5
$\begingroup$

The answer to this physics se question, established that 2.2Mev (Mega electron volts) of energy is emitted as a gamma ray photon when the hydrogen atom nucleus oka protium absorbs a slow moving free neutron causing the formation of a deuterium nucleus. Deuterium is heavy hydrogen with both a proton and neutron in the nucleus and weighs twice as much as regular hydrogen.

Deuterium is a more stable nucleus than protium, meaning it is more energetically favored over protium and neutron seperate. This is due to the strong nuclear force binding nucleons together. The mass of the deuteron is less than the sum of the proton and neutron. The deficit of mass in the reaction is released as energy; in this case in emission of the gamma ray photon of 2.2MeV. This is what happens in nuclear fusion where lighter element nuclei merge together into a heavier element nuclei and releasing a great deal of energy.

Now this reaction cited is a very special case because we don't need millions of degrees temperature to cause the fusion since there is no Coulomb repulsion. A slow neutron strikes a regular protium hydrogen nucleus (just a proton) and we have 2.2MeV of energy. This happens at room temperature with no energy consuming magnetic fields used to contain high temperature plasma as is done with current fusion reactors. Granted we are not getting 17.59MeV that is released when tritium and deuterium are fused, but we are not requiring 25MeV of input power to cause and maintain said reaction, resulting in net loss of power.

In the neutron-protium fusion we have a net 2.2MeV energy output with little or no energy input. My main question is why this energy is not being harnessed for power generation? 2 somewhat challenging but not insurmountable reasons come to mind.

For one, we need free neutrons. These can be obtained from a fission nuclear reactor, cyclotrons, alpha radiating radioactive isotopes with beryllium targets, and others.

If we want to avoid the fission reactor as the neutron source, we can use various other methods. This physics se question probes an exothermic beryllium target process. I had in mind a cyclotron to mildly accelerate alpha particles (helium nuclei) to the appropriate energy level to cause neutron emission when striking a beryllium target. The beryllium target would be encased in liquid or super-critical hydrogen to maximize neutron capture.

My estimation is that the energy output from the n + p = d reaction will supply enough power for the neutron generating cyclotron and then some for useful output power. My reasoning for this is the alpha particles don't need that much energy for this purpose. The process in and of itself is exothermic so we are getting free neutrons plus energy. We are also not concerned with the enormous energy required to confine a plasma.

The other hurdle is harnessing the energy from the emitted gamma ray. This SE question explores the gamma ray energy conversion topic. Then we have to figure out a way to use some of the output energy that was harnessed from the gamma ray to generate more neutrons to "keep the fire going".

While this process is indeed challenging, I think a working system can be developed for a fraction of what is being spent on ITER, the most expensive science project in history to investigate surplus energy from fusion.

$\endgroup$
  • 2
    $\begingroup$ Neutrons cannot be accumulated or harnessed in a volume in numbers that will allow enough energy to be produced , they go through everything and leave. you cannot have a plasma of neutrons and protons. $\endgroup$ – anna v Nov 24 '17 at 5:23
  • $\begingroup$ It's getting the free neutrons that's the problem. If you have to build a fission reactor to generate the free neutrons doesn't that rather defeat the object? $\endgroup$ – John Rennie Nov 24 '17 at 6:05
  • $\begingroup$ Wait, why do alpha-emitters generate thermal neutrons? $\endgroup$ – Jon Custer Nov 24 '17 at 16:02
  • $\begingroup$ Duplicate(?): physics.stackexchange.com/questions/293802/… $\endgroup$ – dmckee Nov 24 '17 at 18:47
  • $\begingroup$ @JonCuster Alpha emitters mostly do not generate neutron (thermal or otherwise) directly but they are often used to initiate $A(\alpha,n)X$ reactions on light nuclei (as in a Americium-Beryllium source). $\endgroup$ – dmckee Nov 24 '17 at 18:50
4
$\begingroup$

Most stable nuclei have a "neutron separation energy" of eight or ten MeV --- that's the minimum cost to produce a neutron, before you factor in any inefficiencies. You can't get ahead spending 10 MeV to free a neutron, having the neutron deposit its 1-2 MeV kinetic energy in some sort of thermalizing moderator so that it can actually capture on hydrogen, and then capturing the 2 MeV photon.

There are some exceptions. For instance, if you put refined uranium in a big pile, neutrons come out of the pile for free. But each fission event releases 200 MeV and only a couple of neutrons that you could use for your purpose; collecting the 4-6 MeV from putting those neutrons on hydrogen is basically rounding error.

There's also the $\rm {}^9Be(\alpha,n\gamma)$ reaction which you mention. However that reaction is not very efficient. In a high-quality AmBe or PuBe source, the number of alphas that undergo $(\alpha,n\gamma)$ on beryllium is about sixty per million. And the americium or plutonium for those sources comes from ... reactor neutrons on uranium. You don't win there.

Let's suppose that you accelerate some helium nuclei onto a beryllium target to set off this $(\alpha,n\gamma)$ reaction. You magically overcome all the technical problems somehow every 3 MeV alpha particle turns into a 10 MeV beryllium disintegration, all of which energy you capture usefully. That's a terrible operating configruation for a power supply, where 30% of the energy (in the wildly optimistic case) has to be renivested in the operation of the power plant. Compare to a campfire (also a pretty inefficient power source) where you find the fuel on the ground, light it with a tiny match, and it keeps you warm for hours.

$\endgroup$
  • $\begingroup$ Thank you for illustrating the energy inefficiencies in both non-fission and fission neutron generation. What about neutrons generated from nuclear waste? I believe most of the long term waste is not really waste at all but dispersed plutonium since most of the other radioactive waste has a rather short half-life. Currently it is too expensive to re-process and concentrate the Pu into usable fuel. Should I update the question with this consideration or just ask another question? $\endgroup$ – 0tyranny 0poverty Nov 29 '17 at 14:25
  • 1
    $\begingroup$ Even if neutron emission were a major component of the radiation from nuclear waste (it isn't), a more efficient way to extract the energy would be to put the waste in enough shielding to absorb all the radiation and collect the heat. Which is actually a thing. $\endgroup$ – rob Nov 29 '17 at 15:04
  • $\begingroup$ Re, "if you put refined uranium in a big pile, neutrons come out of the pile for free." Um, it's not "free" if you count the cost of dollars in building the facility where that process can be implemented on a commercial scale. Nor is it "free if you count the cost to humanity (and all of life on Earth) of dealing with various by-products that would be produced. $\endgroup$ – Solomon Slow Nov 29 '17 at 16:21
  • 1
    $\begingroup$ @jameslarge I didn't mean that assembling a pile of uranium, or disposing of it, is without cost; I meant that you don't have to continually supply energy to the reactor in order for the reaction to continue. $\endgroup$ – rob Nov 29 '17 at 16:42
2
$\begingroup$

Your estimation that the reaction creates enough power to power an neutron producting cyclotron is incorrect. It costs far more energy than $2.2~\rm MeV$ per produced neutron, and so even if you could perfectly efficiently extract the produced energy you still wouldn't be producing any net energy. The main problem is that most alpha particles scatter without actually producing a neutron, and so all the energy spent accelerating those alphas is wasted.

$\endgroup$
0
$\begingroup$

Actually, we do on a rather large scale... in light water reactors. This is the only abundant supply of free (not bound to some nucleus) neutrons we have today. The cooling water absorbs some of the neutrons escaping the chain reaction and then absorbs the heat from the gamma-ray dissipation. Profit! (Probably way less than 1% of the total reactor heat output).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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