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Wave-Particle Duality

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    At what Kinetic energy will a particle's debroglie wavelength equal its Compton Wavelength


    2. Relevant equations
    DeBroglie
    λ = h/mv

    Compton
    λ = h/mc

    3. The attempt at a solution

    Setting the two equations equal to eachother, I got v = c, then said KE = (1/2)mc^2, but somehow that just doesn't sound right. What do you think?
     
  2. jcsd
  3. Oct 19, 2011 #2

    vela

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    Nope, that's not right, as you suspected. The DeBroglie wavelength is actually [itex]\lambda = h/p[/itex] where p is the momentum of the particle. You need to use the relativistic expression for the kinetic energy to get the correct answer.
     
  4. Oct 19, 2011 #3
    So, it's when p = mc, so would you use E = Sqrt((pc)^2+(mc^2)^2) = sqrt((mc)^2+(mc^2))?
     
  5. Oct 19, 2011 #4

    vela

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    Essentially, yes, but you need to get the algebra right. You have [itex]p = mc[/itex] so [itex]pc = mc^2[/itex] and
    [tex]E=\sqrt{(pc)^2+(mc^2)^2} = \sqrt{(mc^2)^2+(mc^2)^2} = \cdots[/tex]Also, remember E gives the total energy, not the kinetic energy.
     
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