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I thought that carbon with its even number of protons and neutrons has no magnetic spin or moment and yet I learn that neutrons are scattered quite effectively by carbon atoms.

I understood that neutrons interacted with matter through the strong force and through their magnetic moment so why should they be scattered by carbon?

Thank you in advance!

Edit: I think am getting confused between scattering due to path difference and scattering due to the magnetic moment. I think scattering can be considered as having an incoherent and coherent aspect. A nucleus with no magnetic moment such as carbon 12 generates no incoherent scattering but it a neutron 'wave' could be scattered by the spatial structure of matter, the coherent aspect (like Youngs slits). Is that it?

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    $\begingroup$ Welcome to our community, I suspect someone here may know the answer but this question is probably more suited for the physics stack exchange. Maybe someone else can weigh in. +1 anyways though as I am curious. $\endgroup$ Commented Oct 8, 2020 at 1:00
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    $\begingroup$ I agree with @TristanMaxson, this is a physics question. In the case of a macroscopic sample, you would not end up with C12 only, some C13 will be present (otherwise, carbon NMR would not be a thing). $\endgroup$
    – TAR86
    Commented Oct 8, 2020 at 4:06
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    $\begingroup$ +1. Welcome to our community, and thank for contributing your question here! We hope to see much more of you !!! $\endgroup$ Commented Oct 8, 2020 at 4:29
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    $\begingroup$ I think the argument can be made that this is a matter modeling question since neutron scattering is a technique used extensively in matter modeling and materials science. $\endgroup$ Commented Oct 8, 2020 at 16:44
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    $\begingroup$ I agree with @taciteloquence. The question is on topic here. $\endgroup$ Commented Oct 8, 2020 at 16:57

2 Answers 2

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Your assumption that neutron-nucleus scattering is primarily via magnetic interaction is not correct. Neutron-nucleus scattering is primarily from the strong interaction, which is (as its imaginative name suggests) stronger than electromagnetism. While neutrons can interact by scattering magnetically from electrons, as currently suggested by another answer, the neutron-electron interaction is much smaller than the neutron-nuclear interaction.

Most nuclei have a “neutron separation energy” of order $\sim 8\rm\,MeV$: if a free neutron were to settle down and be trapped inside of a nucleus, that’s the amount of energy it would have to release. Nuclei are smaller than atoms, and the nuclear interaction is short-range. So to a free neutron, a crystal lattice looks more or less like empty space with a network of delta-function-like attractive potentials at the locations of the nuclei.

When you solve the Schrödinger equation for a square well,

$$ \left(\frac{\hat p^2}{2m} + V(\vec r)\right)\psi = E\psi $$

you find that a free particle has more momentum (and therefore shorter wavelength) inside of the well than outside. For thermal-or-cold neutrons the difference is enormous: the free particle has milli-eV kinetic energy and angstrom-scale wavelength, while the piece of the wavefunction within the nucleus is mega-eV above the bottom of the well and has femtometer-scale wavelength.

If you go through the freshman-quantum exercise of matching the wavefunction at the boundary of the square well, so that it is continuous and smooth, with spherical symmetry, you find something interesting. For nearly all widths and depths of the square well, the free part of the wavefunction will have a zero-crossing somewhere near the nuclear radius. (I remember a good illustration of this in Golub’s textbook.) You can produce the same external wavefunction if, instead of a finely-tuned attractive well, you imagine an infinite, repulsive barrier whose radius is the location of this zero in the wavefunction. The radius of this made-up repulsive potential is called the “scattering length,” and the whole model is called the Fermi pseudopotential. The scattering length is directly related to the scattering cross section.

If you look at a table of scattering lengths and cross sections (like this one, which used to be available as a little booklet that fit in your pocket), you’ll see that all of the even-even nuclei (like $\rm{}^{12}C_6$) have zero incoherent scattering cross section, because even-even nuclei have no spin to exchange with the neutron. But all of the coherent scattering cross sections are a a few barns, because all of the coherent scattering lengths are a few femtometers. (There are a couple of low-mass nuclei where the scattering length is negative, because the zero in the wavefunction shows up on the wrong side of the origin, but even then the length scale is femtometers.)

Contrast this with the neutron-electron scattering length of $10^{-3}$ femtometers: the electrons are basically invisible, compared to the nuclei.

Even for spin-incoherent scattering, it’s not right to think of the neutron-nucleus interaction as magnetic. The energy scales are wrong. The nuclear magneton $e\hbar/2m_N$ is much smaller than the Bohr magneton $e\hbar/2m_e$, but incoherent neutron-nucleus scattering lengths are still mostly femtometer-ish. Incoherent neutron scattering comes from spin exchange mediated by the strong force.

So, that’s the answer to your title question. Neutron scattering from carbon-12 isn’t about magnetism at all. Neutron scattering from carbon-12 is about the neutron looking down into the abyss, wondering what its life would be like as part of a carbon-13, deciding that there is no place for it there, but having its long-distance behavior changed by the experience.

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  • $\begingroup$ that's a brilliant answer, or at least I liked it very much. Thank you. $\endgroup$
    – Naz
    Commented Jan 7, 2022 at 14:15
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There are a few possibilities (the relevant importance of each probably varies depending on the type of scattering and the material).

Even for pure C12, neutrons can scatter off unpaired electrons, since many allotropes have unpaired valence electrons. Neutrons can also scatter off the nucleus itself by nuclear forces. Graphite is used in some nuclear reactors as a neutron moderator precisely because it scatters neutrons, slowing them down so they have a better chance of causing causing fission when they hit a Uranium nucleus. See Wikipedia for Neutron cross section.

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    $\begingroup$ +1 for the earnest attempt, and a speedy answer which may help the question show up in the HNQ list, which is exactly what Tristan and TAR86 seem to be suggesting is necessary, based on their comments! $\endgroup$ Commented Oct 8, 2020 at 16:57
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    $\begingroup$ Thank you for the replies. I proceeding slowly. I note the use of C as a moderator (Chernobyl) but I thought that was a collisional effect where energetic neutrons collide with C until they take on its temperature. I don't think it's use as a moderator implies that it's predominant effect is in scattering neutrons though that depends on one's definition of scattering. I was thinking in terms of the magnetic moment, and the spatial structure of matter. I think the C nucleus has no magnetic moment and hence it exhibits only coherent scattering due to spatial structure - bonding. $\endgroup$
    – Naz
    Commented Oct 8, 2020 at 19:38
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    $\begingroup$ @Naz, I'm not sure what you mean by this distinction. The reason C is a good moderator is because it scatters neutrons well, but doesn't adsorb them (and C is relatively light, so it comes away from the collision with a fair amount of KE). For magnetic scattering (i.e. using neutron scattering to look at magnetic structure), I think the neutrons are usually scattering off the electron spins/orbital angular momentum. $\endgroup$ Commented Oct 15, 2020 at 17:02
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    $\begingroup$ A placeholder comment until I can write an answer: it's incorrect that low- energy neutron scattering is primarily from electrons in any material. The strong interaction wins. $\endgroup$
    – rob
    Commented Oct 16, 2020 at 19:46
  • $\begingroup$ @rob, which statement is incorrect? Looking forward to your answer! $\endgroup$ Commented Oct 19, 2020 at 9:29

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