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Magnetism of the Localized Electrons on the Atom 1. The hydrogenic atom and angular momentum 2. The many-electron atom 3. Spin-orbit coupling 4. The Zeeman interaction 5. Ions in solids 6. Paramagnetism Comments and corrections please: [email protected] Dublin January 2007 1 1. The hydrogenic atom and angular momentum Consider a single electron in a central potential. A hydrogenic atom is composed of a nucleus of charge Ze at the origin and an electron at r,!,". First, consider a single electron in a central potential "e = Ze/4!e 0r z 2 2 2 ! = - (" /2m)# - Ze /4!e 0r l -e r In polar coordinates: ! #2 = #2/#r2 +(2/r)#/#r + 1/r2{#2/#!2 + cot!$/#! + (1/sin2!)#2/#2"} Ze "l The term in parentheses is -l2. Schrödinger’s equation is ! % = E% x The wave function % means that the probability of finding the electron in a small volume dV ar r is %*(r)%(r)dV. (%* is the complex conjugate of %). Eigenfunctions of the Schrödinger equation are of the form %(r,!,") = R(r)&(!)'("). ( The angular part &(!)'(") is written as Ylml(!,"). The spherical harmonics Ylml(!,") depend on two integers l, ml, where l is $ 0 and |ml| % l. '(") = exp(iml") where ml = 0, ±1, ±2 ....... The z-component of orbital angular momentum, represented by the operator lz = -i"#/#", has eigenvalues <'| lz |'> = ml". &(!) = Plml(cos!), are the associated Legendre polynomials with l $ |ml|, so ml = 0, ±1, ±2,....±l. Dublin January 2007 2 z The square of the orbital angular momentum l2 has eigenvalues l(l+1)". The orbital angular momentum has magnitude )[l(l+1)]" and its projection along z can have any value from -l" to +l". The quantities lz and l2 can be measured simultaneously (the operators commute). In the vector model, The total angular momentum is a vector which precesses around z. m l" )[l(l+1)]" Spherical harmonics. s p d f Y00 = )(1/4!) Y10 = )(3/4*) cos ! Y20 = )(5/16*)(3cos2! - 1) Y1±1 = ±)(3/8*) sin ! e±i" Y2±1 = ±)(15/8*) sin! cos! e±i" Y2±2 = )(15/32*) sin2! e±2i" Y30 = )(7/16*)(5cos3! - 3cos!) Y3±1 = ±)(21/64*)(5cos2! - 1)sin! e±i" Y3±3 = ±)(35/64*) sin3! e±3i" Y3±2 = )(105/32*) sin2!cos! e±2i" Dublin January 2007 3 Spherical harmonics. n =1 l=0 s n =2 l=1 ml =0 p ml =±1 n =3 l=2 ml =0 ml =±1 d ml =±2 Dublin January 2007 4 The radial part of the wavefunction R (r) •The radial part R(r) depends on l and also on n, the total quantum number; n > l; hence l = 0, 1, ......(n-1). 0.5 10 2 ! |Rnl (! )| 0.4 R(r) = Vnl(Zr/na0)exp[-(Zr/na0)] 0.3 "2/me2 = 1. Here a0 = 4!+0 = 52.9 pm is the first Bohr radius, the basic length scale in atomic physics. 2 V10 0.6 21 20 0.2 32 31 0.1 30 43 42 41 40 0.0 0 5 10 15 20 25 30 ! = r / a0 The energy levels of the 1-electron atom are E = -Zme4/8h2+02n2 = -ZR0/n2 The quantity R0 = me4/8h2+0 = 13.6 eV is the Rydberg, the basic energy in atomic physics. For the central Coulomb potential "e, the potential energy V(r) depends only on r, not on ! or ". E depends only on n. Dublin January 2007 5 The three quantum numbers n, l, ml denote an orbital, a spatial distribution of electronic charge. Orbitals are denoted nx, x = s, p, d, f for l = 0, 1, 2, 3. Each orbital can accommodate up to two electrons with spin ms = ±1/2. No two electrons can be in a state with the same four quantum numbers (Pauli exclusion principle). The hydrogenic orbitals are listed in the table n l ml ms No of states 1s 1 0 0 ±1/2 2 2s 2 0 0 ±1/2 2 2p 2 1 0,±1 ±1/2 6 3s 3 0 0 ±1/2 2 3p 3 1 0,±1 ±1/2 6 3d 3 2 