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Wave properties of matter

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the statement below made by a student: "Muons have a higher mass than electrons, but because the energy, E, is related to the wavelength by E=hc/[tex]\lambda[/tex], muons that have the same kinetic energy as electrons will also have the same wavelength."

    Do you agree or disagree with this statement? Explain your reasoning.

    2. Relevant equations

    [tex]\lambda=h/p[/tex] (de Broglie wavelength)

    3. The attempt at a solution

    The statement seems wrong to me. If you substitute in for [tex]\lambda[/tex] in the first equation, you get cp, but kinetic energy is p^2/(2m) and those two can't be the same (solving for c gives c=v/2).

    I'd never seen the first equation before, but looking in my textbook it looks like E is the change in energy of an atom when a photon is absorbed or emitted and I don't know how you could apply it to an electron/muon (can you?)
     
  2. jcsd
  3. May 18, 2008 #2

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    [itex]\lambda[/itex] is different for different objects, so why should E be the same?

    E is the total relativistic energy, not p^2/2m. Also, E = hf, where f is the frequency associated with the de Broglie wave.

    Just do the simple math. Also, reading up on de Broglie wavelength would be a good idea.

    EDIT:
    -----

    Are you talking about the muons and the electrons having the same energy, by any chance? This is not mentioned in the problem, so I assumed not.

    Also, [itex]E=hc/\lambda[/itex] is valid only for massless particles which travel at speed c.
     
    Last edited: May 18, 2008
  4. May 18, 2008 #3
    >[itex]\lambda[/itex] is different for different objects, so why should E be the same?
    The question is assuming you have two particles with the same kinetic energy.

    >For high speed particles, E is the total relativistic energy, not p^2/2m.
    Ok, but I still doubt that cp=Ek for an electron. Is that wrong?

    >Also, E = hf, where f is the frequency associated with the de Broglie wave.
    And if f=v/[tex]\lambda[/tex], then E = vp. But v can't be c for an electron, so the equation can't work here (?)
     
  5. May 18, 2008 #4
    Sorry, you posted while I was replying. Yes, that's the problem:
    "muons that have the same kinetic energy as electrons..."

    >Also, is valid only for massless particles which travel at speed c.

    I'm guessing this is essentially the answer to the question
     
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