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Homework Help: Percent uncertainty of momentum

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    The position and momentum of a 1KeV electron are simultaneously determined. If its position is located to within 1Angstrom (Δx), what is the PERCENTAGE of uncertainty in its momentum?


    2. Relevant equations
    ΔxΔp >= h(bar)/2


    3. The attempt at a solution
    Δx = 1A = 1x10^-10m
    Δp>= h(bar)/2Δx
    Δp = 1.055x10^-34(kg/m^2)/s
    -------------------------
    2(1x10^-10m)
    = 5.275x10^-25 kgm/s

    how do I get the percentage of uncertainty?
     
  2. jcsd
  3. Mar 2, 2008 #2

    Shooting Star

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    Homework Helper

    ([itex]\Delta[/itex]p/p)100.

    Use the energy to find p.
     
  4. Mar 3, 2008 #3
    Thanks!

    Here's what I did...
    (I don't know how to use latex =/ )

    KE = 1/2 mv^2
    1 x10^3 eV = 1.602 x10^-16 J
    1.602 x10^-16 J = 1/2 (9.1 x10^-31 kg) (v^2)
    Sqrt[(2 (1.602 x10^-16 J))/(9.1 x10^-31 kg)] = v
    v = 1.88 x10^7 m/s
    p = mv
    p = (9.1 x10^-31 kg) (1.88 x10^7 m/s)
    p = 1.708 x10^-23 Kgm/s
    Δp/p = .031 = 3.1 %

    or click the attachment=)
     

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