2
$\begingroup$

In recoil proton scintillators, the hydrogen acts as a neutron converter to produce protons when a neutron strikes a H atom. But what happens to the electron in H atom? Where does it go? Does it remain a free electron?

$\endgroup$
  • $\begingroup$ Naively, I would think that it would be captured by another molecule in the scintillator, making that molecule an ion. But scintillator physics is well outside of my expertise. $\endgroup$ – Michael Seifert Apr 12 '19 at 19:25
  • $\begingroup$ Actually, if you have a hydrogen atom, and it collides with a neutron to produce a proton and an electron, doesn't the proton and the electron recombine to produce hydrogen again? $\endgroup$ – Cinaed Simson Apr 12 '19 at 21:34
  • $\begingroup$ @CinaedSimson - indeed, with time, some electron will combine most likely. The chance it is the original electron is pretty small, but that will depend on the recoil energy. $\endgroup$ – Jon Custer Apr 12 '19 at 21:51
  • $\begingroup$ Is there photo multiplier tube on the end of scintillator? How do they detect an event electronically? $\endgroup$ – Cinaed Simson Apr 12 '19 at 22:35
  • $\begingroup$ @CinaedSimson yes there is a photomultiplier at the end of the scintillator. But what would the electron combine with? The recoil proton assists in producing a photoelectron when it interacts with a scintillation molecule $\endgroup$ – Betsy Apr 12 '19 at 22:44
2
$\begingroup$

Scintillators aren't generally made of pure hydrogen; the hydrogen is either covalently bonded to some large molecule or otherwise bound up in the crystal structure of the solid.

If incoming radiation, such as a neutron, knocks a proton free, the proton generally leaves its associated electron behind. So organic scintillators that react this way would have a large neutral molecule turn into a slightly-less-large molecule with net negative charge, due to the missing proton. A crystalline scintillator which underwent proton recoil (though I'm not sure there is such a thing) would, for some amount of time, have an extra electron at the site of the recoil. The relaxation of both of these processed back to equilibrium is generally what produces the scintillation light which actually reaches your detector.

$\endgroup$
  • $\begingroup$ Thank you, that makes it clear! Also, say I have a non-scintillating hydrogenous neutron converter like PMMA attached to a non-hydrogeneous scintillating crystal, the PMMA would develop a net negative charge over time as it gets irradiated with neutrons, right? $\endgroup$ – Betsy Apr 16 '19 at 21:58
  • $\begingroup$ To develop a net negative charge, your neutrons would have to eject protons from thw scintillator entirely. That depends on where in the scintillator the neutron captures occur and the range of the struck protons. Unless your scintillator is very thin, I wouldn't expect it to be an issue --- but it's really a question to be answered by modeling. $\endgroup$ – rob Apr 16 '19 at 23:23
1
$\begingroup$

Given that the ionisation energy in a solid is of the order of a few eV and that the energy of the incoming neutron (and recoil proton) is of the order of several MeV, there is a lot more than the initial electron to worry about. The recoil proton causes a shower of ionised particles in the detector. It is the re-combination of these millions of electron-ion pairs that produces the photons that we "see" with the photo-multiplier tube.

So I don't think you can say for any certainty what happens to the original electron. It just jumps around in the lattice until it finally finds an orbital to call home.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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