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Uncertainty question

  1. Dec 6, 2005 #1
    In an atom, an electron is confined to a space of roughly 10^-10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty in its momentum?


    delta(x)delta(p)>h (this is how it was derived in the previous parts to the problem)
    Ok so i know px = mvx and m = 9.11X10^-31 kg and that delta px = 10^-10(px). I guess where i am stuck is, what do i do with the velocity term?
     
  2. jcsd
  3. Dec 6, 2005 #2

    Galileo

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    You are not asked about the velocity. You are asked about the uncertainty in momentum. So you got [itex]\Delta x \Delta p \geq \hbar[/itex] (or without the bar, depening on what version your book has).

    So you know [itex]\Delta x[/itex], that was given and you are asked the minimum value [itex]\Delta p[/itex] can take without violating the inequality. You should be able to tell that immediately from the equation. Or least least without reference to other relations.
     
    Last edited: Dec 6, 2005
  4. Dec 6, 2005 #3
    am i missing some or would it be 1.05e-34J/10e-10?
     
  5. Dec 6, 2005 #4

    Galileo

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    That's right. Simply [itex]\Delta p_{min}=\hbar/\Delta x[/itex]. What would you be missing?
     
  6. Dec 6, 2005 #5
    well the answer should be in kg*m/s thats why i feel i am missing something, but its probably just a conversion or something?
     
  7. Dec 6, 2005 #6

    Galileo

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    The answer is in kg m/s, since that is the unit of momentum. The units of h is Js. Since J=Nm=kgm^2/s^2 you can work it out.
     
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