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Electron in a TV

  1. Jul 23, 2014 #1
    Electrons leave the cathode of a tv tube at essentially zero speed and are accelerated toward the front by 10, 000 V potential. At what speed do they strike the screen? Express this value also as a fraction of the speed of light.

    I found-
    vi=0 m/s V=10 000 V q=-1.6 X 10^-19 C (not sure if it's negative)
    m= 9.1 X 10^-31 kg vf= ?(speed of light)
    v(light)= 3.00 X 10^8 m/s


    Delta PE= qV=(-1.6 X 10^-19 C) ( 10 000 V) = -1.6 X 10^-15 J
    And then --> Delta KE= -Delta PE --> 0.5mv^2=1.6 X 10^-15
    vf= SqRt{ 2(1.6 X 10^-15)/9.1 X 10^-31 = 59 295 868= 5.9 x 10^7 m/s

    Fraction of speed of light --> vf/v(light)= 0.197

    I am unsure if I am doing this properly-Just looking for some confirmation
     
  2. jcsd
  3. Jul 23, 2014 #2

    ehild

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    It looks correct.

    ehild
     
  4. Jul 23, 2014 #3

    rude man

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    You might have included relativistic effects. At 0.2c the error is about 2%. So admittedly it's not much.
     
  5. Jul 23, 2014 #4
    Awesome-
     
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