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Why is the Emission Spectrum the same every time?

  1. Dec 31, 2005 #1
    I'm confused

    So the atoms becaome excited by absorbing energy, like from a hot flame, so this energy makes an electron (or is it all electrons) in the outer shell (or is it all the shells) move to a higher energy level and when the electron(s) return to its(their) ground state, they give off energy in the form of electromagnetic radiation, which scientists use to ID elements. So the spectrum of an element must always be the same if they can use it like a finger print for that element.

    The thing that i don't get is the thing that makes the spectrum the same each time. i mean if you have an atom, and you give it energy, and the electron in the outer shell jumps to a higher shell, if you keep giving energy won't it just keep going higher? and if it can go to different shell won't it give out different wavelengths of light as it jumps back, since the distance of the jump back is different? what about loosing the electron after it has enough energy?

    I googled it but i keep getting the statement that the "Each atom's atomic emission spectrum is unique" not the reason why its always the same
     
  2. jcsd
  3. Dec 31, 2005 #2
    Happy New Year
     
  4. Dec 31, 2005 #3

    James R

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    No, because the electron only absorbs energy in quantum "chunks". For absorption to occur, the energy has to be just right. And once an electron is at a higher energy level, it only stays there for a short time before falling back to a lower level again, releasing light. The chances of it being excited to an even higher level in the available time are quite small under most conditions.

    Yes, it will. But bear in mind that you're usually not exciting one atom but millions and billions. Each atom does its own thing, so that different atoms emit different frequencies of light. In combination, all of the characteristic emission lines of the element are covered.

    That can happen. In that case, the atom becomes ionised.
     
  5. Dec 31, 2005 #4

    GCT

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    atoms have distinct emission spectrums and the energy levels involve uv/vis portion of the spectrum. The electronic transitions are more or less probable under different conditions. Molecules may be IR active due to vibrational and rotational levels, the result is bands (due to limits of instrumentation as well as the closeness/overlaps in spectrum) rather than the linewidths seen with atomic spectra. Despite the complexity, each distinct atom/molecule, have probabilities associated with such transitions, and the spectrum is often unique (depending on the method of analysis), so there's no sense in wondering whether such additional contributions to the spectrum, as you mentioned, will completely obscure uniqueness.
     
  6. Dec 31, 2005 #5

    GCT

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    The peak wavelengths will be the same each time when taking into account the natural linewidth-without instrumental and environmental deviations. The actual spectrum however, will not be exactly the same, in fact one will need to take into account several factors which may have contributed to a wavelength shift such as complexities associated with sample preparation. The intensities will differ according to the precision and accuracy of the method....the key here though to "uniqueness" is in reference to the intrinsic electronic energy levels and the theoretical energy required for such transitions.
     
  7. Jan 12, 2006 #6
    Thanks its a lot clearer in my mind

    But it still doesn't quite link in with everything else i know

    As far as i know atoms are transparent, so how could light be reflected from them? like a leaf, we see it as green because green light is reflected from it, and some gets absorbed and some goes through. So where does it get absorbed? how is it reflected? what determines what gets reflected and what doesn't?
     
  8. Jan 12, 2006 #7

    Gokul43201

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    To help picture the scenario in the original question, think of each different atom as a ladder with a specific set of spacings between the rungs on it. Different ladders have different arrangements of rungs. So the set of numbers representing all the distances between rungs is unique to a given ladder. In much the same way, the energy differences between different (not necessarily adjacent) levels is a set of numbers unique to a particular atom.

    'Transparency', 'absorption' and 'reflection' are not terms that you can use in the context of single atoms. They refer to phenomena involving photons and a large number of atoms (specifically, their electrons). Absorption happens when the energy of a photon is use to excite a mode of atomic (or molecular) vibrations. The energies (frequencies) that are absorbed are the energies of the allowed vibrational modes of the atoms/molecules. The energies (or frequencies or wavelengths or colors) that are not absorbed are typically reflected.
     
