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Electron flow through a wire time

  1. Mar 10, 2007 #1

    fsm

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    1. The problem statement, all variables and given/known data
    The electron drift speed in a 1.00 -mm-diameter gold wire is 5.50*10^-5 m/s
    How long does it take 1 mole of electrons to flow through a cross section of the wire?

    2. Relevant equations
    I=n*q*V*A

    3. The attempt at a solution
    I=(5.9*10^28)*(1.6*10^-19)*(3*10^-5)*(.0005^2*pi)
    I=0.222 col/sec

    I believe my answer is correct. The problem is I never had chem so I don't know how to convert to a mole. Any help?
     
  2. jcsd
  3. Mar 10, 2007 #2

    hage567

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    I'll answer the mole part: 1 mole of electrons would be 6.022x10^23 electrons. What's a "col/sec"?
     
  4. Mar 10, 2007 #3

    fsm

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    It's a couple of things. Its a rate the charge moves through a wire. Isn't this also the electron current? Or I can convert it to the electron current?
     
  5. Mar 10, 2007 #4
    An amp is charge/time. What unit is amps or charge per time in? Har har:biggrin:
     
  6. Mar 10, 2007 #5

    hage567

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    Ah, coloumbs/sec. I usually see that as C/s. Didn't recognize what you were getting at there.
     
  7. Mar 11, 2007 #6

    fsm

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    Guess I'm not getting it. So do I convert my answer to electron current and then divide by 6.022x10^23?
     
  8. Mar 11, 2007 #7

    hage567

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    I'm not sure what you mean by convert to electron current. To me, that's what you already have. Think of the definition of current (charge/time). Start with: How many coloumbs would be in one mole of electrons?

    By the way looking at your first calculation I=(5.9*10^28)*(1.6*10^-19)*(3*10^-5)*(.0005^2*pi)
    Isn't the drift speed 5.50x10^-5 m/s?
     
  9. Mar 12, 2007 #8

    fsm

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    I used the example my teacher provided I just transposed some numbers. I see now he used what he called the formal definition I=n*A*V_d*q. Then he made a conversion to -e/sec by dividing 1.6*10^-19. Which cancels out the charge. Seeing this I now understand he was just using the formula i=n*A*V_d where little i is the electron current. Recalculating I get about 2.55*10^18 -e/sec. The answer is wanted in a day. Do I first convert to -e per day then I divide by 6.02*10^23?
     
    Last edited: Mar 12, 2007
  10. Mar 12, 2007 #9

    hage567

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    If the point of the question is to get the number of moles per day, then I would think your reasoning is correct. But I don't understand how it got to this from the original question you posted, which to me is asking something different. So I'm not sure what to say about that.
     
  11. Mar 12, 2007 #10

    fsm

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    This problem is a bugger. I'm still getting the wrong answer. I know its me and cant figure it out. I even tried to do the same type of problem in my book and got that wrong too. Please any help.
     
  12. Mar 12, 2007 #11

    hage567

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    I'm not sure what you're trying to accomplish anymore, is it still the question in your original post? I don't understand where this e- per day thing is coming from. Perhaps if you post your entire calculation, we can see if you are making some kind of math error. What is the right answer supposed to be?
     
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