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I have been taught that the mass of an atom is concentrated of an atom, this can be proved by invoking the fact that very few particles are deflected. However, unless I'm mistaken the deflection is primarily due to electrostatic repulsion. Therefore, couldn't we instead say that the charge density varies with radius such that it's mostly in the Centre, whereas the mass density is very low, and spread out through out the atom and get the same result?

I'd really appreciate any help

Note:- Rutherford's gold foil experiment = Geiger–Marsden experiments

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  • $\begingroup$ may I know why this question has received a downvote? $\endgroup$
    – math and physics forever
    Commented Sep 7, 2022 at 17:25
  • $\begingroup$ I'll make corrections, if the downvoter can point out any mistakes in my post $\endgroup$
    – math and physics forever
    Commented Sep 7, 2022 at 17:26

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Electric charge does not exist independent of particles with mass. This means that "charge density" will match "mass density" in Rutherford's scattering model. Furthermore, the alpha particles being used as probes in that experiment were highly energetic, which meant they had enough energy to penetrate the electron cloud surrounding the nucleus and "see" the bare nucleus as a scattering target. All of this is covered in Rutherford's own report on the experiment, which you should read.

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