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Kinetic Energy (in eV) of He with given change in potential

  1. Jun 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Point A is at a potential of +230 V, and point B is at a potential of -140 V. An -particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An -particle starts from rest at A and accelerates toward B. When the -particle arrives at B, what kinetic energy (in electron volts) does it have?

    hence:
    Change in potential=370 Volts


    2. Relevant equations
    Change in potential= -work/q

    and

    1e- moving through one volt of potential= 1.6e-19 Joules



    3. The attempt at a solution

    He= 2 protons = 2 (1.6e-19 coulombs)= 3.2e-19c

    (370 Volts)(3.2e-19c)= 1.184e-16

    Of course, this isn't right or I wouldn't be posting this. My ideas for where I've gone astray:

    -the charge (in coulombs) of He is different than 3.2e-19
    -The answer asks for the kinetic energy, but I thought eV was a measure of Potential Electric Energy?
     
  2. jcsd
  3. Jun 30, 2009 #2

    LowlyPion

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    Welcome to PF.

    If they want the answer in eV ...

    ... isn't it simply 2*370 eV ?
     
  4. Jun 30, 2009 #3
    Thanks for the welcome.

    Isn't 370 volts just the difference in potential from A-->B ?
     
  5. Jul 1, 2009 #4

    LowlyPion

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    Yes.

    And if W = KE = q*ΔV

    So ... q = 2 electrons worth and ΔV = 370 V ...
     
  6. Jul 1, 2009 #5
    I had tried that originally, but it didn't work. Here's how:
    2(1.6e-19)*(370)=1.184e-16


    Thanks for the help
     
  7. Jul 1, 2009 #6

    LowlyPion

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    That answer is in Joules, not eV.
     
  8. Jul 1, 2009 #7
    Ohhhh, got it. Much more simple than I made it out to be.

    Thanks again
     
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