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Is there at least an approximation of the decrease in strong nuclear attraction vs distance from the center of the nucleus?

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  • $\begingroup$ The force actually increases up to a distance because of the strange way quarks + gluons form chains. $\endgroup$ – Brandon Enright Jul 21 '14 at 4:27
  • $\begingroup$ Just to add to the above comment. For distances larger than the nuclear scale, the potential is modelled by the Yukawa potential. $\endgroup$ – suresh Jul 21 '14 at 5:20
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You should read the article in wikipedia on nuclear force.

Various models exist that describe the behavior of nuclear forces, which are the result of a spill over of the strong force, the force that exists within the proton and the neutron.

From the link

force between nucleons

Force (in units of 10,000 N) between two nucleons that experience the nuclear force, as a function of distance. In the graph that falls below the horizontal axis, the spins of the particles (which must be different, such as a neutron and proton) are aligned, and they become bound with a negative "binding energy," which becomes maximal at a distance a little more than 1 fermi (Fm) at the minimum of the region shown. Particles much closer than this optimal distance experience a repulsive force. Particles farther than the distance of highest attractive force still experience a smaller attractive potential (Yukawa potential), but it falls at an exponential function of distance.

Repulsive forces exist because of charge distributions ( quarks within the nucleons are charged) and the Pauli exclusion principle.

More complicated nuclei are a many body problem and models exist that fit the measurements and predict behaviors, like the shell model, or the liquid drop model.

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