Proton - neutron fusion?

Question

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.

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. — Emilio Pisanty, Nov 20 '16 at 17:24
have a look at https://en.wikipedia.org/wiki/Chronology_of_the_universe#Hadron_epoch , at that time the energies and densities are such that neutrons can meet protons :) — anna v, Nov 20 '16 at 17:52
@annav Thanks. I recalled being aware of that reaction in the early universe, but I couldn't find a reference to it. — garyp, Nov 20 '16 at 17:59
@ACuriousMind a deuteron (which is to say, the unique bound state of one proton and one neutron). — Logan M, Nov 21 '16 at 05:36
@ACuriousMind Sometimes a deuteron is written with capital $D$, just like the deuterium atom (a deuteron nucleus with one electron in orbit). — Jeppe Stig Nielsen, Nov 21 '16 at 10:07

score 27 · Accepted Answer · answered Nov 20 '16 at 17:54

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.

What about in a light-water reactor? I imagine the relative contribution to sensible energy is very small, but any idea what the contribution is - i.e. of n,p reaction - given so much water (2 protons ea molecule) in presence of decent neutron flux? — jpf, Mar 24 '21 at 23:36

Poutnik · Answer 2 · 2018-02-04T08:48:34.780

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% )

score 1 · Answer 3 · answered Jul 23 '17 at 18:42

1

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.

answered Jul 23 '17 at 18:42

Tamas

19

3

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. – dmckee --- ex-moderator kitten Jul 23 '17 at 19:35
2

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. – rob Jul 26 '17 at 06:16
1

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. – rob Jul 26 '17 at 06:16
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?) – Tamas Aug 23 '17 at 20:07

score 0 · Answer 4 · edited Mar 24 '21 at 15:13

The neutron source would then be the $$^9\text{Be} + \alpha \to \,^{12}\text{C} + \text{n}$$ reaction. You would need the normal Beryllium-$9$, an alpha emitter and water, which could simultaneously serve as a moderator (slowing down the neutrons a little) and a target (where the neutrons could then fuse with the protons). Old fuel rods could perhaps be used as alpha emitters. The Uranium-$238$ contained in them (the non-fissile one) is an alpha emitter. One would get the nominal Carbon-$12$, deuterium but unfortunately also decay products of Uranium.

The reactor would have to be designed in such a way that the uranium decay products, the carbon and the heavy water could be removed from the reactor and new materials could be spat out. This would require a medium level of effort for radiation protection.

The question is whether it is then so productive in terms of energy production. $2.2$ MeV per proton-neutron fussion. The alpha particle of Uranium-$238$ has $4.2$ MeV. Ideally, this is then braked down to $2.6$ MeV. There is then a certain probability that a neutron with $740$ keV will be produced, which in turn must reach thermal velocities ($100$ meV range). And then, with a certain probability, there is again this proton-neutron fusion.

That means, if such a reaction chain runs ideally, we get $$2.2 \text{ MeV} + (4.2-2.6) \text{ MeV} + (740-0.0001)\text{ keV} = 4.5 \text{ MeV}.$$ Average energy consumption per household is $22,400$ kWh per year. This means an average power of $2.56$ kW. This would require $3.56\times 10^{15}$ perfect reaction runs. This means that $1.4\times10^{-6}$ grams of Uranium, $0.05$ grams of Beryllium and $0.0059$ grams of Hydrogen would be consumed per second.

To calculate the costs for this, I found the following values for Uranium and Beryllium: Uranium $19.45$ euros per gram and Beryllium between $0.30$ and $1.50$ euros per gram depending on the quality. This means that the main cost is the Uranium. A kilowatt hour would therefore cost about $4$ cents. In terms of generation costs, this would lie between a nuclear power plant and a coal-fired power plant. If I have calculated correctly. Because then all sorts of things like the EEG surcharge and the profits of the corporations and others are added to the electricity prices, we then pay around $9$ times this price. So a proton-neutron reactor for the home might even be worthwhile.

Okay, that was the ideal case. Of course, the disposal of the radioactive material costs money. The deuterium could possibly even be sold. Then, of course, you need a bit more material, e.g. more water as a moderator. And not every reaction chain runs ideally.

score 0 · Answer 5 · answered Mar 24 '21 at 21:55

Small mistake in my answer

"This means that 1,4e-6 grams of uranium, 0,05e-6 grams of beryllium and 0.0059e-6 grams of hydrogen"

And i would add, that this cold nuclear fusion induced by radioactive decay will not result in a nuclear catastrophe. It is not a self-reinforcing process. In nuclear power, the more nuclei that are split, the more neutrons are released that potentially split nuclei again. Here, radioactive decay does not accelerate when there is more fusion.

Proton - neutron fusion?

5 Answers5

Linked