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De Broglie Wave Energy

  1. Mar 8, 2013 #1
    Problem:

    a. Calculate the energy in eV of an electron with a wavelength of 1 fm.

    b. Make the same calculation for a neutron.

    Solution (so far):

    a. λ=h/p=(hc)/(pc)=(1240 MeV fm)/(pc)=1fm

    so, pc=1240 MeV

    E=√[(pc)^2+E_0^2]

    =√[(1240 MeV)^2+(.511MeV)^2]

    ∴E=1.24 GeV

    This is the same answer as the back of my book, so I'm assuming this is the correct method of solution. However, I do the same thing for the neutron and my answer does not agree.

    b. E=√[(1240 MeV)^2+(940 MeV)^2]

    ∴E=1560 Mev

    My book says the correct answer is 616 MeV.

    I don't see how an energy like that is even possible. Solving the following for pc,

    E^2=(pc)^2+E_0^2

    pc=√[E^2-E_0^2]

    When you plug in the "correct" answer of E=616 MeV you get,

    pc=√[(616 MeV)^2-(940 MeV)^2]

    You certainly cannot take a square root of a negative number and get a meaningful answer. Any suggestions?
     
  2. jcsd
  3. Mar 8, 2013 #2

    rude man

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    Fact: when we talk of a particle with x eV of energy we are talking about its KINETIC energy, not its total energy.

    So - compute the neutron's total energy E, subtract rest energy E_0, which gets you the K.E. , and you get what?
    (Hint: the advertised answer).
     
  4. Mar 8, 2013 #3
    Rude man,

    You will always catch my stupid mistakes, and for that I thank you.

    It hadn't occurred to me that the kinetic energy is what was being asked for.
     
  5. Mar 8, 2013 #4

    rude man

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    Don't worry about it, it's just the convention!
     
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