Langevin Theory Final PDF

The Langevin theory of paramagnetism Recall that Susceptibility 𝑀 𝑥= 𝐻  𝑛𝜇2 𝑁𝜌 𝜇2 𝐶 𝜒𝑉 = = = 3𝑘𝑇 𝐴 𝑘𝑇 𝑇 Classical theory of Paramagnetism Curie found that the mass susceptibility𝜒𝑚 was independent of temperature for diamagnetic mater...

The Langevin theory of paramagnetism Recall that Susceptibility 𝑀 𝑥= 𝐻  𝑛𝜇2 𝑁𝜌 𝜇2 𝐶 𝜒𝑉 = = = 3𝑘𝑇 𝐴 𝑘𝑇 𝑇 Classical theory of Paramagnetism Curie found that the mass susceptibility𝜒𝑚 was independent of temperature for diamagnetic materials, but that it varied inversely with the absolute temperature for paramagnetic materials: The Langevin theory Qualitatively, his theory of paramagnetism is simple He assumed a paramagnetic to consist of atoms, or molecules, each of which has the same net magnetic moment 𝜇, because all the spin and orbital moments of the electrons do not cancel out. In the absence of an applied field, these atomic moments point at random and cancel one another, so that the magnetization of the specimen is zero. ❑ When a field is applied, there is a tendency for each atomic moment to turn toward the direction of the field; ❖ if no opposing force acts, complete alignment of the atomic moments would be produced and the specimen as a whole would acquire a very large moment in the direction of the field. But ❖ thermal agitation of the atoms opposes this tendency and tends to keep the atomic moments pointed at random. The result is only partial alignment in the field direction, and therefore a small positive susceptibility. The effect of an increase in temperature is to increase the randomizing effect of thermal agitation and therefore, decrease the susceptibility. Quantitatively 1) We consider a unit volume of material containing N atoms, each having a magnetic moment 𝜇. 2) Let the direction of each moment be represented by a vector, and let all the vectors be drawn through the center of a sphere of unit radius.  In the absence of a field, the number of 𝜇 vectors passing through unit area of the sphere surface is the same at any point on the sphere surface. 𝑑𝑁 ∝ 𝑑Ω 𝑑𝑁 ∝ 2𝜋 sin 𝜃𝑑𝜃 for a sphere of unit radius (solid angle) When a field is applied, 𝜇 vectors all shift towards the direction of the field. The potential energy of each atomic moment in the field is 𝐸𝑝 = −𝜇 𝐻𝑐𝑜𝑠 𝜃  𝑀 = 𝑁.