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Homework Help: Physics Electron Velocity - HELP!

  1. May 3, 2009 #1
    Hey Guys, so here is the problem.

    1. The problem statement, all variables and given/known data

    Find the speed an alpha particle requires to
    come within 3.4 × 10−14 m of a gold nucleus.
    Coulomb’s constant is 8.99 × 109 N · m2/C2,
    the charge on an electron is 1.6 × 10−19 C,
    and the mass of the alpha particle is
    6.64 × 10−27 kg. Answer in units of m/s.

    2. Relevant equations
    F=ma=mv^2/r=k x q1xq2/r^2

    3. The attempt at a solution

    v=sqrt(8.99x10^9 x (2x79x(1.6x10^-19)^2))/(6.64x10^-27x3.4x10^-14))

    I get 1.269e7, which seems like the correct answer, but my homework server is not accepting it. Anything that yall may see that is a small error would be greatly appreciated. The only thing that I am wondering about this is that the a=mv^2/r, the r is not the same as the r^2 in the other part of the equation? I am not really sure, but if someone could fill me in, I would love it! Thanks
  2. jcsd
  3. May 3, 2009 #2


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    Homework Helper

    Welcome to PF.

    Doesn't the kinetic energy have to be what overcomes the electrostatic potential?

    So aren't you going to need to consider

    1/2*m1*v2 = W = q1*ΔV = q1*k*q2/r
  4. May 3, 2009 #3
    Okay, so I understand correctly, what you are saying is that kinetic energy will have to be stronger than the electrostatic, i.e. it will have to do work. So the second part of the equation that you provided, after the W, would seem to be the correct structure. I am assuming that q1 would then be the alpha particle charge, i.e. 2 electrons. Are you saying that I would cancel that out? I feel like I would have to cancel out the gold charge since we are calculating the velocity for the alpha. Am I correct to assume that?
    Thank you!
  5. May 3, 2009 #4


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    Homework Helper

    Actually it will be the product of the 2 charges - the alpha particle and the nucleus - won't it? It is the basis of the repulsion to be overcome - the protons of the alpha and the 79 protons of the gold.

    My notation was intended to convey that the kinetic energy of the mass m1 must overcome the work of moving the charge q1 against the potential field from infinity to the nucleus of the gold q2.
  6. May 3, 2009 #5
    yeah, that's right. I just did 1/2mv^2=kq1q2/r and it all worked out.

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