  9. Jan 12, 2006 #8
    So the atoms of the red piece of paper i'm holding absorb all light apart from red? but it can't since there are specific energies that can be absorbed, unless the paper has paint on it which is made of a combination of atoms which absorb all the colours apart from red, which sound highly unlikely, if they did absorb all these wavelengths wouldn't it start glowing as the electrons start returning to their ground shells? unless red it the colour which is being absorbed and given out again and the red being given out somehow has a greater intencity than the rest of the light makeing it look red, but then if it was analised the full spectrum would be there, wouldn't it? or do these atoms convert somehow change the blue light to red by changeing the frequency or wavelength, but i doubt that, the absorbed energy would have to be given out at some time or another making it glow

    But it still doesn't explain why its reflected, what makes that photon, change directon, oh and if it is changeing direction wouldn't it have to at one point be stationary? kind of like bounceing a ball off the floor
     
  10. Jan 12, 2006 #9
    The changing directions part is handled by conservation of momentum. The photon is not stationary at any point. It is absorbed ("destroyed") and thus excites an atom. The excited states have finite lifetimes and thus a new photon is emitted ("created") so that conservation of energy and momentum are satisfied.
     
  11. Jan 12, 2006 #10

    Gokul43201

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    Something (your piece of paper, say) looks red because more of the higher frequencies (yellow, green, blue,...) are absorbed by the material and converted into heat (vibrations). This doesn't mean that all frequencies but red are absorbed - only that there is less red light absorbed than other frequencies. If you analyze the spectral composition of the "red" light that is reflected by your piece of paper, you'll most likely find it has some general distribution of wavelengths, but those near red are the most intense (the reflection spectrum has a maximum near red).

    Secondly, recall what I said about absorption. It has to do with exciting molecular vibrational modes, not electronic modes. When electronic modes are excited, you have fluorescence or phosphorescence.
     
  12. Jan 12, 2006 #11

    ok, there are three posible situations with light and matter:
    1) they dont interact with each other - this happens when the energy of the photons doesnt correspond to the energy gaps of the matter.
    in this situation, the light simply passes through the matter.
    2) the photon's energy correspond to phonon (in the case of solid state) or molecular vibration\rotation (in liquid\gas states) energy of the matter.
    in this case the matter absorbs light and heats up.
    3) the photon's energy correspond to the energy gaps in the atom, exciting it.
    in this case the atom will re-emmit the photon in a random direction after a short period of time - and you can call it reflection (because its re-emmited in random direction and phase, the reflection wont be mirror-like, just like you see with your red paper)

    although the spectrum is descrete, it can be very dense at some ranges, so it would seem that only red is not absorbed by the red paper.
     
    Last edited: Jan 12, 2006
  13. Jan 12, 2006 #12

    GCT

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    I just took I. analysis and I think I remember a special case where the relatively faster rate of internal conversions prevent fluorescence/phosphorescence almost completely. Besides aren't there electronic transitions that are do not correspond to the visible range?
     
  14. Jan 12, 2006 #13

    GCT

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    Yeah, I remember that if there are many superimposed energy levels, one will observe no fluorescence/phosphorescence at all due to eventual internal conversion to ground state. I'm interested to know more of the details of your view regarding absorption applied to electronic transitions.

    Or is it that I misunderstood your meaning by "electronic modes?"
     
  15. Jan 13, 2006 #14

    Gokul43201

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    Yes, it is possible to have absorption as a result of an electronic excitation, followed by internal conversion. Ignoring the mechanism of the process, the net effect is that much (and not necessarily all) of the photon's energy went into exciting a vibrational (I use this term generally to include rotations as well) mode.

    In general, it is possible to relax to the ground state through the numerous possible combinations of electronic and vibrational (both individual and collective modes) relaxations. In the long-term though, the energy pumped into the system must be released as radiation. (Even vibrational modes eventually will decay radiatively - this is nothing but cooling.)

    Relaxations that are mostly electronic (ie : fluorescence, phosphorescence) are, relatively speaking, rare. And yes, you can very well have fluorescence at energies higher or lower than the visible range. There's also the extremely low probability (at typical excitation intensities) of fluorescing at frequencies higher than the exciting frequency.

    Naturally, it's much easier to analyze and calculate probability amplitudes for the various relaxation modes for a simple, pure system. For something like paper though, it's a lot easier to just do the experiment !
     
  16. Jan 13, 2006 #15

    GCT

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    are you talking about Raman, anti-stokes?
     
  17. Jan 13, 2006 #16

    GCT

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    Actually, Raman is in reference to virtual states (different phenomena completely), your "antistokes fluorescence" then I suppose is in reference to excitation from a vibrational/rotational level to a higher electronic state(because a vibrational mode would result in relaxation) , then emission towards the ground state or any of the lower vibrational states....I guess this is what you're referring to.
     
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