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Rutherford Experiment, energy and acceleration.

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations

    If i can make that out its Ep=kq1q2/r where k is coloumbs constant

    3. The attempt at a solution

    Im a little confused how do i know the radius of the alpha particles?

    I was thinking its conservation of energy, Ep+Ek=Eka then try to solve for a that way, but would Fnet be 0 on this alpha particle?
     
  2. jcsd
  3. Jan 27, 2013 #2

    tms

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    Right.
    Assume it's a point.
    That's the way to start, but it's a two-step process; you don't get [itex]a[/itex] directly.
    Of course not. If the net force were 0 then there would be no acceleration.
     
  4. Jan 27, 2013 #3

    gneill

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    Staff: Mentor

    Well, you don't know and don't need their radius. You will need to look up their mass and charge though. The radius being referred to in the potential energy equation is the radial distance of the alpha particle from the center of the gold atom nucleus.

    Oh yes, you'll need to know the charge on the gold atom nucleus, too!
    Conservation of energy will certainly play a role here. Assuming that the alpha particle heads directly towards a gold nucleus, at what point in its trajectory do you think it'll experience maximum acceleration?
     
  5. Jan 27, 2013 #4
    Is the conservation set properly? IF it is, than in the Ep+Ek, what do i use for r?, or is Ep ignorable? Mass of an alpha particle and the charge are, 6.65*10^-27kg and 2e, and gold is what 79 protons, how do i know the charge of that?

    It would be at its maximum right before it hits, just as the question suggest wouldnt it be?
     
  6. Jan 27, 2013 #5

    gneill

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    What's the charge on one proton? What's the charge due to 79 of them?

    The alpha particle is unlikely to have enough energy to enter (hit) the gold nucleus. It'll approach and rebound due to electrostatic repulsion. The thing you'll have to determine is, how close will it get with its given initial KE? What'll be the net force acting on it when it reaches that closest approach?
     
  7. Jan 27, 2013 #6
    Oh, since its a positive, that meets a positive atom?, but isnt the nucleus neutral, of any atom? It would be 79+ i would assume. Wouldnt net force just be Force electric, i suppose gravity is too but it is negligible, there was no magnets what so whatever from what i remember from the rutherford experiment.
     
  8. Jan 27, 2013 #7

    gneill

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    The nuclei of all atoms (except for antimatter atoms, of course) is positively charged. It's the electrons surrounding the atom that give the atom a net neutral charge. The electric force is the only one you need be concerned with here.
     
  9. Jan 27, 2013 #8
    how do i calculate, Fe, i dont see a formula i can use, V=E/d and E=V/d only work for parallel plates right. and i dont know r, the standard Fe=Eq, applies, but both of you said to use conservation of energy, at the moment i cant see to factor it in
     
  10. Jan 27, 2013 #9

    gneill

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    The problem statement tells you how to determine potential energy for two charges separated by distance r, and as for force, Coulomb's Law comes to mind...
     
  11. Jan 27, 2013 #10
    But you said you dont need to know r, and coulomb's law involves r as well Fe=kq1q2/r^2?
     
  12. Jan 27, 2013 #11

    gneill

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    Staff: Mentor

    I said you don't need to know the radius of an alpha particle. The radial distance of the alpha particle from the center of the gold nucleus is a different matter -- you will want to calculate a particular value.
     
  13. Jan 28, 2013 #12
    Can you give me some sort of a hint?, its not centripetal since it bounces straight back, besides no mag fields, also the is r in the Ep formula, the radius of an alpha particle, or the radius between the two particles?
     
  14. Jan 28, 2013 #13

    gneill

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    There is NO alpha particle radius ANYWHERE in this problem. All r's are the distance between charges (alpha particle and nucleus).

    Use conservation of energy to determine the closest distance that the alpha particle can approach to the nucleus.
     
  15. Jan 28, 2013 #14

    tms

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    You start with what you know. You know the initial kinetic energy, and the initial potential energy can be set to zero. You know the potential energy due to the electric field at the point of closest approach, and you know what the kinetic energy will be there, too (assume the alpha goes straight in and straight out). Write down the expressions for the initial and final energy, and equate them.
     
  16. Jan 28, 2013 #15
    Ek+Ep = Ekf +Epf
    1*10^-12 + 0 = 1/2mv^2 + Kq1q2/r ?
     
  17. Jan 28, 2013 #16

    tms

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    Don't use numbers at this stage (except for 0); do everything symbolically. It will be easier to understand.

    What is the kinetic energy at the point of closest approach? (This time give a number).
     
  18. Jan 28, 2013 #17
    I dont know, 0?
     
  19. Jan 28, 2013 #18

    tms

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    You surely do know. Throw a ball straight up. What is its kinetic energy at the highest point of its trajectory?
     
  20. Jan 28, 2013 #19
    zero right, since at the top as it comes back down in that instant the velocity is zero?

    Ek+ 0 = 0 + Ep ? find r through that and use that r in Fe=kq1q2/r^2 ? using charge of alpha particle and charge of gold nucleus, as the other poster stated?
     
  21. Jan 28, 2013 #20

    tms

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    Right. Although you should be stating these things, not asking them.
     
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