Zero-spin

Nuclear magnetic resctnance involves the transitions between energy levels of the fourth quantum number, the spin quantum number, and only certain nuclei whose spin is not zero can be studied by this technique. Atoms having both an even number of protons and neutrons have a zero spin for example, carbon 12, oxygen 16 and silicon 28. 

Carbon 12, the most abundant naturally occurring isotope, has zero spin and thus cannot be studied by NMR. On the other hand, its isotope carbon 13 has an extra neutron and can be its low natural occurrence (1.1%) nevertheless makes the task somewhat difficult. Only pulsed NMR can be utilized. 

The notation used for the Slater Condon rules will be the same as used in the text (a.) zero (spin orbital) differenee ... 

A nucleus with non-zero spin acts as a magnetic dipole, giving raise to a vector potential Aa. 

Optically detected magnetic resonance (ODMR) has yielded valuable information about dynamics of long-lived pholoexcitations of conjugated polymers. The technique relies upon the paramagnetic interaction of excitations with an applied magnetic field. For a particle with non-zero spin, placed in a magnetic field, the Hamiltonian is ... 

Spin 1, Mass Zero Particles. Photons.—For a mass zero, spin 1 particle, the set of relativistic wave equations describing the particle is Maxwell s equations. We adopt the vector 9(x) and the pseudovector (x) which are positive energy (frequency) solutions of... 

The state of a particle with zero spin s = 0) may be represented by a state function (r, t) of the spatial coordinates r and the time t. However, the state of a particle having spin 5 (5 7 0) must also depend on some spin variable. We select for this spin variable the component of the spin angular momentum along the z-axis and use the quantum number ms to designate the state. Thus, for a particle in a specific spin state, the state function is denoted by (r, ms, t), where ms has only the (2s + 1) possible values —sh, (—s + )h,... 

The spin of is as expected. Deuterium (2H) could have spin 0 or 1, depending on the relative alignment of the proton and neutron spins. It is observed to be 1. 4He has zero spin. From these observations a set of empirical rules specifying the spins of all important nuclei has been derived. 

The NSE principle - as described above - only works if the neutron spin is not affected by the scattering process (some exceptions hke complete spin-flip could be tolerated and would just replace the n-flipper). A related problem occurs if the scattering nuclei have a non-zero spin and their scattering power depends significantly on the relative orientation of nuclear and neutron spin. 

The Slater Condon rule for zero (spin orbital) difference with N electrons in N spin orbitals is ... 

The only term depending upon % is the last on the left hand side and this is just the spin-orbit couphng. If this term is dropped x-independent solutions, corresponding to zero spin-orbit couphng, are obtained. However, the large relativistic shifts, the mass velocity and Darwin terms are retained. In Fig. 3, for example, the relativistic levels remain in place but each of the spin-orbit split pairs is replaced by the average energy level... 

A/=+l. (One way of thinking about this is that the photon has zero spin and one unit of angular momentum. Conservation of spin and angular momentum then produces these rules.) For a sodium atom, for example, the 35 electron can absorb one photon and go to the 3jo level. (There is no restriction on changes of the principal quantum number.) The 3s electron will not, however, go to the 3d or 4s level. Figure 8.1 illustrates allowed and forbidden transitions. 

Figure 6.10 In the absence of spin polarization, which corresponds to the ROHF picture, there is zero spin density in the plane containing the atoms of the methyl radical. Accounting for spin polarization, which corresponds to the UHF picture, results in a build-up of negative spin density (represented as a shaded region) in the same plane...

If a nucleus with a non-zero spin number, which can be compared to a small magnet, is exposed to a magnetic field 7i0, with an angle 6 with the spin vector, fl and Bq will become coupled. This coupling modifies the potential energy E of the nucleus. However, fl will not necessarily align itself in the direction of the external field, in contrast to the action of a compass. 

A nucleus, represented by z%, will have a non-zero spin number / giving an NMR signal as long as the number of protons Z and neutrons A are not both even numbers. For example, H, j-C, F and 15P all have a spin number / = 1/2 while ]H (deuterium, D) and N have 7=1. However, nuclei such as C, 2He, gO, f Si and cannot be studied by NMR. In fact, more than half of the stable nuclei known (at least one isotope per element) yield NMR signals. However, sensitivity varies enormously depending on the nucleus. Hence the proton, also known as H, or the nucleus 19F, are easier to detect than 13C, which is thousands of times less sensitive than the proton because of its weak natural isotopic abundance. 

For dissimilar pairs, the parameter ys equals zero and we have Eq. 5.36. Like pairs of zero spin are bosons and all odd-numbered partial waves are ruled out by the requirement of even wavefunctions of the pair this calls for ys = 1. In general, for like pairs, the symmetry parameter ys will be between -1 and 1, depending on the monomer spins (fermions or bosons) and the various total spin functions of the pair. A simple example is considered below (p. 288ff.). If vibrational states are excited, the radial wavefunctions xp must be obtained from the vibrationally averaged potential, Fq(R). The functions gf(R) and gM(R) are similar to the pair distribution function, namely [294]... 

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