## Main Question or Discussion Point

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:
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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mathman
What is "a negative charged radius"?

phyzguy
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.

This makes sense to me.

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.

phyzguy
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.

I'm not an expert here, but I think the answer is yes. The equation is just a simple fitting function.
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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