Monday, November 24, 2008
90 Degree Pulses for I = n/2 Quadrupolar Nuclei in the Solid State
The 90 degree pulse for an I = n/2 quadrupolar nucleus in the solid state depends on the strength of the rf pulse with respect to the quadrupolar frequency. If the strength of the pulse is much greater than the quadrupolar frequency, the pulse is non-selective and excites all transitions equally. If however it is much less than the quadrupolar frequency, then the pulse is selective to the central (m = 1/2 - m = -1/2) transition. The duration of the pulse producing a maximum signal is shorter for selective vs. non-selective pulses at a similar power level. In solution, where the quadrupolar interactions is averaged by random isotropic molecular motion or in the solid state, if the symmetry around the I = n/2 nucleus is cubic, the quadrupolar frequency is small with respect to the strength of the rf pulses and the pulses are non-selective. When the symmetry around the I = n/2 nucleus in the solid state is non-cubic, the quadrupolar frequency is significant and the pulses are very often selective to the central transition. This is illustrated in the figures below for the 23Na MAS spectrum of a mixture of NaCl (cubic) and Na2SO4 (non-cubic). The first figure shows the 23Na MAS spectrum labelling each component of the mixture. The second figure shows the effect of increasing the pulse duration. One can clearly see that the 90 degree pulse for NaCl is close to twice that of Na2SO4.
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5 comments:
Hi,
I would like to understand how a quadrupolar frequency effects the intensity evolution. Does the quadrupolar interaction prevent inefficient flipping of spins when pulses are applied? I guess I don't understand how the coupling between the quadrupole moment and EFG would effect the 90 or 180 degree time from a qualitative perspective for say two systems that have different quadrupolar frequencies.
Any advice?
Thomas
Dear Thomas,
Thank you for your question. The answer has to do with the fact that for small quadrupolar coupling constants an rf pulse is nonselective. It excites all of the transitions. For large quadrupolar couplings a typical rf pulse is selective only towards the central transition, the satellite transitions being too far off resonance to be excited. A detailed answer to your question is given in Melinda Duer's book, "Solid-State NMR Spectroscopy" chapter 5 section 5.5.2, page 249. I hope this helps.
Glenn
Hi Glen,
I'm just curious to know how the 90 Degree Pulses for I = 1 (2H) Quadrupolar Nuclei in the Solid State will look like?? Also, non-selective and selective pulse excitation effect will come into the picture?
Unknown,
2H is a spin I=1 nuclide with a very small quadrupole moment. As a result its two transitions are very close in frequency, typically tens on kHz apart. The frequencies of the transitions depend in the orientation of each crystallite in the sample. In polycrystalline samples, the transitions overlap to produce a powder spectrum. Since the transitions are very close in frequency (and overlap in the case of polycrystalline samples), the 90 deg pulse for one transition is generally the same as that for the other transition. If one had a single crystal sample producing a sharp doublet spectrum, one could use a low power selective pulse to selectively excite one of the transitions. In a polycrystalline sample one could use a low power selective pulse to irradiate the "shoulder" of a Pake doublet to selectively excite one transition.
Glenn
Thank you for this very clear and concise explanation. We've been banging our heads against the wall over some recent 27Al data, and now have a much better understanding of what's going on. You do a tremendous service to our community Glenn! Happy Holidays!
Nicole
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