𝜇ҧ We wish to calculate the number of moments (𝑑𝑁) inclined at an angle between (𝜃 𝑎𝑛𝑑 𝜃 + 𝑑𝜃) to the field H.  The potential energy of a moment is 𝐸 = −𝜇𝐻𝑐𝑜𝑠 𝜃 Let 𝑁 𝜃 𝑑𝜃 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑚𝑎𝑘𝑒 𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝜃 𝑎𝑛𝑑 𝜃 + 𝑑𝜃 with the applied field. 𝑁 𝑡𝑜𝑡𝑎𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑝𝑒𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 −𝐸 The Boltzmann factor = 𝑒 = 𝑒𝜇𝐻𝑐𝑜𝑠𝜃/𝑘𝑇 describes the relative probability of 𝑘𝑇 an atomic moment 𝜇 to make angle 𝜃 𝑎𝑛𝑑 𝜃 + 𝑑𝜃 with 𝐻 i.e. inside a solid angle (𝑑Ω) (between 𝜃 + 𝑑𝜃 ) 𝑁 𝜃 𝑑𝜃 ∝ 1) 𝑑Ω (= 2π sin 𝜃𝑑𝜃) 2) 𝑒 𝜇𝐻𝑐𝑜𝑠𝜃/𝑘𝑇 ❑ the proportionality constant is 𝑛𝑜  𝑢𝐻𝑐𝑜𝑠𝜃 𝑁(𝜃)𝑑𝜃 = 𝑛𝑜 2𝜋 sin 𝜃𝑑𝜃 𝑒 𝑘𝑇 (1) 𝜋 𝑁= ‫׬‬0 𝑁𝜃𝑑𝜃 𝜋 𝑀 = 𝑁.𝜇ҧ 𝑀 = න 𝑁 𝜃 𝑑𝜃 𝑢𝑐𝑜𝑠𝜃 (2) 0 Substitute by 1𝜋in 2 𝑢𝐻𝑐𝑜𝑠𝜃 𝑀 = න 2𝜋𝑛𝑜 sin 𝜃𝑑𝜃 𝑒 𝑘𝑇 𝜇𝑐𝑜𝑠𝜃 0  𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 𝑀 = 2𝜋𝜇𝑛𝑜 න sin 𝜃 𝑒 𝑘𝑇 𝑐𝑜𝑠𝜃𝑑𝜃 0 𝑁 To eliminate 𝑛𝑜 multiply by ( ) 𝑁 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 𝑁𝜇2𝜋𝑛𝑜 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑐𝑜𝑠𝜃𝑑𝜃 𝑀= 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 2𝜋𝑛𝑜 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑑𝜃 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 𝑁𝜇 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑐𝑜𝑠𝜃𝑑𝜃 𝑀= 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑑𝜃 Solving the integration gives 𝟏 𝑴 = 𝝁𝑵𝑳 𝜶 𝒘𝒉𝒆𝒓𝒆 𝐋 𝜶 = 𝒄𝒐𝒕𝒉𝜶 − 𝜶 Solving the Langevin integral 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 𝑁𝜇 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑐𝑜𝑠𝜃𝑑𝜃 → 𝑁𝑈𝑀𝐸𝑅𝐴𝑇𝑂𝑅 𝐼= 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑑𝜃 → 𝐷𝐸𝑁𝑂𝑀𝐼𝑁𝐴𝑇𝑂𝑅 𝜇𝐻 𝑡𝑎𝑘𝑒 ∝= , 𝑥 = cos 𝜃 , 𝑑𝑥 = −𝑠𝑖𝑛𝜃𝑑𝜃 𝑘𝑇 𝜃: 0 → 𝜋 𝑎𝑛𝑑 𝑥 = 1 → −1 −1 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 = න 𝑥𝑒 ∝𝑥 (−𝑑𝑥) 1 −1 ∝𝑥 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 =න 𝑒 (−𝑑𝑥) 1 Use integration by parts න 𝑓 𝑥 𝑔′ 𝑥 𝑑𝑥 = 𝑓 𝑥 𝑔 𝑥 − න 𝑓 ′ 𝑥 𝑔 𝑥 𝑑𝑥 −1 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 = න 𝑥𝑒 ∝𝑥 (−𝑑𝑥) 1 𝑓 𝑥 = 𝑥, 𝑔′ 𝑥 = 𝑒 𝛼𝑥 𝑒 ∝𝑥 𝑓′ 𝑥 = 1, 𝑔 𝑥 = ∝  ‫‪−1‬‬ ‫‪1‬‬ ‫𝑥𝑑 𝑥∝ 𝑒𝑥 ‪−1‬׬ = )𝑥𝑑‪1 𝑥𝑒 ∝𝑥 (−‬׬ = 𝑟𝑜𝑡𝑎𝑟𝑒𝑚𝑢𝑁 ‬ ‫‪1‬‬ ‫‪𝑥𝑒 ∝𝑥 1‬‬ ‫𝑥∝ 𝑒‬ ‫[ = 𝑟𝑜𝑡𝑎𝑟𝑒𝑚𝑢𝑁‬ ‫]‬ ‫‪− න 1.