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Magnets, Metals and Superconductors Tutorial 1 Dr. Abbie Mclaughlin G24a.

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Presentation on theme: "Magnets, Metals and Superconductors Tutorial 1 Dr. Abbie Mclaughlin G24a."— Presentation transcript:

1 Magnets, Metals and Superconductors Tutorial 1 Dr. Abbie Mclaughlin G24a

2 1.Determine the ground state configuration and predict the effective magnetic moment for the following Ln 3+ ions. Gd 3+, Er 3+ L SymbolSPDFGHIK mjmj Gd 3+ = f 7 S = 7/2, L = 0, J = 0 i.e. is spin only! eff = 7.94 Term symbol = 8 S 0

3 mjmj Er 3+ = f 11 The term state symbol is written 2S+1 L J. Hunds Rules: For less than half-filled shells, the smallest J term lies lowest; for more than half filled shells the largest J lies lowest. S = 3/2 L = m j = 6, J = 15/2, 13/2, 11/2, 9/2 Term symbol = 4 I 15/2 g j = 6/5 µ eff = 9.58 L SymbolSPDFGHIK

4 2 Exam question 1 (2004) (b). The gradient =1/C. This can be used to determine S from the equation: C = Ng 2 µ B 2 S(S+1)/3k The value of can be determined from the 1/ vs T plot. This gives an indication of the strength and nature of the interactions between neighbouring molecules.

5 b) Ferromagnetic exchange: S T = nS nS = 10 x 2 = 20. =41 B 3a. 2 cyclic-[Fe(OMe)(OAc)] 10 Fe 2+ d 6 S = 2 n = 10 a) Antiferromagnetic exchange S T = 0 or ½ depending on whether there is and even or odd number of electrons. There are 10 antiferromagnetic S = 2 ions, S T = 0

6 c) Non interacting (high temperature limit). S = 2, n = 10 = 15.5 µ B. 3b) [Cu 3 OCl 4 (Mepy) 4 ]Cu 2+ d 9, S = 1/2, n = 3 a) Antiferromagnetic exchange: S T = 0 or ½ depending on whether there is and even or odd number of electrons. There are 3 antiferromagnetic S = 1/2 ions, S T = ½. µ eff = 1.73 µ B

7 c) Non interacting (high temperature limit). S = 1/2, n = 3 = 3.00 µ B b)Ferromagnetic exchange: ST = nS nS = 3 x 1/2 = 3/2. = 3.87 µ B.

8 4. Determine eff per mole of Cu 2 (OAc) 4.2H 2 O. Apply a diamagnetic correction to and redetermine eff. Does it make a difference? =1.7 µ B per mole of dimer Diamagnetic correction (for dimer) Cu = -11 X 10 -6, OAc = -30 X 10 -6, H 2 O = -13 X Overall correction = (X ) = -168 X M = dia + para para = M - dia =1.2 x X = x

9 eff = µ B per mole of dimer Uncorrected = 1.7 µ B = per mole of dimer 0.1 difference. Its important to correct if you want to be accurate.

10 2 Exam question 1 (2004) (a). What is meant by Curie behaviour? Give reasons why paramagnetic materials may deviate from Curie behaviour and explain what additional information can be extracted from such deviations. The magnetic susceptibility, (M/H) is dependent on 1/T. =C/T. As the temperature increases the increase in thermal energy gives rise to greater randomisation of the spin orientation and hence a smaller induced magnetisation. The Curie constant C, comprises a series of fundamental constants and S, the spin quantum number. Thus from a plot of 1/ Vs T the value of S can be determined form the gradient.

11 Paramagnetic materials may deviate from Curie behaviour if: a) there are local ferromagnetic or antiferromagnetic interactions between spins. The materials can then be described as Curie Weiss paramagnets. = C/(T- ) When > 0 it indicates ferromagnetic interactions; if = 0 we have ideal Curie behaviour and if < 0 then it indicates antiferromagnetic interactions. can be determined from a plot of 1/ vs T, which should be linear with an intercept on the T axis equal to. The larger the value of the greater the interaction between spins on neighbouring molecules.

12 Paramagnetic materials may deviate from Curie behaviour if: b) If the material shows Van Vleck behaviour. This occurs when there is thermal population of excited states whose magnetic behaviour is different to that of the ground state. For example Eu 3+. The ground state term is 7 F 0 hence the predicted µ eff is 0 B.M. Observed values are typically in the range B.M. at room temperature, although the value decreases upon cooling. In the case of Eu 3+ the separation of the ground state 7 F 0 and the first excited state is ca. 300cm -1. At room temperature there is enough thermal energy for the 7 F 1 state to be partially populated.

13 On cooling the 7 F 1 state becomes depopulated and the magnetic moment approaches 0 B.M. as T approaches 0 K when all the ions are in the 7 F 0 state. However a second effect (temperature independent paramagnetism, TIP) is required to rationalize the data satisfactorily.


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