0,±1,±2 ±1/2 10 4s 4 0 0 ±1/2 2 4p 4 1 0,±1 ±1/2 6 4d 4 2 0,±1,±2 ±1/2 10 4f 4 3 0,±1,±2,±3 ±1/2 14 •The Pauli principle states that no two electrons can have the same four quantum numbers. Each orbital can be occupied by at most two electrons with opposite spin. Dublin January 2007 6 2. The many-electron atom Hartree-Foch approximation, each electron sees a different self-consistent potential Vl(r). ! E Coulomb interactions between electrons n= n 0 6s 6p 4p 4s 6d 1 2 3 4 5 6 1s 2s 3s 4s 5s 6s 2p 3p 4p 5p 6p 3d 4d 5d 6d 4f 5f 6f 5g 6g 6f 4 3 3d 3p 3s 2 2p 2s 1s l s p d f Dublin January 2007 7 Dublin January 2007 8 Example: The carbon atom 1s22s22p2 There are no options for the first four electrons. There must be a pair of them with opposite spin in each s orbital. However, there are 15 ways of accomodating two electrons in the p orbitals. The 15 states fall into three groups (terms). Notation: States with L = 0,1,2,3,4,5,6 are S,P,D,F,G,H 2S+1X Dublin January 2007 9 Term 1S 3P 1D S J L S (ML, MS) 0 1 2 0 1 0 (0,0) (1,1)(1,0)(1,-1)(0,1)(0,0)(0.-1)(-1,1)(-1,0)(-1,-1) (2,0)(1,0) (2,1)(2,0)(0,0)(-1,0)(-2,0) L The terms are widely separated in energy; one is the lowest Addition of L and S in the vector model Finally we need to couple the spin and orbital angular momentum to form a resultant J. J = L + S L-S % J % L+S Hund’s rules; A prescription to give the ground state of a multi-electron atom 1) First maximize S for the configuration 2) Then maximize L consistent with that S 3) Finally couple L and S; J = L - S if shell is < half-full; J = L + S if shell is > half-full. In the example, S = 1, L = 1, J = 0. The ground state of carbon is 3P0, which is nonmagnetic (J = 0). General notation for multiplets is 2S+1XJ where X = S, P, D ...... for L = 0, 1, 2 ..... In spectroscopy, the energy unit cm-1 is used. Handy conversions are:1 eV , 11605 K and 1 cm-1 , 1.44 K Dublin January 2007 10 Some examples: Fe3+ 3d5 S = 5/2 L = 0 -----| ooooo J = 5/2 6S Ni2+ S=1 -----|...oo J=4 3F 3d8 L=3 5/2 4 Nd3+ 4f3 S = 3/2 L = 6 ---oooo |ooooooo 4I J = 9/2 9/2 Dy3+ 4f9 S = 5/2 L = 5 -------|..ooooo 6H J = 15/2 15/2 Dublin January 2007 11 3. Spin-orbit coupling This relatively-weak relativistic interaction is responsible for Hund's third rule. In the multi-electron atom, the spin-orbit term in the Hamiltonian can be written as ! so = /L.S / is > 0 for the first half of the 3d or 4f series and < 0 for the second half. It becomes large in heavy elements. / is related to the one-electron spin-orbit coupling constant 0 by / = ±0/2S for the first and second halves of the series. The resultant angular momentum (see above) is J =L±S The identity J2 = L2 + S2 + 2 L.S is used to evaluate Hso.The eigenvalues of J2 are J(J + 1) "2 etc, hence L.S can be calculated. 1 3+ Spin-orbit coupling constants in the 3d and 4f series L ion 4f Ce 920 4f2 Pr3+ 540 4f3 Nd3+ 430 3d1 Ti3+ 124 4f5 Sm3+ 350 3d2 Ti2+ 88 4f8 Tb3+ -410 3d3 V2+ 82 4f9 Dy3+ -550 3d4 Cr2+ 85 4f10 Ho3+ -780 3d6 Fe2+ -164 4f11 Er3+ -1170 3d7 Co2+ -272 4f12 Tm3+ -1900 3d8 Ni2+ -493 4f13 Yb3+ -4140 Dublin January 2007 12 9 L S J 8 7 6 5 4 3 2 1 0 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Application of Hund’s rules to the trivalent rare-earth ions. Dublin January 2007 13 Magnetic properties of free atoms: only the elements marked in bold are nonmagnetic, w ith J = 0 Dublin January 2007 14 4. The Zeeman interaction The magnetic moment of an ion is represented by the term m = (µB/") (L + 2S) The Zeeman Hamiltonian for the magnetic moment in a field B applied along z is –m.B ! Z = (µB/")(L + 2S).B z The vector model of the atom, including magnetic moments. First project m onto J. J then precesses around z. We define the g-factor for the atom or ion as the ratio of the component of magnetic moment along J in units of µB to the magnitude of the angular momentum in units of ". g = -(m.J/µB)/(J2/") = m.J/J(J + 1)"µB.e but S m J S L m.J = (µB/"){(L + 2S).