# What happens to molecules when a neutron is captured?

What happens to a molecule during neutron capture? Does the process release enough energy to break the molecule bond? In fission this is obviously the case, since the compound nucleus forms and then splits with enough energy to split the nucleus let alone the molecule it was in. But what happened in a thermal capture event?

For example, consider water molecules with 2 Protium and 1 Ogygen being bombarded with thermal neutrons. After a sufficient amount of time, one of the Protiums absorbs a neutron and becomes deuterium. What happens to the water molecule? Does it become HDO or does the D bond break and it becomes a free D+ ion and OH pair until grabs onto something most likely another H2O molecule or adheres to the wall of the container?

Does the behavior change as you go up the chart of the nuclides? For example, does a Carbon, Iron, and Uranium atom before differently? Assuming of course the Uranium atom doesn't fission.

• – Ben Crowell Jan 22 '18 at 19:32
• – Ben Crowell Jan 22 '18 at 19:44

Welcome to physics.SE! The basic intuition is that nuclear energy scales are MeV, while chemical energy scales are eV, so we do not expect molecular bonds to survive any nuclear process. There are cases where this intuition can be violated, as in some cases of beta decay, but in general it's correct.

A thermal neutron has a negligible amount of energy on the nuclear scale. The energy released in neutron-proton fusion is 2.2 MeV, and this is released by emission of a gamma ray. Because the neutron and the gamma have low momentum, I believe the recoil energy of the deuterium nucleus will be on the eV scale, so it may or may not be enough to break the bond, depending on the neutron's energy. The gamma has a strong and intensely oscillating electric field, so I would expect it to ionize the hydrogen on the way out, but I don't know how big the probability is.

More generally, neutron capture often leads to further nuclear reactions or decays, such as neutron or proton emission, induced fission, and alpha, beta, or gamma decay. In most cases these are virtually guaranteed to have enough energy to break bonds and ionize atoms (and they may also transmute the atom into another element).

• Indeed, the fact that molecular tritium can sometimes survive beta decay is part of the basis for the proposed relic neutrino experiment PTOLEMY. – probably_someone Jan 22 '18 at 19:58
• "-1". ... the neutron and the gamma have low momentum, I believe the recoil energy of the deuterium nucleus will be on the eV scale .... Why would you believe this? – A.V.S. Feb 2 '18 at 6:11
• " The gamma has a strong and intensely oscillating electric field," this is not correct for single photons of gamma ray energy. Any electric and magnetic field information is in the photon wavefunction cds.cern.ch/record/944002/files/0604169.pdf and can contribute with scatters/interactions as it is going out. The wave is a probability wave, not a change of electric fields in the gamma path. – anna v Feb 2 '18 at 6:36

Neutron capture on proton produces 2.22 MeV gamma ray. Its emission produces recoil of the resultant deuteron. Let us check, what the kinetic energy of the deuteron would be as a result of this recoil. The momentum of deuteron would be equal to the momentum of gamma ray $p_\text{D}=E_\gamma/c$ (assuming slow neutron). Since 2.2 MeV is much smaller than the rest energy of deuteron $m_\text{D} c^2 \approx 1.9 \,\text{GeV}$ we could use nonrelativistic expression for the kinetic energy: $$E_\text{D}=\frac{p_\text{D}^2}{2\,m_\text{D}}=\frac{E_\gamma^2}{2 m_\text{D} c^2}=\frac{(2.2\,\text{MeV})^2}{2 \cdot 1.9\,\text{GeV}}\approx 1.3\,\text{keV}.$$ This energy is much larger than the chemical bond in a water molecule (about 5 eV) and so deuteron would break away.

Incidentally, this figure also tells us that there is little point in trying to slow neutrons too much prior to the capture: the net result would be the same.

Additionally, the electron of hydrogen atom has a good chance to stay with the deuteron. Indeed, the change of momentum needed for the electron is $p_\text{D}\frac{m_e}{m_D}$. This is less than a characteristic momentum in the ground state: $p_0=\hbar / a_0$ ($a_0$ is the Bohr radius), and so the deuteron could impart this momentum without losing the electron.