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What is the mechanism of cold neutron radiative capture by proton? I have seen some experimental papers on studies of parity violation via this reaction, but I haven't seen a fundamental description of the process or references on one. The old nuclear theory books seem to either omit this topic completely, or consider thermal neutron capture cuz of absence of data for cold neutrons at that time. Any links, references, ideas are highly appreciated.

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  • $\begingroup$ The first thing to understand is what "cold" means: that the neutrons are moving slower than the speed that corresponds to kinetic energy around the thermal energy $k_B T$. I could take a guess at the meaning of "radiative" in this context, but it would only be a guess. $\endgroup$ – dmckee --- ex-moderator kitten Apr 27 '17 at 22:08
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Your treatments for thermal neutron capture are also valid for cold neutron capture. A thermal neutron's wavelength is roughly an angstrom, which is already many orders of magnitude longer than the nucleus, and the energy is much lower than any resonances in any ground-state nucleus --- let alone the deuteron, which doesn't have any bound excitations at all. Changing the neutron energy by a factor of ten by looking at neutrons from a cold hydrogen moderator rather than a room-temperature water moderator only changes the wavelength by a factor of three, which doesn't have much impact on the arguments.

You can see this in the publicly-accessible cross-section data: below the energy of the first nuclear resonance, neutron capture cross sections are proportional to $1/\sqrt E = 1/v$, which means the probability of capture is proportional to the neutron's "dwell time" in the vicinity of the nucleus.

For a reference you may not have considered, have a look at Byrne's Neutrons, Nuclei, and Matter, 1994.

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Anytime (at low energies) you get neutron and proton close enough to feel the strong interaction, you can describe it as a scattering state. There is a resonance in $p-n$ system at very low energy with very high cross section (it is actually a virtual state of the deuteron there).

This scattering state can continue either as elastic scattering (neutron is scattered with some pattern) or there can an electromagnetic process take place and irradiate 2.23 MeV energy by gamma and the final state will be the ground state of the deuteron.

The is lot of information on the net... e.g. https://journals.aps.org/prc/abstract/10.1103/PhysRevC.11.103 is discussed in https://journals.aps.org/prc/abstract/10.1103/PhysRevC.73.044002

There may be a technical problem having a free proton and a free neutron.

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  • $\begingroup$ Wait, but cold neutron capture is $np \rightarrow d\gamma$. It is not a scattering. Why are you making this parallel is not obvious to me, sorry. $\endgroup$ – MsTais Apr 28 '17 at 20:21
  • $\begingroup$ But it also could be. This is how it can be described. Approach, scattering state, transition operator, final state (whatever it is). $\endgroup$ – jaromrax Apr 28 '17 at 20:30

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