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Emission spectrum

  1. Aug 21, 2010 #1
    Hi all

    I'm unsure about something and so would like to pose the following question.

    When an electron moves to a lower state it emits energy in the form of a photon. But in order to get excited into this higher state it must first receive a minimum amount of energy. Is the energy received originally not always transferred in the form of a photon?
  2. jcsd
  3. Aug 21, 2010 #2
    As I know electron while changing it's states always receives or emits only EM energy (photons)!
  4. Aug 21, 2010 #3
    and not only electron-all the electric charges receives or emits EM(photons) energy while changing it's speed.
  5. Aug 21, 2010 #4
    So if all energy is always transferred via photons why do we see a distinct emission spectrum for hydrogen and not photons in all the other frequencies?
  6. Aug 21, 2010 #5
    it's because of the object you Chose!
    How lasers work?they are getting energy from the light (electrons get to higher states) and when you turn off the light electrons get to lower states by turn-emitting EM waves in the frequencies we see,but there are lot's of objects that emits EM waves in not visual for us frequencies !!
  7. Aug 21, 2010 #6
    It is not.

    In an ordinary neon light for example, the process begins with an electrical discharge between electrodes, creating fast moving electrons that collide with gas atoms.
    The collision transfers energy to the gas atoms, which are left in an excited state. They subsequently decay, emitting UV light. (The UV is converted by a secondary energy transfer with the coating on the tube.)
  8. Aug 21, 2010 #7
    But as I know lasers emits not only UV light Sir!?
  9. Aug 21, 2010 #8
    In the Helium Neon laser, the process again begins with collision of electrons with the Helium gas.
    This excites helium from the ground state to the 23S1 and 21S0 excited states. Further collision of the excited helium atoms with the ground-state neon atoms results in transfer of energy to the neon atoms, exciting neon electrons into the 3s2 level. When that level decays under stimulated emission it emits the familiar laser light.
  10. Aug 21, 2010 #9
    Yes I know the process but read attentively please. I just said that not only UV light emits, but 632,8nm EM wave too which is in the red portion of the visible spectrum.
  11. Aug 21, 2010 #10


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    The initial energy can be anything. In your typical fluorescent lights, you have an electron source (usually via thermionic emission). These electrons are accelerated due to some potential. But because the tube is filled with gas (typically mercury), these electrons collide with these gas molecules and cause them to be excited to a higher energy. So here, the initial energy was in the form of the kinetic energy of the electrons. It doesn't always have to be a photon.

  12. Aug 21, 2010 #11
    Thanks very much for your answers.

    In that case where the source is a photon do we see absorption lines in the spectrum because
    A: The photon that was previously heading towards us is temporarily absorbed by the atom and then emitted in a random direction, causing a drop in the average number of photons reaching us from the source of that frequency.
    B: Sometimes a 2nd amount of photon energy will hit the excited electron before it drops to its base state and move it to a higher orbit, the electron may then drop 2 or more shells and release a higher frequency photon - so even if by pure chance it still heads towards us it will not be of the original frequency?

    In the case of heat energy is it the collisions of the atoms in the source striking against other atoms which transfers energy?

    And finally...

    When two atoms bond (such as h2o) they release energy, from which part of the atom does this energy come?

    Thank you all so much!
  13. Aug 21, 2010 #12


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    Well, on one level, energy is transmitted either through photon emission, or 'non-radiatively' through collisions of atoms, molecules and ions. On another level, the collisions occur through electromagnetic force, which is mediated by photons. So in the end it's just photons really, and the distinction is between real photons and the virtual, force-mediating ones that aren't detected.

    To answer the second question, it's A. When you look at absorption, you're looking at the amount of light lost due to scattering. When you look at an emission spectrum, the detector is at a 90 degree angle to the exciting source, so you only see the scattered light.

    When two atoms bond and release energy, almost all that energy comes from the change in (kinetic/potential) energy of the electrons.
  14. Aug 21, 2010 #13
    When you introduce enough energy which causes a particle collision and a bonding between carbon and two oxygen atoms we essentially have fire. The energy released from the bonding then applies a force to another atom causing that to fire off and collide, and so we have a chain reaction.

    However as these atoms fly around they would sometimes hit other atoms to which they do not bond so this would surely cause less CO2 bonding over time, so what is it that keeps a fire burning as long as you don't run out of oxygen and fuel?
  15. Aug 21, 2010 #14


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    You're going to have to rephrase this. I can't make any sense of it.
  16. Aug 22, 2010 #15
    As long as the temperature remains sufficient, the reaction will continue. Why would there be less bonding over time?
  17. Aug 22, 2010 #16
    Some of the temperature radiates into the air and heats things around the fire (such as me, watching it) so the amount of heat energy within the fire is decreasing. Where does the extra heat energy come from?
  18. Aug 22, 2010 #17
    Franck and Hertz accelerated electrons in a dilute gas of mercury vapor until at about 4.9 volts, the electrons lost a lot of energy in the inelastic collisions with the mercury atoms. The inelastic collisions were exciting the mercury atoms. This won the Nobel Prize in 1925:


    Bob S
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