3 ms·

It generates high-speed neutrons, helium cores, photons, fissile fragments, and in rare circumstances free protons. Of those, only the protons have an electric

It generates high-speed neutrons, helium cores, photons, fissile fragments, and in rare circumstances free protons. Of those, only the protons have an electric charge you could use in any form of generator… but I’m not aware of any form of reaction that predominantly creates protium. The fragments are also charged, but the implication that you’re using fission means they’re in the middle of a block of uranium and won’t keep their speed for long. The others ignore electromagnetic fields for the most part, and will fly around until they smash into something and go goooong like the world’s smallest bell. Or smash through something, perhaps; it’s a chain reaction after all. This mostly just makes other stuff move about. Repeat a few dozen times, and you’ve got heat.

adrian_b 2y ago
The helium nuclei a.k.a. alpha particles are also charged, like the free protons. However, the alpha decay of some product of the fission reactions does not change the total charge of the fissile material, because the emission of a helium nucleus with a double positive elementary charge leaves a heavy nucleus with a diminished nuclear charge, by those two elementary charges. Only when the alpha decay happens to occur close to the surface of the material, the positive helium nuclei may escape from it and they could land on a collecting electrode, making that electrode positively charged and leaving the fissile core negatively charged. However such a process would extract only a negligible part of the energy produced by fission. Even if the alpha extraction could be enhanced somehow, the decay energy of the fission products is small in comparison with the energy produced by the initial fission of the uranium nuclei. Extracting directly from the fissile material the nuclei generated by fission would be much more difficult than extracting the helium nuclei. The Darpa project may succeed to stimulate the creation of some electric generators that could deliver additional energy from a fission reactor, by direct electric charge separation, but that would remain a small part of the total fission energy, most of which will still have to be extracted through thermal methods, like today.
oneshtein 2y ago
2D + 3He → 4He + 1p + 18.3 MeV 3He + 3He → 4He + 2 1p + 12.86 MeV

adrian_b 2y ago

The helium nuclei a.k.a. alpha particles are also charged, like the free protons. However, the alpha decay of some product of the fission reactions does not change the total charge of the fissile material, because the emission of a helium nucleus with a double positive elementary charge leaves a heavy nucleus with a diminished nuclear charge, by those two elementary charges. Only when the alpha decay happens to occur close to the surface of the material, the positive helium nuclei may escape from it and they could land on a collecting electrode, making that electrode positively charged and leaving the fissile core negatively charged. However such a process would extract only a negligible part of the energy produced by fission. Even if the alpha extraction could be enhanced somehow, the decay energy of the fission products is small in comparison with the energy produced by the initial fission of the uranium nuclei. Extracting directly from the fissile material the nuclei generated by fission would be much more difficult than extracting the helium nuclei. The Darpa project may succeed to stimulate the creation of some electric generators that could deliver additional energy from a fission reactor, by direct electric charge separation, but that would remain a small part of the total fission energy, most of which will still have to be extracted through thermal methods, like today.

oneshtein 2y ago

2D + 3He → 4He + 1p + 18.3 MeV 3He + 3He → 4He + 2 1p + 12.86 MeV