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Electrons leaving a cathode

  1. Jan 6, 2007 #1


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    I am not sure if this is the right place to post this question but here it is any way...

    Electrons leave the cathode of a TV tube at essentially zero speed and are accelerated toward the front by 10,000v potential. At what speed do they strike the screen? Express this value also as a faction of the speed of light.

    OK so here is what I did does it look right?

    PE1 + KE1 = PE2 + KE2
    KE1 = PE2 = 0
    PE1 = KE2

    PE = 1/2QV
    KE = 1/2mv^2


    1/2QV = 1/2mv^2

    Rearrange for velocity

    v = sqroot(QV/m)
    v = sqroot(1.6 x 10^-19 x 10,000 V / 9.11 x 10^-31)
    v = 4.2 x 10^7 m/s

    c 3.00 x 10^8 m/s

    Therefore the speed of the electron equals approximately 2/15c

    Does this look right or am I way off base?????
    thanks for your help....
  2. jcsd
  3. Jan 6, 2007 #2


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    1. The correct place to post this would be the Homework and Coursework subforum (near the top of the main page). Keep that in mind for your next thread.

    2. Recheck the formula you've used for PE - there's an error in it.

    3. If you haven't yet had any Relativity, then this approach is correct. If you fix the error in 2 you should be good.

    4. I haven't actually checked the numbers.
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