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Rutherford Scattering Help

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey I am revising for my physics test and am having a bit of trouble understanding rutherford's alpha scattering experiment. My teacher gave us a powerpoint but it does not seem to help me, more so it confuses me. I am a good physics student and not understanding this is hard for me.

    If you could help me by answering these questions I would be ecstatic, thanks in advance guys:

    why do alpha particles scatter?
    - How do they scatter?
    - force involved?
    - how this force works?
    - what if the metal wasn't gold?
    -would this have an affect?

    If you want to throw in some things you think might help I am willing to give anything a try, thanks again
  2. jcsd
  3. Mar 10, 2012 #2


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    repulsive electrostatic forces between the +ve alpha particles and the +ve nucleii

    repulsive electrostatic forces between the +ve alpha particles and the +ve nucleii

    repulsive electrostatic forces between the +ve alpha particles and the +ve nucleii

    the electrostatic forces between the +ve alpha particles and the +ve nucleii are repulsive

    Most metals are significantly more massive than alphas. Rutherford assumed the gold nucleii were essentially fixed. Of course, if the metal nucleii are of similar mass to the alphas then that doesn't hold true any more. If you shine alphas on to beryllium, for example, you'll end up with a stream of neutrons coming out of the metal!!

    just remember that repulsive electrostatic forces between the +ve alpha particles and +ve nucleii results in scattering (called 'Coulomb scattering' as it is the 'Coulombic' force... did I mention that is the electrostatic force :smile: )
  4. Mar 10, 2012 #3
    thanks alot for that i needed it. I can't believe its all because of the same force. I didnt really understand the bit about the metals though. Is it that if the metal he used was more proton rich there would be greater deflection angles and less proton, smaller deflection? because i used an applet just before and it appeared that as the number of protons decreases so does the deflection angles
  5. Mar 10, 2012 #4


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    The comment on the different metals was to do with the atomic mass - it is like a ping-pong ball hitting a bowling ball, it'll just bounce off. But if it were another ping pong ball then both would shoot off from the collision point.

    I'd expect the electric field around a naked gold nucleus would, indeed, be stronger. However, it is strong enough around any nucleus that the alphas would not reach the nucleus (unless* they have enough energy to penetrate the Coulomb barrier - an MeV at least, in the case of gold that'd be many many MeV I am sure)

    *[- or if there is some quantum tunnelling event, but I'll not complicate this answer with that]

    However, in real life you'd also contend with the fact that the alpha is flying through an electron soup before it reaches the nucleus, and the fewer protons the fewer electrons. I confess I've not crunched those numbers or thought about it much, so I'm not sure enough to comment on what the variations of results would be if you used a metal foil of different atomic mass.
  6. Mar 10, 2012 #5
    ok i get it now thanks again
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