‬‬ ‫𝑥𝑑‬ ‫∝‬ ‫‪−1‬‬ ‫‪−1‬‬ ‫∝‬ ‫‪𝑥𝑒 ∝𝑥 1‬‬ ‫‪𝑒 ∝𝑥 1‬‬ ‫[=‬ ‫]‪] −[ 2‬‬ ‫‪∝ −1‬‬ ‫∝‬ ‫‪−1‬‬ ‫∝ 𝑒‪1‬‬ ‫∝‪𝑒 −‬‬ ‫∝‪𝑒 +∝ 𝑒 −‬‬ ‫=‬ ‫‪− −1‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪− 2‬‬ ‫∝‬ ‫∝‬ ‫∝‬ ‫∝‬ ‫∝‪𝑒 ∝ 𝑒 −‬‬ ‫∝‪𝑒 +∝ 𝑒 −‬‬ ‫=‬ ‫‪+‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪− 2‬‬ ‫∝‬ ‫∝‬ ‫∝‬ ‫∝‬ ‫‪1‬‬ ‫=‬ ‫∝[‬ ‫] ∝‪𝑒 ∝ + 𝑒 −∝ − 𝑒 ∝ − 𝑒 −‬‬ ‫‪∝2‬‬  1 ∝𝑥 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 = ‫׬‬−1 𝑒 (𝑑𝑥) 𝑒 ∝𝑥 1 1 ∝ = [ ] = [𝑒 − 𝑒 −∝ ] ∝ −1 ∝ 1 ∝ −∝ ∝ −∝ 2 [∝ 𝑒 + 𝑒 − 𝑒 − 𝑒 ] 𝐼= ∝ 1 ∝ [𝑒 − 𝑒 −∝ ] ∝ −∝ ∝ 1 ∝ 𝑒 +𝑒 𝐼 = [ ∝ −∝ − 1] ∝ [𝑒 −𝑒 ] 𝑒 ∝ + 𝑒 −∝ 1 𝐼= ∝ −∝ − [𝑒 − 𝑒 ] ∝ 1 𝐼 = 𝑐𝑜𝑡ℎ ∝ − ∝  1 𝑀 = 𝑁𝜇(coth 𝑎 − ) 𝑎 𝑁𝜇 = 𝑀𝑜 The Langevin theory leads to two conclusions: For small a, L(a)=a/3 𝑁𝜇𝑎 𝑁𝜇2 𝐻 𝑀= = 3 3𝑘𝑇 The Langevin theory leads to the Curie law For small a, L(a)=a/3 𝑁𝜇𝑎 𝑁𝜇2 𝐻 𝑀= = 3 3𝑘𝑇 𝑀 𝑁𝜇2 𝐶 𝜒 = 𝜒𝑉 = = = 𝐻 3𝑘𝑇 𝑇 𝑁𝜇2 The constant 3𝑘 is called the Curie constant. 𝜒𝑉 𝑁𝜇2 The mass susceptibility 𝜒𝑚 = = 𝜌 3𝜌𝑘𝑇 Where 𝜌 is the density. ❑N is the number of( atoms /volume), 𝑁𝐴 (atoms/mole), A (mass/mole) 𝐴𝑁 Therefore 𝜌 = 𝑁 𝐴 𝑁𝐴 𝜇 2 𝑚3 Therefore, 𝜒𝑚 = 3𝐴𝑘𝑇 the units are the reciprocal of density≡ 𝑘𝑔  Susceptibility continued 𝑀 𝑁𝜇2 𝑁𝐴 𝜌 𝜒𝑉 = = 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 (𝑁 = ) 𝐻 3𝑘𝑇 𝐴 𝑁𝜇2 𝑁𝐴 𝜌 𝜇2 𝐶 𝜒𝑉 = = = 3𝑘𝑇 3𝐴 𝑘𝑇 𝑇 The curie constant depends on : 1. the type of material (𝜇) 2. The structure of matter (N) Quantum theory of paramagnetism The main conclusions of the classical theory are modified by quantum mechanics, but not radically so. We will find that quantum theory greatly improves the quantitative agreement between theory and experiment without changing the qualitative features of the classical theory. The central postulate of quantum mechanics is that the energy of a system is not continuously variable In quantum theory, 𝜃 is restricted to certain definite values 𝜃1 , 𝜃2 ,... , and intermediate values are not allowed. ❑ This restriction is called space quantization, and is illustrated schematically in Fig Assumptions ❑ That there is no orbital contribution to the moment, so that J =S. The orbital moment is, in such cases, said to be quenched. ❑ This condition results from the action on the atom or ion considered by the electric field, called the crystalline or crystal field, produced by the surrounding atoms or ions in the solid. ❑ This field has the symmetry of the crystal involved. ❑ Thus the electron orbits in a particular isolated atom might be circular, but when that atom forms part of a cubic crystal, the