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Rutherford scattering calculation

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data

    The fraction of [tex]6.0 MeV[/tex] protons scattered by thin gold foil, of density [tex]\rho=19.3 g/cm^{3}[/tex], from the incident beam into a region where scattering angles exceed [tex]60 degrees[/tex] is equal to [tex] 2.0 x 10^{-5} [/tex]. Calculate the thickness of the gold foil using the result of the previous problem.

    2. Relevant equations
    The result I got for the previous problem was a formula for the number of particles scattered thru and angle of [tex]\theta[/tex] or greater and is:

    [tex]N(\theta)= (\frac{1}{4\pi\epsilon_{0}})^{2} \pi I \rho t (\frac{zZe^{2}}{Mv^{2}})^{2} cot^{2}(\frac{\theta}{2}) [/tex]

    3. The attempt at a solution

    So I took [tex]\frac{N(\theta)}{I}[/tex] which is equivalent to removing the I from the above equation and attempted to solve for t, which is the thickness...

    [tex]t=\frac{2x10^{-5}}{(\frac{1}{4\pi\epsilon_{0}})^{2} \pi (19.3 g/cm^{3}) (\frac{79(1.602x10^{-19}coul)^{2}}{2(6 MeV)})^{2} cot^{2}(\frac{\pi}{6})}[/tex]

    problem is that I keep getting an answer in m/kg, so I must be missing some term, though I can't figure out what it is...

    any help is very much appreciated.
  2. jcsd
  3. Oct 20, 2007 #2


    User Avatar

    Staff: Mentor

    Absorber thickness is sometime measured in kg/m2 or gm/cm2 such that dividing by density in kg/m3 or gm/cm3 yields a thickness in m or cm, respectively.

    Make sure the relationship between N and I is correct, and check one's units on the various terms.
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