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In reviewing some problems in an elementary book, I ran across a reference to the reaction $p+n\rightarrow d$ + "energy".

Is that possible? I don't see any reason why not, but I don't find any mention of this reaction at all using Google. It seems to me that the "energy" would have to be a combination of deuteron kinetic energy and a gamma.

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  • $\begingroup$ It's probably possible - but keep in mind that free neutrons are unstable, and they decay into a $p^+e^-\bar\nu_e$ triplet within about fifteen minutes, so your problem is going to be getting hold of the neutron to begin with. That's what makes this reaction so rare w.r.t. $pp→de^+$ fusion in the Google results. $\endgroup$ – Emilio Pisanty Nov 20 '16 at 17:24
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    $\begingroup$ have a look at en.wikipedia.org/wiki/Chronology_of_the_universe#Hadron_epoch , at that time the energies and densities are such that neutrons can meet protons :) $\endgroup$ – anna v Nov 20 '16 at 17:52
  • $\begingroup$ @annav Thanks. I recalled being aware of that reaction in the early universe, but I couldn't find a reference to it. $\endgroup$ – garyp Nov 20 '16 at 17:59
  • $\begingroup$ Kind of a silly question, but: What is "$d$"? $\endgroup$ – ACuriousMind Nov 21 '16 at 3:20
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    $\begingroup$ @ACuriousMind a deuteron (which is to say, the unique bound state of one proton and one neutron). $\endgroup$ – Logan M Nov 21 '16 at 5:36
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Of course the reaction is possible. It doesn't even require special environmental conditions. Having no charge the neutrons don't need to overcome a strong Coulomb barrier to interact with atomic nuclei and will happily find any nuclei that can capture them at thermal energies. KamLAND (for instance) relies on this reaction as the delayed part of the delay-coincidence in detecting anti-neutrino events in the detector. In the mineral oil environment of KamLAND the free neutrons have a mean lifetime around $200 \,\mathrm{\mu s}$.

Neutron capture even on a proton releases 2.2 MeV. Chlorine, boron and gadolinium are all better neutron capture agents than hydrogen bearing molecules like water and oils, and captures to those absorbers release even more energy per event.

So why isn't everyone jumping around cheering for room temperature fusion and prognosticating a beautiful future full of safe and abundant energy?

Because there is no adequate supply of free neutrons. With their roughly 15 minute beta-decay lifetime there is no naturally occurring reserve and you can't store them in any case.

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I would like to oppose dmckee's statement, there are very good and stable neutron sources in fission power plants. Referring to garyp's question, if the p + n - D + "energy" proceeds as an exoterm process so the proper conditions are at hand in small experimental nuclear power plants. It is true, the neutron is stable only for a relative short time, but this time would be sufficient to execute a fusion reaction. I suppose, it would be an interesting experiment to combine a fission process with a fusion process in an experimental power plant by modifying the adsorbing medium to produce deuterium having additional thermal energy. It could contribute to the energy of the cooling medium. The evolved deuterium nucleus could be used to further experiments in D,T reactions as well.

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    $\begingroup$ Moderated neutrons in a nuclear pile do capture, and do contribute their capture energy toward the heat evolved by the plant. But that merely reduces a possible source of inefficiency in the conversion of available fissile energy to a useful form. You can also buy commercial off the shelf neutron generators which work by artificially induced fusion, but you have to put more energy into the generator that you can—even in theory—recover from the neutrons. $\endgroup$ – dmckee Jul 23 '17 at 19:35
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    $\begingroup$ Quantitatively, a uranium fission reaction releases about 200 MeV. Recovering a 2 MeV photon from neutron capture on water instead of, say, and 8 MeV photon from neutron capture on cadmium is pretty deep in the weeds of the thermal engineering. $\endgroup$ – rob Jul 26 '17 at 6:16
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    $\begingroup$ As far as extracting the deuterium, transmutation has its own challenges. For my thesis experiment, looking at a weak-interaction effect in $np\to d\gamma$, we needed to observe about $10^{18}$ neutron captures to get the statistical sensitivity we wanted; we therefore made a few micrograms of deuterium mixed in with perhaps a kilogram of ordinary hydrogen, using one of the most intense neutron beams in the world for more than a year. Natural deuterium is already more common than this. $\endgroup$ – rob Jul 26 '17 at 6:16
  • $\begingroup$ Referring to rob's experiment I would like to ask about the neutron's energy range coming from the neutron generator. As it is well known, the hydrogen is a good medium for the absorption of neutrons, but I suppose, it must be an optimal energy level - perhaps around the thermal region, - where the neutron capture through fusion could reach a maximum and the D-yield of the system is much higher than in rob's experiment. (Is there any possibility to read the publication about the experiment?) $\endgroup$ – Tamas Aug 23 '17 at 20:07
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The p + n --> D and D + n --> T reactions are the basis of the difference between light and heavy water used as the moderator in fission reactors. Because the former reaction has bigger cross section and steals more neutrons from the chain reaction neutron economy.

Because of the optimal proton/neutron mass ratio, light water would be better moderator than heavy water, as it faster slows down neutrons to the thermal speed. It would be, but is not, as this effect is overruled by higher neutron absorption.

The result is, that while heavy water moderator allows to build fission reactors with unriched uranium, reactors with light water moderator require enriched uranium. ( typically 2-3 % of 235U, instead of the natural 0.7% )

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