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Electron emmission from orbitals

  1. Oct 30, 2006 #1

    TeX

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    Hello.

    Is there a way of telling from which electron orbital of unstable radioactive isotope an electron will be emmitted (if we take for example that decay mode includes electron emmission)? Because different atoms have different electron orbital configurations (1s, 2p, ..) and there can be limited number of electrons in each orbital, plus considering different energies in orbitals, I was wondering, can something like this be predicted? Is this a known thing for every unstable isotope decay mode involving electrons (from which orbital it is more likely for electron to be emmitted)?
    If so, than is there some electron bouncing effect possible because of other orbitals surrounding the orbital from which electron is emmitted (if such geometric configuration of orbitals exist)? If we would have, say, a perfect crystal cube structure of pure unstable radioactive isotope atoms, and there would be deacy mode involving electron emmission, would there be more dense electron emmission detected on some sides of cube because it would be known from which orbitals there is more chance of emmission?

    Thanks to anyone with more indepth knowledge of particle physics for clarifying this to me. Also, if something like this asked in my questions exists, can someone point me to some literature (books, articles, ...) or Internet sites which explain this specific subject in more detail?

    thanks
     
  2. jcsd
  3. Oct 30, 2006 #2
    I am not an expert on this, but I think that the phenomenon you are talking about is half and half. It can be easy to predict which electron will jump if the the next orbital is full (and you find the electron's spin to see which one will jump into the next slot). Largely, quantum numbers can help you "predict" which electron will jump.

    In repsonse to your second question, (which I think I understand what you mean) electrons don't just bounce off other orbitals until they "find" the right one. When a spot opens, they take a quantum leap; that's it. No boucing between orbitals (that would burn energy and get the electron no where).

    Finally, for an atom that is unstable. I am, once again, no expert on chemisty. However, I think, logically, that the outer-most orbitals would loose electrons first. Therefore, depending on an electron's quantum numbers (seeing which electron is furthest from the nucleas), it is sort of possible to gain probabilities of whether it will leave the atom or not. Now, it's only a matter of chance.
     
  4. Oct 31, 2006 #3

    jtbell

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    Staff: Mentor

    Are you referring to nuclear beta decay? The electron emitted in nuclear beta decay is not one of the atom's orbital electrons. It is created inside the nucleus, in the decay process: neutron --> proton + electron + antineutrino.
     
  5. Oct 31, 2006 #4

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    In case of so called "internal conversion" (transition between states of the same nucleus), if I understand this correctly, an excited nucleus transfers energy to orbital electron and ejects it. I don't know tough, if this happens in any of decay modes of unstable radioactive isotopes. Also, what happens if we have, say, a molecule AB, of which atom A is unstable radioactive isotope, atom B is stable isotope and then decay of A happens. Would in case of this "internal conversion" decay mode electron be ejected from atom A orbital which forms a bond with atom B, or would it be some other orbital of atom A. In such case I tought, since atom B is completely different atom then A, such electron ejected from atom A would not be capable of penetrating into bonding atom B, but would bounce off from electrons orbiting around atom B nucleus (since electrons have same charge). No matter what happens to such AB molecule after such decay process of A atom (for example A atom decays into daughter atom and still stays connected to atom B, or, A atom completely desintegrates leaving atom B alone, and so on ...), wouldn't this be a case of possibility of predicting in which direction there is more chance of electron being emmitted?
     
  6. Oct 31, 2006 #5

    jtbell

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    OK, I didn't realize that you were asking about internal conversion. What makes internal conversion possible is that the QM wavefunction of an atomic electron can overlap the nucleus. That is, at any point in time there is a small probability that the electron is located inside the nucleus. The larger this probability is for a particular orbital, the more likely it is that an electron in that orbital can undergo internal conversion.

    Nuclear physics isn't my specialty, but according to the Wikipedia article on internal conversion, K shell (1s) electrons are most likely to undergo internal conversion because the 1s wavefunction has the highest probability of being inside the nucleus, and this sounds reasonable to me.

    The bonds between atoms in a molecule are formed by the outer electrons, which surely have a very much smaller probability of being inside the nucleus than the inner-shell electrons. I haven't tried to look up actual wave functions to test this, but I expect that this "nuclear overlap" probabilty would be utterly insignificant for an outer-shell electron, even if not strictly zero.
     
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