Electric Field Gradient in Aluminium Due to Vacancy and Muon * M
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Electric Field Gradient in Aluminium due to Vacancy and Muon * M. J. Ponnambalam University of the West Indies Z. Naturforsch. 47a, 45-48 (1992); received July 29, 1991 The electric field gradients (EFG) in aluminium due to a monovacancy and the interstitial muon are evaluated. The valence effect EFG qv is calculated using perturbed electron density ön(r) values obtained from density functional theory in an analytic expression which is valid at all distances from the impurity. The size effect EFG qs is evaluated using a new oscillatory form for the near neighbour (nn) displacements. The numerical values of qs are computed using fractional nn displacements available in the literature. For the total EFG good agreement with experiment is obtained without the use of any parameter. Key words: Nuclear Quadrupole Interaction, Electric Field Gradient 1. Introduction they used the following oscillatory form for the dis- placement of the host ions: In a cubic crystal, the electric field gradient (EFG) u(r) = A cos (2/c r + </>) r/r3, (2) vanishes at the site of any ion due to cubic symmetry F [1]. The presence of an impurity in a cubic host locally where kF is the Fermi wave vector while A and </> are destroys the cubic symmetry and thus produces a non- constants. vanishing EFG at the site of the host near neighbours In the case of a muon and a vacancy in Al and Cu, (nn) - thus leading to a non-vanishing nuclear quadru- Tripathi et al. [8] and Schmidt et al. [9] have carried pole interaction of the host nuclei. Conventionally, for out a lattice sum in the evaluation of qs, using avail- computational convenience the EFG is split into a) the able lattice displacements. Recently, using KKR- valence effect EFG qv, arising from the perturbation Green's function method, Dederichs et al. [10] have <5n(r) of the ambient electron density, and b) the size evaluated the EFG's in Cu alloys. They find that much effect EFG qs, arising from the strain on the host nn of the EFG is caused by the d-electrons of the Cu ion due to the misfit of the impurity ion in a host ion site. under consideration. Asymptotic and pre-asymptotic solutions for qv In this paper, we present our results on the EFG in have been obtained [2, 3]. Sagalyn and Alexander [4] Al due to a mono-vacancy and a muon in the octa- proposed a new method to evaluate qv using integra- hedral interstitial. Unlike Cu alloys, wherein the EFG tion and a special geometry. In addition, they evalu- is complicated by the d-electrons of Cu [10], Al is a ated qs using the point ion model of Faulkner [5] and simple metal with completely filled inner shells. the expression Hence, qv is evaluated using the analytic expression of u(r) = Dr/r3 (1) PJ [7] and perturbed electron density ön(r) values ob- tained from density functional theory. A new, oscilla- for the displacement of the host ions. Here, D is a tory form for the nn displacement is used in the alge- constant. This formalism has been used by others with braic solution of qs, while its numerical value is some modifications [6]. obtained using available nn displacements. Good A solution for qv, valid at all distances from the agreement is obtained with experiment without the impurity, was given by Ponnambalam and Jena (PJ) use of any parameter. The details of the Theory are [7], who showed that their results went over to the presented in Section 2. This is followed by Results and earlier asymptotic and pre-asymptotic ones at the ap- Discussion in Section 3 and the Summary in Section 4. propriate limits. In addition, in the evaluation of qs, * Presented at the Xlth International Symposium on 2. Theory Nuclear Quadrupole Resonance Spectroscopy, London, United Kingdom, July 15-19, 1991. Reprint requests to Dr. M. J. Ponnambalam, Physics Dept., Let V(r) be the electrostatic potential at the site of UWI, Kingston 7, Jamaica (West Indies). any host ion surrounding the impurity at the origin in 0932-0784 / 92 / 0100-0045 $ 01.30/0. - Please order a reprint rather than making your own copy. 46 M. J. Ponnambalam • Electric Field Gradient in Aluminium due to Vacancy and Muon a cubic host. If V^ is its derivative with respect to x, the nn displacements are taken to be radial. Since and Xj, then different nn's may move differently (i.e. some inward and some outward), (1) is