Understanding Rutherford's Alpha Scattering Experiment

AI Thread 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. The scattering occurs as these forces cause deflections when alpha particles approach the nucleus. If a different metal than gold is used, the mass and charge of the nucleus can affect scattering angles; lighter nuclei may result in greater deflections, akin to a ping-pong ball colliding with another ping-pong ball. The strength of the electric field around a nucleus influences whether alpha particles can penetrate the Coulomb barrier, which requires significant energy. Overall, understanding the forces involved and the properties of the target metal is crucial for grasping the experiment's implications.
DJ-Smiles
Messages
46
Reaction score
0

Homework Statement


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
 
Physics news on Phys.org
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
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top