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I am finding some conflicting information which is resulting in some confusion when teaching the topic of nuclear physics to high school students.

The residual strong nuclear force has an attractive range of around 0.7 - 3fm.

A nucleon has a diameter somewhere in the range of 1-2fm.

According to my understanding, nucleons will be in equilibrium when the residual strong nuclear force is zero for a neutron-neutron or neutron-proton interaction and slightly further apart for a proton-proton interaction (since the RSNF has to now be as attractive as the electrostatic force is repulsive). This means about 0.7 fm for n-p and n-n or 0.8fm for p-p. In all cases this is (substantially) less than the sum of the radii of the two nucleons.

Does this imply that there is significant overlap of the nucleons? I know that the hard sphere idea is not good at this level as we are well into the quantum realm. I assume that quantum fuzziness is to blame, in which case what is really meant by the radius of a nucleon.

Thanks Martyn

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  • $\begingroup$ I can think of at least one example (deuteron) where the proton and the neutron orbitals fully overlap. $\endgroup$
    – fraxinus
    Dec 22, 2023 at 7:48

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The nuclear diameter is measured in several ways. One is elastic electron scatter which measure the charge distribution, showing that the RMS charge radius is 0.84 - 0.87 fm.

As far as the residual strong nuclear force goes, it is really complicated. It has serval central force term, both repulsive and attractive, the are attributed to various meson exchanges. Spin-orbit coupling, spin-spin, and helicity terms are not perturbative...they're strong. There's also a tensor force, which is non-central, and iso-spin terms that are significant. Finally, there are three and four body interactions that cannot be described as $N$-body central-potential interactions. I would consider these very difficult to explain at the high-school level.

Note also that protons and neutrons don't retain their identity in a nucleus, which has to do with quantum addition of isospin. It makes those pictures of two colored marbles in a ball more deceptive than descriptive.

Oh, and the nuclear shapes are constrained by their spin (really orbital angular momentum), so protons and alpha particles are absolutely spherical (up to tiny symmetry violations that have not been measured), while deuterons are a mix of at least a spherical part and a prolate or oblate part.

And I forgot about the "Exchange term", which has to do with the Pauli exclusions principle.

See http://www.scholarpedia.org/article/Nuclear_Forces for further discouragement.

Really, don't attempt a quantum description of the nucleus until you've mastered the hydrogen atom, which is already riddled with semi-classical misinterpretations from over simplification.

Nevertheless, there is hope. The semi-empirical mass formula (https://en.wikipedia.org/wiki/Semi-empirical_mass_formula) provides a quantitative and easily understood description of what's going on.

Note that there is a charge term, but I have never seen it considered as a modification to the $pp$ vs $nn$ short range interaction; rather, the electrostatic energy of the nucleus goes as $Z^2$, and for large $Z$, that can overcome nuclear binding effects--so it's a global consideration.

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  • $\begingroup$ Thanks for the excellent response. As I supposed it is more complicated than I need to get into. $\endgroup$
    – MartynW
    Dec 21, 2023 at 21:25
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    $\begingroup$ "for further discouragement" +1 $\endgroup$
    – fraxinus
    Dec 22, 2023 at 7:46
  • $\begingroup$ @MartynW See the world famous "Argonne V18 Potential" for more, e.g: nukephysik101.wordpress.com/2022/02/07/… and phy.anl.gov/theory/research/av18 ...the latter link shows data from years of experiments to figure it out. Meanwhile Wigner figured out the basic hydrogen atom from symmetry...I don't think he even needed to solve the Schodinger eq... $\endgroup$
    – JEB
    Dec 22, 2023 at 16:57

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