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Four Protons Are Fused Into An Alpha Particle, Energy Released

  1. May 15, 2013 #1
    (answered)Four Protons Are Fused Into Alpha Particle, Energy Released

    Answered: I wasn't supposed to multiply by c2. Thank you ehild

    I apologize in advance if I posted this in an incorrect format or in the wrong section. I read the rules and I believe this is the correct format/place.

    1. The problem statement, all variables and given/known data
    The proton-proton reaction that takes place in the Sun has the net effect of converting four protons (Mp= 1.007825 u) into alpha particles (Ma= 4.002603 u). The resulting "released" energy is shared between the kinetic energy of the alpha particles, positrons, neutrinos and gamma ray photons. Assume we could eventually design a 1.00 GW (billion watt) fusion power plant using this same reaction. Also assume that 12.5% of the "released" energy (kinetic + photon) can be converted into electrical power.
    How many alpha particles would need to be fused in the reactor every second?


    2. Relevant equations
    binding energy:
    [STRIKE]Eb=c2 * (Mnucleons-Mparticle)*931.494MeV[/STRIKE]
    Eb=(Mnucleons-Mparticle)*931.494MeV/u


    3. The attempt at a solution
    Eb = c2 * (4*Mp-Ma)=2.2058*1024eV =3.85408*105j

    Eb*12.5% = 4.8115*104j

    1*109W / Eb = 2.0783*104s-1

    The correct answer is: 1.87*1021

    Thank you in advance for any help anybody can provide.
     
    Last edited: May 16, 2013
  2. jcsd
  3. May 15, 2013 #2

    ehild

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    In what units of mass have you calculated with?


    ehild
     
  4. May 15, 2013 #3
    For the problem the mass of a proton(1.007825u) and alpha particle(4.002603u) is given in atomic mass units(u=931.494MeV). I converted to joules before doing any calculations involving power.
     
  5. May 16, 2013 #4

    ehild

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    Why did you multiplied energy by c^2?

    ehild
     
  6. May 16, 2013 #5
    Well, I think you just found my mistake.

    Just crunched the number without the c^2 and it came out correct.

    Thank you.
     
  7. May 16, 2013 #6

    ehild

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    Yes, taking the proton mass in kg-s, m(proton)c2/(1.6 10-19(J/eV) )≈931 MeV.

    ehild
     
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