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Neutral hydrogen and Bremsstrahlung

  1. Apr 18, 2004 #1

    hellfire

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    An electron which is decelerated emits a photon (Bremsstrahlung). When neutral hydrogen is formed putting together one proton and one electron (a free electron is bound by a proton), does the electron decelerate and does it emit a photon?

    Thanks.
     
  2. jcsd
  3. Apr 18, 2004 #2

    mathman

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    When an electron is bound in a hydrogen atom, it is subject to the rules of quantum mechanics, which, among other things, permits electron energy changes to take place in discrete (quantum) amounts (this gives rise to its spectrum). At its lowest level, it just stays at constant energy. There are a lot more details to this subject - look it up with google, or get a book on elementary quantum theory.
     
  4. Apr 18, 2004 #3

    hellfire

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    I already took a look in google, of course, but I did not found an answer. I know that an electron bounded in an atom follows some rules, but my question is whether the electron suffers a deceleration in the time from beeing free to the time when it is bound or trapped by the proton. Lets say, it is not a question how the energetic transitions of the electron inside the atom are related to emission of absorption of radiation, but a question whether radiation plays a role between the electron beeing outside and inside of the atom. I assume that the electron will have a change in its dynamical behaviour...?

    Regards.
     
    Last edited: Apr 18, 2004
  5. Apr 18, 2004 #4

    Nereid

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    If an electron, originally in the lowest bound state in an H atom, is 'booted out', to become a 'free' electron, then it needs to be 'given' energy - e.g. by a photon - at least equal to the "Lyman limit". If the reverse happens, do you think a photon will be emitted?

    If a cloud of H atoms is between you and a distant, point source of copious photons, from gammas to radio, what will you observe in the spectrum of this bright, distant point source? The cloud of H atoms will reach some kind of equilibrium wrt the distant source of photons of many energies; what sort of equilibrium? Why don't you 'see' LyA radiation coming from the point source (caution, trick question)?
     
  6. Apr 19, 2004 #5

    hellfire

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    I am not sure about the answer, Nereid, this is why I asked. On the other hand it is nice that you step into the thread, because my question has ultimatively to do with cosmology…

    Coulds of neutral Hydrogen may absorb high energy photons and get ionzed. As far as I know, when electrons recombine with protons, then they have to step down on energy levels and an emission takes place. But I am not sure of this.

    May be my question was not correctly formulated: what I want to know is whether the formation of neutral hydrogen generates radiation. Independently whether Bremsstrahlung takes place or not (it seams it does not).

    The main reason what makes me doubt is that during the cosmological recombination epoch a lot of radiation should have been generated due to the formation of neutral hydrogen. I assume that this emitted radiation was not necessarily in thermal equilibrium with the rest of the photons at 3000° K.

    Since this radiation was exactly at the last scattering surface it should be visible as a clear anisotropy now. I do not remember to have seen this point mentioned in the references I read about the CMB.

    I apologize for the somehow chaotic way of asking this, starting this ambiguous thread. So, may be this thread should be moved to cosmology.

    Regards.
     
  7. Apr 19, 2004 #6

    ZapperZ

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    This may not exactly answer your question, but if this process is anything similar to an inverse photoemission (IPES) process, then yes, you will get photons being emitted as the free electron (initially having some energy above the vacuum state) make a transition into one of the unoccupied bound state. In fact, one of the specific technique that is used in IPES is called the Bremsstrahlung Isochromat Spectroscopy (BIS).

    This is a well-known experimental technique in condensed matter physics to study various properties of solids.

    Zz.

    After I posted this, I realized that I was sitting on this info - inverse photoionization, which I think it exactly what you're looking for. Try this:

    http://doc.cern.ch/tmp/convert_SCAN-0004050.pdf
     
    Last edited: Apr 19, 2004
  8. Apr 20, 2004 #7

    hellfire

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    Thanks ZapperZ. In the meanwhile I think I have the answer now, which seams rather elemental. Recombination produces emission, since electrons have to step down energetic levels. It seams that some of this should be visible also in the CMB (it is called 'recombination lines').

    Regards.
     
    Last edited: Apr 20, 2004
  9. Apr 20, 2004 #8

    Nereid

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    I got "Document not found" when I clicked on the link; am I too late?
     
  10. Apr 20, 2004 #9

    Nereid

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    You seem to be OK with the answers so far to your questions, but are interested in, and maybe a little puzzled by, what we think was going on in the universe around the time of last scattering.

    In particular, you seem to feel there should be some sort of anisotropy in the CMBR, in some way an imprint of photons emitted when protons and electrons combined to form neutral H atoms.

    While I'm not sure I understand your question, I'd like to go back to the cloud of neutral H atoms and the intense source of photons - of all energies. Some H atoms will be ionised, so we - located on the other side of the cloud from the point source - should see a Lyman absorption spectrum of H in the light from the point source. But then, as you say, the electrons will recombine, emitting photons ... with the Lyman H spectrum! So how come there are still absorption lines? Well, the emission is isotropic, but the ionising radiation is not! (Lots of caveats, as always.)

    At the time of last scattering, there wasn't an 'outside' to the universe, nor a 'point source', so the geometry which allows us to see absorption lines isn't the same.

    Does this help?

    BTW, no apologies needed; if you don't think about things and ask questions, how will you ever learn anything? :smile:
     
  11. Apr 21, 2004 #10

    ZapperZ

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  12. Apr 21, 2004 #11

    hellfire

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    Nereid, I think I understand the example of the neutral Hydrogen cloud, but I am puzzled with your comment regarding CMB.

    In the time between my first question till now, I found out the following:

    During recombination, electrons are trapped by protons and form neutral hydrogen. The electrons step down to the lowest energy level. This produces the emission of a photon, which has enough energy to ionize a hydrogen atom again.

    Recombination takes place anyway, since: (a) the emitted photons are redshifted and also (b) other energetic transitions take place which produce photons with lower energy.

    Both kind of photons do not fit into the blackbody spectrum of the CMB (this is formed by the original radiation of the plasma) and must be somehow visible in a sort of anisotropy.

    These references (I hope they are significative):

    http://www-ra.phys.utas.edu.au/~rdodson/all_together_now/node253.html
    http://arxiv.org/abs/astro-ph/0003227

    explain that these photons should be redshifted now almost to the radio (or low microwave). They are called ‘recombination lines’.

    Regards.
     
    Last edited: Apr 21, 2004
  13. Apr 21, 2004 #12

    ZapperZ

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    You need to be a bit careful here. What you have just described involved TWO separate transitions.

    1. Free electron gets captured by proton. This involves a transition from an energy level ABOVE the vacuum state to a bound state. This can involve the emission of photons also. The bound state need not be stable, it can be a meta-stable state.

    2. A second transition to the ground state - this can again involve another emission of photon. But note that the photon emitted here does NOT have enough energy to ionize another atom. A transition from one bound state to the next bound state within a H-atom is always less than 13.6 eV. So you will never have enough energy from the emitted photon to reionize another H atom.

    So it appears that if there is any ionization taking place, it will have to come from photons from the first transition process. Depending on how energetic the free electron was before capture, it's energy is certainly high enough above the vacuum state to cause the emission of a photon larger than 13.6 eV if it is captured and settled to a deep-enough metastable state.

    Zz.
     
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