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Scattering formulas, what's the point?

  1. Oct 3, 2013 #1


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    I am currently redaing Goldstein's Classical Mechanics (3rd ed. 2002, with Poole and Safko). The sections 3.10-11 deal with scattering, of the type originally studied by Rutherford, and which led him to formulate his "planetary system" atom model.

    Rutherford fired a beam of positively charged alpha particles against a thin gold foil, and could explain the results of the scattering of the particles with the model that the atom consists of a (relatively) heavy positively charged nucleus with light negatively charged electrons orbiting it at a (relatively) large distance. The deflections take place if and when the particles come close to the nucleus. The more general problem is to calculate the deflection pattern if a beam of particles is fired against an originally fixed center of force.
    Goldstein makes great pains to calculate the so called cross section for scattering in a given direction, first assuming that the target (in Rutherford's case the nucleus) remains fixed all the time, and then when the target moves in reaction to the collision.

    I don't see the point with these detalied calculations. It seems to me that the formulas will not apply in a real experiment such as Rutherford's:

    1. In the calculations, it is assumed that the center of force is originally fixed. But, unless it is held fixed in a lattice or something like that, this can only be true before the interaction with the first particle in the beam which hits it. When the second, third etc. particle in the beam hits the center of force, it will move in reaction to prior collisions, in a chaotic manner making it unmanageable to calculate the scattering angles, and hence the cross section for scattering to be measured.

    2. In Rutherford's experiment, why don't the particles in the beam interact with the electrons, and become deflected by such interactions?

    3. Why don't the particles in the beam interact with each other, and become deflected by such repulsions?
  2. jcsd
  3. Oct 3, 2013 #2
    1. The target atoms are fixed to a lattice and there are many more target atoms than incoming particle so the chances of two incoming particle hitting the same atom is negligible

    2. The electrons are a bunch of light weights that scatter out of the way of the incoming particles with negligible effect on them

    3.The beam density is too low so the particles in the beam are too far away from each other and their mutual interactions is also negligible.
  4. Oct 3, 2013 #3


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    Ok, thanks Dauto. One more thing, though:

    But the mass of an electron does not matter for the electrostatic force it exerts upon a particle, only its charge matters for this, and this charge is the same as the charge of a much heavier proton (although with opposite sign). On the other hand, the nucleus of a gold atom (which Rutherford used) contains 79 protons, so its charge is 79 times the charge of an electron and exerts a much greater force upon the incoming particle than any electron...
  5. Oct 3, 2013 #4
    The electrons scatter OUT OF THE WAY, so they are never close enough to the incoming particles to have a measurable effect on them
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