I suppose that method has the advantage of actually directly observing a single election's charge, but I'm not sure how reasonable it is. Being a mathematician, I probably like the abstract existence proof because it is more elegant and, statistically, airtight. – Ryan Reich Dec 27 '17 at 17:17