Fusion of He, Li, Be in Z-pinches?

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When you do a Z-pinch and deuterium is present in the plasma there will be fusion; now I wonder about other elements or isotopes. If, say a Lithium wire, is imploded in a Z-pinch will there be fusion and bursts of neutrons? Or if you use a mixture of He and Li or whatever light elements? References?

I searched for this on Google and in the university computers but found nothing useful, therefore I ask here.

S.A.
 
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If temperature and pressure are sufficient, you'll get every possible fusion reaction. Deuterium/tritium has by far the best fusion cross-section at reasonable conditions, but it is not the only combination where fusion happens.
 
mfb said:
If temperature and pressure are sufficient, you'll get every possible fusion reaction. Deuterium/tritium has by far the best fusion cross-section at reasonable conditions, but it is not the only combination where fusion happens.

Yes, I know, but I was after references to specific scientific articles on the subject. What decade of the 20th century would be best to narrow my search to?
 
There are several fusion reactions under investigation for fusion reactors. For tokamaks, deuterium (d) and tritium (t) will be required for ignition. If a sustained fusion is achieved deuterium burning d+d will occur, generating either t+p (H1) or He3 + n. He3 + d is a nice neutron free reaction, but requires higher plasma temperatures than d+t. Finally Be11 + p --> 3 He4 has been discussed as an aneuronic (no neutrons) reaction. That might be a great source for a z-pinch--if you use a Be11 wire in a chamber where the walls are coated with hydrogen the you can get a "pure" pinch with no high z impurities.

What about lithium? Technically lithium does not participate in fusion, since all the resulting particles are lighter than the lithium. (Yes, this sort of applies to the Be11 reaction above as well...) However, many operational fusion plant designs use (high energy) neutrons in Li6 + n --> He4 + t, to "breed" tritium. (And use up the neutrons from d+t and/or d+d). Can you get enough tritium that way? Good question. Since d+d yields 50% tritium, getting some d+d burning becomes a necessity. Li7+p --> 2He4 is another aneutronic reaction, but the relative cross-sections favor Be!+p instead, or He3+p, if you have a source of He3

Please remember that in a working fusion reactor, the actual reaction will include just about anything possible. Even with d+t which is requires relatively low energies, the actual plasma will have some twists and pinches where most of the t+d reactions occur, along with d+d, He3+d, t+t and t+p. (The He3 comes from the other 50% of d+d reactions.)
 
eachus said:
Since d+d yields 50% tritium, getting some d+d burning becomes a necessity.
It is neither necessary nor plausible for foreseeable fusion reactors. At D+T reaction conditions, D+D is negligible.
Li7 + n -> He4 + T + n is an alternative - reducing the energy of the neutron, but breeding tritium without using it completely.
There are also some neutron multiplication reactions like Be9 + n -> 2 He4 + 2n.
 
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In tokamaks, it is true that very little d+d burning will take place at d+t temperatures. But there are many other proposed fusion reactors such as the levitating dipole and Elmo bumpy torus, where the plasma density, and thus temperature varies significantly. This is where ignition comes into play. Once ignition occurs, the plasma continues to heat up until the (usually RF or neutral beam) pumping can be turned off. Do I think that tokamaks will be profitable fusion power generators? Never have. On the other hand, several early fusion experiments would be immediately practical if scaled up from lab dimensions to a practical power plant. (They had loss modes such as the magnetic mirror machine, where one of the parameters is a dimension of the machine. Make a magnetic mirror long enough, and losses through the end cap are insignificant. Stellerators grow even better. A critical loss parameter is the number of times an ion circles the machine between collisions.)
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...

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