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Heisenberg's uncertainty principle question

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    The position and momentum of a 1.00 KeV electron are simultaneously determined. If the position is located to within 0.100 nm, what is the percentage of uncertainty in its momentum?

    (Arthur Beiser - Concepts of modern Physics, 3rd Part exercise 33)


    2. Relevant equations
    [tex]Δx\cdotΔp≥\frac{\hbar}{2}[/tex]


    3. The attempt at a solution

    [tex]Δx\cdotΔp≥\frac{\hbar}{2}\rightarrow \; Δp≥ \frac{\hbar}{2\cdotΔx}[/tex]
    Given that Δx=0,100 nm, we can evaluate the uncertainty in its momentum Δp.
    The momentum of the particle should be of roughly equal magnitude, and evaluating the fraction [tex]\frac{Δp}{p}%[/tex] gives me 98%, where the answer is 3,1 %. What did I do wrong?
     
  2. jcsd
  3. May 30, 2012 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    What did you plug for p?

    It should be p=1[Kev] /c where c is the speed of light.
     
  4. May 30, 2012 #3
    Yep.

    [tex]K=\sqrt{m_{0}^{2}c^{4}+p^{2}c^{2}}≈pc[/tex]
    so
    [tex]p=\frac{1,6\cdot10^{-16} J}{3\cdot10^{8} \frac{m}{s}}=0,533 \cdot 10^{-24} kg\cdot\frac{m}{s}[/tex]
    and because [tex]Δp=\frac{1,05\cdot10^{-34}J\cdot s}{2\cdot 10^{-10}m}=0,525\cdot 10^{-24} kg\cdot \frac{m}{s} [/tex]
    we get
    [tex]\frac{Δp}{p}≈0,98[/tex]

    :(
     
  5. Jun 3, 2012 #4
    Could I get a bumpity-bump 'cause these exercises are crucial for the finals?
     
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