Understanding Rutherford's Alpha Scattering Experiment

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SUMMARY

Rutherford's alpha scattering experiment demonstrates that alpha particles scatter due to repulsive electrostatic forces between positively charged alpha particles and positively charged nuclei. This scattering, known as Coulomb scattering, occurs because the electrostatic force is repulsive. If a metal other than gold is used, such as beryllium, the results change significantly; lighter nuclei can lead to different scattering behaviors, including the emission of neutrons. The strength of the electric field around the nucleus also plays a crucial role in determining the deflection angles of the alpha particles.

PREREQUISITES
  • Understanding of Coulomb's Law and electrostatic forces
  • Familiarity with atomic structure and nuclear physics
  • Knowledge of particle interactions and scattering phenomena
  • Basic grasp of energy levels and quantum tunneling
NEXT STEPS
  • Study Coulomb's Law in detail to understand electrostatic interactions
  • Explore Rutherford's original experiment and its implications in nuclear physics
  • Investigate the effects of different atomic masses on scattering outcomes
  • Learn about quantum tunneling and its relevance in nuclear reactions
USEFUL FOR

Students preparing for physics exams, educators teaching nuclear physics concepts, and anyone interested in the fundamental principles of particle interactions and scattering experiments.

DJ-Smiles
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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
 
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why do alpha particles scatter?
repulsive electrostatic forces between the +ve alpha particles and the +ve nucleii

- How do they scatter?
repulsive electrostatic forces between the +ve alpha particles and the +ve nucleii

- force involved?
repulsive electrostatic forces between the +ve alpha particles and the +ve nucleii

- how this force works?
the electrostatic forces between the +ve alpha particles and the +ve nucleii are repulsive

- what if the metal wasn't gold?
-would this have an affect?
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!

If you want to throw in some things you think might help I am willing to give anything a try,

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: )
 
thanks a lot 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
 
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.
 
ok i get it now thanks again
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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