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Proton Velocity (potential/field/charge)

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    (background info)
    The source of the sun's energy is a sequence of nuclear reactions that occur in its core. The first of these reactions involves the collision of two protons, which fuse together to form a heavier nucleus and release energy. For this process, called nuclear fusion, to occur, the two protons must first approach until their surfaces are essentially in contact.

    Assume both protons are moving with the same speed and they collide head-on. If the radius of the proton is 1.2*10^-15m, what is the minimum speed that will allow fusion to occur? The charge distribution within a proton is spherically symmetric, so the electric field and potential outside a proton are the same as if it were a point charge. The mass of the proton is 1.67*10^-27kg.


    2. Relevant equations



    3. The attempt at a solution

    qV=(1/2)mv^2
    V=kq/r
    kq^2/r=(1/2)mv^2
    plugging in for the numbers gives me 1.5*10^7

    The answer should be 7.58*10^6
     
  2. jcsd
  3. Feb 25, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    How close will the proton's centers be to each other when their surfaces just meet?

    How many particles will share the required kinetic energy? How will it be distributed between them?
     
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