As far as I have understood neutron nuclear reactions, if a fast neutron gets captured by a nucleus, since the kinetic energy has to go somewhere, the newly formed nucleus is radioactive and must loose this energy (by gamma emission or decay).

What happens in the case of cold (meaning kinetic energy below 0.025 eV) neutrons? Does irradiation of a stable element by cold neutrons produce radioactive isotopes? If yes, only theoretically or have there been experiments?


3 Answers 3


Every neutron capture causes a nuclear transmutation. The “neutron separation energy” of the daughter nucleus is generally released promptly, usually as a cascade of gamma rays. The ground states of the daughter nuclei may be stable or unstable. The new unstable nuclei are referred to a “neutron activation products”; material which has become radioactive following exposure to neutrons is said to have been “activated.”

Some practical examples:

  • High-density polyethylene, whose chemical formula is long chains of $\rm CH_2$, mostly captures neutrons by hydrogen to deuterium, with some fewer captures changing carbon-12 to carbon-13. Neither capture product is radioactive, so clean HDPE is not activated by neutron beams.

  • Aluminum has only one stable isotope. Neutron capture on aluminum forms aluminum-28, which beta decays to silicon with half-life of a couple minutes. Neutron-activated aluminum has no detectable activity after an hour or so.

  • If you were building an experiment that used cesium iodide detectors to look at the prompt gamma rays from neutron capture, but your shielding design were flawed and neutrons entered your detector crystals, you would also detect radiation from the decays of iodine-128 (25m), the isomer cesium-134m (2.9h), and the ground state cesium-134 (2y). Boy howdy, was that an expensive mistake. Not that I’m bitter.

  • Cold neutrons in the air undergo a nucleon transfer reaction with nitrogen,

    $$\rm ^{14}N + n \to p + {}^{14}C$$

    which has biological consequences.

  • If you own a smoke detector with an americium ionization source, that americium was produced by repeated neutron capture on uranium in a reactor core.

In general, neutrons whose kinetic energy is below the energy of any nuclear resonance have cross section

$$ \sigma = \sigma_\text{thermal} \sqrt{\frac{E_\text{thermal}}{E}} $$

where $E_\text{thermal} \approx \rm \frac{1}{40}\,eV$ is the kinetic energy associated with “room temperature.” Nuclear resonances bottom out in the kilo-eV range, so the approximation is good for basically all milli-eV neutrons. The energy dependence is often referred to as a “one over vee” cross section, referring to the neutron velocity $v = \sqrt{2E/m}$. A hand-waving interpretation of a $1/v$ cross section dependence is that the probability of neutron capture is proportional to the neutron’s “dwell time” in the vicinity of the nucleus.

The smallest neutron separation energies are mega-eV, so the kinetic energy of a milli-eV neutron is completely negligible in a capture reaction.

  • $\begingroup$ So you say, HDPE is not activated by cold neutrons? It should be composed of 1,1% C-13 and 0,01% H-2 (Deuterium) (each per element per mol), so C-13 and H-2 do not have any neutron cross section, right? But N-14 does, you said. Is there a table or database where I can look up $\sigma_{thermal}$ for different isotopes? $\endgroup$
    – iblue
    Commented Feb 12, 2022 at 20:44
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    $\begingroup$ @iblue Ask and receive. www-nds.iaea.org/exfor/endf.htm $\endgroup$
    – Dan
    Commented Feb 12, 2022 at 22:15
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    $\begingroup$ @iblue I have a paper copy of this little pocket-sized book, but from a computer it might be easier to use the National Nuclear Data Center. A useful homework problem would be to compute whether the neutron activation due to capture on C-13 or on H-2 is more important. In practice, everything that comes out of the neutron beam has to live in a designated cabinet until the radiation control team has cleared it; HDPE is usually is inactive on its first test. $\endgroup$
    – rob
    Commented Feb 12, 2022 at 22:34
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    $\begingroup$ Well, @TLW, americium-241 is formed from U-238 by capturing three neutrons. But that has to happen in a reactor core, where, as you say, there are "enough" neutrons. The question is about cold neutrons, which in practice have to be extracted from some moderator in a way that reduces the neutron flux quite a bit. In the experiment I mentioned, we stopped a microgram of neutrons in a kilogram of liquid hydrogen, transmuting one nucleus per billion; that was at one of the world's most intense cold neutron beams, and it took two years of beam time. Chemically significant transmutation is hard. $\endgroup$
    – rob
    Commented Feb 15, 2022 at 5:16
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    $\begingroup$ @TLW Compare my experiment’s $10^{-9}$-ish beam-induced transmutation to the natural deuterium abundance of $10^{-4}$. For the case of HDPE, consider also that any incidental tritium is quite long-lived. Furthermore the range of the low-energy tritium $\beta$ in HDPE is much shorter than the range of neutrons which form it, so any tritium activation tends to be self-shielded. As to whether to worry about any incidental carbon-14, I invite you to do the homework problem I suggested in a previous comment. $\endgroup$
    – rob
    Commented Feb 15, 2022 at 14:47

