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Which Energy does E signify in E=hf.

  1. Nov 9, 2009 #1
    We have the famous Einstein's Relation E = hf.
    He originally used this relation for photons of light only.
    But de-broglie expanded the concept and postulated that every matter exhibits a wave phenomenum (called matter wave) which satisfies these two relations
    1.[tex]\lambda[/tex] = h / P
    2. E = hf
    It is by using these two relations, we have derived schrodinger's Equation (concepts of modern physics, Arthur Bieser, 6th edition).
    What I am wondering is, is that Energy E, only the kinetic Energy of the particle or does it also takes into account its rest mass energy m0c2
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2

    Dale

    Staff: Mentor

    I think it is the total energy including both rest and kinetic. Otherwise you could probably detect the waving of large objects like baseballs.
     
  4. Nov 9, 2009 #3

    Born2bwire

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    Gold Member

    As I recall Schroedinger's equation does not incorporate the rest mass, m_0c, because it is non-relativistic. Its energy values are the same as the classical analogues, the kinetic and potential energies. If you use a relativistic wave equation, like the Klein-Gordon or Dirac, then you will get the relativistic energy form. This can be most explicitly seen in comparing the energies of the Schroedinger and Kein-Gordon for the free particle.

    For deBroglie, it may be instructive to see what form the energy has if we were to use the relativistic momentum in the calculation and compare the resulting energy to the Klein-Gordon energy (or whatever equivalent relativistic energy) of a free particle. As I recall, the deBroglie relation is correct for non-relativistic free particles. I do not know if we can extend it to relativistic particles.
     
  5. Nov 10, 2009 #4
    Even for non-relativistic cases (i.e. v <<c ), we can't deny the presence of rest mass energy. Its present even for 0 velocity. If we are going to change an electron to a wave, thereby completely vanishing its particle existence, then I don't see any reason why you could ignore its rest mass energy.

    if any theory considers the total energy is just its kinetic energy (excluding the rest-mass energy), in the light of relativity, that theory is simply wrong.
     
  6. Nov 10, 2009 #5
    E = hf is not Einstein's but Plank's.

    De Broglie wave-length does not include the rest mass. It is a wave of a free particle. The electron diffraction corresponds to the De Broglie wave: lambda = h/mv.
     
  7. Nov 10, 2009 #6

    Born2bwire

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    No, we can certainly ignore the rest mass for non-relativistic systems. The entire success of Newtonian physics is based upon this. Schroedinger's equation is non-relativistic, it does not account for rest mass at all. You need to move to a relativistic form like Klein-Gordon, Dirac, or better yet quantum field theory. The rest mass is not a real factor for many problems because we are mainly interested in the change in kinetic energy. Many situations we can apply an arbitrary constant to the kinetic energy, which is what the rest mass will do, and still get the right physics back out. The greatest advantage of a relativistic quantum theory is the incorporation of the equivalence principle. We can now talk about the creation and annihilation of particles from the energy contained in fields.
     
  8. Nov 10, 2009 #7
    I am sorry for that mis-reference.

    If it doesn't include rest-mass, how much correct am I to use it in Schroedinger's Equation.
     
  9. Nov 10, 2009 #8
    You are completely correct to use it in the Schroedinger equation. All wave functions - bound states and plane waves are based on this relationship.
     
  10. Nov 10, 2009 #9
    Here is the question I came across, bob.
    Q. What is the Kinetic Energy of the electron whose de-Broglie wavelength [tex]\lambda[/tex] is 2E-12 m?
    So, I solved this this way, since you said E is only kinetic,
    1. p = h / [tex]\lambda[/tex]
    2.E = K.E. = p2 / 2m

    But what Arthur Beiser does in his book (concept of modern physics, sixth edition, Pg.103) is quite different:
    1. p = h / [tex]\lambda[/tex]
    2. K.E. = E - E0 = sqrt( E02 + (pc)2) - E0

    He got an answer of 292 Kev but I got 376 Kev.
    So, Who is correct???(Although I am almost certain the big Arthur is correct compared to a little me; I am simply asking to know, why?)
     
    Last edited: Nov 10, 2009
  11. Nov 10, 2009 #10
    When the electron (or neutron) is non-relativistic, both formulae give the same KE. By the way, non-relativistic neutrons (ultra cold neutrons) follow the formula lambda=h/p.

    When the electron is relativistic, the KE is expressed via p differently, as in the book. The electron rest mass is about 512 KeV, so your electron approaches relativistic velocities.
     
  12. Nov 10, 2009 #11
    I am extremely sorry for my lack of quick understanding ability; but I need to ask this final question as a check.
    Now, I am asked: The de-borglie frequency of an electron is 1.5 E 20 Hz, what is its kinetic Energy?
    Should I do,
    A.
    1. E = K.E. = hf = Ans

    OR
    B.
    1. E = hf
    2. K.E. = E - E0 = Ans.
    (I think my text book would prefer doing B, just like in my previous post)
     
  13. Nov 11, 2009 #12
    It Seems that I asked (childishly)more questions than was morally allowed. Sorry for bothering, bob.
     
  14. Nov 11, 2009 #13
    E=hf only for photons.

    KE=KE(p) for a particle. lambda=h/p for a particle. Find p, express f via labmda and use KE(p) in the relativistic formulation. If p is small, you will obtain the numerical value close to KE = p2/2. If p is relativistic, you will obtain relativistic KE (see the book formula).

    P.S. I did not understand what de Broglie frequency mean. If it is the factor in KE = hf, then you have the answer (A).
     
  15. Nov 11, 2009 #14
    The de-broglie matter-waves have wavelength lambda= h / p, and frequency f = v / lambda.
    Thank you for all the help you have provided.
    So, I am walking away with this knowledge
    What's the Energy of a particle whose de-broglie frequency is 2E20 Hz and velocity is 2E8. m/s?
    Ans:
    A. 1. E = hf = Ans (wrong, we can't use E = hf for electrons)
    Right Answer:
    B. 1. first, lambda = v / f.
    2. p = h / lambda
    3. K.E. = sqrt((m0c2)2 + (pc)2) - m0c2
    4. E = E0 + K.E. = Ans

    Thank you once again for all the help.
     
    Last edited: Nov 11, 2009
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