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Neutron capture cross section of tritium is zero

  1. Sep 30, 2012 #1
    I wish to verify a couple of assertions. From the NNDC and other references, it appears that the neutron capture cross section of 3H (Tritium) is zero to a first approximation. This holds true for all neutron energies, including cold neutrons. These statements are equivalent to asserting that 4H ("quadrium" !?) does not exist. As a result, all questions about the behavior of putative "quadrium" are not legitimate physics questions. Is this correct?
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  3. Sep 30, 2012 #2


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    According to the references at Wikipedia, it exists as a very short-living nucleus. It should be possible to produce 4H with 3H+n, but deuterium is easier to accelerate and the single proton is probably a cleaner signature.
  4. Sep 30, 2012 #3
    Huh. Thanks. I'm not quite sure what to make of that, because it decays by ejecting the neutron again, typically in less than a millionth of a femtosecond. This is going to look almost like a peculiar kind of inelastic collision...neutron in, neutron out. I suppose it might be important to include this in some kind of very detailed behavioral model. But the 4H doesn't exist in any useful sense. Thanks again for your response.
  5. Sep 30, 2012 #4


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    Looking at (n,total) vs (n,elastic) cross-sections for H-3 down to 1E-5, they are essentially identical, so the absorption cross-section is neglible, i.e., H-4 effectively doesn't exist, i.e., the binding energy is neglible. Elastic scattering is the more likely outcome.
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