1
$\begingroup$

I teach IB physics and recently attempted explain to a student where the Fermi Radius formula comes from in response to a question. For those unaware the fermi radius formula is shown below.

$$ R = R_0 A^{1/3} $$

Where $R_0$ is the Fermi Radius (1.20 fm), $R$ is the nuclear radius of an arbitrary nucleus, and $A$ is the mass number of the nucleus.

Post Submission Note: It seems the terms "Fermi Radius" is not common parlance in the physics world. I had never heard it until teaching IB physics but assumed that's because my education was chemical engineering and not physics. In the IB syllabus they give this formula and called $R_0$ the Fermi Radius which is stated as equal to 1.20 fm.

I assumed the relation came from the constant nuclear density relationship as shown below.

In the below work the H subscript stands for Hydrogen and variables missing a subscript are an arbitrary nucleus for which you wish to calculate the radius (R) for and know its mass number (N).

Assuming all nuclei have equal density you get....

\begin{align} \rho_H &= \rho \\ \\ \frac{N_H}{\frac43\pi R_H^3} &= \frac{N}{\frac43\pi R^3} \\ \\ \frac{N_H}{R_H^3} &= \frac{N}{R^3} \end{align}

Solving for $R$ and substituting the mass number of hydrogen (1) and Hydrogen's nuclear radius (or proton radius which equals 0.84 fm) in we get

$$ R = R_H \left(\frac{N}{N_H}\right)^{1/3} = (0.84\mathrm{\,fm}) \left(\frac{N}{1}\right)^{1/3} $$

$$ R = (0.84\mathrm{\,fm}) N^{1/3} $$

But according to the Fermi Radius formula the initial radius should 1.20 fm. I expected a little variance, but 50% is quite large and I don't see why I should be this far off. I'm mostly looking for an answer that would be useful for a high school student and not an overly technical one of that's possible. Any answers are appreciated though.

Final Note: Since "Fermi Radius" seems to not be common terminology it's possible this is just a mistake in the IB curriculum I guess.

$\endgroup$
5
  • 2
    $\begingroup$ What exactly do you mean by the "Fermi radius"? $\endgroup$
    – AfterShave
    Apr 28, 2022 at 1:42
  • $\begingroup$ Could you be a referring to the radius of the so-called "Fermi sphere"? $\endgroup$
    – joseph h
    Apr 28, 2022 at 2:01
  • $\begingroup$ I didn't major in Physics so all I know for sure is that this is what the constant R0 is referred as in the IB syllabus. I was under the impression this was common terminology. But hopefully my post makes it clear that it seems to be a radius related to the smallest nuclei possible, or at least I think, which should mean it equals the radius of a proton but since it's not I'm confused. From what I just searched only the Fermi sphere seems unrelated to this. $\endgroup$
    – Eric Craig
    Apr 28, 2022 at 2:07
  • $\begingroup$ Well your assumption that all nuclei have equal density isn't true, hence the discrepancy in the final result. The internal structure of an atomic nuclei is a very complicated subject and to make an inaccurate model you'd have to consider many factors. With that said, your rough analysis should work fine as an order of magnitude analysis. $\endgroup$
    – AfterShave
    Apr 28, 2022 at 2:23
  • $\begingroup$ I have multiple college textbooks that all say that nuclear density is is NEARLY constant so I'm pretty sure the starting assumption is correct. You maybe be emphasizing the nearly part but since nuclei are all composed of protons and neutrons which have very little difference in mass and volume I don't think a 50% variance can be explained by the slight variations. $\endgroup$
    – Eric Craig
    Apr 28, 2022 at 2:37

2 Answers 2

5
$\begingroup$

The proton is a complicated object and doesn’t have a single “radius” like a ball of yarn does.

Your value for the “proton radius”, $r_p \approx 0.8\,\rm fm$, is the proton’s “charge radius,” a parameter that describes the distribution of electric charge in the region surrounding the center of a proton. If you were to imagine a rigid, hard-shelled proton, there are several different ways you might distribute its electric charge throughout its volume and its immediate external environment, which would give you different reasonable definitions of the “charge radius.” For example,

  • You might imagine that the proton, like a conducting sphere, has all of its charge on its surface. In that case the “charge radius” and the radius of the sphere would be the same.

