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Electrons in a TV tube accelerated from rest.

  1. Oct 1, 2008 #1

    _F_

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    1. The problem statement, all variables and given/known data
    Electrons in a TV tube are accelerated from rest through a 25-kV potential difference. With what speed do they hit the TV?

    V = 25*10^3 J/C


    2. Relevant equations
    I'm not really sure here. I'm thinking you have to use Newtonian mechanics somewhere here.
    So,
    x(t) = (1/2)(a_x)t^2 + (v_0)t +x_0
    (v_f)^2 = (v_0)^2 + 2(a_x)d

    You also need electomagnetic equations.
    So,
    delta-V = delta-U/q = -E*delta-r

    3. The attempt at a solution
    This is where I have trouble. I don't even know where to start. Any pointers in the right direction (that aren't too vague!) would greatly be appreciated.
     
  2. jcsd
  3. Oct 1, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi _F_! Welcome to PF! :smile:

    Hint: volts = energy/charge (as you can see from the J/C units).

    So what KE does each electron have after going through a 25kV potential difference? :smile:
     
  4. Oct 1, 2008 #3

    _F_

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    Re: Welcome to PF!

    Okay, so K = (1/2)mv^2

    We have volts in J/C, so to get a value of just energy would we multiply V*q?
    If so:
    V*q = (2.5*10^3 J/C)(1.6*10^-19 C) = 4*10^-16 J.

    So K = 4*10^-16 J,
    Then
    K/[(1/2)m] = v^2 => (4*10^-16 J)/[(1/2)(9.1*10^-31 kg) = 8.79*10^14 J/kg

    So v^2 = 8.79*10^14 J/kg
    Then, v = sqrt(8.79*10^14 J/kg) = 2.9*10^7 m/s

    But this is not what the back of my book says. In the back, the answer is 9.37*10^7 m/s.
     
  5. Oct 1, 2008 #4

    tiny-tim

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    Hi _F_! :smile:

    Hint: check your units … you're out by a factor of √10. :wink:
     
  6. Oct 1, 2008 #5

    _F_

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    J = kg*m^2/s^2

    V = [J/C]
    Vq = [J]
    K = [J]
    v^2 = 2K/m = [m^2/s^2]
    v = [m/s]

    Where did I go wrong?
     
  7. Oct 1, 2008 #6

    Dick

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    V is given to be 25*10^3 J/C. You used 2.5*10^3.
     
  8. Oct 1, 2008 #7

    _F_

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    Ack! Thanks. Such a stupid mistake. :uhh:
     
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