Thursday, 8 October 2020

particle physics - What is the percentage of useful energy do we get from matter-antimatter annihilation?


This is a theoretical question since we haven't made enough antimatter to try it in reality of course. But I am asking about the physics part in this.


Also, by "useful energy" I mean the energy we are able to use either as a heating source for something like a nuclear reactor, or as energy for an explosion like nuclear explosions.


If I am not mistaken, a large part of the energy we get from the annihilation is in the form of neutrinos, which we for some reason can't consider them useful energy. So now, if we subtract the energy of the neutrinos, is it safe to consider the rest as useful energy as I explained ?


Please try to be as simple as possible because I don't speak English very well.



Answer



I found a interesting paper on the use of antimatter for rocket propulsion from NASA that addresses this subject. Your question has a pretty complicated answer.


I think you will find more formation than you wanted reading that article. In the report it says about an electron-positron collision:




His results indicate that the neutrinos carry off ~22% of the available pion energy (~1248 MeV) whereas the muons retain ~78%. The unstable muon, having an average energy of ~300 MeV, also decays (in ~6.2μs) into an electron, or positron, and two neutrinos as shown in Table 2. The energy appears to be about equally distributed among the three particles with the neutrinos carrying off ~2/3 of the available energy. Ultimately, the electrons and positrons can also annihilate yielding additional energy in the form of two 0.511- MeV gamma rays. The neutrinos are considered to be massless and move at essentially the speed of light. They are extremely penetrating and rarely interact with matter. Under vacuum conditions the various muon and electron neutrino particle–antiparticle pairs carry off ~50% of the available annihilation energy following a p–p reaction.



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