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B About Laser Production

  1. Oct 30, 2017 #1
    I want to ask several questions regarding laser production:

    1. To produce laser, electron needed to be moved from ground state to excited state. Does ground state mean the innermost shell (n = 1) and excited state mean any other shells (n = 2, n = 3, etc)? Can laser be produced if electron is moved from, let say, n = 3 to n = 5 (in certain atom)? Can we use any atoms to produce laser?

    2. Let assume the electron moves from n = 1 to n = 2. Spontaneous emission occurs when this electron goes back to n = 1 naturally. Stimulated emission occurs when this electron is in metastable state and incoming photon triggers this electron to de-excites.
    What is metastable state? Is this only the term used to say that the electron stays in n = 2 longer than it should be naturally or maybe there is certain "place" in n = 2 that refers to metastable state? Does all atom have this metastable state?

    3. Photon that triggers stimulated emission is not absorb by the electron, so how can it trigger the electron to de-excite? Does it hit the electron (just like when photon hits electron in n = 1 and electron absorbs energy from the photon to move to n = 2)?

    4. Let say the energy difference between n = 1 and n = 2 is 10 eV. Can stimulated emission occur when photon having energy 20 eV tries to trigger the electron?

    5. How can the two photons of laser have same direction?

    Thanks
     
  2. jcsd
  3. Oct 31, 2017 #2

    mfb

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    Lasers are typically solid state devices or at least molecules where the shell concept doesn't apply to the relevant energy levels. Ground state is simply a state that is normally occupied, and excited state is a state that is normally not occupied.
    For practical reasons, the spontaneous transition from the excited state to the ground state shouldn't be too fast. You don't want this spontaneous transition. If the excited state is long-living, it is called metastable.
    I don't think classical analogies work here. You can calculate that it works in quantum mechanics.
    No. At least not if you want a proper laser.
    Where do you see an issue with that?
     
  4. Nov 1, 2017 #3
    So it is not correct if I say n = 1 is the shell consists of max 2 electrons and n = 2 is the shell consists of max 8 electrons and electron is excited from n = 1 to n = 2?

    Why can metastable state occur? Why can there be an exicted state that is short-living and long-living?

    I am not going into formula right now. I just want to understand it qualitatively or maybe intuitively. Usually photon hits electron and the electron absorbs that photon energy then moves to higher energy level corresponds to energy absorbed. If when stimulated emission happens the photon is not absorbed, what does actually happen? How can the photon trigger stimulated emission without transferring its energy to electron?

    I don't understand why the two photons are always in same direction. Why don't they move in opposite direction or making certain angle towards each other?

    Thanks
     
  5. Nov 1, 2017 #4

    Drakkith

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    No. Most lasers do not work this way. Solid state lasers, for example, work by exciting electrons in a doped semiconductor whose doping is chosen to provide specific energy bandgaps. Electrons fall from a state at the top of the bandgap to the ground state, releasing a photon in the process.

    Since solid state lasers deal with electrons within a bulk material, their energy levels aren't atomic shells. Instead, the energy levels arise from the combined bonds formed between the huge number of atoms.

    Other lasers work slightly differently, but most involve transitions between energy states in molecules, not single atoms.

    If you're not willing to go into formulas and math then I'm not sure there's any way to explain it.

    Again, if you're not willing to get into the details of the math then I don't think you can hope to understand it. The best you can do is accept that it happens.

    Some things can't be understood without learning the math, and I think atomic and subatomic physics is one of those things.
     
    Last edited: Nov 1, 2017
  6. Nov 1, 2017 #5
    I haven't learnt any formula related to this. Can you tell me the formula or maybe post a link so that I can look it up first before asking question here. I really don't have idea where to start.

    And this includes the reason why the two photons of laser always have same direction?

    Thanks
     
  7. Nov 1, 2017 #6

    Drakkith

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    I'm afraid I don't have any good links for you. The only thing I can suggest is to pick up a book on lasers or semiconductor physics that's in-depth enough to cover the math. MFB or someone else may have so good links though.
     
  8. Nov 1, 2017 #7

    Henryk

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    Interaction between the photon and the electron is via "dielectric dipole moment of the transition". There is a quantum formula to calculate this. The key is the initial state of the electron before interaction with the photon. If the electron is in the lower state, the photon gets absorbed. If the electron is in the upper state, then, you have stimulated emission. Actually, the probabilities for both processes are exactly the same. Therefore, for laser operation, you need population inversion: that is more electrons in the upper state than in the lower state. Only in this case you get net amplification of the light - laser operation.

    The key is to get the population inversion. The first laser (ruby) actually used 3 level system. The actual energy levels are the levels of chromium ions: dopant in the crystal lattice of aluminium oxide. The electron was excited to a third level, then spontaneously decayed to a slightly lower metastable level. That's how the population inversion was achieved.
    Since then, other ways of achieving population inversion were invented. Semiconductor laser works by injecting electrons and holes across a p-n junction.
    Dye laser work by pumping dye molecule with high intensity light, etc.

    This is by the basics of the emission: the emitted photon is exactly the same as the incident one, that is same energy, same direction.

    The reason the photo is not absorbed is that the electron is already in the upper energy state, hence it cannot absorb the photon. It boils down to having the population inversion.
     
  9. Nov 3, 2017 #8
    Thank you very much for the help
     
  10. Dec 17, 2017 at 12:04 AM #9
    In the standard texts, the interaction of atoms with electromagnetic radiation is usually done using perturbation theory on an oscillating field (aka the dipole approximation). This leads to "selection rules" where electrons can only make transitions to other states if its angular momentum quantum number, l, changes by 1. Such was derived in Sakurai's textbook using an analysis on the properties of parity (space inversion operation, x->-x) of the perturbation.

    You can put electrons in a metastable state by pumping it from, say an n=1, l=0 state to an n=3, l=1 state. It will spontaneously decay into the n=2, l=0 state (the metastable state), after which it is stuck, because there is no l=1 state for n=0. However, the "selection rules" arise only from the previously mentioned simplified model, which doesn't take into consideration, say, the weak interaction, which violates parity invariance. The weak interaction will actually allow electrons in the n=2, l=0 state to decay into the n=1, l=0 state (violating the selection rules), but at a larger time scale. Hence, the term "metastable." It is stable in the atomic time scale, but not truly stable universally.
     
    Last edited: Dec 17, 2017 at 12:12 AM
  11. Dec 17, 2017 at 12:12 AM #10

    mfb

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    The weak interaction has nothing to do with this. It is way too weak to be relevant in electronic transitions.

    The photon can carry orbital angular momentum or the process can happen via the emission of more than one photon. The decay via a different state (like n=2, l=1) is possible as well.
     
  12. Dec 17, 2017 at 12:19 AM #11
    How would an electron decay directly from an n=2,l=0 to an n=2,l=1 state if the principle quantum number stays the same (implying no energy change), unless we are also considering fine/hyperfine structure transitions?
     
  13. Dec 17, 2017 at 12:47 AM #12

    mfb

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    Fine structure is the key point. The energies are not the same, due to the Lamb-shift l=0 has a slighty higher energy.
     
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