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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
    light?

    Known
    (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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