The neutron capture reaction 3He(n,p)3H is very useful for neutron detection as the Q value of ~700keV is converted to kinetic energies in the produced p and 3H. These charged products can then ionize gas mixed in with 3He in a gas chamber and then be detected. The cross-section for this reaction is massive at ~kilobarns for thermal neutrons (~2000m/s).

My question is: why does the reaction 3He(n,gamma)4He not occur very often? The Q-value for this is ~20MeV! This is because the 4He nucleus is so tightly bound. A quick literature search for this puts the cross-section at ~microbarns.

Am I missing something? Is there some selection rule I'm not aware of?

  • $\begingroup$ I've found in Krane - Introductory Nuclear Physics on the section on neutron capture the following: "...gamma emission [after neutron capture] is the most probable decay process. (Charged-particle emission is inhibited by the Coulomb barrier and unlikely to occur in any but the very lightest nuclei.)" I don't fully know how to interpret this. $\endgroup$
    – Kent
    May 5, 2014 at 22:09
  • $\begingroup$ The Coulomb barrier is a term used in Gamow's model of charged-particle decay. For example, see figure 14.9 here $\endgroup$
    – rob
    May 6, 2014 at 3:22

1 Answer 1


It's because the lowest-energy excited state in $^4$He is at 20.2 MeV and has spin-parity $0^+$. (I think that this link selects $^4$He from the chart of isotopes; click "list of levels" or "level scheme.") A $0^+\to0^+$ transition cannot emit a photon, because the photon must carry off one unit of angular momentum. That state is just wide enough to overlap with the energy of an unbound $^3$He and neutron at rest.

You get the same sort of effect in $^6$Li(n,$\alpha$)$^3$H. (Though tonight I can't find an associated excitation in $^7$Li, grrr.) For this reason, enriched lithium is also used to dope scintillators for neutron detection or for neutron shielding where it's important to have a low gamma background.

All the heavier nuclei have internal transitions and emit 1–10 MeV of photons after neutron capture. Usually the photons come from a cascade of internal transitions and don't have anything like a line spectrum.

The $^3$He nucleus has a fairly large magnetic moment and can pick up a nuclear spin from an electron-polarized alkali vapor, like optically pumped rubidium or potassium. Since the $^3$He+n capture passes through a $0^+$ state, polarized $^3$He will prefer to absorb one neutron spin state. This is a common method for polarizing beams of thermal or cold neutrons.

  • $\begingroup$ That's a great explanation. Thanks! I found 7Li levels on p.4 of tunl.duke.edu/nucldata/fas/07_1966.pdf. The mass difference between 6Li + n and 7Li gives 7.2MeV of excitation energy. Which of these states are preferentially excited? The 1st excited at 0.5MeV does allow gamma emission. There's one at 4.6MeV, which gives the alpha & triton, and another at 7.5MeV that gives an additional n emission channel. $\endgroup$
    – Kent
    May 7, 2014 at 16:59
  • $\begingroup$ I was explaining to someone that the large singlet, and near zero triplet, n+3He capture cross-section is due to the Pauli exclusion principle. Is this the correct "deeper" interpretation of why ground and first two excited states have J=0? Also, I'm looking at the table and asking why for the higher energy levels where J=1 so that photon emission IS possible, but it doesn't occur. $\endgroup$
    – Kent
    May 7, 2014 at 17:11
  • $\begingroup$ Nice find. The 6Li(n,X) reactions are numbers 4–8 in the list on pages 4–6. The (n,n) is a scattering reaction for eV neutrons and faster. (n,$\gamma$) has cross section 45 mb, compared to 1 kb for (n,$\alpha$). If you look at the states' energies and their widths it should be pretty clear that only the 7.5 MeV state has very much overlap with the 6Li and free neutron at rest. $\endgroup$
    – rob
    May 7, 2014 at 17:14
  • $\begingroup$ There could be some isospin argument against any low-lying $1^-$ states in $^3$He, but I don't know it. I have used the exclusion principle argument myself, no shame in that. Thermal neutrons can't access the higher-energy states because they don't have enough energy. $\endgroup$
    – rob
    May 7, 2014 at 17:19
  • $\begingroup$ Great discussion! "There could be some isospin argument against any low-lying 1− states in 4He" (and not 3He, right?). I agree that it's not possible to excite the 2- or 1- states of 4He with 3He+n with thermal neutrons. But suppose that we had neutrons with ~3-5MeV or use other reactions to get to those states. Why do they not emit a gamma? In the table you posted, it only shows n & p emission branches. $\endgroup$
    – Kent
    May 7, 2014 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.