orbits might become elongated along three mutually perpendicular axes because of the electric fields created by the adjoining atoms located on these axes. ❑ In any case, the orbits are in a sense bound, or “coupled,” rather strongly to the crystal lattice. ❑ The spins, on the other hand, are only loosely coupled to the orbits. Thus, when a magnetic field is applied along some arbitrary direction in the crystal, the strong orbit–lattice coupling often prevents the orbits, and their associated orbital magnetic moments, from turning toward the field direction, whereas the spins are free to turn because of the relatively weak spin–orbit coupling. ❑ The result is that only the spins contribute to the magnetization process and the resultant magnetic moment of the specimen; the orbital moments act as though they were not there. ❑ Quenching may be complete or partial. Back to Quantum theory of paramagnetism Assuming that g and J are known for the atoms involved, we can proceed to calculate the total magnetization of a specimen as a function of the field and temperature. The procedure is the same as that followed in deriving the classical (Langevin) law, except that: 1. The quantized component of magnetic moment in the field direction (𝜇𝐻 = 𝑔𝑀𝐽 𝜇𝐵 ) replaces the classical term 𝜇 cos 𝜃. 2. A summation over discrete moment orientations replaces an integration over a continuous range of orientations. The potential energy of each moment in the field H is 𝐸𝑝 = −𝑔𝑀𝐽 𝜇𝐵 𝐻 The probability of an atom having an energy 𝐸𝑝 is proportional to 𝑒 −𝐸𝑝 /𝑘𝑇 = 𝑒 𝑔𝑀𝐽 𝜇𝐵𝐻/𝑘𝑇 Recall that using the classical treatment we obtained 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 𝑁𝜇 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑐𝑜𝑠𝜃𝑑𝜃 𝑀= 𝜋 𝑢𝐻𝑐𝑜𝑠𝜃 ‫׬‬0 𝑒 𝑘𝑇 sin 𝜃 𝑑𝜃 Using the quantum mechanical treatment: If there are 𝑁 atoms per unit volume, σ 𝒈𝑴𝑱 𝝁𝑩 𝒆𝒈𝑴𝑱 𝝁𝑩𝑯/𝒌𝑻 𝑴=𝑵 σ 𝒆𝒈𝑴𝑱 𝝁𝑩 𝑯/𝒌𝑻 Where the summations are over 𝑀𝐽 and extend from −𝐽 𝑡𝑜 + 𝐽. Example The chromium ion Electronic Configuration 𝐾𝐶𝑟(𝑆𝑂4 )2. 12𝐻2 𝑂 The only magnetic component is 𝐶𝑟 3+ ion. For the free state ion 3 3 2 𝐽 = 2 , 𝐿 = 3, 𝑆 = 2 which leads to 𝑔 = 5 Example continued 𝐾𝐶𝑟(𝑆𝑂4 )2. 