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Maximum size of a nucleus

  1. Feb 20, 2008 #1
    If a free proton is pushed to within 10^-13 centimeters of a nucleus, it overcomes the repulsion interactions and is locked into the nucleus.

    Is there a known limit to the size of a nucleus? Is there a limit to how many protons we can add to a nucleus?

    Thanks for your help

    John
     
  2. jcsd
  3. Feb 20, 2008 #2
    Really interesting question...
    I think that accordingly to our present knowledge the answer is in the periodic table..

    regards
    marco
     
  4. Feb 20, 2008 #3
    Hm, I remember I took a coarse about this once. There is a limit size or in other words number of protons (mixed with a special fraction of neutrons) in the nucleus. Nucleus with Z>109 are instable and decay very quickly into lighter atoms, also radiating [tex]\gamma[/tex]-rays.

    Since protons and neutrons are made up of 3 quarks each, you could think of the nucleus as a "quark-liquid" (like [tex]H_2O[/tex] molecules in a small water drop). Each quark carries electric charge and Coulomb repulsion/attraction will occur (mainly repulsion). Phenomenologically one introduce also an attractive force between the quarks like:

    [tex]U_{i,j} = + constant\mtimes\mid\vec{r}_i-\vec{r}_j\mid[/tex]

    The balance between attractive quark force and repulsive coulomb interaction will determine the size of the nucleus. If you try to move quarks from each other to far, the energy will be so high that this energy could create particle-antiparticles and will decay.

    The idea is that in large enough "quark-soup" there could be effectively to large quark-quark distances involved (attraction energy increases above [tex]mc^2[/tex]), so that the nucleus will decay rapidly.


    I hope it give you something,
    Per
     
    Last edited: Feb 20, 2008
  5. Feb 20, 2008 #4
    As per said, it matters how long you want the nucleas to stay togever without decaying. As shown in periodic table, the biggest nucleus discovered so far is Ununoctium, but it stays togever for very little time and decays. So in theory you could have probably alot more than there is currently in periodic table, but it would last for such a short time, it would be hardly any use to us.

    Tachyon.
     
  6. Feb 20, 2008 #5
    Thank you for your helpful replies.

    Can I put it crudely to see if I am getting the right idea.

    Is it assumed to be a build up of the electromagnetic repulsion interaction that eventually defeats the strong nuclear attraction interaction?

    In a manner of speaking, does the strong force break because the volume added to a nucleus by the addition of a proton exceeds the overall volume increase of the nucleus. The disproportionality eventually leading to an unstable nucleus?

    I am looking into the implications that the relative atomic mass has for my enquiry, thanks for the pointers to the periodic table.

    Thanks again for your valued help.

    John
     
  7. Feb 23, 2008 #6
    If it helps further, there are theoretical sizes of nuclei that are (potentially) extremely stable elements; element 114 (more specifically the 286(?) isotope), is a theoretical size that physical chemists and nuclear physicists have been trying to create for a little while now. Although I have to admit that I don't really understand why certain superheavy elements are so stable, from the way it was explained to me it seems that the right combination of neutrons and protons in the right pattern could potentially produce nuclei quite large, but only of specific sizes. Glenn Seaborg won a Nobel Prize for his work in this, and there is an article in Nature about element 114:

    http://www.nature.com/nature/journal/v280/n5723/abs/280543a0.html
     
    Last edited: Feb 23, 2008
  8. Feb 24, 2008 #7
    Yes, the electromagnetic repulsion will eventually overcome strong nuclear force: nuclear force has low reach, so it acts only between neighbour nucleons. Consequently it's binding energy increases only lineary with the number of nucleons A.
    Electrostatic energy is proportional to e^2/r, so it will increase as Z^2/A^(1/3). (radius of nucleus increases as A^(1/3)). Z is the number of protons, which is aproximately proportional to the number of nucleons (A), so the (positive) electrostatic energy will increase aproximately as A^5/6, which is faster than strong force energy (A).

    You might ask why we can't have a pure nevtron atom (to get rid of electrostatic repulsion). The reason is Pauli exclusion principle, which forces fermions (nevtrons, protons) to ocupy higher energy states as their density increases (this energy has a minimum when the numbers of nevtrons (N) and protons (Z) are equal). This energy is high enough to forbid pure nevtron atom, but can't always force Z=N, since it has to compete with electrostatic energy.

    There are also some other effects involved: I advice you to search for "semi-empiric mass formula" (for example on wikipedia), which uses liquid drop model to estimate binding energies of atoms (as a function of numbers N and Z).
     
    Last edited: Feb 24, 2008
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