Nuclei with Other Spins

The general result for the nuclear charge radius and the Darwin-Foldy contribution for a nucleus with arbitrary spin was obtained in [9]. It was shown there that one may write a universal formula for the sum of these contributions irrespective of the spin of the nucleus if the nuclear charge radius is defined with the help of the same form factor for any spin. However, for historic reasons, the definitions of the nuclear charge radius are not universal, and respective formulae have different appearances for different spins. We will discuss here only the most interesting cases of the spin zero and spin one nuclei. 

The case of the spin zero nucleus is the simplest one. For an elementary scalar particle the low momentum nonrelativistic expansion of the photon-scalar vertex starts with the term, and, hence, the respective con- 

---

An isolated nucleus with a spin number other than zero behaves like a small magnet of magnetic moment fl (J T ) where ... 

A nucleus with a spin 1= 0 has no NMR spectrum, it is NMR silent, and does not couple to other nuclei. Important examples are C, 0, and S. We are all familiar with the consequences from C-NMR. Here the predominant isotope is C, an NMR silent nucleus. The NMR active nucleus C has a natural abundance of only 1.1%. Coupling to H is observed and usually deactivated by broadband decoupling. Couphng to carbon is not observed, except in the case of C enrichment, due to probability constraints. 

If the X nucleus is x, a p electron of the bonding pair will tend to be found nearby (because that is energetically favorable for it). The second electron in the bond, which must have a spin if the other is p, will be found mainly at the far end of the bond (because electrons tend to stay apart to reduce their mutual repulsion). Because it is energetically favorable for the spin of Y to be antiparallel to an electron spin, a Y nucleus with p spin has a lower energy than a Y nucleus with a spin ... 

If an unpaired electron is affected by a nucleus with a spin quantum number other than 1/2, a given nucleus will produce more than two lines. For example, a deuterium nucleus ( H) has / = 1 so that M/ = 1,0, or —1. Deuterium atoms will produce a spectrum with three lines. However, unlike the spectrum of hJ, the three lines will be of nearly equal intensity, since there is one state with each value of M/ and the states are nearly equally populated. A nucleus with 7 = 3 would produce seven lines of equal intensity. 

The above relationships emerge from the fact that the protons and neutrons in the nucleus possess spins. Protons will thus form pairs with other protons in the nucleus with opposite spins. Similarly, neutrons will pair with other neutrons in the same nucleus but with opposite spins. In nuclei which have even numbers of protons and neutrons, all the spins will be paired and the spin number / will be zero. However if there is an odd number of either protons or neutrons, the spin quantum number I will have a quantized value of i, 1, etc. If the sum of protons and neutrons is even, I will be zero or a multiple of 1. If the sum is odd, I will be an integral multiple of j. The nuclei with which the organic chemist is most frequently concerned are and both of which have a spin quantum number of Other elements which may be of interest are and P which also have / = i while deuterium, and have a spin quantum number of 1. 

The behavior of a multi-particle system with a symmetric wave function differs markedly from the behavior of a system with an antisymmetric wave function. Particles with integral spin and therefore symmetric wave functions satisfy Bose-Einstein statistics and are called bosons, while particles with antisymmetric wave functions satisfy Fermi-Dirac statistics and are called fermions. Systems of " He atoms (helium-4) and of He atoms (helium-3) provide an excellent illustration. The " He atom is a boson with spin 0 because the spins of the two protons and the two neutrons in the nucleus and of the two electrons are paired. The He atom is a fermion with spin because the single neutron in the nucleus is unpaired. Because these two atoms obey different statistics, the thermodynamic and other macroscopic properties of liquid helium-4 and liquid helium-3 are dramatically different. 

For liquids, the dominant relaxation mechanism is the nuclear-nuclear dipole interaction, in which simple motion of one nucleus with respect to the other is the most common source of relaxation [12, 27]. In the gas phase, however, the physical mechanism of relaxation is often quite different. For gases such as the ones listed above, the dominant mechanism is the spin-rotation interaction, in which molecular collisions alter the rotational state of the molecule, leading to rotation-induced magnetic fluctuations that cause relaxation [27]. The equation governing spin-rotation relaxation is given by... 

Other good news comes in the shape of the 13C nucleus having a spin quantum number of /2. This means that 13C signals are generally sharp as there are no line-broadening quadrupolar relaxation issues to worry about and we don t have to deal with any strange multiplicities. 

In the case of iron, magnetism is due to the unpaired electrons in the 3d-orbitals, which have all parallel spin. These electrons interact with all other electrons of the atom, also the s-electrons that have overlap with the nucleus. As the interaction between electrons with parallel spins is slightly less repulsive than between electrons with anti parallel spins, the s-electron cloud is polarized, which causes the large but also highly localized magnetic field at the nucleus. The field of any externally applied magnet adds vectorially to the internal magnetic field at the nucleus. 

NMR observes the chemistry of only the proton nucleus (though it can observe many other nuclei independently). This means that hetero and metallic chemistry cannot be observed directly. Thus, sulfur, nitrogen, oxygen, and metals cannot be directly analyzed by NMR, though secondary correlations can be obtained from the proton chemistry of the sample. In combination with electron spin resonance (ESR) analyzers that can operate in the fringe fields of the NMR magnet the presence of paramagnetic metals and free radicals can be quantified. 

Briefly, the physical basis of NMR is as follows. A nucleus possessing a spin quantum number I other than zero is endowed also with spin angular momentum, p, given by... 

Some examples to illustrate the use of this spin system notation to distinguish between first- and second-order systems and to explain the concept of magnetic inequivalence will now be discussed. Because is the only spin-1/2 nucleus with 100% natural abundance that forms a wide variety of inorganic ring systems, most of the examples are taken from phosphorus chemistry (for other examples, see Chapter 11). 

The spin of the electron has not been considered in this theory, but it hardly can introduce any appreciable change. We completely agree with Van Vleck s opinion that this theoretical value is more reliable than the result of the extremely difficult experimental determination of the index of refraction. The fact that for helium the experimental value c — 1 = 0,000074 is very different from the above value, is not consistent with a model in which one electron is close to the nucleus, the other farther away. It can be regarded as a further confirmation of Heisenberg s theory assuming equivalent orbits for the two electrons. 

Remember that a proton, or any other nucleus with spin I — /2, can have (2/ + 1) = 2 orientations in an applied magnetic field. The two possible orientations are aligned with (m/ = + and against (w/ = — /2) the applied field. Remember also that the frequency v of an NMR transition is proportional to the magnetic field experienced by the nucleus,... 

---