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Beta Positive Decay reactor

  1. Jun 2, 2012 #1
    Okay, I have had a idea for some time and I want to see if it will work.

    first you get Magnesium 24 launch protons at 20 Mev then a decay happens. Now you have Na22 sodium 22 beta positive decays into a anti electron of 1 Mev with its counter particle they annihilation into 2 gamma rays which produce heat due to electrical resistance now you use that pressure to make electricity. you lose all but 1/20th but you have a battery of 80,000 j/s for 2.6 years or 28 Tj/kg over half life then it halves at a slope but i do not want to calculate that. Mass into energy in a small area you have a long source of energy for robots and stuff.

    EQ = ((6.022 x 10^23/22000) = 1 mev = 1mev *( 6.022 x 10^23/22000) * 1..6 10^-19 = 80,000 (j/s) 1/20th, small container
    super conducting accelerators-------> decay force ----------> anti-matter annihilation------> Heat or electricity.

    alot of energy in a Small container more than every oil barrel on the world.


    That is what I was thinking, Is it possible?
     
    Last edited: Jun 2, 2012
  2. jcsd
  3. Jun 2, 2012 #2

    Astronuc

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    This is not practical considering the energy required to produce 20 MeV protons, and the energy losses from scattered protons.

    Mg24(p,p'd)Na22 is not practical.
     
  4. Jun 2, 2012 #3
    Would you be able to propel the positively charged deutons at 20 mev the energy could be less, due to 30T for little energy due to superconductive linear accelerators.

    less energy by my resistance more by magnetic push
    1/R=A R is lower A > than Rcu, energy to push it
    but still long periods of energy loss 80,000/s isnt bad for long periods like a 2.6 year battery for 1 kg of fuel, 28 Tj.
    GJ -------> tj
     
    Last edited: Jun 2, 2012
  5. Jun 2, 2012 #4

    Vanadium 50

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    It takes at least 20 MeV of energy to accelerate a particle to 20 MeV. Since you get less energy back than this, it's not a viable source of energy.
     
  6. Jun 2, 2012 #5
    its 20 Mev per 1 mev in packet of 1 kg a small amount for a 2.6 yr 80 kj /s or 200Tj for 20 Tj 1 kg enough for a robot or space shuttle

    80,000 per second for 2.6 yr is the key. 20 per 1 (6.3*10^23)/(1.6^-19) = Mev Mj = 80 kj/s for 16 Mj/s over 2.6 years in 1 small packet of 1 kg. storage not creation for long distances. EE mass of protons(1/20000)C^2 = (E/dt(1/2 Decay)
     
    Last edited: Jun 2, 2012
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