"Reversal" of Nuclear Decay in Beta Emitters

  • #1
Hello all,

I've got a question on nuclear decay "reversal" in beta emitters.

I've been researching the Cowan-Reines experiment, which used neutrinos to convert protons into neutrons. Recently, I found out that the particle which hits the proton need not necessarily be a neutrino in order to induce a proton-to-neutron conversion. It looks like an energy of 1.29 MeV is enough to convert a proton into a neutron.

That being said, there's a couple of questions that come to mind:

Say that we have a beta emitter which decays to a stable isotope. We use an accelerated electron of 1.29 MeV to convert one of the protons in the nucleus into a neutron.

But does this mean that the isotope will undergo the beta decay again, with the same energy value? Or does there need to be additional energy in order to not just convert a proton to a neutron, but to produce an electron as well?

Answers and Replies

  • #2
If you can get the reverse reaction with your electron beam then the produced nucleus can do beta decay again, sure. If your electron beam doesn't provide enough energy for a nuclear reaction then there won't be one.
  • #3
Meir Achuz
Science Advisor
Homework Helper
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"We use an accelerated electron of 1.29 MeV to convert one of the protons in the nucleus into a neutron."
The process is e +p --> n +neutrino. A neutrino must always be involved. 1.29 MeV is the minimum possible energy, but the actual energy is always higher. Any beta decay energy is determined by the energy difference of two nuclear levels.
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  • #4
If it's a proton in a (non-trivial) nucleus then the required energy can be lower or higher. It's no longer sufficient to treat the proton as isolated particle, the binding energy of the parent and possible daughter nucleus has to be considered. In some cases there is not even an energy threshold.
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  • #5

this is the state-of-the-art theoretical framework of electron capture

Converting a proton into a neutron, or viceversa, involves weak interaction. As its own name says, it is not likely to happen. Reaction rate is given by

r = n1 * n2 * <sigma*v>

n1, n2 = density of two interacting particles
<sigma*v> = cross section averaged by Boltzmann distribution (it is a function on the temperature)

If cross-sections is low, you will need huge densities.

Coming back to shooting high energy electrons to nuclei... the most likely interaction will be elastic scattering, which was worth a Nobel prize