I'm an electrical engineer. Throughout my education, I was taught that current always has a return path. This understanding applies to everything.

Batteries use an electrolyte to strip electrons from the positive electrode and move them to the negative electrode. The potential is then derived when the electrons want to return back to the positive electrode.

In a generator, a magnet passes by a coil of wire. The motion of the magnetic field causes the electrons in the wire to move to one side. This then causes an abundance of electrons on one wire and a deficit on the other. Again this sets up a difference of potential because the electrons want to get back to the positive electrode.

A hydrogen fuel cell has hydrogen on one side of a membrane and oxygen on the other. The hydrogen wants to combine with the oxygen but when it passes through the membrane the electrons are stripped off. This causes an abundance of electrons on one side and a deficit on the other.

An HB11 fusion reactor bombards a pellet of boron 11 with hydrogen atoms. When a collision occurs between a hydrogen and a boron atom the result is two helium atoms with no electrons. The helium atoms form the positive electrode due to the lack of electrons. Where do the electrons come from?

That is really where my confusion comes from. I suspect that the helium will rob electrons from whatever is around. That seems like a problem. What am I missing?

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    $\begingroup$ Any needed electrons come from the surroundings, and, ultimately from a ground connection. And it had better make 3 alphas in total. $\endgroup$ – Jon Custer Apr 5 at 14:33
  • $\begingroup$ @JonCuster Would this eventually create a deficit of electrons? $\endgroup$ – vini_i Apr 6 at 12:12
  • $\begingroup$ Does a power plant run out? (OK, kinda snarky.) Somewhere at the fusion facility you start with a bottle of H2 gas and a chunk of B11, both charge neutral. To accelerate the hydrogen you dissociate it, ionize it (strip an electron off), and send it through some kind of ion accelerator. You hope it smashes into a B11 nucleus and you make the 3 alphas. In the reactor you are only missing the one electron from the H, so you let the electrons you stripped from it in the first place go to the reactor to recombine with some positive ion. $\endgroup$ – Jon Custer Apr 6 at 13:11

All nuclear reactions preserve charge, so there will be electrons produced as well as alpha particles. The electrons often called beta particles, and may be omitted from the graphics because the energy of the beta is going to be much less than the energy of the alpha. I assume there will be an electric field applied to the target, which will drive the alphas one direction and the betas another.

Charged particles (alpha and beta) tend to react with just about everything in the system, so there will be a lot of losses in the system. The interactions will eventually be transformed to heat, which will reduce the efficiency of the system. Not all of the alpha and betas will be converted directly to electricity. This is one problem with the design.

Another problem with the design is that the cross section for the (p,B11) reaction is fairly small. There will be other reactions going on. How do you get the densities high enough to have enough reactions of interest?

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