(L + S)} J2 = J(J + 1)"2; (µB/"){(L2 + 3L.S + 2S2)} (µB/"){(L2 + 2S2 + (3/2)(J2 - L2 - S2)} (µB/"){((3/2)J2 – (1/2)L2 + (1/2)S2)} (µB /"){((3/2)J(J + 1) – (1/2)L(L + 1) + (1/2)S(S + 1)} Jz = MJ" hence g = 3/2 + {S(S+1) - L(L+1)}/2J(J+1) Dublin January 2007 15 The g-factor is also the ratio of the z-component of magnetic moment (in units of µB) to the z component of angular momentum (in units of ") mz/Jz = m.J/J2 = gµB/ " The magnetic Zeeman energy is EZ = –mz B. This is –(mZ /Jz).(Jz B) = (gµB&/")JzB Hence EZ = -gµBMJB MJ -5 / 2 Note the magnitudes of the energies involved, with g = 2 and µB = 9.27 10-24 J T-1. The splitting of two J=5/2 adjacent energy levels is gµBB. For B = 1 T, this is only ' 2 10-23 J, equivalent to 1.4 K. [kB = 1.38 10-23 J K-1] -3 / 2 -1 / 2 1/2 3/2 5/2 In a large field at low temperature the moment is gµBB saturated at the value -gµBJ Bohr magnetons. The effect of applying a magnetic field on an ion with J = 5/2. Susceptibility gives meff2 = g2µB2J(J+1) Dublin January 2007 16 Co2+ free ion Energy levels of Co2+ ion, 3d7. Note that the Zeeman splitting is not to scale. Dublin January 2007 17 Data on rare-earth ions Dublin January 2007 18 Data on 3d ions Dublin January 2007 19 Summary: For free ions: ( Filled electronic shells are not magnetic. A - and a . electron is paired in each orbital ( Only partly-filled shells may possess a magnetic moment ( The magnetic moment is related to the total angular momentum by m = g(µB/ ")J, where J is the total quantum number given by Hund’s rules. The third point has to be modified for ions in solids ( Orbital angular momentum for 3d ions is quenched. The spin-only moment is m = gµBS, with g =2. ( Certain crystallographic directions become easy axes of magnetization – magnetocrystalline anisotropy. Dublin January 2007 20 5. Ions in solids When an ion is embedded in a solid, the Coulomb interaction of the charge distribution of the ion with the potential 1cf created by the surrounding charges is the crystal-field interaction. ! cf = ( 1cf(r)2(r) d3r Dublin January 2007 21 The Hamiltonian now has four terms ! = ! 0 + ! so+ ! cf+ ! Z Typical magnitudes of energy terms (in K) !0 ! cf ! Z in 1 T 3d 1 - 5 104 102 -103 1 - 104 1 1 - 6 105 1 - 5 103 !3 102 1 4f ! so ! so must be considered before ! cf for 4f ions, and the converse for 3d ions. Hence J is a good quantum number for 4f ions, but S is a good quantum number for 3d ions. The 4f electrons are generally localized, and 3d electrons are localized in oxides and other ionic compounds. Dublin January 2007 22 The most common coordination for 3d ions is 6-fold (octahedral) or 4-fold (tetrahedral). Both have cubic symmetry, if undistorted. The crystal field can be estimated from a point-charge sum. q 5.1 One-electron states Octahedral and tetrahedral sites. To demonstrate quenching of orbital angular momentum, we consider the l = 1 states %0, %1, %-1 corresponding to ml = 0, ±1. %0 = R(r) cos ! %±1 = R(r) sin ! exp {±3"} The functions are eigenstates in the central