In the NYT article about the new breakthrough in fusion research, it says …
Blockquote Out flowed a flood of neutron particles — the product of fusion — which carried about 3 megajoules of energy, a factor of 1.5 in energy gain.
That doesn’t seem exactly right to me. The basic idea is fusing two deuterium ions into one helium atom, which wouldn’t leave any leftover neutrons to flow out, right? I’ve always thought the energy produced would just be electromagnetic radiation, like heat for example. However, tritium is also involved somehow, so maybe there are extra neutrons and those are carrying kinetic energy?
The fuels for the fusion reaction in the NIF are deuterium (one proton, one neutron) and tritium (one proton, two neutrons), also known as D-T fusion. The resulting products are an alpha particle (essentially ionized helium, two protons, two neutrons) with 3.52 MeV of kinetic energy, and a free neutron with 14.06 MeV of energy. D-D fusion is also possible but requires much higher temperature and pressure conditions, and only produces ~4 MeV of kinetic energy, so it is a much less desirable reaction from a yield standpoint.
They are fusing a deuterium atom (1 proton and 1 neutron) and a tritium atom (1 proton and 2 neutrons) into regular helium (2 protons and 2 neutrons) so there is a neutron left over. This combination is used because other hydrogen isotopes are even more difficult to fuse.
There are drawbacks to D -T fusion, notably the extra neutron that has to be contained. Also, tritium is extremely rare and must be produced in regular nuclear reactors.
D-D fusion will also produce a neutron (although at significantly lower energy of 0.82 MeV) approximately half of the time.
Well, reactions with 6Li, which unfortunately is a pretty rare isotope of lithium, so it’s not anything like free. There are also the proliferation concerns; tritium is used to boost the yield of nuclear fission weapons, and so even though a fusion reactor doesn’t make plutonium it is still a concern for wider access to nuclear weapon materials.
Given m_{\text{He}_3} = 2.8084 GeV and m_\text{n} = 0.9396 GeV, [and a yield of 3.27 MeV] we see that the majority of the kinetic energy should be carried off by the neutron
I stand corrected; the 3He has the 0.82 MeV and the neutron is 2.45 MeV. Still way lower energetic yield than D-T fusion. Note that this is also a problem with supposed “aneutronic” D-3He fusion because a significant number of the fusion reactions will be D-D reactions with that neutron being produced ~50% of the time, so while there are fewer and less energetic neutrons it will still produce a lot of neutrons and at a threshold power density that is almost two orders of magnitude higher than D-T fusion.
I can’t get to DPRK’s link for some reason even though “Down or Not” sites say it’s up. Even so, I think I’m getting it. In the D-D fusion, half of the He produced is a two proton, one neutron isotope, so free neutrons are produced, just not as many as in the D-T fusion. In either case, they’re moving fast so can be used as an energy source.
What does it mean that the neutrons must be contained? They must be contained in order to get usable energy from them, or they must be contained because free neutrons are dangerous? If the latter, then how are they dangerous?
Neutrons are highly penetrating because they don’t interact electrodynamically and can radiologically activate elements within the body into unstable isotopes, such that even isotopes emitting alpha particles (which are essentially harmless outside the body as they are stopped by the epidermis layer) can do internal damage resulting in genetic damage, cancer, or at high enough levels radiation sickness leading to systemic organ failure as evidence by the horrific death of Hisashi Ouchi.
Those high energy neutrons are problematic in all sorts of ways. They have to be contained so that the energy can be extracted. But neutron bombarding is really disruptive to the containing material. Neutrons can be captured by a nucleus, which will change the isotope and may change the element via neutron capture and beta decay. This may negatively impact the structural strength of the containment. Neutrons can also dislocate atoms in the containment, again affecting structural strength and possibly storing energy in the material (Wigner effect). This energy may need to be released by annealing, but again, this may have implications for the structural strength of the containment. Finally, a free neutron that is slowed down in the containment without capture will decay into a proton and a beta particle (electron). This will contaminate the containment with hydrogen, which affects structural strength.
Getting fusion going and sustained is just one problem to solve - actually extracting the energy and designing containment to do so safely and sustainably is another problem looking for solutions.
IIRC, this was one of the criticisms of “cold fusion”. The experimenters claimed to be seeing excess heat they attributed to D-D fusion - heavy water electrolyzed to D and O and the D was infusing the metal lattice, hence being forced together by the characteristics of the metal lattice. However, there was no measurable neutron emissions consistent with the alleged energy, so unlikely (impossible) D-D fusion was happening.
Because they have no charge, neutrons are less influenced by the materials they pass through, are harder to stop (by colliding with atomic nuclei) and so penetrate further. Collisions can cause some atoms to split or absorb the neutron, thus creating other elements (split) or isotopes (extra neutron) that may be unstable and so radioactive.
Castle Bravo proved that 7Li fuses extravagantly under proton (free hydrogen nucleus) bombardment into a bunch of useless 4He and a crapton of energy. Not applicable here, but nice to cheaply boost a thermonuclear weapon.