Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

There are some nice pictures on the web showing the counter-spiralling paths of an electron positron pair produced in a bubble chamber with a uniform magnetic field, for example:-

e+ e- pair production

Would it be practical to produce a cloud chamber and Helmholz Coils capable of imaging such events in an ordinary domestic garage? What sort of source would I need? How often might such events be detected?

share|improve this question

3 Answers 3

Basics

You need to

  1. be able to generate a pair-creation event and
  2. be able to image it well enough to know what it was.

Getting a pair conversion event

Pair creation calls for the highest energy gamma you can get and as much mass in the chamber as you can arrange.

The odds of getting a pair-conversion event are graphed in figure 31.17 of the 2013 Review of Particle Physics chapter on the passage of radiation through matter, but it doesn't hit 10% in air until somewhere between 6 and 10 MeV. While this process can happen as soon as you get energies above $2m_e \approx 1.022 \,\mathrm{MeV}$, you need some energy left over to give the created particles some momentum. The mass attenuation length is graphed in figure 31.16 of the same reference, but for a few MeV gammas you are looking at distances around 20 cm.

Being able to tell that you got it

To be able to tell that you got a pair conversion event you need long enough tracks to convince yourself that you have a isolated "vee" and to tell that the tracks curve in opposite directions. That cares for a long enough propagation distance for the created problem it be line-like and to see a non-trivial curvature. For arguments sake let's say that a one-tenth radian curve is good enough. That means getting tracks that are at least one-tenth their radius of curvature.

The radius of curvature of a particle in a magnetic field is given by $$ r = \frac{p}{qB} \,.$$ (Note that I don't write $p=mv$ because the pair may be at least moderately relativistic. Just stick to momentum.)

Engineering concerns

You need to pick a source of gamma rays, and you want it as energetic as possible. For really energetic gammas you need an accelerator based system, but this make the project many times as big as it started. I'm going to assume you will use a radiological source, despite the low energy and unpredictable timing. Cobalt-60 would be enough if you have patience, but I'd suggest Thorium-232 if you can get it (you'll actually be taking advantage of the high energy gamma from the daughter isotope Thallium-208).

Using radiological sources means that you need a continuous data-acquisition systems of some kind--say digital video.

Finally you have to chose the strength of the magnetic field, and that depends on the expected momentum of your pair, which depends on the energy of your source.

share|improve this answer
    
For the curious, the market rate for 1 $\mu\mathrm{Ci}$ of Co-60 (which seems to be the most one can have without a license, but the NRC's rules are too convoluted for me to be sure what they say) seems to be about \$80. –  Chris White Jul 24 at 0:51
    
Co-60 Th-232 Tl-208.... I really hope this guys garage is no where near mine, this sounds like safety perils abound... A particle accelerator sounds no better since these are very high energy photons. –  Dan S Jul 24 at 1:53
    
@DanS You can buy well sealed sources CoTS (in fact, there is more paperwork to get access to raw radio-isotopes than for the sealed sources). And the quantities needed are pretty low. That $\mu\mathrm{Ci}$ that Chris mentions is about $40000\,\mathrm{Bq}$ which is plenty for what the OP desires. The OP will want to store them in a source safe or something equivalent, but they pose no danger to someone outside the garage. –  dmckee Jul 24 at 1:56
    
@dmckee Great answer thanks but I am weak on relativistic momentum, I can work out "Newtonian momentum" given:- Me, E and v=c but I guess your p is derived differently. How? I need to know B to see if I can build a sustained and strong enough magnetic field with 30amp limit to garage power. –  steveOw Jul 24 at 19:30
    
@steveOw You can't just set $v=c$ and use the Newtonian formula. If you know the (total) energy and mass of a particle than $(m c^2)^2 = E^2 - (pc)^2$. If you only know the kinetic energy $T$ of a particle then $E = T + mc^2$ and proceed as above. Note that particle physicists and particle physics references normally set $c=1$ and quote energies, masses and momenta in multiples of electron-volts. –  dmckee Jul 24 at 19:34

These electron-positron pairs are created by gamma rays. I don't know anything about how to make a cloud chamber, but detecting cosmic gamma rays at the surface of the Earth is very very rare. The atmosphere is very opaque to gamma rays (Source). Cosmic gamma rays burst are commonly detected on satellites orbiting the Earth, but very few make it to the surface.

These images of pair productions are likely from gamma rays created at particle accelerators.

share|improve this answer
    
Thanks. I read that gamma rays can be produced naturally by lightning but I guess the kit would have to be pretty close to catch any. I could erect a conductor but I dont think the neighbours would be too happy. –  steveOw Jul 24 at 19:35

I don't think it would be practical at all: purely technical problems aside, I am not sure you would be able to ensure safety - we are talking about ionizing radiation, remember?

share|improve this answer
    
Thanks. Yes I wouldnt be very happy having it around unless it came with very good safety instructions. –  steveOw Jul 24 at 19:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.