I am thinking if it's possible to detect Cherenkov radiation in water caused by dissolved natural potassium (KOH) in it without using photo-multipliers.

The main question is does 1 beta particle generate 1 photon, or multiple? Specifically, how many photons of visible light should I expect from 1.33 MeV electron in water?

I know that blue light photon has energy of 2.76 eV, but I have no idea about efficiency of this process.


The Frank-Tamm formula previously mentioned by another user is indeed the correct one to use, however actually implementing it in order to come up with a final estimated number of blue photons is not trivial.

The standard simulation tool preferred by working physicists to perform these sorts of calculations is a library of C++ classes called Geant4: http://geant4.cern.ch/. It is a gargantuan package which simulates far more types of interactions than just Cherenkov radiation.

The teeny tiny little subsection which handles Cherenkov radiation can be found here: http://www-geant4.kek.jp/lxr/source/processes/electromagnetic/xrays/src/G4Cerenkov.cc. The code is written and maintained by and for the professional academic physics community, which means that if you can get it running and actually use it to produce a comprehensible output, the physics results themselves are very likely to be validated and error-free.

  • $\begingroup$ Thanks for the link to software. I realized that it's probably easier to do an experiment. As a result I was unable to see any cherenkov radiation caused by natural potassium. Probably this requires more advanced setup (like cooled sensors, many hours of explosure...) $\endgroup$ – BarsMonster Jul 29 '13 at 9:53

I don't quite understand what role KOH plays in your setup, but you can calculate the intensity of Cherenkov radiation using the Frank-Tamm formula (https://en.wikipedia.org/wiki/Frank%E2%80%93Tamm_formula ).

  • $\begingroup$ Role of KOH - it's the source of 1.33 MeV electrons, due to decay of K-40. Thanks for the formula - I would probably unable to use it directly (that would need precise data on permeability and index of refraction across frequencies), but apparently it shows that energy is released as particle travels through water, so it means multiple photons would be emitted. The question is what is rough estimation on number of photons per 1.33 MeV electron, and which fraction of these are visible light? $\endgroup$ – BarsMonster Jul 27 '13 at 15:22
  • $\begingroup$ @BarsMonsters: You can assume that permeability equals 1 and the index of refraction in water is constant in the visible range (about 1.33 - en.wikipedia.org/wiki/List_of_refractive_indices ). Then you can integrate over frequency (within the visible range) and over the angle (mostly around the Cherenkov cone). Sorry, I am too lazy to do that:-) $\endgroup$ – akhmeteli Jul 27 '13 at 15:47
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    $\begingroup$ I've tried to simulate cherenkov radiation using this formula, but figured out that it's harder that I've thought. Energy of the particle drops as it moves, and to calculate how fast it drops - we need to calculate integral across all frequencies, including deep UV. In order for this integral to be finite - index of refraction must be a function which is less than 1 for high frequencies. But it is clear now, that multiple photons are emitted, and their total energy is significant part of 1.3MeV. The last question remains is how much is visible light and how much is UV... $\endgroup$ – BarsMonster Jul 27 '13 at 17:49
  • $\begingroup$ @BarsMonster: In the first approximation, you can assume that the energy of the electron is constant. I may be wrong, but it seems to me it makes no sense to calculate electron energy losses using the Frank-Tamm formula, as, I suspect, ionization losses may be much higher. However, I did not check that. $\endgroup$ – akhmeteli Jul 27 '13 at 18:13

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