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Cant seem to figure this out

  1. Aug 20, 2006 #1
    in a rutherford gold experiment, alpha particles with sharge +2e and a kinetic energy of 7.7 Mev were beamed at gold foil. the nuclius of a gold atom contains 79 protons, giving it a charge of +79e. What is the closest distance that an alpha particle can get to a gold nucleus when it approaches head on ?

    ok i used the equation En=k^2 x e^4 me . Z^2
    2xh^2 h^2

    i am getting n^2= -9.07x10^21

    i know this is wrong and am getting quite frustrated
    if some on could help me it would be great
  2. jcsd
  3. Aug 20, 2006 #2
    can some one tell me if i am eve using the right formula\
  4. Aug 20, 2006 #3


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    We don't know the distance of the alpha particle source with the gold sheet?
  5. Aug 20, 2006 #4

    Andrew Mason

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    I am not sure what you are doing here. Can you explain what the principle is here? How is the kinetic energy of the alpha particle related to the closest distance between the alpha and the gold nucleus? What is the kinetic energy of the alpha particle when it is at the closest distance? Where did the energy go?

  6. Aug 22, 2006 #5


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    Staff: Mentor

    Adding to AM's questions, what is the electric potential energy of the alpha particle at its closest approach, and how is it related to the alpha particle's kinetic energy?
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