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Apologies this is probably a stupid idea but I am curious and my knowledge of physics is limited as I am 14. So I was wondering if we could use particle accelerators to achieve nuclear fusion. I have found a few other posts about this but they seem to be about unnecessarily high energy accelerators when only approximately 1 MeV seems to be necessary to reach the fraction of the speed of light for fusion.

So it seems that there are two forces involved there is the electromagnetic which causes the protons to repel preventing fusion and has an infinite range. Then there is the strong force which is approximately 137 times stronger than electromagnetic but has a limited range of a femtometer and is necessary for fusion. So basically all that is required is to overcome the Coulomb barrier which requires the protons to be travelling at around 0.05-0.07c and requires approximately 1 MeV according to my research and calculation. This means that 18.6 MeV profit per collision (19.6 MeV per fusion reaction) so that means that over 5.1% of the nuclei need to fuse to gain energy. This seems to be a major flaw as this seems unlikely to happen.

However I was wondering if this could be overcome by using a 1 MeV linear accelerator to accelerate the hydrogen ions at a target which is essentially a very long and dense highly pressurised chamber of deuterium gas where it is almost certain for collisons to occur due to the high amount of atoms?

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marked as duplicate by John Rennie, ACuriousMind♦, Kyle Kanos, Martin, Qmechanic♦ Jul 12 '15 at 17:30

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    $\begingroup$ Sure, you can. That is how new isotopes are created, for instance. Look up "Farnsworth fuser. Google "home made fusion reactor". Check out the images, too: they're pretty. Causing fusion to happen in the lab is almost easy. That's never been the problem. The problem has always been getting more energy out then you put in. $\endgroup$ – dmckee♦ Jul 12 '15 at 3:53
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    $\begingroup$ possible duplicate of Could we use particle colliders as fusion generators? $\endgroup$ – John Rennie Jul 12 '15 at 10:22
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In your scheme, other processes, such as Coulomb scattering, are much more probable than fusion, so you will have net energy loss, as dmckee noted.

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  • $\begingroup$ What about if slightly higher MeV is used? This means that a denser longer target to produce a profit but could compensate $\endgroup$ – user43087 Jul 12 '15 at 12:47
  • $\begingroup$ @user43087: As far as I know, that won't help, moreover, I am not sure you won't make it worse this way. Energy losses for deceleration radiation and ionization will prevail. $\endgroup$ – akhmeteli Jul 12 '15 at 16:41

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