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No, there are still plenty of electrons. You start with hydrogen (one proton, one electron) and boron (five protons, five electrons). You end with three heliums
by DennisP 5y ago
No, there are still plenty of electrons. You start with hydrogen (one proton, one electron) and boron (five protons, five electrons). You end with three heliums (each with two proton, two electrons).
The energy isn't extracted because of a build-up of charge. What happens is that each nuclear reaction is a tiny explosion. The helium nuclei shoot out really fast, and they're positively charged. You just need them to push through a magnetic field, and then you're getting electricity, like any generator moving magnets through a coil.
- pfdietz 5y agoThat is not HB11's proposed direct conversion scheme. HB11 proposes to have the alpha particles go up a large potential gradient, in effect charging a capacitor. I don't think their scheme works, btw. Having the ions move against a magnetic field doesn't change their energy. What would be needed for that scheme is to subject the ions to a changing magnetic field to produce an electric field that reduces their energy. Helion proposes something like this.
- DennisP 5y agoInteresting, thanks. Would it be possible to use Helion's scheme for HB11 (assuming they manage net power)?
- pfdietz 5y agoSince I also don't think their central idea works (it's off by a factor of 100) it hardly matters. pdf: https://arxiv.org/ftp/arxiv/papers/2012/2012.14533.pdf https://arxiv.org/ftp/arxiv/papers/2012/2012.14533.pdf
- DennisP 5y agoOk fine, that's a separate issue, maybe you're right. But also maybe there's a chance that there's a way to make it work (as your link suggests at the end). It seems to me that the initial impetus, a small explosion of charged particles, is the same for both HB11 and Helion. So it doesn't seem like it matters if HB11's energy extraction doesn't work, as long as Helion's does.