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Why can't a compact 10 MeV electrostatic particle accelerator not be built?

Physics Asked by user11377 on May 28, 2021

I’ve been thinking about this problem for a while, since I want to see if a compact accelerator driven nuclear reactor can be built for powering aircraft and rockets. The reaction I am considering for the nuclear reactor is the following chain:

He-4 + Be-9 + 10 MeV-> C-12 + n

n + Li-7 -> Li-8

Li-8 -> Be-8 + e- + 17 MeV

Be-8 -> 2 He-4 + 0.032 MeV

The reactor would need to accelerate helium nucleii to 10 MeV to reach a good capture/scatter cross section ratio. Electrostatic acceleration is the best bet for an airborne system since it doesn’t need heavy electromagnets or its associated cooling systems. If placed inside a pressurized air vessel held at 3 atm of pressure, the accelerator should be able to hold 10 MV of voltage. The real area to work on is how to make the accelerator as compact as possible, since that drives weight, as a longer accelerator means more volume to pull to vacuum, and more material to resist compressive forces.

So I had an idea and didn’t see any evidence that it had been tried. The idea is this: instead of accelerate the helium nucleii between two plates straight, have it bounce between plates to gain the energy, similar to the set-up in a photomultiplier, except instead of having the nucleii impact the plates, charge the plates enough to reflect the nucleii. This would reduce the acceleration distance, since the path now becomes compressed into a zig-zag shape. The only problem I see if with bremsstrahlung, since each reflection will be a sizeable acceleration. However, for 10 MeV for helium nucleii, the Lorentz factor shouldn’t be very high, and since the power emitted varies by the Lorentz factor to the sixth power compared to acceleration to the third power, the losses should be small.

So here is my question to the group: Has this been tried? Is there some factor I am overlooking that will cause this to not work?

*The length I am aiming for is 5 centimeters, instead of the usual 2-3 meters.

One Answer

In the bouncing you do not consider that the alpha will be attracted to the ground plate as much as being repelled by the positive plate , with all the probabilities of loosing any energy gained when bouncing off it.

Also I do not see the analogy with a photomultiplier, which works by electrons hitting metal and releasing many more electrons.

Here you have a reaction that releases a neutron, (which cannot be focused} to hit a Li7 and make Li8 which decaying will give an extra 7 MeV over the the energy spent of 10 MeV to initiate the reaction, and finally one more alpha than the incoming one.

In a photomultiplier the number of electrons in the multiplication is unlimited.

The basic stop is that neutrons cannot be focused therefor their probability of hitting a Li7 will be very low. They will leave the reactor and contaminate radioactively the surroundings.

Whether a 5 cm gap can hold 10 Megavolts with the vacuums technically available is another weak point. An accelerator expert should answer this.

It is a no go proposal, imo.

Answered by anna v on May 28, 2021

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