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Rutherford Experiment

  1. Apr 18, 2007 #1
    1. The problem statement, all variables and given/known data

    In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle, very far from the gold foil, with an initial kinetic energy of 3.6 MeV heading directly for a gold atom (charge +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus.

    2. Relevant equations

    Force:
    mv^2=kZe^2/r

    k=8.99x10^9 N^2/C^2
    Z= # protons
    e=1.6x10^-19 C
    3. The attempt at a solution

    From the question, I'm assuming that the gold and the alpha particle never collide. Thus, I figured that by Newton's third law, I can deduce the distance (r) between them. First, I did 3.6MeV*2 to find mv^2 (since KE=1/2mv^2 and I know that all KE is converted to PE). I got 7200 eV. Then, I plugged in 79 for Z on the other side of the equation and tried solving for r. I came up with 2.525E-30 m. However this answer is not correct. Had I missed an important step? Thanks in advance for your help!
     
  2. jcsd
  3. Apr 18, 2007 #2

    cepheid

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    Hi,

    Conceptually you've got this question: (assume that all of the particle's initial kinetic energy is converted into electrical potential energy). All of your problems stem from silly mistakes here:

    1. As you correctly stated in the body of the solution, this equation has nothing to do with forces, but rather is an equation of energies.

    2. So, the left hand side has mv^2 = 2*KE. That's all well and good, but how did you get rid of the 1/2? Answer: by multiplying BOTH sides of the equation by two. I don't see that on the right hand side.

    3. You appear to have forgotten that your particle is an alpha particle with charge +2e. Your equation for the Coulomb potential energy is correct, but it applies to a particle of charge e in the vicinity of a nucleus of charge Ze. Remember the most general form that applies to the potential energy of a system of any two charges:

    [tex] U = \frac{kq_1q_2}{r} [/tex]

    Thus, for this problem you would have kZe(2e)/r = 2kZe^2/r

    Let me know if these corrections do the trick.
     
  4. Apr 18, 2007 #3

    cepheid

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    Ok, here's a question for you: how the hell does 3.6 MILLION electron volts times two equal 7200 eV?

    Also, did you convert everything to SI units when doing your calculation?
     
  5. Apr 19, 2007 #4
    Oops! For some reason, I thought 3.6MeV was 3.6 x 10^3 eV, but its 10^6!!


    However, I tried what you proposed in your first post, and I understand the reasoning...but I'm still not arriving at the correct answer. Here's what I'm doing:

    Total E= 1/2mv^2 - k(q1)(q2)/r

    mv^2=2k(q1)(q2)/r

    KE = 1/2mv^2
    3.6MeV * 2 = 7.2 x10^6 eV

    r= 2k(q1)(q2)/7.2 x10^6 eV
    = [(2)(8.99x10^9)(79)(2)(1.6x10^-19)^2]/[7.2 x10^6 eV]
    = 1.01008 x 10^-32 m

    which is Incorrect. :confused:
     
  6. Apr 21, 2007 #5
    *bump for viewing*
     
  7. Apr 21, 2007 #6

    hage567

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    Convert the kinetic energy from eV to Joules.
     
  8. Apr 21, 2007 #7
    Thank you! I got the answer :)
     
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