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While studying NMR theory, my textbook explained that only nuclei with odd mass numbers are NMR active because they have non-integer spin quantum numbers and nuclei with an even mass number and atomic number have I= 0 spin.

This doesn't make much sense to me. NMR works by generating a magnetic field which interacts with spinning charges (like a proton or electron) that also generate a magnetic field. So how come the nucleus of Carbon-12 is "invisible" to NMR?

In other words, if we treat each proton as miniature bar magnet that is capable of flipping between alpha and beta states (parallel and antiparallel), what is stopping the protons from "reversing spin" in a carbon-12 nucleus?

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    $\begingroup$ Across the entire periodic table, nuclear spin values ranging from I = 0 to I = 8 in ½-unit increments can be found. Protons and neutrons each have net spins of ½, but this derives from spin interactions among the elementary particles (quarks) of which they are composed. As a result of this complexity, no simple formula exists to predict I based on the number of protons and neutrons within an atom. mri-q.com/predict-nuclear-spin-i.html $\endgroup$ – Wolphram jonny Jun 17 '16 at 19:48
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    $\begingroup$ What you're neglecting is that the nucleons also interact with each other, organising themselves so that their magnetic moments effectively cancel out. Consequently, you need to treat the entire nucleus as a single entity. $\endgroup$ – lemon Jun 17 '16 at 19:49
  • $\begingroup$ @lemon in that case, how come C-13 has I=0.5 and H-2 has I = 1? Both have one "unpaired" proton... $\endgroup$ – Nova Jun 17 '16 at 19:53
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    $\begingroup$ @Nova I should also add that due to the dependence of nuclear force on isospin deuterium is only stable when the neutron-proton spins are parallel, hence $I_z = 1/2 + 1/2 = 1$. $\endgroup$ – thodic Jun 17 '16 at 20:08
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    $\begingroup$ @Nova As your book explains all fermions (particles with half integer spin) behave like small magnets and will precess in a magnetic field. This is due to their spin not their charge (see en.wikipedia.org/wiki/Neutron_magnetic_moment). $\endgroup$ – thodic Jun 17 '16 at 21:15
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The interaction between nucleons in a nucleus is very strong so the energy changes associated with flipping a spin are very high. The first few energy levels of a carbon nucleus can be found in this document. The energy spacing between the ground state $J=0$ and the first excited state $J=2$ is 4.44 MeV. This isn't much by the standards of modern accelerators, but it is far, far too big for the transition to be caused by an external magnetic field.

So as far as NMR is concerned the carbon atom behaves as an elementary particle of spin zero with no internal excitations.

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