12𝐻2 𝑂 The only magnetic component is 𝐶𝑟 3+ ion. For the free state ion 3 3 2 𝐽= ,𝐿 = 3, 𝑆 = which leads to 𝑔 = 2 2 5 The magnetic moment per molecule ( chromium ion is given by) M = 𝑁𝑔𝐽𝜇𝐵 𝐵 𝐽, 𝑎′ 𝑀 = 𝑔𝐽𝐵(𝐽, 𝑎′ )𝜇𝐵 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 If we assume the orbital component is entirely quenched this gives: 3 𝐿 = 0 𝑎𝑛𝑑 𝑔 = 2 𝑎𝑛𝑑 𝐽 = 𝑆 = 2  The maximum moment in the field direction (𝜇𝐻 ) is 3 given by 𝑔𝑀𝐽 𝜇𝐵 = 2 𝜇𝐵 = 3𝜇𝐵 2 And the effective moment is 𝜇𝑒𝑓𝑓 = 𝑔 𝐽(𝐽 + 1)𝜇𝐵 3 5 = 2 ( )( )𝜇𝐵 2 2 = 15𝜇𝐵 = 3.87𝜇𝐵 The wide disparity between the classical theory and experiment is evident at large (H/T) Upper curve The upper curve is obtained using the classical approach (𝐽 = ∞). Langevin theory is applied 3 The maximum moment in the field direction is given by 2 𝜇𝐵 = 3𝜇𝐵 2 And the effective moment is 𝜇𝑒𝑓𝑓 = 𝑔 𝐽(𝐽 + 1)𝜇𝐵 3 5 = 2 ( )( )𝜇𝐵 2 2 = 15𝜇𝐵 = 3.87𝜇𝐵 ( does not agree with experiment)  The lower curve The Brillouin function is applied. 3 2 𝐽= , 𝑔= 2 5 Does not agree with experiment The middle curve 2 Using 𝑔 = 2 𝑛𝑜𝑡 5 i.e. The orbital contribution is quenched L=0 rather than 3. 3 𝐽=𝑆= 2 The curve agrees well with experiment.  The wide disparity between the classical theory and experiment is evident at large (H/T) At higher temperatures, larger fields are required to achieve saturation ( large values of a)  Example 𝐾𝐶𝑟(𝑆𝑂4 )2. 12𝐻2 𝑂 This substance follows the Curie law exactly, with a molecular Curie constant (Curie constant per mole) 𝐶𝑀 of 1.85. a) Find the effective magnetic moment in Bohr magnetons. b) Calculate the maximum component of the moment. K= Solution a) Find the effective magnetic moment in Bohr magnetons. 𝑁𝐴 𝜇 2 𝑚3 𝑐𝑚3 Recall that, 𝜒𝑚 = 3𝐴𝑘𝑇 Therefore, the units are the reciprocal of density≡ 𝑘𝑔 𝑜𝑟 𝑔 To obtain the Curie constant per mole 𝐶𝑀 we multiply 𝐶𝑚 by A 𝑁𝐴 𝜇𝑒𝑓𝑓 2 𝐶𝑀 = = 1.85 3𝑘 3𝑘𝐶𝑀 1/2 𝜇𝑒𝑓𝑓 = ( ) 𝑁𝐴 3(1.38 × 10−16 1.85 ) 1/2 𝜇𝑒𝑓𝑓 =( ) 6.02 × 1023 𝜇𝑒𝑓𝑓 = 3.57 × 10−20 𝑒𝑟𝑔/𝑂𝑒 3.57 × 10−20 𝜇𝑒𝑓𝑓 = = 3.85𝜇𝐵 0.927 × 10−20  continued b) Calculate the maximum component of the moment. Knowing that 𝜇𝐻 = 𝑔𝐽𝜇𝐵 We need to calculate J We can do this only if we assume that spin moments only are involved ( orbital motion is quenched). Taking 𝑔 = 2 Using 𝜇𝑒𝑓𝑓 = 𝑔 𝐽(𝐽 + 1)𝜇𝐵 = 3.85𝜇𝐵 3 This gives 𝐽 = 1.49 = 2 𝜇𝐻 = 𝑔𝐽𝜇𝐵 = 𝑔 1.49 𝜇𝐵 = 2.98𝜇𝐵 The magnetic susceptibility ( for small a’) 𝑀 = 𝑁𝑔𝐽𝜇𝐵 𝐵(𝐽, 𝑎′ ) For very small 𝑎′ 𝐽+1 ′ 𝑀 = 𝑛𝑔𝐽𝜇𝐵 𝑎 ( ) 3𝐽 𝜇𝐻 𝐻 And 𝑎′ = and 𝜇𝐻 = 𝑔𝐽𝜇𝐵 𝑘𝑇 𝑔𝐽𝜇𝐵 𝐻 𝐽 + 1 𝑀 = 𝑛𝑔𝐽𝜇𝐵 ( )( ) 𝑘𝑇 3𝐽 2 𝐽+1 𝑀 = 𝑛𝑔 𝐽 ( )𝜇𝐵 2 H 3𝑘𝑇 𝑛𝑔2 𝐽(𝐽 + 1)𝜇𝐵 2 H 𝑀= 3𝑘𝑇 Using 𝜇𝑒𝑓𝑓 = 𝑔 𝐽(𝐽 + 1)𝜇𝐵 𝑁𝜇𝑒𝑓𝑓 2 𝐻 𝑀= 3𝑘𝑇 𝑀 𝜒= 𝐻 Curie Weiss Law The Langevin theory of paramagnetism, which leads to the Curie law, is based on the assumption that the individual carriers of magnetic moment (atoms or molecules) 1. do not interact with one another. 