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De Broglie Wavelenght of a 5.5Mev

  1. Aug 24, 2007 #1
    Calculate the de Broglie wavelenght of a 5.5 Mev alpha particle emitted from Am (241) nucleus, could this particle exist with inside the Am (241) nucleus (diameter = 1.6x10^-14m)?


    the wavelenght:
    Lambda=h/p = (hc) / (sqrt[2mc^2)xeV)
    = 1240 / sqrt(2x3727.38x10^6x5.5x20^6)

    ???
     
  2. jcsd
  3. Aug 24, 2007 #2
    Well...

    The alpha particle is nonrelativistic since KE/mc^2 = 10^-3, so you can use the classical formula for momentum p = sqrt(2mKE).

    Use this momentum to find the deBroglie wavelength.
     
  4. Aug 24, 2007 #3
    Use the uncertainty principle to find out the spatial resolution of a particle with this momentum.

    (hands wave around)

    x ~= h/(4*pi*p)

    Is this x larger than the diameter of the nucleus?
     
    Last edited: Aug 24, 2007
  5. Aug 24, 2007 #4
    I find 6x10^-15 m for lambda, so the diameter of Am is 1.6x10^-14, so the wave length is less than the diameter, is that mean the wavelength can be exist inside the Am nucleus?
     
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