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Electron Charge to Mass Ratio

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Electrons from a heated cathode are accelerated through a potential difference of 25.0 V. Calculate the speed of the electrons escaping through a hole in the anode.

    2. Relevant equations


    v = (2eV/m)^(1/2)

    3. The attempt at a solution

    v = (2*25*1.759*10^11)^(1/2) = 2.97*10^6 m/s

    but this answer is wrong. why?
     
  2. jcsd
  3. Sep 30, 2008 #2

    Redbelly98

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    What are the units for each number you use in the calculation? Perhaps the units are not consistent.
     
  4. Oct 1, 2008 #3
    v = (2* 25V * 1.759*10^11 C/kg) ^(1/2) = 2.97*10^6 m/s

    (VC/kg)^(1/2) = m/s

    or

    VC/kg = m^2/s^2

    is that true?
     
  5. Oct 1, 2008 #4

    Redbelly98

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    Okay, I agree with your answer, including the number of significant figures. I wasn't familiar with the value of e/m.

    Hmmm ... since the calculated answer is 2.9655... *10^6 m/s, maybe try 2.96*10^6 m/s. They might have done some intermediate rounding off that changed their final answer slightly.
     
  6. Oct 2, 2008 #5
    that still turned out to be wrong. maybe i am using the wrong formula. this problem deals with a cathode and anode from a helm's coil. any suggestions?
     
  7. Oct 2, 2008 #6

    Redbelly98

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    I'm unfamiliar with a helm's coil (or do you mean Helmholtz coil?).

    Your formula is correct for electrons with zero, or at least negligible, initial kinetic energy at the cathode.
    (Substitute eV = ½ m v2 in the equation to verify it, if you wish.)

    Does anything in the problem statement, or in the text book, indicate a nonzero initial speed for the electrons? If not, I am stumped ... does somebody else have any ideas?
     
  8. Oct 4, 2008 #7
    thank you so much! i had been typing the answer wrong. the computer did not want the answer in scientific notation. so your advice was correct. thank you
     
  9. Oct 4, 2008 #8

    Redbelly98

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    Your welcome!

    Computers often use a different form of scientific notation. Instead of typing 2.97*10^6, it would be 2.76e6

    You might try that out on the next problem, just to see if the computer recognizes it.
     
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