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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. $\endgroup$ – Tristan Maxson Oct 8 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 Oct 8 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 !!! $\endgroup$ – Nike Dattani Oct 8 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. $\endgroup$ – taciteloquence Oct 8 at 16:44
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    $\begingroup$ I agree with @taciteloquence. The question is on topic here. $\endgroup$ – Nike Dattani Oct 8 at 16:57
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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! $\endgroup$ – Nike Dattani Oct 8 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 Oct 8 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. $\endgroup$ – taciteloquence Oct 15 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 Oct 16 at 19:46
  • $\begingroup$ @rob, which statement is incorrect? Looking forward to your answer! $\endgroup$ – taciteloquence Oct 19 at 9:29

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