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The Brogile wavelength of an Alpha particle

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    An alpha particle of energy 5.78 MeV is emitted from a particular nucleus.
    Calculate the de Broglie wavelength of the alpha particle. How does it compare
    with the nuclear diameter, which is known to be approximately 2x10−14 m.


    2. Relevant equations

    E=h/p and E=hf

    3. The attempt at a solution

    I was thinking of using E=hc/lamda but this would be wrong because alpha particles don't travel at c.
     
  2. jcsd
  3. May 23, 2009 #2

    rock.freak667

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    The energy of 5.78MeV is entirely kinetic.
     
  4. May 24, 2009 #3
    so how do i do the question?
     
  5. May 24, 2009 #4

    rock.freak667

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    Use Ek=1/2mv2
     
  6. May 24, 2009 #5
    well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?
     
  7. May 24, 2009 #6

    rock.freak667

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    find v, then use

    [tex]\lambda = \frac{h}{mv}[/tex]

    I am sure you can look up the mass of an alpha particle. I am not sure how else you can solve this problem without my suggestion though.
     
  8. May 27, 2009 #7
    how do you get lambda = h/mv
    i know E = hc/lambda = h/mv
    so, if i combine both equations, shouldnt i suppose to get lambda = cmv ?
    did i miss something or what.
     
  9. May 27, 2009 #8
    p = h/lamda
    lamda = h/p=h/mv

    You can't use E = hc/lamda because alpha particles don't travel at the speed of light.
     
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