It is easy to construct a home-made cloud chamber. In observation, one finds that the directions of the trajectories are quite random. Does this mean that all the particles detected are secondary cosmic rays?
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$\begingroup$ For the most part primaries don't make it to elevations where humans live. Remember that there are around 10,000 kg of atmosphere above each square meter of surface. That's a lot of shielding. $\endgroup$– dmckee --- ex-moderator kittenCommented Feb 28, 2017 at 15:41
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It turns out there is a lot of interesting physics involved in explaining this.
Cosmic rays are about 90% protons and 10% alpha particles (with small amounts of other objects). However these mostly get stopped by collisions in the upper atmosphere. The collisions mostly create muons, so the cosmic rays we see at ground level are mostly muons. Strictly speaking these are already secondary cosmic rays, so I guess that would make the tracks you observe in the cloud chamber tertiary rays, but lets overlook this nicety for now.
The tracks you see in the cloud chamber start when a high energy cosmic ray muon collides with and ionises an air molecule. The collision ejects an electron from the molecule and it's that electron that leaves the track. So the track starts at the site of the muon-molecule collision and fades out after a few cm as the electron loses energy.
So the question is why the high energy muon doesn't leave a track while the low energy electron does. And the answer is (as I've implied by mentioning the energy) that in general the energy loss per unit length is higher for low energy particles like the electrons that it is for high energy particles like the muons i.e. the electrons interact more strongly with the air in the chamber than the muons do.
The energy loss per unit length is described by the Bethe formula. Rather than just quote the complicated and largely incomprehensible formula let me just show the graph of energy loss as a function of energy from that Wikipedia article:
The energy of the muons created by the cosmic rays is around 4GeV at the Earth's surface so that's off the right hand end of the graph meaning that those muons interact only weakly. Offhand I don't know the energy of the electrons that leave the tracks, but I would guess they are around $1-10$ MeV and therefore they interact far more strongly.
answered Feb 28, 2017 at 16:17
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1$\begingroup$ Very interesting answer, and very well formulated, thanks John ! $\endgroup$– ChamCommented Feb 28, 2017 at 16:25
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$\begingroup$ But why cannot the proton travel down to the sea level? It is less than 10 times heavier than muon. $\endgroup$ Commented Mar 4, 2017 at 1:38
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$\begingroup$ @Jiang-minZhang: there is 100km of atmosphere in the way! $\endgroup$ Commented Mar 4, 2017 at 6:01