According to P.J. Hore in the Oxford Chemistry Primer "Nuclear Magnetic Resonance" p57, 'the dipolar coupling, modulated by molecular motions, causes nuclear spins to experience time-dependent local magnetic fields which, if they contain a component of the NMR frequency, can induce the radiationless transitions which return the spins to equilibrium'.

What does Hore mean by the local magnetic fields containing a component of the NMR frequency?

What is the mechanism of the radiationless transition? I don't understand any explanation that I've read in textbooks.


1 Answer 1


From Einstein's law for spontaneous emission that relates the lifetime of the excited state, τ, to the frequency of the transition, $\nu$:

$τ ∝ \frac{1}{\nu^3}$

since in NMR spectroscopy we deal with radiofrequencies lifetimes may reach values as large as 108 s (which means years).

Fortunately there exist nonradiative relaxation mechanisms, with decay times much shorter than radiative ones, which allow us to produce a NMR spectrum. This non-radiative mechanism are induced by fluctuating magnetic fields oscillating at the same $\omega_0$ of the nucleus that has to relax.

There are several proecesses, both intramolecular and intermolecular, that contribute to spin-lattice relaxation. The principal contributor is dipole-dipole interaction. The spin of an excited nucleus interacts with the spins of other magnetic nuclei that are in the same molecule or in nearby molecules. These interactions can induce nuclear spin transitions and exchange. For carbon nuclei, relaxation is fastest if hydrogen atoms are directly bondend.

The key point to understand is that in order to induce the transition needed for spin-lattice relaxation, the fluctuations of the magnetic field have to be at the Larmor frequency of the nucleus you want to relax, in the same way that the external $B_1$ fields has to oscillate at the Larmor frequency to cause the excitation.


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