potential V(r) but they are not eigenstates of ! cf. Suppose the oxygens can be represented by point charges q at their centres, then for the octahedron, ! cf = eVcf = Dq(x4 +y4 +z4 - 3y2z2 -3z2x2 -3x2y2) where D ' e/4pe oa6. But %±1 are not eigenfunctions of Vcf, e.g. (%i*Vcf%jdV) 4ij, where i,j = -1, 0, 1. We seek linear combinations that are eigenfunctions, namely %0 = R(r) cos ! (1/*2)(%1 + %-1)= R’(r)sin!cos" (1/*2)(%1 - %-1)= R’(r)sin!sin" = zR(r) = pz = xR(r) = px = yR(r) = py Dublin January 2007 23 Note that the z-component of angular momentum; lz" = i"#/#" is zero for these wavefunctions. Hence the orbital angular momentum is quenched. The same applies to 3d orbitals; the eigenfunctions there are dxy = (1/*2)(%2 - %-2) = dyz = (1/*2)(%1 - %-1) = dzx = (1/*2)(%1 + %-1) = dx2-y2 = (1/*2)(%2 + %-2) = = d3z2-r2 = %0 R’(r)sin2!sin2" R’(r)sin!cos!sin" R’(r)sin!cos!cos" R’(r)sin2!cos2" R’(r)(3cos2! 5 1) The three p-orbitals are degenerate in a cubic crystal field, whether octahedral or tetrahedral, whereas the five d-orbitals split into a group of three t2g and a group of two eg orbitals ' ' ' ' ' xyR(r) yzR(r) zxR(r) (x2-y2)R(r) (3z2-r2)R(r) t2g orbitals eg orbitals dx2-y2, dz2, eg dxy,dyz, dzx px,, py, pz oct / tet 6Dq 10Dq dxy,dyz, dzx oct t2 t2g dx2-y2, dz2, e tet Notation; a or b denote a nondegenerate single-electron orbital, e a twofold degenerate orbital and t a threefold degenerate orbital. Capital letters refer to multi-electron states. a, A are nondegenerate and symmetric with respect to the principal axis of symmetry (the sign of the wavefunction is unchanged), b. B are antisymmetric with respect to the principal axis (the sign of the wavefunction changes). Subscripts g and u indicate whether the wavefunction is symmetric or antisymmetric under inversion. 1 refers to mirror planes parallel to a symmetry axis, 2 refers to diagonal mirror planes. Dublin January 2007 24 s, p and d orbitals in the crystal field Dublin January 2007 25 As the site symmetry is reduced, the degeneracy of the one-electron energy levels is raised. For example, a tetragonal extension of the octahedron along the z-axis will lower pz and raise px and py. The effect on the d-states is shown below. The degeneracy of the d-levels in different symmetry is shown in the table. Px,py pz The effect of a tetragonal distortion of octahedral symmetry on the one-electron energy levels. The splitting of the 1-electron levels in different symmetry . l Cubic Tetragonal Trigonal Rhombohedral s 1 1 1 1 1 p 2 3 1,2 1,2 1,1,1 d 3 2,3 1,1,1,2 1,2,2 1,1,1,1,1 f 4 1,3,3 1,1,1,2,2 1,1,1,2,2 1,1,1,1,1,1,1 Dublin January 2007 26 One-electron energy diagrams b1g 0 : e 1/2: 1/20 a1g 3/590 90 b2g 2/590 e 8 7 1/38 eg 2/37 6 4 e a1 field-free ion octahedral Oh tetragonal D4h trigonal C3v Dublin January 2007 monoclinic C2 27 z z Jahn Teller Effect x x x y y x x z x z z 8 8 Dcfse E 90 4 4 Dublin January 2007 •A system with a single electron (or hole) in a degenerate level will tend to distort spontaneously. The effect is particularly strong for d4 and d9 ions in octahedral symmetry (Mn3+, Cu2+) which can lower their energy by distorting the crystal environment. This is the Jahn-Teller effect. If the local strain is +, the energy change 9 E = -A+ +B+2, where the first term is the crystal-field stabilization energy Dcfse and the second term is the increased elastic energy. The J-T distorsion may be static or dynamic. 