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Relationship of velocity of an electron/velocity of proton to mass ratio

  1. Feb 5, 2008 #1
    Relationship of velocity of an electron/velocity of proton to....mass ratio

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

    An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed ve, while the proton acquires a speed vp.


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    (d) What is the algebraic expression for the ratio ve/vp of the speed of the electron to that of the proton? Express your answer in terms of the mass me of the electron and the mass mp of the proton. (Answer using m_e for me and m_p for mp.)

    2. Relevant equations

    SO I need to write an equation relating Velocity of electron/ velocity of proton...I understand that much...but I am unsure where to start.

    I think I need the equation 1/2mv^2 = -q (charge)deltaV(voltage)


    3. The attempt at a solution

    by using the equation above I solved and got sqrt(2qdeltaV/m) ...I am not sure where to go next...? Can anyone help me out?
     
  2. jcsd
  3. Feb 5, 2008 #2

    Tom Mattson

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    Yes, that's what you want to use.

    Actually, you should get sqrt(-2qdeltaV/m). Then if q<0 (like it is for an electron) then deltaV>0 and the radicand is positive. And if q>0 (like it is for the proton) then deltaV<0, and again the radicand is positive. Just find expressions for the speeds of the proton and the electron, and take their ratio and you'll have it.
     
  4. Feb 5, 2008 #3

    Hootenanny

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    So you have used conservation of energy to almost correctly [note the sign correction] solved for the velocity of a generic particle of charge q and mass m (traveling at non-relativistic speeds). Let me write your equation out in a form that may be easier to work with;

    [tex]v^2 = -2q\Delta V \cdot \frac{1}{m}[/tex]

    So now you want that ratio [itex]v_e/v_p[/itex]. So you can now write two equations;

    [tex]v_e^2 = -2q_e\Delta V \cdot \frac{1}{m_e}[/tex]

    [tex]v_p^2 = -2q_p\Delta V \cdot \frac{1}{m_p}[/tex]

    Can you take the next step?

    Edit: Tom beat me to it!
     
    Last edited: Feb 5, 2008
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