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Homework Help: Electrostatics Problem

  1. May 27, 2008 #1
    [SOLVED] Electrostatics Problem

    I've been trying to solve this problem for the longest time, but it doesn't seem to give enough information.

    1. The problem statement, all variables and given/known data
    If an alpha particle is fired straight at a nucleus of a gold atom, (q = 79 e) at a velocity of 3.3 x 10^4 m/s, how close will it get to the gold nucleus?

    mass of alpha particle = 6.64 x 10^27 kg
    e = 1.6 x 10^-19 C
    (I looked this up on the internet, it's not given in the question, but: charge on alpha particle = +2e)


    2. Relevant equations

    I have v1 and v2 (I'm assuming it's zero), but that's pretty much it. I can't calculate the force between these charges because I don't know their distance.

    I also tried: W = Delta Ek
    W = (0.5)(m)(v)^2
    W = (0.5)(6.64x10^-27)(3.3x10^4)^2
    W = 3.62x10^-18 J
    F * d = 3.62x10^-18

    But, as I said, I can't calculate the force, and the "d" in that formula gives me the distance it traveled, not how close it got to the nucleus.

    3. The attempt at a solution

    I'd really like to post an attempt, to show that I truly tried this problem, but I'm really stumped. I have thought of using the electrostatic force equation, but that didn't work out. I'm really not sure where to start.

    Could someone please tell me the general direction I should go? I don't need an actual answer, I just want some guidance. Thanks.
     
    Last edited: May 27, 2008
  2. jcsd
  3. May 27, 2008 #2

    marcusl

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    You should look up the charge of a gold nucleus, then go back to Coulomb's law--specifically, the energy between two charged particles. Think about how the energy changes as you bring the two particles towards each other from infinity.
     
  4. May 27, 2008 #3
    Well I know the charge on the gold nucleus is 79e. (1.264 x 10^-14 C)

    The alpha particle is repelled by the nucleus, so the electric potential energy increases as it is brought towards the nucleus.

    E = k q1 q2 / r
    E = 3.65 x 10^-26 / r

    For the force,

    F = 3.65 x 10^-26 / r^2.

    But I still don't see how this can get an answer.
     
  5. May 27, 2008 #4
    Is it because all of the kinetic energy is converted into electric potential energy? If that is the case, I guess I could isolate for r. =P
     
  6. May 27, 2008 #5

    Redbelly98

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    Yes.
     
  7. May 27, 2008 #6
    Thanks!
     
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