Actually the neutron capture cross section is typically higher for slower neutrons (that's why you need a moderator in a nuclear reactor that slows down the hot neutrons released in the fission reactions). For the nuclear reactions it does not matter energy wise whether the neutrons are thermal or cold (a few eV are irrelevant at the energy scales we talk about).

A neutron captured by a nucleus releases binding energy due to the strong nuclear force. The initial configuration is typically an excited state, often there is $\gamma$-emission as the nucleus relaxes to its ground state. An unstable nucleus may well be the result.

As the neutron content of the nucleus increases, it may no longer be stable against $\beta$-decay, so an unstable nucleus may be the result.

Wikipedia has a helpful article: https://en.wikipedia.org/wiki/Neutron_capture it includes a nuclide map with the cross sections for thermal neutron capture and reaction examples.


Things are more complicated than your question implies.

As Sebastian Riese mentions, higher energy usually means lower cross section. There is a lot of complexity. St some intermediate energies there are resonances. It is getting far afield from your question. But let us set that aside. Let us suppose there is an interaction.

It will depend on the target nucleus. This will give you many possible results.

In the following you should take it as understood that many of these reactions can produce some extra gammas along the way. If a reaction puts a nucleus into an elevated energy state, an isomer, then it can give off a gamma to relax back to the ground state. Even with this warning I am glossing over a lot of details.

For some isotopes, it will be possible for the incoming neutron to produce fission. The result is often two fragments of the original nucleus, plus some few neutrons. The number of neutrons is random and usually in the range of 1 to 3, depending on the type of nucleus and the energy of the neutron. The fragments are very often radiocative since they will very likely have too many neutrons to be stable. There are several possible radioactive decays they can follow, depending on what isotope they are. They can release a neutron. They can release an alpha particle. They can beta decay. Some can do an electron capture. And lots of them will release gammas on the way to some other decay. Fission fragments are a "soup" of many different types of radiation at many different energies and many different half lives.

Many isotopes can capture an incoming neutron. Depending on the energy of the incoming neutron, this can happen a few ways.

If there is an energy state available for the resulting nucleus, it is possible for it to simply absorb the neutron. Much more usual is what is called an n-gamma reaction. The neutron gets absorbed and the nucleus immediately releases a gamma to allow it to go into one of its available energy states. This can result in a new nucleus with one more neutron. This new isotope can be stable or it can be radioactive. This will depend on the starting nucleus and the energy of the incoming neutron.

For an example, consider iron. Natural iron has four isotopes: Fe54, Fe56, Fe57, and Fe58. Fe56 is by far the most common. But for fun, suppose the target nucleus is Fe54, which is 5.85% of naturual iron. So it catches a neutron and becomes Fe55. Fe55 can do electron capture with a half life of 2.73 years, and become Mn55, which is stable.

There are other things the incoming neutron can do. One important reaction is called "spalling." The incoming neutron can cause neutrons from the target nucleus to get kicked off. This is called an n-2n reaction. It can happen in lead. The result is, lead can be a poor choice for shielding against neutrons since it can result in more neutrons than you started with. They will be at lower energy than you started with, but more of them. And in some situations, that is worse than no shielding at all.

As to the lead nucleus, lead has four stable isotopes: Pb204, Pb206, Pb207, and Pb208. So if the spalling moves from one to the other of these it is still stable. PB205 can electron capture (possibly after releasing some gammas) with a half life of 17.3 million years. And PB203 can electron capture to Tl203 with a half life of 51.9 hours, and Tl203 is stable.

There are some other reactions that can happen, depending on the nucleus involved. But it is already quite complicated.

So, to sum up. It depends on the energy of the incoming neutron and the type of nucleus it is hitting. It can produce a new stable isotope, or it can produce a radioactive isotope through one of several reactions. And it can release several different types of radiation.

  • $\begingroup$ AFAIK, neutron-induced fission is only likely to occur with heavy nuclei. I doubt it happens with cold neutrons hitting light nuclei. $\endgroup$
    – PM 2Ring
    Commented Feb 12, 2022 at 15:09
  • $\begingroup$ BTW, you have a few typos. $\endgroup$
    – PM 2Ring
    Commented Feb 12, 2022 at 15:09

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