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Proton accelerated thru potential difference - find final speed

  1. May 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A proton is accelerated from rest through a potential difference of 107 V. Calculate the final speed of this proton.

    2. Relevant equations
    I'm not really sure which equations to use, but I figured I give all I tried...
    and I think this applies...v(f)^2= v(i)^2 + 2ad

    3. The attempt at a solution

    I really can't figure out what equation i should use. What I did was..
    1)used the equation V = Ed
    2) becomes V= Kc(q/d^2)d
    3) becomes V=Kc(q/d)
    4)107= 8.99x10^9 (1.602x10^-19)/d
    5)Solved for d and found it was 1.34598 x 10^-11
    6) used the equation v(f)^2= v(i)^2 + 2ad
    7) became v = sqrt (2a (1.34598x10^-11)
    ....but I don't know what "a" is so I can't solve. Any help would be greatly appreciated. I gave this a shot and really couldn't figure it out. Thanks, everyone.
  2. jcsd
  3. May 2, 2007 #2
    Try energy method.
    Find relationship between initial energy, and final energy.
    What is the initial energy?, and what's the final energy?
  4. May 2, 2007 #3
    Yes, by far the easiest method is to just use energy conservation.

    Proton KE = potential difference
  5. Feb 9, 2008 #4
    Don't forget that 1eV = 1.60E-19 J

    Since the potential difference is given (107 V), you have to find the kinetic energy that it would take for that to happen. Set up a ratio like this: (1 eV)/(1.60E-19 J)=(107 V)/(x). Solve for x. This value, x, is the kinetic energy (in J) that this would take.

    Now, use 1/2mv^2=KE. Since you have the KE to be x, solve for v.

  6. Oct 4, 2009 #5
    what is "m"?
    what is "sqrt"?
  7. May 29, 2011 #6
    m: meters
    sqrt: square root of
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