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Heisenbergs uncertainity principle for an electron

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The speed of an electron is measured to within an uncertainty of 2e4 m/s. What is the size of the smallest region of space in which the electron can be confined?


    2. Relevant equations
    Diracs Constant=ΔxΔp
    p=mv

    3. The attempt at a solution
    what I did was (Diracs Constant)/mv = Δx, (1.054560653e-34)/((9.109e-31)(2e4))=6nm this is this incorrect but I dont understand why?
     
  2. jcsd
  3. Sep 27, 2012 #2
    Hi,
    I am not sure, I may be wrong.
    Use Heisenberg uncertainty principle:
    [tex]\Delta x \Delta p ≥ \hbar/2.
    [/tex]
    The given value is not the speed of electron. It is the uncertainty of speed.
    Got it?
     
  4. Sep 27, 2012 #3
    Hi. Not value but only order of value has meaning in this situation. Order of nano meter seems fine. What is the 'correct' answer you have got?
     
  5. Sep 27, 2012 #4
    Sweet springs the correct answer is 5.8nm what I get is 6nm I know it's so close but it's not what I'm getting. Rajini given the uncertainty of speed how then would I find speed go the electron so I can find p.
     
  6. Sep 27, 2012 #5
    Ok I'm starting to think I got the right answer after using wolfram I get 5.78 I am starting to think my calculator has some kind of rounding algorithm
     
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