Neutron capture cross section of tritium is zero

[pdxjjb]

Main Question or Discussion Point

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?

 

[mfb]

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.

 

[pdxjjb]

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.

 

[Astronuc]

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.