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Collision of electons

  1. Jun 13, 2012 #1
    I got the concept of spin quantum number and got enlightened of the fact that it has got nothing to do with the movement of electron around the nucleus and that it is concerned only with the rotation of electrons about their own axis. I thank 'pf mentor' for giving me clarity about that. But I still dont get why two electrons present in the same orbital dont collide each other.. I didn't understand the example of surface of a ball and orbital. So plz dont mind clarifying my doubt..
  2. jcsd
  3. Jun 13, 2012 #2
    First of all, there is no end-all analogy for the quantum world. As soon as you understood the first watered down analogy I gave and found the flaw in it, you'd be right back at the doorstep looking for the next one. And so on, and so on, until sooner or later my analogy would be 'electrons act like electrons because electrons act like electrons.' And you would say, "What a douchebag, he just wasted all this time to say nothing." There is no way to completely discuss the quantum world using English, if there was, there would be no Physicists, only English majors (and what a sad world that would be.) That being said, here we go.

    We don't think of electrons orbiting the nucleus anymore. Electrons exist in orbitals, probability clouds, around the nucleus. The word orbital means 'similar to an orbit.' It's not an orbit, it will never be an orbit, and no matter how closely you look at it or hard you think about it: it's not an orbit. We observe electrons in the orbitals using statistical methods so we wouldn't know when and where they collide or if they collide at all. The question 'do electrons in the same orbital collide?' is nonsensical. There is no classical interpretation of the quantum mechanical model of atomic electrons. The QM model is the only tool we have to understand atomic physics, so why would it matter if they collided in the first place?
  4. Jun 14, 2012 #3
    I do know that in practical two electrons in the same orbital do not collide with each other because if that happens atom wouldnt be stable. But how can they exist without colliding one or the other time.. I'm sure there is an answer for it and hope I get it...
  5. Jun 14, 2012 #4


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    They are not billard balls which could collide in a classical sense.
    Their interaction is already part of their wave functions in the atom.
  6. Jun 14, 2012 #5


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    Like charges repel. The electrons try to position themselves around the atom the furtherest they can get from each other (there are more rules to this but I can't remember them - the electrons are attracted to the positively charged protons - but not to each other). The closer the electrons are to each other the greater the repulsion - Colombs law.

    The electrons arrange themselves into shells. If the shells need more electrons to be stable, the atom will react with another chemical if it gets the chance. Oxygen's outer shell needs two electrons to be stable - and Hydrogen has a single electron, but needs another one for it's shell to be stable - so they react and produce H20 -Two Hydrogens and one oxygen = water.
  7. Jun 14, 2012 #6


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    In QM, electrons with opposite spins actually *do* collide with each other, if you want to use these terms. The pair distribution function,
    g(x,x') = \frac{\langle \psi^\dagger(x)\psi^\dagger(x')\psi(x')\psi(x) \rangle }{\langle \psi^\dagger(x)\psi(x)\rangle \;\langle \psi^\dagger(x')\psi(x')\rangle },
    with psi being the field operators, describes the probability of finding two electrons at spin-positions x and x' simultaneously, divided by the probability of finding them at positions x and x' individually. This function can be evaluated in practice in molecules and other correlated electronic systems (say, the uniform electron gas) and turns out to *not* be zero in general for x1 and x2 at the same position, but with opposite spin.

    The reason this is not associated with an infinite Coulomb energy is that the volume element in which the energy is infinite is also infinitesimally small. This just results in so called derivative-discontinuities in the wave function.
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