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Measuring the wavelength and position of a photon

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A wavelength measurement gives Lambda= 6000 x 10^-10m with an accuracy of one part in a million. What is the minimum uncertainty in the position of the photon?


    2. Relevant equations
    deltaP*deltaX > h/4pi

    p = h/lambda


    3. The attempt at a solution
    In order to solve, I know deltaP has to be expressed in terms of deltaLambda. Delta lambda is (6000 x 10^-10)(1 x 10^-6) = 6 x 10^-13.

    delta P = dp/dLambda * deltaLambda

    I'm not sure how to solve this relationship to get an expression for deltaLambda, to plug back into the original

    deltaX > h/(4pi)(deltaP)

    Once I get the expression for momentum in terms of lambda, I think I just sub it back in and solve for deltaX.

    Any help? or have I done anything wrong so far?
     
  2. jcsd
  3. Oct 14, 2007 #2

    Astronuc

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    Staff: Mentor

    Ok - One has [itex]\Delta\lambda[/itex], from which one can obtain [itex]\Delta[/itex]E.

    What is the relationship between E and p for a photon?
     
  4. Oct 14, 2007 #3
    Well, E = hc/[tex]\lambda[/tex]

    From de Broglies wavelength equation,
    [tex]\lambda[/tex] = h/p

    so here [tex]\Delta[/tex]E=[tex]\Delta[/tex]pc

    So to find the uncertainty in position, I have [tex]\Delta[/tex][tex]\lambda[/tex], change that to [tex]\Delta[/tex]E. From there I can get [tex]\Delta[/tex]p, and

    [tex]\Delta[/tex]x = h/4[tex]\pi[/tex][tex]\Delta[/tex]p

    Where [tex]\Delta[/tex]x is the minimum uncertainty in position?

    Does that make sense?
     
  5. Oct 14, 2007 #4

    Astronuc

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    Staff: Mentor

    Yes. That would do it.
     
  6. Oct 14, 2007 #5
    Thanks for the help!
     
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