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I De Broglie Momentum doesn't seem in Agreement with p=mv

  1. Jan 22, 2017 #1
    Hi there.

    So I had this lab last week about De Broglie hypothesis. In a simulation, we plugged in the electron velocity and the computer gave back a beautiful wavefunction, from wich I can measure the wavelength. So here I have an electron going at 0.6 m/s with a wavelenght of 0.00060606 meters. When we calculate the momentum, we have two choices, either p=mv or p=h/λ. I respectively found 5.466E^(-31) and 1.093291E^(-30). The link ? A factor of 1/2 between the two values. I did this with many other wavelengths and velocities. What is the point I am missing ? Shouldn't the two equations give the same answers ?
     
  2. jcsd
  3. Jan 22, 2017 #2

    PeterDonis

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    What was the wavefunction?
     
  4. Jan 22, 2017 #3
    A classical sine wave, we were only considering the problem in one dimension with only one velocity at a time. We then went further with wavepackets, but I had problems way before going into that. I measured myself the wavelenghts. It's a Matlab simulation that our Prof gave us, it seemed legit, that's why I ask myself these questions. Here is a link to see the distribution. Don't be scared because there are some french words, the math is the same https://docs.google.com/document/d/1k0BKPiMQSJ7jxplDCH7-yf0z7-08c3mhDWBgW8cWzVU/edit?usp=sharing
     
    Last edited: Jan 22, 2017
  5. Jan 22, 2017 #4

    PeterDonis

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    Did you measure a full wavelength (sine function going from 0, to 1, to 0, to -1, to 0) or just a half wavelength?
     
  6. Jan 22, 2017 #5
    A full wavelength. Actually, I counted 33 λ over 20 mm, so there is the value I gave you.
     
  7. Jan 22, 2017 #6
    If you are sure those 2 equations are coherent together, I'll just assume I'm the cause of the error and maybe I'll try to talk to my teacher, I just didn't want to look silly in front of him.
     
  8. Jan 22, 2017 #7

    Mentz114

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    If you used software to do the calculations, then you should send a bug report to the makers.:wink:
     
  9. Jan 22, 2017 #8

    PeterDonis

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    I get the same thing from the image at the link. But the image doesn't give the formula that was used to calculate the wave given the electron's velocity. What formula was used?
     
  10. Jan 22, 2017 #9
    I don't know, it is not given. The more velocities we were plugging, the more the distribution was narrowing but no equation was given. It's a lab, we had to "measure" ourselves I suggest.
     
  11. Jan 22, 2017 #10
    Yes, I tend to believe I did the error before doing so, but I'll surely do it if I'm sure there is an error.
     
  12. Jan 23, 2017 #11

    PeterDonis

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    That makes me wonder, since the obvious way to calculate the wavelength from the velocity is by equating the two expressions for momentum, i.e., ##m v = h / \lambda##. So the fact that there is a factor of 2 error makes me suspect that there is an error in the formula somewhere, and the fact that it isn't given makes me suspect that it has not been carefully checked.
     
  13. Jan 23, 2017 #12
    Alrigth, thanks guys \
     
  14. Jan 23, 2017 #13
    Hey, finally it turned out the distribution was squarred, because it's a probability distribution. When you square a sine function, the frequency double, hence the wavelenght becomes the half.
     
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