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Electromagnetic radiation

  1. Apr 19, 2008 #1
    A question that I reckon will be fairly simple for you guys. Electromagnetic radiation originates from electrons going down different energy levels, correct? But then what determines the frequency and wavelength of that radiation?
     
  2. jcsd
  3. Apr 19, 2008 #2
    I could be completely off-base, but I'm inclined to imagine the orbit of the affected electron as a disk rolling away from the atom in all directions simultaneously at the speed of light. The wavelength is then the speed of light divided by the period of the affected orbit .

    Regards,

    Bill
     
  4. Apr 19, 2008 #3

    ZapperZ

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    This is not entirely accurate.

    While you do get EM radiation upon an atomic transition, this isn't the only means to create it. Any accelerating charge can produce the same thing. A charge oscillating produces EM radiation as well.

    In an atomic transition, the "frequency" f corresponds to the energy of the photon, i.e. E=hf. When you create EM radiation via accelerating charges, the frequency corresponds to the cyclic frequency of that charge.

    Zz.
     
  5. Apr 19, 2008 #4
    Thanks a ton, are these three (atomic transition, accelerating charge and oscillating charge) all possible 'creators' (excuse my English) of EM radiation or are there even more?

    Kind regards,

    Erwin
     
  6. Apr 19, 2008 #5
    I agree.

    An electron orbiting an atom might as well be an oscillating charge to a distant observer.

    With respect to atomic transition, does the energy of the photon not necessarily equal the change in energy of the orbiting electron??

    Regards,

    Bill
     
  7. Apr 21, 2008 #6
    yes but an electron doesn't draw any particular orbit around the nucleus in an atom actually, it is rather "moving around" than anything else. Moreover, an atomic transition (for an non hydrogenoid atom) changes the whole electronic configuration and this is quite complicated to discribe with simple "images" in reality.
     
  8. Apr 21, 2008 #7
    What do you mean by "moving around"?

    The electron orbits each have their own period, and re-orient (e.g. "move around" on their shell) to maintain the equilibrium of forces over time.

    Yes - for an atom with more than one electron, the loss of one electron affects what passes for a state of equilibrium among all that remain.

    Regards,

    Bill
     
  9. Apr 21, 2008 #8

    ZapperZ

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    But this isn't your ordinary "orbit" as in the classical sense, because that would cause it to radiate in a stable orbit. We see no such thing in the ground state of an atom. So already such classical picture doesn't work.

    There is a FAQ on this in this General Physics sub-forum.

    Zz.
     
  10. Apr 21, 2008 #9
    I didn't mean to imply that it was. Surely, the intersection of a cone and an electron shell would not fit the classical sense.

    I didn't say the orbits were "stable" either - I claimed that they re-orient with respect to time. In this scenario, the charge need not apear at two points on a line every half period (oscillate).

    Regards,

    Bill
     
  11. Apr 21, 2008 #10

    ZapperZ

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    I have no idea what you just said here.

    Recall that you responded to my "oscillating charge" post by writing this:

    So you certainly were equating an atomic orbital with an oscillating, or accelerating charge, which is incorrect.

    If you have a published physics paper that is describing what you just wrote here, please post it. Otherwise, I've never heard of such a thing and it is purely speculative. And I believe you know already what that implies.

    Zz.
     
    Last edited: Apr 21, 2008
  12. Apr 21, 2008 #11
    As far as I know there is no evidence that such shells exist, this is much more the contrary actually.
    In principle the electron in the ground state for example can be found in the whole volume around the center of the nucleus with a given probability, I'm very curious to know how you retrieve this result within your "shell theory".
     
  13. Apr 21, 2008 #12

    Andy Resnick

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    There are lots of ways to generate radation: transitions between (electronic, vibronic, rotational) bound states produces emission with a discrete spectrum. Accelerating charged particles produces continuum radiation (oscillating is a form of acceleration). Scattering also produces continuum radiation.
     
  14. Apr 21, 2008 #13

    Andy Resnick

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    I think you are using the concept of electron orbitals a little too literally.
     
  15. Apr 21, 2008 #14
    I've read the whole thread and I'm surprised no one has given what I would consider
    a clear answer. The Schroedinger picture of the hydrogen gives a very clear picture of the energy level transition in which the production of radiation is accoreding to Maxwell's equations.

    Look at the transition from the 2p (excited) to the 1s (ground) state. Each of those states has a charge distribution (or "cloud") which is stationary with time. But the phase of each charge cloud has a complex frequency, different for each state. When we picture the atom in a mixture of those two states, the phases interfere with a beat frequency equal to the difference of the two frequencies. As a result, the location of the charge cloud oscillates about a central plane. As a small dipole antenna, we would classically calculate that it radiates energy at a certain power level. So it would take a certain amount of time to shed the excess energy from the exicted state to the ground state. In fact, this is the same time spoken of in quantum mechanics as the lifetime of the excited state; or from another point of view, it tells us the spectral linewidth for that transition.

    I don't believe there are any instances of radiation in quantum mechanics that can't ultimately be analyzed as a case of oscillating charge (or magnetic moment).
     
  16. Apr 21, 2008 #15
    It is simply my interpretation of an electron with a precessional orbit, and how it might apply to what we were discussing.

    However, since you seem to be of the opinion that I am breeching forum guidelines, I will cease contributing to this thread to avoid further transgression.

    Regards,

    Bill
     
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