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Simple Scattering

  1. Dec 27, 2003 #1
    As every physics major knows, elastic scattering from a hard sphere of radius a can be modeled by the following potential:
    U(r) = 0 for r>a,
    and infinity for r<a.

    Consider the effective potential l^2/(2mr^2) + U(r). And consider scattering with impact parameter b less than a. Where is the turning point of the motion?
    Then, how would you use this line of reasoning to show that the differential cross section is Sigma(theta) = a^2/4?

    I know how to find the differential cross section geometrically, by finding b(theta), but I don't know how to do this with effective potentials.

    This isn't homework. It's a problem from pa.uky.edu/~cvj/PHY404G/ 1998 final exam.

    Last edited: Dec 27, 2003
  2. jcsd
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