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Atomic Physics problem.

  1. Dec 30, 2011 #1
    I am going to try traduce the problem in the best way i can.



    What is the minimum distance that protons of a 1 MeV bundle that collides in a gold leaf
    can approach to the atom core.

    Solution: 1.14 x 10-13 m




    I am posting these problem because i do not have any idea how to start it. So please, do not delete this post because i do not have any attempt to solution. I searched in lot of places and i am not seeing what is the formula to resolve this problem.
     
  2. jcsd
  3. Dec 30, 2011 #2
    How about energy conservation !!!
    initial KE of proton is known ... final KE of system is 0
     
  4. Dec 30, 2011 #3
    And what is the point of know the kinetic energy of proton?
     
  5. Dec 30, 2011 #4
    What? i cant get you ...
     
  6. Dec 30, 2011 #5
    I am not understanding why i need to now KE (i think is kinetic energy) of proton?
     
  7. Dec 30, 2011 #6

    SammyS

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    What do you understand about this problem?

    Can you calculated the velocity of a 1MeV proton?
     
  8. Dec 31, 2011 #7
    I know how to calculate a velocity of 1MeV proton by the equation E = 0.5mv^2 (i think)

    but why should i need to know the velocity?
     
  9. Jan 1, 2012 #8
    You don't need velocity ... You just need Kinetic energy ... (but yes do convert it into SI)

    Then use energy conservation ... for the system ...

    PS: Happy New Year
     
    Last edited: Jan 1, 2012
  10. Jan 1, 2012 #9

    ehild

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    Imagine you give a push to a book lying on the table. It moves towards the edge but slows down along its path. If the speed is lost completely, the book does not get closer to the edge any more. The minimum distance the book approaches the edge of the table is the distance of the point where it stopped.

    ehild
     
  11. Jan 1, 2012 #10
    So, ok we know that

    1Mev = (1*10^6eV) * (1.6 * 10^-19 C) = 1.6 * 10^-13 J

    We know that in final point KE of proton is 0.


    Ok So ΔK = 1.6 * 10^-13 J KE conservation.

    W, as we know = ΔK

    so F = W*d

    F = mg (i think) = proton mass * gravity acceleration = 1.64*-26N (this because of imagining the push)


    d = F/W = 1.02*-13???

    Is that correct?? because i think not
     
    Last edited: Jan 1, 2012
  12. Jan 2, 2012 #11
    Well you cant use mg here because mg is not the repelling or working force here.its the coulomb force doing the work.
    And you cant use coulomb force directly too because its not constant but keeps changing.

    That is why is suggested using energy conservation, ie

    Kinitial + Uinitial = Kfinal + Ufinal

    But i just noticed ... initial distance b/w proton and gold is not specified ...
    either you left some part of question or i'm missing out on some concept ...
     
  13. Jan 2, 2012 #12

    ehild

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    The proton starts from a greater distance, where the Coulomb force of the gold nucleus is screened by the electron cloud, so the starting potential energy is zero, like in infinity. The potential energy changes mainly when the proton is near to the nucleus, closer than the radius of the first electron shell.

    ehild
     
  14. Jan 2, 2012 #13
    But then there's very less energy and its coming from a very huge distance. I dont think closost distance would be of order of 10-13m
     
  15. Jan 2, 2012 #14

    Curious3141

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    This problem is a simple application of the conservation of energy. Initially, the proton has a KE of 1MeV and an Electrical PE of 0. Finally, the proton has a KE of 0 and an Electrical PE defined by:

    [tex]PE = \frac{q_1q_2}{4{\pi}{\epsilon_0}r}[/tex]

    where q1 is proton charge (equal to e, the electron charge), q2 is the positive charge of the gold nucleus (depends upon the number of protons in it, which can easily be looked up). [itex]\epsilon_0[/itex] is the permittivity of vaccuum, which can also be easily looked up. r is the minimum closing distance (which is to be determined).

    That expression has to be equated to 1MeV. I would advise against converting to J (joules), because one of the 'e's can easily be cancelled out.

    This problem is simple, and there are tons of simplifying assumptions, but all the same, I wonder if we're entirely justified in completely neglecting the electron cloud?
     
  16. Jan 2, 2012 #15

    ehild

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    The proton comes from an accelerator and the gold leaf is a few mm or cm from the accelerator exit. Outside the gold film, the Coulomb potential is zero, so the total energy of the proton is kinetic.
    At the closest approach, the total energy is electric potential energy.
    We suppose that no dissipative interaction With the electrons takes place, as the atom is almost empty. Then the total energy transforms into the electric potential energy, and it results in the closest approach of about 10-13 m.

    http://physicaplus.org.il/zope/home/1223030912/god_particle_en/?skin=print [Broken]

    The image shows the typical sizes of atom and nucleus.


    sat117002_0401.gif

    ehild
     
    Last edited by a moderator: May 5, 2017
  17. Jan 2, 2012 #16


    Hi ehild
    Nice figure!! Did you make that? on which software?
     
    Last edited by a moderator: May 5, 2017
  18. Jan 2, 2012 #17

    ehild

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  19. Jan 2, 2012 #18
    Ok people thanks a lot for the help!
     
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