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Final Velocity of Positron through Potential Difference

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    If a positron (or electron antiparticle) beam is accelerated across a potential of 20 kV, find the final velocity v of the particles. Do this problem TWICE, once using MKS units (J for energy) and a second time using "modern" units (eV for energy). Use the following values for the positron's mass: 9.11 *10-31 kg or 0.511 MeV/c^2.

    2. Relevant equations
    E = gamma * m * c^2
    E = (V2 - V1)*q

    3. The attempt at a solution
    I set the two equations above equal to each other. Gamma is a function of v, which ends up being the only unknown. Solving for v using units of Joules and kg gives an imaginary solution, however. I'm not sure how else I would go about doing this.
     
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  3. Feb 4, 2017 #2

    kuruman

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    How jeronimo3000 and welcome to PF.

    Why do you think you need relativity to solve this problem?
     
  4. Feb 4, 2017 #3
    Well, I guess because we're just coming off of chapters about relativity haha. Also, typically all of the particles we have worked with in class are moving at relativistic speeds. The answer using (V2-V1)*q = 0.5*m*v^2 gives 1.12 * 10^-7. That seems very small.
     
  5. Feb 4, 2017 #4
    Whoops, I was using the "modern" mass units and Joules for energy. MKS units gives 8.38 * 10^7 m/s. That's 0.28c, which is fast enough to start considering relativistic effects, right?
     
  6. Feb 4, 2017 #5

    kuruman

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    Maybe. Let's do this using relativity and then compare the two calculations. One of the relevant equations that you listed is E = γmc2. What does the symbol "E" represent?
     
  7. Feb 4, 2017 #6
    E is the total energy of the particle. The energy from its mass and then it's kinetic energy as well.
     
  8. Feb 4, 2017 #7

    kuruman

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    Correct. It is the total relativistic energy. What is another way of writing this total relativistic energy after the particle has accelerated through the potential difference?
     
  9. Feb 4, 2017 #8
    √[ (pc)^2 + (mc^2)^2 ]

    where p is the momentum of the particle
     
  10. Feb 4, 2017 #9

    kuruman

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    This is no different from E = γmc2. Let's consider your earlier thought.
    Where does the kinetic energy, which needs to be added to the rest mass energy, come from?
     
  11. Feb 4, 2017 #10
    The kinetic energy comes from its velocity, which comes from the particle being accelerated through the potential difference.
     
  12. Feb 4, 2017 #11

    kuruman

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    Correct. No potential difference, no velocity, no kinetic energy. How then do you think the kinetic energy is related to the potential difference?
     
  13. Feb 4, 2017 #12
    In general, kinetic energy is 0.5*m*v2. The energy the particle gains from going through the potential difference, ΔV*q, should equal the particle's kinetic energy after it's gone through the potential difference. Does 0.5*m*v2 hold even at relativistic speeds? If so then I definitely understand the problem. I just thought all of that stuff got wonky and changed at those speeds.
     
  14. Feb 4, 2017 #13

    kuruman

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    Actually, this is not true in general. It is true only when the speed is much less than c. In general, the kinetic energy for a free particle is the total energy less the rest mass energy. However, energy conservation demands that if a particle starts from rest and falls through a potential difference, thus gaining kinetic energy, the change in potential energy must equal the change in kinetic energy. Do you see where this is going now?
     
  15. Feb 4, 2017 #14
    I understand that the changes must be equal, but I'm not sure how to equate them mathematically. The potential energy is ΔV*q, right? If 0.5*m*v2 doesn't hold when speeds are near the speed of light then what would the correct equation for kinetic energy be?
     
  16. Feb 4, 2017 #15
    Wait, looked through my equations. Relativistic total energy is gamma*m*c^2, and relativistic KE = E - mc^2. So (gamma - 1)*mc^2. Let me try that and see what I get.
     
  17. Feb 4, 2017 #16
    Aha! That produced 8.14*107 m/s. That looks much better!
     
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