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Finite nucleus magnetic interactions

Here, h r) is a one-particle Dirac Hamiltonian for electron in a field of the finite size nucleus and y is a potential of the interelectron interaction. In order to take into account the retarding effect and magnetic interaction in the lowest order on parameter a (a is the fine structure constant), one could write [23]... [Pg.233]

Magnetic interactions, finite mass of the nucleus, special relativity, to name a few. [Pg.157]

Unlike the electric quadrupole moment, contributions to the magnetic interactions with the nucleus can come from (electron) spinors with j = 1/2, and thus the magnitude of the finite nucleus effect is likely to be larger than for the electric quadrupole moment. [Pg.255]

Before formally developing the tensor it is perhaps worthwhile to discuss the various types of interactions which contribute to it. The coupling between nuclear and electron magnetic moments are conveniently divided into those which are isotropic and those which depend on orientation. The former is the result of the impaired electron having a finite probability of being at the nucleus. This type of interaction is termed the contact interaction, and is described by the constant,... [Pg.336]

In actual practice the unpaired electron is not free. It is generally associated with one or more nuclei, which may have a nuclear spin magnetic moment. This moment generates a magnetic field at the location of the unpaired electron, due to the so-called contact or Fermi hyperfine interaction (the electron has a finite probability of penetrating to the atomic nucleus) and to the through-space dipolar interaction between nuclear and electronic magnetic spin moment, represented by... [Pg.100]

When full, isotropic averaging is achieved, most of the interactions between a nucleus and its environment are averaged to zero. One exception to this is the amount by which the applied magnetic field is modified by the electrons circulating around the nucleus (shielding), which is averaged to a finite (non-zero) value. This value depends on the... [Pg.57]

We next consider the effect of finite nuclear size on the nuclear spin Hamiltonian. The electric moments were derived by considering the Coulomb interaction of the nuclear charge density, expanded in a multipole series, with the electrons. By analogy, the magnetic moments are derived by considering the Gaunt interaction of the nucleus with the electrons. It is at this point that we must consider, at least as a formal entity, the nuclear wave function, and from it obtain a nuclear spin density that interacts with the electron spin density. [Pg.253]


See other pages where Finite nucleus magnetic interactions is mentioned: [Pg.249]    [Pg.252]    [Pg.397]    [Pg.131]    [Pg.34]    [Pg.73]    [Pg.131]    [Pg.182]    [Pg.10]    [Pg.90]    [Pg.224]    [Pg.67]    [Pg.257]    [Pg.68]    [Pg.209]    [Pg.60]    [Pg.279]    [Pg.58]    [Pg.29]    [Pg.43]    [Pg.217]    [Pg.57]    [Pg.314]    [Pg.177]    [Pg.225]    [Pg.57]    [Pg.58]    [Pg.339]    [Pg.342]    [Pg.912]    [Pg.3338]    [Pg.5]    [Pg.2450]    [Pg.253]    [Pg.587]    [Pg.320]    [Pg.191]    [Pg.246]    [Pg.711]    [Pg.510]    [Pg.512]    [Pg.252]   
See also in sourсe #XX -- [ Pg.255 ]




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