28 dx2-y2, dz2 eg dxy, dyz, dxz t2g 3/59o dxy, dyz, dxz t2 2/59c 2/59t 9o E 9t 9c 3/59t 3/59c 2/59o e t2g dx2-y2, dz2 dxy, dyz, dxz eg dx2-y2, dz2 cubic tetrahedral spherical Dublin January 2007 octahedral 29 T1g 5.2 Many-electron states A2g T1g P In insulators, the electrons in an unfilled shell interact strongly with each other giving rise to a series of sharp energy levels which are determined by the action of the crystal field on the orbital terms of the free atom. The spacing of theses levels may be determined by spectroscopy, and the crystal-field determined. T2g F T1g T2g T1g A2g E Orgel Diagrams !0 These diagrams show the effect of a cubic crystal field on the Hund’s rule ground state term. Since a half-filled shell has spherical symmetry, the cases dn and d5+n are equivalent. Also, since a hole is the absence of an electron, the cases dn and d10-n are related. !0 0 d3, d8 octahedral (d2, d7 tetrahedral) d2, d7 octahedral (d3, d8 tetrahedral) Eg T2g E D S A1 A1 T2g Eg !0 ; Do 0 Dt < 4 9 d , d octahedral (d1, d6 tetrahedral) d5 octahedral or tetrahedral Dublin January 2007 0 !0 d1, d6 octahedral (d4, d9 tetrahedral) 30 High-spin and low-spin states An ion is in a high-spin state or a low spin state, depending on whether the Coulomb interaction U leading to Hund’s first rule (maximize S) is greater or less than the the crystal-field splitting D. Consider a 3d6 ion such as Fe3+. eg eg eg t2g t2g eg D U D t2g U t2g U > D, gives a high-spin state, S = 2 e.g. FeCl2 U < D, gives a low-spin state, S = 0 e.g. Pyrite FeS2 Dublin January 2007 31 Tanabe-Sugano diagrams These show the splitting of the ground state and higher terms by the crystal field. The high-spin < lowspin crossover is seen. Diagrams shown are for d-ions octahedral environments. d6 1 3 3 3 3 I G F P 3 T2 g 3 A1g 3 T1g 3 Eg 3 Eg ; (t2 g ) (eg )3 H E 5 3 D 1 A1g (t2 g ) 6 3 T2 g 3 T1g 3 T2 g 3 T1g 5 T2 g ; (t2 g )(eg ) 2 crystal field splitting Dublin January 2007 32 d5 Matching the optical absorption spectrum of Fe3+ - doped Al2O3 with the calculated Tanabe-Sugano energy-level diagram to determine the cubic crystal field splitting at octahedral sites. Dublin January 2007 33 d2 d7 Note the similarities between the Tanabe-Sugano diagrams for d2 and d7. The differences are associated with the possible low-spin states for d7 (e.g. Co2+) Dublin January 2007 34 d3 For Cr3+ in Al2O3, the cf parameter Dq/B is 2.8 Dublin January 2007 35 d3 For Cr3+ in Al2O3, the cf parameter Dq/B is 2.8 Dublin January 2007 36 5.3 Single-ion anisotropy The electrostatic interaction of the ionic charge distribution 21(r) with the potential 1cf created by the rest of the crysta gives rise to the crystal field splittings. It is also the source, via spin-orbit coupling, of magnetocrystalline anisotropy. E = = 1cf 2(r)dr where 1cf(r) = -(e/4*+0) = {2(R) / |R - r|} dR The anisotropy energy is therefore Ea(r) = -(e/4*+0) = {2(r,!") 2(R ) / |R - r|}dr dR Both the charge distribution 2(r) and the potential 1cf(r) can be expanded in spherical harmonics. Using the Wigner-Eckart theorem, it is possible to write the corresponding crystal-field Hamiltonian in terms of angular momentum operators Jx, Jxy Jz J2 which is a particularly useful way to find the energy-levels (eigenvalues). The Hamiltonian matrix is written in an ML or MJ basis for the 3d transition elements or 4f rare earths respectively. In concise form ! cf = >>n=2,4,6>m=-nn Bnm Ô nm Crystal field parameters !n?rn@Anm Stevens coefficients Stevens operators Crystal field coefficients In