2016
DOI: 10.1016/j.adt.2016.02.003
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Dependence of nuclear quadrupole resonance transitions on the electric field gradient asymmetry parameter for nuclides with half-integer spins

Abstract: Allowed transition energies and eigenstate expansions have been calculated and tabulated in numerical form as functions of the electric field gradient asymmetry parameter for the zero field Hamiltonian of quadrupolar nuclides with I = 3/2, 5/2, 7/2, and 9/2. These results are essential to interpret nuclear quadrupole resonance (NQR) spectra and extract accurate values of the electric field gradient tensors. Applications of NQR methods to studies of electronic structure in heavy element systems are proposed.

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Cited by 6 publications
(4 citation statements)
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“…Here, i and j are labels for each of the spin states, i.e., 1 / 2 , 3 / 2 , 5 / 2 , 7 / 2 . More interestingly, numerical solutions (i.e., Δ E ij as well as each E i ) have been tabulated for all η Q values (0 ≤ η Q ≤ 1). Reciprocally, all the NQR frequencies can be obtained from C Q and η Q . Since antimony possesses two quadrupolar isotopes, a spin- 5 / 2 and a spin- 7 / 2 , five different pure NQR frequencies are available (excluding overtones).…”
Section: Resultsmentioning
confidence: 99%
“…Here, i and j are labels for each of the spin states, i.e., 1 / 2 , 3 / 2 , 5 / 2 , 7 / 2 . More interestingly, numerical solutions (i.e., Δ E ij as well as each E i ) have been tabulated for all η Q values (0 ≤ η Q ≤ 1). Reciprocally, all the NQR frequencies can be obtained from C Q and η Q . Since antimony possesses two quadrupolar isotopes, a spin- 5 / 2 and a spin- 7 / 2 , five different pure NQR frequencies are available (excluding overtones).…”
Section: Resultsmentioning
confidence: 99%
“…The EFG principal component V 33 of a spin- 3 / 2 nuclide can be determined (in SI units) with eq : , where ν is the frequency of the transition at zero field. Assuming that η = 1 based on the 35 Cl and 37 Cl NMR line shapes, the value of V 33 obtained using eq is 0.577 au.…”
Section: Discussionmentioning
confidence: 99%
“…0 ≤ η ≤ 1 and the larger η becomes, the more mixing there is between sublevels. The Hamiltonian (1) can be diagonalized to find the eigenvalues and eigenvectors, which are shown here for the ground I g = 5/2 and excited (isomer) state I e = 3/2 [45][46][47][48]. Figures 17 and 19 show the energy of the split ground and excited states where C g/e = eQ g/e V zz /(4I g/e (2I g/e −1)).…”
Section: Discussionmentioning
confidence: 99%