Since alpha-particles (used in Rutherford's experiment) have positive charge, they probably should attract electrons (and stop being alpha particles). Was it the case in Rutherford's experiment? Did the foil become slightly positively charged (ionized)?

  • $\begingroup$ the energy of particles involved does not permit such interactions which can generate stripping of electrons..my guess..as time interval must be pretty small leading to energy transfer $\endgroup$
    – drvrm
    Commented Sep 10, 2018 at 12:28

1 Answer 1


The phenomenon that occurs in the experiment is known as Rutherford Scattering. That second word is rather important - the $\alpha$-particles scatter and don't stop in the gold foil. You start with a tightly collimated beam of $\alpha$s and observe that, after you put the foil in the path, the beam gets slightly diffused in the forward direction. What is surprising is that you get a significant amount of back-scatter - that is, $\alpha$s coming back from the foil.

In either event, the $\alpha$s don't stay in the foil so it doesn't "charge up".

For the pedants: That's not to say there isn't any local ionisation. An $\alpha$-particle is highly ionising and will strip electrons readily from atoms in the foil as it passes through.

  • $\begingroup$ I know they mostly penetrate the foil. But still electromagnetic interaction should be rather strong, unless there is stronger force between nucleus and electrons. This is because according to nuclear model $\alpha$ particle can approach electron at much closer distance than it is between electrons and nucleus. So, your "For the pedants" section is affirmative answer, right? $\endgroup$
    – user168013
    Commented Sep 10, 2018 at 13:23
  • $\begingroup$ @rus9384 As the $\alpha$s pass through the foil, they lose energy by knocking electrons off the atoms as they pass. This is what I meant by *local ionisation*. Note that this doesn't "charge up" the foil since the electrons don't leave the foil and quickly re-combine with ions in the lattice. Eventually, if the foil is thick enough, the $\alpha$ will lose all its energy and stop. Then you would have a net charge of +2 in the foil. But this wouldn't make for a very good Rutherford Scattering experiment. For that, the foil has to be very thin - precisely so that they can get out again. $\endgroup$ Commented Sep 11, 2018 at 6:15
  • $\begingroup$ Why are they knocking electrons off instead of attracting them as opposite charges should? Is there another force? $\endgroup$
    – user168013
    Commented Sep 11, 2018 at 6:29
  • $\begingroup$ Note that sending an ion beam at a foil will charge it up unless it is grounded and conducting. Some is the net charge from the beam, some is secondary electron emission (which you can use to image the sample) where electrons do, indeed, leave the surface. To get accurate measurements of beam current you need secondary suppression, typically about 300V to redirect secondaries back to the sample. $\endgroup$
    – Jon Custer
    Commented Sep 11, 2018 at 16:14
  • $\begingroup$ the paper shows no grounding of the foil web.ihep.su/dbserv/compas/src/geiger13/eng.pdf $\endgroup$ Commented Sep 24, 2018 at 16:42

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