oversimplified. This problem eq? =V -ö V2V/3. (3) j ij ij is overcome by the oscillatory form of u(r) in (2), which v Diagonalizing qjj, the principal components of q are reduces to (1) if (2/cFr + 0) = O. obtained. When V(r) and hence the perturbed electron Now, in a simplified model, the displacement of an density ön(r) is approximated to be spherically sym- nn is influenced by the difference in the sizes and metric, the principal components of <jv are parallel (||) charges of the impurity and host ion. In the case of the and perpendicular (1) to the direction connecting the latter, one is aware that the impurity-host charge dif- impurity to the host and are given by the following ference tends to be screened out by the re-arrangement analytic expressions [7]: of the conduction electrons as given by the perturbed electron density <5n(r). Hence it is not unreasonable to 8 n ön(r) + expect the displacement field to exhibit the Friedel 2 {Zeff-Z(r)} 4nr oscillations of ön(r). Consequently, the nn displace- v ments are assumed to be of the form = -2q[ = -2ql\ (4) u(r) = A cos(2/cFr + 4>) (9) where ot(kF) is the Bloch enhancement factor which takes into account the departure of the electronic where A and 0 are constants while kr is the Fermi wavefunctions from plane wave character as well as wave vector. The choices of 2 kF as the scaling factor anti-shielding effects [2] and of r4 in the denominator (instead of the r3 in (2)) have no fundamental justification except that now Z(r) = ]ön(r')dv' and Zeff = Z(oo). (5) |i/(r)| is of the same form as the asymptotic ön(r) [2]. o Using (7), (8), and (9) in (6), the following result is The components qfj of the size effect EFG, linear in obtained for the pth nn: the strain components e^, are given by [4, 7] ^ qh = Lp[Su - f (1 -öu)] (2Su - 6xf xj/dj), (10) 3 e qfj = i - Fj 2) ^e!! - jX + 2 (1 - <y F44 gy, 9j/2Z a[ fp(\ + dp/3). (11) Here, dp is the position of the pth nn with fractional where FJ/s are components of a fourth rank tensor. displacement / , Z is the host valency and aL the For an FCC host, these are given, in the point ion lattice constant. It should be noted that in [7] Lp in- model as [11] volves cos (2 /cF r + 0) which is evaluated using the 3 F11-F12=-3F^=9Ze/d 1 = n]/2Ze/al, (7) equilibrium conditions of elasticity theory. Herein, however, Lp is re-expressed in terms of fp whose values where Z is the host's valency, d is the first nn distance x are available from lattice statics calculations. and a the lattice constant. Equation (7) with Z=1 L Diagonalizing qfj in (10), the principal components agrees with the result of Faulkner [5]. Faulkner's of qs are obtained as result, which was derived for monovalent Cu, has l 2 s been used incorrectly for polyvalent Al by the earlier ql = —4L =-2q l =-2q ± , along [100], groups [6, 7]. In this paper (7), which is valid for poly- 2 4 = L = q{*/2 =-q s/3, along [110], valent hosts like Al, is used. The elastic strain tensor s 2 is given in terms of the nn displacements u(r) as = 8L/3 =-2gi = —2q along [111], j2.463L = 7.389q= -0.881 q2 s, along [211]. 1 / dU: dU: ± (12) u + 2 0X: ÖX, (8) It should be noted that along [211] none of the com- Thus, the computation of qfj requires an analytic ex- ponents of qs are exactly along the parallel direction, pression for u(r). although the first component's direction is fairly close Both lattice statics calculations [12] and EXAFS to that. (In this case one may combine qjj with qfj and measurements [13] indicate that the displacements are diagonalize the resultant matrix. However, this result radial along the symmetry directions [100], [110] and differs very little from that using the approximate one [111]. Due to this and due to simplicity in calculations, in (12).) 47 M. J. Ponnambalam • Electric Field Gradient in Aluminium due to Vacancy and Muon Table 1. The largest component of the total EFG q in nm 3 nn of Al-vacancy are written (in nm 3) as and the asymmetry parameter rj for Al host. q|| =(-1700+1694)- 94= —100, Impurity nn Exp. [18] /P[ll, 17] Theory q\= (850- 847) —188= —185, <7 n <1 1 q\= (850- 847) + 282 = 285; >7 = 0.30. (14) Vacancy 1 -0.0174 285 0.30 280 0.65 2 -0.0093 376 0 - 0 Here, the first terms are contributions from eZeff, the 3 -0.0023 98 0.78 93 - 4 -0.0007 43 0.26 69 - underlined terms are from the spherically symmetric Muon 1 0.0273 166 0 180 0 ön(r) and the last terms are from the size EFG.