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Impact Parameter of Alpha Particle Rutherford

  1. Feb 28, 2005 #1
    In the Rutherford scattering experiment a very thin gold foil target is bombarded with a beam of a particles of known kinetic energy. A detector which can be moved on a circle around the target counts the scattered particles. What was the impact parameter of a 4.57 MeV a particle if it was detected at 52.4 degree angle?

    Im using b= (rmin/2)cot(theta/2) where rmin = Z1Z2e^2/4piEok

    so i have ((2)(79)(1.6*10^-19 c)^2 / (4.57 *10^6 eV)(1.6*10^-9 J/eV) ) * 8.99*10^9 Jm/c ....and got 3.46 *10^-15 is this right so far? ...i got 8.99*10^9 by (1/4piEo) ....thanks
     
  2. jcsd
  3. Feb 28, 2005 #2

    dextercioby

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    The order of magnitude for r_{min} is good,indeed there is fermi/femtometer.Compute "b" now...

    Daniel.
     
  4. Feb 28, 2005 #3
    i then get an answe of 3.52 *10^-15 m ...but this is not the correct answer ... i dont know what i did wrong
     
  5. Feb 28, 2005 #4

    dextercioby

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    Devious...Make sure you insert correct numerical values.The energy (which is supposed to be in the denominator) is 4.57MeV=4.57*1.6*10^{6}*10^{-19}J...

    Daniel.
     
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