A friend of mine was telling me about building a cloud chamber while he was in graduate school. As I understand it, this allows you to "see" interactions caused by high energy particles going through the cloud chamber. This has fascinated me, and I would like to build one with my daughter, but I want to make sure I am able to explain it to her when the eventual questions come. Can someone help me out please? How would I explain a cloud chamber to someone who is a freshman in high school?

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    $\begingroup$ have you looked at the wikipedia article? Any specific questions not answered there? $\endgroup$
    – luksen
    Aug 28, 2011 at 19:40
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    $\begingroup$ Mainly the construction as well as suggested language for a 15 year old. Since I want to make one and then be able to explain it to her. $\endgroup$ Aug 28, 2011 at 19:46
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    $\begingroup$ A cloud chamber operates on the principle that ionized particles can act as seeds for droplets, and that charged particles ionize particles, so that the path of a charged particle is visible as clouds. $\endgroup$
    – Ron Maimon
    Aug 28, 2011 at 20:41
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    $\begingroup$ For me the interesting part of the question is "Why do ionized particles act as nucleation sites for <phase change>, and what are the conditions under which this happens?" I say phase change instead of "cloud formation" because bubble chambers are very similar. $\endgroup$ Aug 28, 2011 at 21:20
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    $\begingroup$ @Larian: Except I don't know any good way to make it obvious that ions are good seeds for phase changes. It's experimentally true, but that's not a reason... $\endgroup$
    – Ron Maimon
    Aug 29, 2011 at 7:26

2 Answers 2


Feynman used to say - if you can't explain something in simple words, such that a child could understand, then you don't understand it either. So here's my take:

A cloud chamber is nothing more than a box where mist is about to form, but not quite yet. There's vapors of stuff (either alcohol, or water, or something else) in it, and the temperature is such that the vapors are almost about to produce mist (or "clouds"). Imagine wetlands or marshes on a cold autumn morning, it's kind of like that - fill a box with that kind of "cold wet air".

Now a charged particle (such as Alpha radiation from a chunk of radioactive ore) zips through the chamber at high speed. It bumps into water (or alcohol) molecules and ionizes them - it creates a trail of ionized molecules marking its path.

Now, the vapors are such that they really want to produce mist; any tiny disturbance is enough to push them over the edge. The trail of ionized molecules is enough to do that - the ions attract a bunch of molecules, the resulting clumps attract even more, and before you know it a droplet of water is formed, then another, and another. Voila, a trail of mist follows the particle.

I could try to describe the construction, but this Instructables page will do it much better:


Basically, you evaporate some alcohol and let it run over a very cold area (cooled by the Peltier elements). Like breath coming out of your mouth in the cold air of winter, the alcohol vapors will tend to produce mist, so some vapors will turn to mist anyway. But the process happens a lot faster when a charged particle zips through the chamber - so, if you place a tiny bit of radioactive material nearby, tiny white trails will seem to come from it and traverse the chamber, because mist tends to form that much better around the ionized trails left by the radiation in its wake.

More designs:




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    $\begingroup$ Feynman used to say - if you can't explain something in simple words, such that a child could understand, then you don't understand it either. Love that quote. +1 $\endgroup$
    – JasonR
    Sep 16, 2011 at 13:15
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    $\begingroup$ Keep in mind, it's not an exact quote, it's a rephrasing, but the spirit is the same. $\endgroup$ Sep 16, 2011 at 20:11
  • $\begingroup$ @FlorinAndrei Watching the video on the first link, I saw that the trail disappear after some distance. What happen with the particle when the trail end's? Does the particle decay in a neutral particle? $\endgroup$ Sep 18, 2011 at 13:36
  • $\begingroup$ @RodrigoThomas Several scenarios: The particle may indeed interact with something in the chamber and change into something else. The natural decay of short-lived particles might be involved. The chamber itself may generate trails only in a limited volume. Or the particles may lose their energy and stop. $\endgroup$ Sep 19, 2011 at 18:58
  • $\begingroup$ I do like this answer as it is fairly complete and a good example of what can be accomplished with qualitative descriptions, but I was particularly interested in a microscopic analysis. $\endgroup$ Sep 20, 2011 at 0:28

The cloud chamber works by producing a super-saturated vapor, as explained by florin. When a charged particle passes by, it ionizes the molecules of the liquid, and these ions become centers for droplets, which condense around the ionization trail. But why are ions such good seeds for droplet condensation?

The reason is just electrostatic-dipole interaction, the cloud-chamber fluids are all dipoles. Choosing unit of energy eV, unit of length 1A, and unit of charge 1e, the Coulomb constant k is 14.4 (eV A/e). The dipole moment of water and alcohol (two common vapors for a cloud chamber) are both about .4 eV A. That means that you have to go out a distance of 13 Angstroms before the thermal energy is comparable to the maximum dipole energy.

Within this region, the statistical equilibrium requires that the first dipole that enters sits on top of the ion, because the potential well is essentially infinitely deep. The ion plus polar molecule will attract another ion, and so on, until a droplet is formed.

Ignoring interaction between the dipoles, the radius of a droplet stabilizes electrostatically approximately when the energy gain from being on the surface of the droplet is equal to about 6kT, the factor of 6 is the approximate entropy gain of being in gas vs. liquid. This happens relatively quickly, so you get a microscopic droplet. But in a supersaturated liquid, there are strong forces already between dipoles which mean that there is very little cost to forming bigger drops. The energy gain just from the inter-molecular forces already balances the entropy loss from leaving the gas phase. The only thing that doesn't balance is the surface tension cost.

The radius of the drop is then approximately determined by the place where the average energy gain for a dipole on the surface is equal to the free energy difference between surface and bulk fluid. When this radius exceeds the critical droplet size for the supersaturated liquid, you get nucleation. Since it is going to be about 10 A, it is already pretty big. There is no chance of producing a 10A drop, containing hundred of molecules, by thermal fluctuations.

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    $\begingroup$ Good explanation overall. There are a few points need clarification though. First, the ionization process by the scattering of liquid molecules by the incoming charged particle is passed over. Presumably, the ions you mentioned are generated from the liquid molecules stripped of their electrons instead of the incoming charged particles and the freed electrons. It is not mentioned. Therefore presumably, the ions are all positively charged, unless the molecule captures extra electrons from the incoming charged particle which is unlikely. $\endgroup$
    – Hans
    Jun 13, 2017 at 23:00
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    $\begingroup$ Second, therefore the ions should repel rather than attract each other, and in turn the last sentence of your third paragraph should more likely read "the ion plus polar molecule will attract another dipole, and so on, until a droplet is formed". I crude picture I have in mind is that one ion resides in the center, the dipoles arrange themselves in spherical layers with the negative ends pointing towards the center and the positive ends pointing away from the center. The number of dipoles in each layer is such that the total negative charge almost cancels the ion charge. $\endgroup$
    – Hans
    Jun 14, 2017 at 0:05
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    $\begingroup$ Because the dipoles are not completely free to move, the cancellation is not complete and the degree of which is described by the electric susceptibility of the vapor. Third, it would be great if you could list some related equations for the thermoelectrostatic interaction of the droplet. In particular how one obtain, for example, 6kT and 10A you mentioned. Fourth, it would be great to provide some references where the detailed computation summarized in your answer is carried out. $\endgroup$
    – Hans
    Jun 14, 2017 at 0:13

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