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Finding the potential difference of an alpha particle.

  1. Apr 18, 2008 #1
    1. Through what potential difference ([tex]\Delta[/tex]V) would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15 x 10^(-15) m [tex]^{238}[/tex]U nucleus?

    2. KE = q[tex]\Delta[/tex]V
    E = [tex]\gamma_{p}[/tex]mc²
    r = [tex]\frac{mv}{qB}[/tex]

    3. Ok, I feel like I am close to what I really need but I am lacking something...
    You can get the charge (q) and mass for both the particle and the U nucleus from the information given, you can also get the radius of the nucleus if needed.

    I have that, I just seem to have two unknowns for every equation I am using, if I solve for u in E = [tex]\gamma_{p}[/tex]mc² I don't know Energy, and if I solve for v or B in
    r = [tex]\frac{mv}{qB}[/tex] I don't know one of them, what am I missing?

  2. jcsd
  3. Apr 18, 2008 #2

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    Where did the B come from?

    If the alpha particle just reaches the surface of the nucleus, its KE is zero, but PE is not.

    But how much KE it should have had to reach that point? Remember conservation of energy.

    Knowing the charge of the alpha particle, what should be ∆V to give that KE at a point far away from the nucleus?

    There, I just solved it for you.
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