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Homework Help: Colliding Protons

  1. Jan 30, 2006 #1
    Hi Guys,

    I was just wondering if you could help me, I seem to be having some problems getting to the correct answer (possibly because of my calculations). If two protons collide head on, both with energy 2keV... show that the closest approach is 7.2 x 10exp-11 cm. Now I know that when the protons are at closest approach then only the potential energy will be considered. Therefore V = q1q2 / r, which for a proton I know is e^2 / r, but for some reason I just cannot get to the right answer and I don't know if it is the way I am using electronvolts and joules. Please help.

  2. jcsd
  3. Jan 30, 2006 #2


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    Staff Emeritus
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    If one is using cgs units, then make sure all units are consistent, e.g. energy = ergs, displacement/distance = cm, charge = esu's.
  4. Jan 30, 2006 #3
    Hi, I'm working in joules and eV...I still can't get it to the right answer.... one question - to change from eV to joules it is simply multiplication of eV and 1.6 x 10exp-19? I haven't used eV for a very long time and can't quite remember how to do it.

    Thanks for your help,

  5. Jan 31, 2006 #4

    Meir Achuz

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    If you use eV for the energy, then you should onlly use one factor of e in the potential energy. That is U=ke/r is the PE for two protons in the unit eV.
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