1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    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


    User Avatar
    Homework Helper

    In what units of mass have you calculated with?

  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


    User Avatar
    Homework Helper

    Why did you multiplied energy by c^2?

  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


    User Avatar
    Homework Helper

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted