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Neutron capture by light nuclei

  1. Mar 4, 2009 #1
    I have a decent understanding of neutron-induced reactions in heavy nuclei (fission, capture), but less so when light nuclei are involved. I do know of Li6 + n --> He4 + T and the Li7 equivalent. Those reactions are more like nucleus splitting (fission).

    My question is: do neutron capture reactions for light nuclei exist? I'm mainly thinking of protium. Would it be possible to produce deuterium by having protons capture neutrons? I read about proton-neutron collisions (as a shielding measure against a neutron flux, slowing the neutrons down), but those didn't involve a forming of a new nucleus. Where do those neutrons actually go after being slowed down? Are they just around until they decay to protons?
  2. jcsd
  3. Mar 4, 2009 #2


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    In protium, a proton can combine with a neutron to form a deuteron in a deuterium atom, and deuteron could capture a neutron and become a triton, which is the nucleus of tritium. The cross-section for n-capture by deuterium is very small, hence the motivation to use heavy water as a moderator in CANDU reactors.

    He-3 is a particular effective capturer (absorber) of neutrons, and the product is He-4.

    B-10 is another effective absorber of neutrons and is used in neutron detectors and control rods in nuclear reactors.

    n(6Li,α)T is a source of tritium.

    Be is a good reflector of neutrons, and at high (fast) energies induces an (n,2n) reaction.

    Protons effectively moderate neutrons since they have approximately the same mass, and by that a neutron may lose up to half of its kinetic energy (by virtue of the conservation of energy and momentum).

    If neutrons are not absorbed, they will eventually decay into a proton, electron and anti-neutrino.
  4. Mar 4, 2009 #3
    Thanks for the reply, much useful information in that.

    About these reactions: something has to be wrong in my calculations. I've calculated that 10B + n --> 11B would be exothermic and yield 11.5 MeV. This seems quite an energy to me. That's approx. 111 TJ/kg. The energy density rivals that of ordinary heavy-nucleus fission. One would just need a neutron flux.

    Can someone correct my calculation, or tell me why isn't this used in energy production?
  5. Mar 11, 2009 #4
    The thermal neutron n + B^10 capture reaction goes to lithium 7 and an alpha particle, plus about 2.3 MeV of ionization energy. The energy release is nearly 100 times less than uranium fission per nucleus, plus the fact that there is no neutron multiplication to sustain the reaction, makes this reaction unsuitable for energy production. I would guess that if you put boron into a reactor, you would poison (quench) the reaction by absorbing neutrons.

    Free protons will capture thermal neutrons to form deuterons, and and emit a 2.2 MeV gamma ray.
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