  • You might imagine that the proton has its charge smeared out uniformly over its entire volume. In that case the “average charge radius” might be somewhat smaller than the radius of the sphere, depending on how you do some integral. (I think the charge radius is the root-mean-square expectation value $\sqrt{\left<r^2\right>}$ taken over the charge distribution, but I haven’t looked that up tonight.)

  • You might imagine that the proton spends part of its time as a neutron orbited by a $\pi^+$ meson, in which case the “average” location of the charge might actually lie outside of the neutron’s radius.

  • You might imagine that the proton is a bound system of two positively-charged up quarks and one negatively-charged down quark, and come up with an argument that the proton’s (overall positive) charge distribution might actually have a negative core, or a negative halo, or both.

Your “Fermi radius” is the typical distance between nucleon centers in a nucleus. This is allowed to be different from the charge radius because electric charge isn’t what holds a nucleus up. The Fermi radius is a feature of the nucleon-nucleon strong interaction, to which the electric repulsion between protons is a small correction. A useful model of the strong interaction is the “Yukawa potential,”

$$ V(r) = \alpha \hbar c \frac{e^{-r/r_0}}{r} $$

where $\alpha$ is a dimensionless coupling constant, and the “range” of the interaction,

$$ r_0 = \frac{\hbar c}{mc^2} $$

depends on the mass $m$ of the particle which mediates the interaction. Electromagnetism has coupling constant $\alpha_\text{e.m.} \approx \frac{1}{137}$ and $r_0\to\infty$, since the photon is massless. The attractive, long-range part of the nuclear interaction is carried by the pion, with $\alpha_\pi \gg \alpha_\text{e.m.}$, mass $m_\pi \approx 140\,\mathrm{MeV}/c^2$, and range $r_\pi \approx 1.3\rm\,fm$. There are heavy mesons which mediate a short-range repulsive force which gives nucleons their “hard core,” but the usefulness of the meson model gets squashy at that point.

That is to say, your value of the “proton radius” comes from electromagnetic scattering, but within a nucleus the Yukawa range of the pion is more important for determining internucleon distances.

I don’t really know whether “Fermi radius” is common terminology or not. Fermi did so much in the early days of nuclear physics that a zillion things are named for him. In fact the femtometer (fm) was frequently called “one fermi” before the standardization of the SI prefixes. Your text may use “Fermi radius” for this distance because real nuclear physicists will really say “the radius of a nucleon is approximately one fermi” in discussions where one significant figure is appropriate.

$\endgroup$
3
$\begingroup$

Why is the Fermi Radius not equal to proton radius?

Very roughly, it's similar to why a structure made of closely-packed spheres has a volume different from a single sphere of the same mass. But there are other complications...

Assuming all nuclei have equal density you get....

This is not necessarily a good approximation. But is perhaps better for larger nuclei.

For example, if we model protons and neutrons as hard spheres, the density of helium would reasonably be expected to be (in some sense) less than hydrogen, because the hard spheres do not pack well.

Of course, protons and neutrons are not really hard spheres, and neither are nuclei. They are a gobbledee-gook of quarks and gluons and all sorts of horrible QCD garbage.

Because of this it is probably most reasonable to say something like: For large enough nuclei the mass is probably going to scale like the radius cubed (whatever that "radius" might be) (based on a simple model like one you presented). But, we really don't know exactly what the scale factor is (and it is very hard to calculate it exactly from first principles). However, we do know that the length scale is about 100000 smaller than the radius of an atom.

So, we could say that, roughly: $$ M \propto A \propto R^3 $$ Or, solving for R: $$ R \propto A^{1/3}\;. $$

There really is no exact proportionality factor that will always work. But we can choose any number that is close to $10^{-15}$ meters. So, go ahead and choose it to be $$ R = R_0 A^{1/3}\;, $$ with the caveat that this is an approximation.

And further, it is not unreasonable to expect this approximate equation to only hold for "large enough" nuclei (here we assume larger nuclei are more amenable to being treated as some kind of homogenous blob). This latter caveat solves the hydrogen nucleus radius issue you have (it is not a large nucleus).

As a final caveat consider also that we are ignoring nuclear binding energies and so on (think Uranium fission and whatnot) that will be important for large atoms but not so much for small ones (like hydrogen).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.