a site with uniaxial anisotropy, the leading term is ! cf = B20 Ô20 The Stevens operator Ô 20 is {3Jz2 - J(J+1)} ! cf = B20 {3Jz2 - J(J+1)} Dublin January 2007 37 9/2 Charge distributions of the rare-earth ions. Those with a positive quadrupole moment (!2 > 0), italic type are distinguished from those with a negative quadrupole moment (!2 < 0) bold type. Note the quarter-shell changes, ±9/2 ±1/2 3+ ±3/2 e.g. Nd J = 9/2 ±5/2 ±7/2 ±1/2 ±9/2 B2 0 < 0 B20 > 0 Dublin January 2007 38 z 6 Paramagnetism 6.1 Langevin theory 2!sin"d" " m H Dublin January 2007 O 39 6.2 Brillouin theory The general quantum case was treated by Brillouin; m is gµBJ, and x is defined as x = µ0mH/kBT. There are 2J+1 energy levels Ei = -µ0gµBMJH, with moment mi =gµBMJ where MJ = J. J-1, J-2, … -J. The sums over the energy levels have 2J+1 terms. Their populations are proportional to exp(-Ei/kT) a) Susceptibility To calculate the susceptibility, we can take x << 1, because the susceptibility is defined as the initial slope of the magnetization curve. We expand the exponential as exp(x) = 1 + x + .., <m> = >-JJgµBMJ(1 + µ0gµBMJH/kBT)/>-JJ(1 + µ0gµBMJH/kBT) MJ Recall >-JJ1 = 2J + 1 -5/2 -3/2 >-JJMJ = 0 J=5/2 -1/2 >-JJMJ2 = J(J + 1)(2J + 1)/3 1/2 Hence <m> = µ0g2µB2HJ(J + 1)(2J + 1)/3(2J + 1)kBT 3/2 The relative susceptibility is N<m>/H, where N is the 5/2 -5/2 3 number of atoms/m . -3/2 Ar = µ0Ng2µB2J(J + 1)/3kBT H -1/2 1/2 This is the general form of the Curie law. Again it can be written Ar = C/T where the Curie constant 3/2 2 2 2 C = µ0Ng µB J(J+1)/3kB or C = µ0Nmeff /3kB where 5/2 meff = gµB)[J(J+1)]. A typical value of C for J = 1, N = 8.1028 m-3 is 3.5 K. Note that results for the classical limit and S = 1/2 are obtained when J < B (m = gµBJ) and J = 1/2, g = 2. (m = µB) . Dublin January 2007 40 Magnetization Curve To calculate the complete magnetization curve, set y = µ0gµBH/kBT, then <m> = gµB#/#y[ln>-JJ exp{MJy}] [d(ln z)/dy = (1/z) dz/dy] The sum over the energy levels must be evaluated; it can be written as exp(Jy) {1 + r + r2 + .........r2J} where r = exp{-y} The sum of a geometric progression (1 + r + r2+ .... + rn) = (rn+1 - 1)/(r - 1) C >-JJ exp{MJy} = (exp{-(2J+1)y} - 1)exp{Jy}/(exp{-y}-1) multiply top and bottom by exp{y/2} = [sinh(2J+1)y/2]/[sinh y/2] <m> = gµB($/#y)ln{[sinh(2J+1)y/2]/[sinh y/2]} = gµB/2 {(2J+1)coth(2J+1)y/2 - coth y/2} setting x = Jy, we obtain <m> = mBJ(x) 1.0 where the Brillouin function BJ(x) = {(2J+1)/2J}coth(2J+1)x/2J - (1/2J)coth(x/2J). 0.8 1/2 Again, this reduces to the previous equations in the limits J < B (m = gµBJ) and J = 1/2, g = 2. 2 0.6 B , 0.4 0.2 Comparison of the Langevin function and the Brillouin functions for J = 1/2 and J = 2. 0 0 Dublin January 2007 2 4 6 8 x 41 7 S = 7/2 (Gd3+) µB per ion Magnetic moment, 6 5 S = 5/2 (Fe3+) 4 3 S = 3/2 (Cr3+) 1.30K 2.00K 3.00K 4.21K Brillouin functions 2 1 0 0 1 2 3 4 -1 BJ /T (TK ) Reduced magnetization curves for three paramagnetic salts, with Brillouin-theory predictions The theory of localized magnetism gives a good account of magnetically-dilute 3d and 4f salts where the magnetic moments do not interact with each other. Except in large fields or very low temperatures, the M(H) response is linear. Fields > 100 T would be needed to approach saturation at room temperature. The excellence of the theory is illustrated by the fact that data for quite different temperatures superpose on a single Brillouin curve plotted as a function of x ' H/T Dublin January 2007 42