Why Do Measured Nuclear Radii Differ from Predicted Values?

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    Nuclear Nucleus Radius
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Discussion Overview

The discussion revolves around the discrepancies between measured nuclear radii and those predicted by the equation r=1.2 fm×A⅓, particularly focusing on the radii of deuterium and tritium. Participants explore the implications of these differences and the concept of a negative charge radius for the neutron, touching on theoretical models and the behavior of nucleons within the nucleus.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants note the equation r=1.2 fm×A⅓ assumes constant nuclear density, but measured values for deuterium and tritium differ significantly, prompting questions about reconciliation of these discrepancies.
  • One participant suggests that the smaller radius of tritium compared to deuterium can be understood through the balance of attractive strong forces and repulsive electromagnetic forces, though this remains a heuristic explanation.
  • There is a discussion about the neutron's negative charge radius, with references to its quark composition and the asymmetric distribution of charge, but participants acknowledge that this is only one of many theoretical models.
  • Some participants express confusion regarding the similarity in radius between neutrons and protons, despite differing interpretations of what "radius" means in this context.
  • One participant emphasizes that the equation is a simple fitting function and may not accurately reflect the complexities of nuclear structure.
  • Another participant introduces recent findings on the charge radii of calcium isotopes, highlighting unique variations that challenge existing theories.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the accuracy of the r=1.2 fm×A⅓ equation compared to measured values, and there are multiple competing views regarding the interpretation of nuclear radii and the behavior of nucleons within the nucleus.

Contextual Notes

Participants note that definitions of "radius" can vary, and the identity of protons and neutrons may change within a nucleus, complicating the discussion. There are also unresolved questions about the implications of the neutron's negative charge radius and the behavior of isotopes.

Nathan Warford
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I have seen numerous sources for radii of atomic nuclii of various elements. One of the most common is the simple equation r=1.2 fm×A, which makes sense if the nuclear density is constant for all elements and all isotopes. However, I've also found a table of measured nuclear charge radii that differ greatly from what that equation suggests they should be:
https://www-nds.iaea.org/radii/
This is probably most apparent with the nuclii of deuterium and tritium, with deuterium having a radius of 2.1421 fm and tritium having a much smaller radius of 1.7591 fm despite having one more nucleon. How can this discrepancy be reconciled?

Also, despite all of my research, I'm having a hard time figuring out what it means for the isolated neutron to have a negative charge radius. Can someone help me decipher that?
 
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What is "a negative charged radius"?
 
Nathan Warford said:
This is probably most apparent with the nuclii of deuterium and tritium, with deuterium having a radius of 2.1421 fm and tritium having a much smaller radius of 1.7591 fm despite having one more nucleon. How can this discrepancy be reconciled?

Well, heuristically, if you think of the attractive strong force as holding the nucleus together against the repulsive electromagnetic force, then it makes sense that tritium would be smaller than deuterium, since you've added one more "attracting" particle without changing the net repulsion. This is why higher-Z nuclei need more neutrons in order to be stable, since it takes more neutrons to overcome the larger repulsion.

Also, despite all of my research, I'm having a hard time figuring out what it means for the isolated neutron to have a negative charge radius. Can someone help me decipher that?

Wikipedia says,

"The best known particle with a negative squared charge radius is the neutron. The heuristic explanation for why the squared charge radius of a neutron is negative, despite its overall neutral electric charge, is that this is the case because its negatively charged down quarks are, on average, located in the outer part of the neutron, while its positively charged up quark is, on average, located towards the center of the neutron. This asymmetric distribution of charge within the particle gives rise to a small negative squared charge radius for the particle as a whole. But, this is only the simplest of a variety of theoretical models, some of which are more elaborate, that are used to explain this property of a neutron.[2]

This makes sense to me.
 
phyzguy said:
Well, heuristically, if you think of the attractive strong force as holding the nucleus together against the repulsive electromagnetic force, then it makes sense that tritium would be smaller than deuterium, since you've added one more "attracting" particle without changing the net repulsion. This is why higher-Z nuclei need more neutrons in order to be stable, since it takes more neutrons to overcome the larger repulsion.
Does this mean that the table is more accurate than the r=1.2 fm×A equation?
Wikipedia says,

"The best known particle with a negative squared charge radius is the neutron. The heuristic explanation for why the squared charge radius of a neutron is negative, despite its overall neutral electric charge, is that this is the case because its negatively charged down quarks are, on average, located in the outer part of the neutron, while its positively charged up quark is, on average, located towards the center of the neutron. This asymmetric distribution of charge within the particle gives rise to a small negative squared charge radius for the particle as a whole. But, this is only the simplest of a variety of theoretical models, some of which are more elaborate, that are used to explain this property of a neutron.
I guess the reason I have a hard time understanding is because most sources say that the neutron and proton have roughly the same radius.

Wikipedia says,
The neutron has a mean square radius of about 0.8×10−15 m, or 0.8 fm

That's a value that I recognize as about the radius of a proton at between 0.84-0.87 fm.
 
Nathan Warford said:
Does this mean that the table is more accurate than the r=1.2 fm×A equation?
I'm not an expert here, but I think the answer is yes. The equation is just a simple fitting function.
I guess the reason I have a hard time understanding is because most sources say that the neutron and proton have roughly the same radius.

Wikipedia says,
The neutron has a mean square radius of about 0.8×10−15 m, or 0.8 fm

That's a value that I recognize as about the radius of a proton at between 0.84-0.87 fm.

You need to define what you mean by radius. They are not billiard balls. The radius will be different depending on what you mean. Also, you need to realize that the proton and neutrons lose their identity inside a nucleus. A nucleus is more of a "soup" of quarks and gluons.
 
phyzguy said:
I'm not an expert here, but I think the answer is yes. The equation is just a simple fitting function.
Thank you for your input. You have been very helpful.
Also, you need to realize that the proton and neutrons lose their identity inside a nucleus. A nucleus is more of a "soup" of quarks and gluons.
Is that so? I thought that only happened at very high temperatures and/or pressures, like in a particle collider or the core of a neutron star.
 
That equation is but a 'broad brush'. Here's a recent look at some detail...
https://phys.org/news/2019-02-puzzling-sizes-extremely-calcium-isotopes.html
quote:
One of the most fundamental properties of the nucleus is its size. The nuclear radius generally increases with the number of proton and neutron constituents. However, when examined closely, the radii vary in unique ways, reflecting the intricate behavior of protons and neutrons inside the nucleus.

Of particular interest is the variation of the charge radii of calcium isotopes. They exhibit a peculiar behavior with calcium-48 having almost the same radius as calcium-40, a local maximum at calcium-44, a distinct odd-even zigzag pattern, and a very large radius for calcium-52. Although the pattern has been partially explained (gray line in the figure), many existing theories struggle to explain this behavior. Below the lightest stable calcium-40 isotope, the charge radius has been known only for calcium-39, due to the difficulty in producing proton-rich calcium nuclei.
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