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Rutherford's gold foil experiment

  1. Aug 27, 2009 #1
    Can anyone explain the experiment? i am having a big doubt in it.
     
  2. jcsd
  3. Aug 27, 2009 #2

    negitron

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    Explain your doubt.
     
  4. Aug 27, 2009 #3

    MATLABdude

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    The OP is probably Indian and means doubt in the sense of 'question' rather than in the more-common (and western) usage of 'disbelief'.
     
  5. Aug 28, 2009 #4

    negitron

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    It doesn't matter; either way, it applies: what's the question, specifically? Sure, I can explain Rutherford's experiment, but without knowing what's giving the OP trouble, it's hard to know which aspect to focus on.
     
  6. Aug 28, 2009 #5
    First of all, if you write out the equations for Rutherford scattering of alpha particles from a point nuclear charge, you get an equation that has an angular dependence something like sin4(θ/2) (my memory fails a little). If it deviates from this at large angles, then the alpha particle is hitting something. I did this experiment in a physics lab many years ago, and my biggest problem was that the gold foil stuck to my fingers.

    Erratum: It should read 1/sin4(θ/2) . See
    http://en.wikipedia.org/wiki/Rutherford_scattering
     
    Last edited: Aug 29, 2009
  7. Aug 29, 2009 #6
    When I was deriving the formula for scattering fast charged particles from atoms, I discovered that, due to motion around the atomic center of inertia, the positive charge in an atom is quantum mechanically smeared, just like the negative (electron) charge but localized in much smaller region. Thus the elastic cross section differs from the Rutherford formula at large angles. It can be observed in specially designed experiments.

    By the way, in a solid state the nucleus QM de-localisation is as large as the lattice step so the elastic cross section differs essentially from the Rutherford formula at large angles. It is the inclusive cross section (elastic + inelastic ones) that is reduced to the Rutherford formula. Normally it is the inclusive cross section that is observed.
     
    Last edited: Aug 29, 2009
  8. Aug 29, 2009 #7

    Born2bwire

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    Oh sure. Whenever we do experiments with lead foil or aluminum foil there are no problems. But the second we use gold foil all the students suddenly get a case of very "sticky fingers" indeed...
     
  9. Aug 29, 2009 #8
    A thin gold foil contains infinite number of atoms.So, by passing alpha rays to this foil , how was Rutherford able to explain the structure of a single atom?
     
  10. Aug 29, 2009 #9
    He did not explain the structure of a single atom, he discovered that the positive charge is entirely concentrated in a heavy particle (nucleus) whereas the negative charge is concentrated in one-charge light electrons. The Rutherford model needed in fact QM to explain the atomic stability.
     
  11. Aug 29, 2009 #10

    uart

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    No. In a thin gold foil the alpha particle might only need pass through several hundred atoms. If these atoms where more or less homogenious then we would expect that every alpha particle would interact with the foil in much the same way as every other alpha particle. But Rutherford instead observed that many alpha particles went straight through with little or no deflection yet a few were greatly deflected and some even rebounded (deflected more than 180 degrees). This provided very good evidence that the atoms were not homogenious and lead to the proposed model of a compact positive nucleus containing most of the mass in only a very small amount of the total atom volume.
     
    Last edited: Aug 29, 2009
  12. Aug 29, 2009 #11
    The Rutherford (backward) scattering is also observed from a thick solid pieces. It is not possible without heavy nuclei. There is a strong Z- and M_a dependence of the scattered flux.
     
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