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Minimum separation, impact parameter, and scattering angle

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Alpha particles of kinetic energy 9.6MeV are incident on a silver foil of thickness 7.0 micro meters. For a certain value of the impact parameter, the alpha particles lose exactly half their incident kinetic energy when they reach their minimum separation from the nucleus. Find the minimum separation, the impact parameter, and the scattering angle.

    K=9.6 MeV, t=7.0 x 10^(-6) meters, density; n=5.68 x 10^28 atoms/m^3
    z=47

    2. Relevant equations

    b=(zZ/2K)(1.44MeV nm)cot([tex]\theta[/tex]/2)



    3. The attempt at a solution

    I can not figure out where to start.



    Thanks for any help,
    Jim
     
  2. jcsd
  3. Oct 14, 2009 #2

    gabbagabbahey

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    Use conservation of energy. If the alpha particle has initial kinetic energy of [itex]K[/itex], what is its kinetic energy at minimum separation from a silver nucleus? What is its potential energy? What does that tell you?
     
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