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Qualitatively, for a plum pudding model atom (where the positive charge of each atom is even diffused throughout the volume of the atom) I think that the deflection of incident alpha particles would be caused by the electrons and not the positive charge at all. Is this correct?

Quantitatively, for a detector at a specific angle with respect to the incident beam, the number of particles per unit area striking the detector is given by the Rutherford formula (source linked here):

$N(\theta)=\frac{N_inLZ^2k^2e^2}{4r^2KE^2\sin^4{(\theta/2)}}$

where $N_i$ is the number of incident alpha particles, $n$ is the number of atoms per unit volume in target, $L$ is the thickness of the target, $Z$ is the atomic number of the target, $k$ is Coulomb's constant, $e$ is the electron charge, $r$ is the target-detector distance, $KE$ is the kinetic energy of the alpha particles, and $\theta$ is the scattering angle.

For a plum pudding atom, would this formula still hold as is or would it need to be modified?

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The thing is, the electrons are known (even at the original time of this experiment) to be

  1. very light compared to a alpha particle
  2. very weakly bound compared to the kinetic energy of the alpha particle

If an alpha hit one head on, the electron would be ejected at high speed but the alpha would be barely deflected.

And at the energies used in the experiment there was a very low limit on the expected deflection due to the broad positive charge distribution.

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  • $\begingroup$ Makes sense. Do you have any comments on the quantitative formulation of the amount of scattering? $\endgroup$ – NeutronStar Feb 23 '14 at 4:45
  • $\begingroup$ Not off the top of my head. That form assumes point charges (or to put it another way, no interpenetration). But it is pretty easy to show that the Po-210 alpha's Rutherford used will penetrate a plum pudding of atomic size, so the expectation will be less scattering to high angles. $\endgroup$ – dmckee --- ex-moderator kitten Feb 23 '14 at 4:48

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