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Let A be the atomic number

The density of a nucleus is computed using the ratio of the number of protons and neutrons, A, to the volume of the nucleus (which at my current level, is assumed to be a sphere)

The math is trivial.

Conceptually, the argument for the central density of a nucleus remaining roughly constant stems from the suggestion (from my text) that each nucleons feels a force only from it's nearest neighbour. As more nucleons are added to the nucleus, the central density increases but still lies within a range sufficient to be classified as being 'constant'.

This does not convince me. Is it necessarily true that a nucleon experiences constant force only from it's nearest neighbour? Suppose there exists a finite number number of nucleons, n, in the nucleus. Each nucleon then experiences a force from (n-1) number of nucleons. The other nucleon experiences also a force from (n-1) number of nucleons. By induction, and since these particles are indistinguishable-protons and neutrons are fermions- is it also not true then that every $$nucleon_i$$ experiences a constant force?

Could someone fill me in conceptually?

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If you were to imagine each nucleon as a hard sphere, then the packed volume of such a sphere will be proportional to the number of nucleons and the sphere will have the same mean density, regardless of the number of nucleons.

Think about the forces experienced by one of these "hard spheres". It feels a strong force from each of its nearest neighbours, but nucleons that are not in "contact" with it exert no influence at all.

This analogy is reasonably appropriate for nucleons bound together, because the strong nuclear force has a short range and becomes very repulsive at small separations. In other words, if we double the equilibrium separation between a pair of nucleons, the strong force is still attractive but so small it can be neglected, whereas if we halve that distance the force become very strongly repulsive.

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  • $\begingroup$ Thank you. I tried to imagine a model of a latice to understand this. I think what confused me is the use of 'nearest' since that by definition implies 'one and only one-unique'. But surely, if a nucleon has nearest neighbor of more than one other nucleon, then the term 'nearest' would not hold. $\endgroup$
    – Physkid
    Commented May 30, 2015 at 0:35
  • $\begingroup$ @Physkid It should be nearest neighbours. Then the BE is just proportional to the number of nucleons. This becomes untrue for nucleons that are not surrounded - hence the modification term that is proportional to the "surface area" of the nucleus. $\endgroup$
    – ProfRob
    Commented May 30, 2015 at 8:27

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