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B Tidal effect in the balance of Coulomb and nuclear forces?

  1. Sep 18, 2017 #1
    In a heavy nucleus, the balance of the Coulomb and nuclear forces is a delicate one. This can be seen in two cases: First, in oblong deformed nuclei, alpha particle emission is more likely to occur at the poles, where the Coulomb barrier is thinner, than at the waist. Second, heavy synthetic isotopes are generally unstable because the attractive nuclear force is inadequate to compensate for Coulomb repulsion.

    Is it possible to have a differential balance of Coulomb and nuclear forces at different ends of an axis that passes through the nucleus? Possible example: subject an atom of a heavy element to a strong magnetic field, and the electron orbitals will alter under the influence of the field, perhaps setting up asymmetric electron screening throughout the nucleus. If this example will not work, a more artificial scenario can be produced. In such a context, would the differential balance of Coulomb to nuclear forces have a destabilizing effect on the nucleus? Might it be possible to alter the rate of spontaneous fission, for example, in an isotope that has this as a decay mode?
     
    Last edited: Sep 18, 2017
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  3. Sep 19, 2017 #2

    mfb

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    We cannot create electric or magnetic fields strong enough to alter nuclear properties in a measurable way. Sufficiently strong fields would have an effect, sure. The typical scale would be 1021 V/m (which doesn't work at all) or 1012 T (where it is unclear if fields that strong exist).
     
  4. Sep 19, 2017 #3
    Thank you, mfb, for the helpful reply. In order to better understand it, I'll ask a further question.

    Consider a p electron orbital, which, roughly speaking, has the geometry of a dumbbell passing through the nucleus with a node at the origin. In the Rydberg case, I understand that it's straightforward to deform such an orbital with a magnetic field attainable in a laboratory. Is a much stronger magnetic field required to deform a p orbital in the ground state? (Significance: when a p orbital is deformed under a magnetic field, it no longer has axial symmetry, and hence there will be an asymmetry in electron density.) If the electron density in the nucleus from the p orbital electron is too small to consider, a similar scenario can be contemplated for an s-orbital electron.

    Or is the point about the strength of magnetic and electric fields needed to alter nuclear properties addressing a different but related matter?
     
    Last edited: Sep 19, 2017
  5. Sep 20, 2017 #4

    mfb

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    The effect of electrons in the nucleus is small, and electromagnetic fields altering this effect a bit will have an even smaller impact.
     
  6. Sep 20, 2017 #5
    There do seem to be some elements where electron capture and presumably electric fields are significant.

    "Around the elements in the middle of the periodic table, isotopes that are lighter than stable isotopes of the same element tend to decay through electron capture, while isotopes heavier than the stable ones decay by electron emission. Electron capture happens most often in the heavier neutron-deficient elements where the mass change is smallest and positron emission isn't always possible. When the loss of mass in a nuclear reaction is greater than zero but less than 2m[0-1e-], the process cannot occur by positron emission but is spontaneous for electron capture."

    A long time ago I read there is one element which is radioactive when ionised but stable when not (or the other way round) but I can't find a reference.
     
  7. Sep 20, 2017 #6

    mfb

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    Both options exist. They are quite obscure, however.

    - Beryllium-7 can only decay via electron capture. Normally there are electrons around to capture, so it is radioactive. Remove all electrons from its environment and it cannot decay any more - it gets stable.
    - A neutral dysprosium-163 atom is stable. If you remove all electrons, it can beta decay, where the electron stays in a low energy level. The energy is not sufficient to have the electron escape or occupy a higher energy level, which would be required for a decay of a neutral atom.
     
  8. Sep 20, 2017 #7
    Those of course are examples of weak decay channels; there are yet others as well: zirconium-89 and strontium-85 (via electron capture) and rhenium-187 (via beta decay). The referenced page agrees with mfb on the matter of strong decays, which is what we're considering here:
    The author explains that the changes are estimated to be very small indeed. But I note the phrases "in principle," and "calculations" and "estimated." What I'm curious about is whether some kind of tidal effect from a differential balance of electron density along an axis passing through the nucleus was contemplated in those estimates and whether there have been any experimental investigations of a possible tidal effect on nuclides that decay via spontaneous emission.

    It's possible the answer to both is "no," which is fine. I was just curious.
     
    Last edited: Sep 20, 2017
  9. Sep 20, 2017 #8

    mfb

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    You are looking at a tiny modification of an effect that is tiny already. Tiny2=negligible.
     
  10. Sep 20, 2017 #9
    I guess I was implicitly querying your assumption (and that of Bill Johnson, the author of the quote I excerpted) about the effect being tiny when asymmetry is involved. I suspect that if the matter has neither been explored theoretically nor investigated empirically, your assumption about a tiny modification of a tiny effect is merely a heuristic or a lucky guess. (Which is not to say that it's a bad heuristic or guess.)
     
    Last edited: Sep 20, 2017
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