2. but are acted on only by the applied field and thermal agitation.  Weiss pointed out that this behaviour could be understood by postulating that the elementary moments do interact with one another. Interaction could be expressed in terms of a fictitious internal field which he called the “molecular field ”𝐻𝑚 which acted in addition to the applied field H. The term “molecular field ”𝐻𝑚 was then modified to be called the “atomic” field.  Weiss assumed that the intensity of the molecular field was directly proportional to the magnetization: 𝐻𝑚 = 𝛾𝑀 𝛾 is called the molecular field constant. The total field acting on the material is 𝐻𝑡 = 𝐻 + 𝐻𝑚 Curie’s Law may be written as: 𝑀 𝐶𝑔 𝜒𝑚 = = 𝜌𝐻 𝑇 𝐶𝑔 is the Curie constant per gram 𝐻 in this expression must now be replaced by 𝐻𝑡  𝑀 𝐶𝑔 𝜒𝑚 = = 𝜌𝐻 𝑇 𝐶𝜌(𝐻 + 𝛾𝑀) 𝑀= 𝑇 𝜌𝐶𝐻 𝜌𝐶𝛾𝑀 𝑀= + 𝑇 𝑇 Rearrange 𝜌𝐶𝛾 𝜌𝐶𝐻 𝑀 1− = 𝑇 𝑇 𝜌𝐻𝐶 1 𝑀= 𝑇 (1 − 𝜌𝛾𝐶 ) 𝑇 𝜌𝐶𝐻 𝑀= (𝑇 − 𝜌𝛾𝐶) 𝑀 Recall that 𝜒𝑚 = 𝜌𝐻 𝑀 𝐶 𝐶𝑔 𝜒𝑚 = = = 𝜌𝐻 (𝑇 − 𝜌𝐶𝛾) (𝑇 − 𝜃) The value of 𝜃 𝑀 𝐶 𝐶 𝜒𝑚 = = = 𝜌𝐻 (𝑇 − 𝜌𝐶𝛾) (𝑇 − 𝜃) Where 𝜃 = 𝜌𝐶𝛾 𝜃 Is a measure of the strength of the interaction because it is proportional to the molecular field constant 𝛾. For substances that obey Curie’s law 𝜃 = 𝛾 = 0 Curie Law versus Curie Weiss Law 𝑀 𝐶 𝜒𝑚 = = 𝜌𝐻 (𝑇 − 𝜃) Positive and negative values of 𝜃 𝑪 𝟏 𝑻 𝜽 𝝌𝒎 = → = − 𝑻−𝜽 𝝌𝒎 𝑪 𝑪 Determine the intercept at T = 0 𝜃 𝑖𝑠 + 𝑣𝑒 𝜃 𝑖𝑠 − 𝑣𝑒 ( i.e. molecular field aids the applied field in ( i.e. molecular field opposes applied aligning the moments and tends to increase field and tends to decrease the the susceptibility) susceptibility) For many paramagnets, 𝜃 is in the order of 10 K or less The quantum relation for Curie-Weiss behaviour Introduce 𝐻𝑚 (= 𝛾𝑀) 𝜒𝑚 = 𝑁𝐴 𝜇 2 3𝐴𝑘𝑇 Again 𝐶𝑔 𝜒𝑚 = (𝑇 − 𝜃) 𝐶 is the Curie constant 𝑁𝐴 𝜇𝑒𝑓𝑓 2 𝑁𝐴 𝑔2 𝐽(𝐽 + 1)𝜇𝐵 2 𝜒𝑚 = = 3𝐴𝑘(𝑇 − 𝜃) 3𝐴𝑘(𝑇 − 𝜃) 𝜌𝑁𝐴 𝜇𝑒𝑓𝑓 2 𝛾 𝜌𝑁𝐴 𝑔2 𝐽(𝐽+1)𝜇𝐵 2 𝛾 As before 𝜃 = 𝜌𝐶𝑚 𝛾 = = 3𝐴𝑘 3𝐴𝑘 Determining the molecular field constant Using the obtained value of 𝜃 (graphically) 𝜌𝑁𝜇𝑒𝑓𝑓 2 𝛾 𝜌𝑁𝑔2 𝐽(𝐽 + 1)𝜇𝐵 2 𝛾 𝜃 = 𝜌𝐶𝛾 = = 3𝐴𝑘 3𝐴𝑘 Or simply use the known value of 𝜇𝑒𝑓𝑓 as we did previously