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Homework Help: Heisenberg indeterminacy principle

  1. Nov 28, 2011 #1
    No object can travel faster than the speed of light, so it
    would appear evident that the uncertainty in the speed of
    any object is at most 3 * 10^8 m s
    (a) What is the minimum uncertainty in the position
    of an electron, given that we know nothing about
    its speed except that it is slower than the speed of

    (Δx)(Δp)=h/4∏ hence Δx=h/4∏(mass of e-)(velocity)
    Mass of e-=9.10x10^-31 kg
    Velocity is equal to 3x10^8 m s
    I plug in my values and came up with 3.049 x10^-59meters
    My answer seems a bit wonky..could you double check..and if wrong give me hints to solve properly...thank you
  2. jcsd
  3. Dec 2, 2011 #2
    There is a relativistic version of the uncertainty principle, ΔxΔv >= m2/(2* E3) where we are using units such that c = h-bar = 1. E = √p2+m2. where m is the rest mass and p is the relativistic momentum p = λmv. p can rise unbounded as v → c and thus so can E. Δv = c = 1, so Δx >= 0. The minimum uncertainty is 0. Or so I'd like to think.
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