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How much energy is released in a matter/antimatter collision?

  1. Jun 24, 2012 #1
    Hi there,

    Let's say a hydrogen atom and an antihydrogen atom collide and annihilate each other. How much energy is actually released during this process?

    I've looked around the interwebs and could not find an answer to my question. Everything I found said that all the mass of both particles is converted into "pure energy" (what does that even mean?) lol.
     
  2. jcsd
  3. Jun 24, 2012 #2
    Energy is released equal to the rest energy of the original two particles. The rest energy of a particle is given by E = mc^2, where m is its mass. In the simplest annihilation process, a particle and its antiparticle are destroyed and in their place appear two photons. Each photon has an energy E = mc^2, where m is the mass of one of the original particles. (This assumes that the original two particles were at rest. If they were moving, their kinetic energy gets added into energy of the two photons).

    "Pure energy" is indeed a meaningless phrase. Generally when people say this they are talking about the annihilation process above that produces photons.
     
  4. Jun 25, 2012 #3

    mfb

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    In the case of proton+antiproton annihilation, I would expect high-energetic pions to carry most of the energy. The total energy (kinetic energy of pions, rest energy of pions, energy of photons, and other particles if they are produced) adds up to the rest energy of the proton plus the antiproton, plus the kinetic energy if the particles did not collide "at rest". This is about ~1860 MeV + kinetic energy.
     
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