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Frequency of the matter wave

  1. Mar 2, 2013 #1
    in my book ,their is a equation

    f= E/h

    what is the E means?
    Total energy????
    Does it includes the rest energy ?
     
  2. jcsd
  3. Mar 2, 2013 #2

    ZapperZ

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    This is a good place for you to learn a bit more on how we do things on here.

    Never, ever simply quote an equation without proper context.

    For example, under what type of discussion was this equation used? Was it trying to explain the energy of photons? Or was it for some other purposes? Have you seen this equation before with regards to photon energy? Etc... etc.... Can you see why simply dumping that equation out of nowhere provides us very little to go on with regards to what YOU are seeing in your text?

    Zz.
     
  4. Mar 2, 2013 #3
    It is not the total energy since it does not include a potential energy term. However it does incluideit does include kinetic energy.

    If, h = Planck's constant and f =frequency of matter wave, then

    E = "inertial energy" = Kinetic Energy + Rest Energy
     
  5. Mar 3, 2013 #4

    Bill_K

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    :confused: Of course it includes both kinetic and potential energy. A stationary state for a particle in a potential well is described by ψ ~ exp(iEt/ħ) where E is an eigenvalue of H = p2/2m + V(x).
     
  6. Mar 3, 2013 #5
  7. Mar 4, 2013 #6
    if the terms includes rest energy
    i will not understand the Schrodinger equation
    the schrodinger equ
    if the terms includes rest energy i will not understand the Schrodinger equation the schrodinger equ for free particle
    [PLAIN]http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/quantum/imgqua/seq1.gif [Broken]
    for free particle U=0 and the second term is eliminated
    and
    http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/quantum/imgqua/seq2.gif [Broken]
    sub ψ into the schrodinger equ for free particle
    and take the derivative
    we got
    (ħ^2)(k^2) / 2m = ħw
    if E= hw = rel. energy
    why can i get (ħ^2)(k^2) / 2m = ħw?
    i mean, (ħ^2)(k^2) / 2m = p^2/2m where p is the momentum
    and E will be equal to non-rel. KE energy only , but there is no rest energy included!
    Plz help =(

    STEPS are included


    reference: Modern physics second edition , international ed, RANDY HARRIS p.110 - 112
     

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  8. Mar 4, 2013 #7

    Actually, i originally think that it refers to KE+rest energy , ie rel. energy
    but what i get from the Schrodinger equation for free particles tells that E= non-rel. KE only

    you can see the reply for more details. Plz help me, i spend a week to find the ans but i still can figure it out
     
  9. Dec 13, 2015 #8
    Isn't the equation E= ħ∨ used for photon

    In my knowledge ,
    we first calculate λ from λ= ħ/p , then use equation v=λ∨ ==> λ=v/∨
    ==>
    v/∨=ħ/p
    ==>
    ∨=pv/h
    ==>
    ∨=mv2/h
     
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