Download presentation

Presentation is loading. Please wait.

Published byPierce Iredale Modified over 2 years ago

1
Magnets, Metals and Superconductors Tutorial 2 Dr. Abbie Mclaughlin G24a

2
1. Explain why dimers of type Cp 2 TiX 2 TiCp 2 (containing 2 bridging X groups) exhibit similar magnetic behaviour to copper acetate, Cu 2 OAc 4.2H 2 O. Values for J for X = F, Cl, Br and I are -31, -110, -138 and -87 cm -1 respectively. Comment on this observation. J is less than 0 for all X. Kinetic superexchange in which antiferromagnetism is stabilised occurs via X ( interactions). As we go from F – Br the Cp 2 TiX 2 TiCp 2 compounds get more covalent. The more covalent the compound the more overlap there is of the Ti and X orbitals and hence greater exchange parameter J. J decreases for X = I -. This is due to a change in bridging angle. Ti 3+ = d 1

3
As X gets larger (as we go from F - - I - ) the bridging angle gets smaller (M-X-M). For F - - Br - this is overcompensated for by the increase in covalency. This is not the case for I -.

4
2. (a) Explain what is meant by the terms correlation length and spin wave. In the ferromagnetic ordered state (below T c ) the magnetisation is always less than the saturation value. This is only obtained at 0 K. We can understand why this happens by considering what is happening to individual spins and using the concepts of correlation length, (x i ) and spin waves. (x i )

5
At T > 0 K there is enough thermal energy for some of the spins to “flip”. As the temperature rises, the number of spin waves increase (more spins are reversed) and Ms decreases, eventually reaching 0 at T c. The correlation length (x i ) is the length of “unflipped spins”. Hence as T increases (x i ) decreases. (b) Estimate the magnetic ordering temperature for [Cp*Fe][TCNE] T c = 4.7 K

6
c)Explain the differences in the shapes of the hysteresis loops recorded at 2 K and 4.7 K? The saturated moment, coercive field and remanent moment are all much smaller at 4.7 K than at 2 K. The saturated moment is larger at 2 K because the correlation length (xi) of the spin waves is increasing. The saturated moment is equal to 2S but this assumes that the spins are all aligned parallel throughout the sample (i.e doesn’t take spin waves into account). The saturated moment increases with decreasing temperature because the length of ordered spins increases and at 0 K the saturated moment = 2S when all the spins are aligned parallel.

7
ii) The coercive field is defined as the reverse field in order to reduce magnetisation to 0 from saturation. i.e. it is the field required to “flip” the spins in order to make the magnetisation 0. The remanent magnetisation is defined as the magnetisation remaining when the field is switched off from saturation.

8
In order to minimise its total free energy, the ferromagnet usually subdivides into domains. Each domain has a net magnetisation in zero field, but the moments of each domain are randomly oriented such that the resultant magnetisation in zero. At T = 2 K there are much larger regions of parallel spins since the thermal energy is less and (x i ) is longer (i.e. the domain sizes are much larger). Hence it is much harder to realign the domain walls and a larger field is required at 2 K. Application of an external field leads to a realignment of the domain walls with the magnetic field. When the applied field is switched off, the spins remain aligned and the sample is said to have become magnetised.

9
At 4.7 K the length of ordered spins is much smaller and it is much easier to realign the domain walls and hence a smaller field is required. Since the domain size is larger at 2 K due to the increase of (x i ) with decreasing temperature more spins remain aligned when the field is switched off and so the remanent magnetisation increases as T decreases.

10
3.Discuss the following: The transition metal oxides MO order antiferromagnetically for M = Mn, Fe, Co and Ni. The strongest exchange interaction is found to be between next nearest neighbour cations spins. The Neel temperature increases across the series Mn-Ni, from 116 to 525 K, but the saturated moment falls from 4.9 to 1.7 µ B. The transition metal oxides MO have the NaCl structure

11
Superexchange leads to a small fraction of the ordered spin being localised on the ligand, e.g. stabilisation of a singlet term by mixing in of a closed shell singlet. This spin is not observed in the measurement of saturated moments. The more mixing there is of the metal (3d) and oxygen (2p) orbitals (i.e. the more covalent the compound is) then the more spin is localised on the ligand. As we go across the transition metal series the effective nuclear charge increases. Hence the cation radius decreases and this results in a stronger covalent interaction between M and O. This results in stronger superexchange and higher transition temperatures. Why does T N increase from 116 to 525 K, but the saturated moment falls from 4.9 to 1.7 µ B as M changes from Mn-Ni, Hence as we go from MnO to NiO and the compounds get more covalent and the measured saturated moment decreases from 4.9 to 1.7 µ B.

12
4. What is spin frustration? How can spin frustration be alleviated. In 3D extended systems which structure types is spin frustration observed? Square and triangular arrangements of ions with equal antiferromagnetic interactions between them. In the square arrangement, all the antiferromagnetic interactions can be satisfied. In the triangular arrangement, we cannot satisfy all interactions at the same time – the structure is frustrated.

13
For frustrated structures, how can we determine which is more frustrated? We can define a degree of frustration, F:- Frustration can be relieved by slight molecular distortions; a small change in bond angle or bond length may weaken an exchange by a few cm-1 – sufficient to induce a preferred spin orientation. It can also be relieved by spin canting (for example canting the spins by 109º 28’ in a tetrahedral complex results in a degree of frustration = 0).

14
Why does canting the spins by 109.8° relieve spin frustration in a tetrahedral complex? If F = 0 then there is no frustration!

15
Frustration is observed in the MO lattices (NaCl type), Rutile lattice, hexagonal tungsten bronze lattice and pyrochlore). There is usually a cubic to rhomobohedral symmetry change at T N. This disrupts the J90º ferromagnetic order and relieves the frustration.

16
This structure contains chains of edge sharing octahedra with direct exchange between M-M centres. However the metals are in a triangular configuration and so there is frustration. Structural distortions are often observed.

17
2D frustration in the HTB structure. The geometry is square planar in the z direction. Low T N s of ~ 100 K. The pyrochlore lattice is basically an array of 3D tetrahedra. Tetrahedra are frustrated so there is frustration in 3 dimensions in this structure type. It is the most frustrated structure in nature and very low T N s are observed (typically 20 K).

18
5. What are the key ingredients for superconductivity? Key Ingredients for superconductivity: At present there is no exact theory as to why the HTSC compounds superconduct at such high temperatures! BCS theory predicted T cMAX = 35 K so clearly something else is going on here. However there are some common structural generalisations that can be made about these layered cuprates. 1. One or more hole doped CuO 2 planes are required. Maximum T c for 3 or 4 adjacent planes. 2. Electropositive cations – Group IIA (Ca, Sr, Ba), Rare earth (La – Lu, Y) enhance Cu-O covalency. (a) Large cations (Sr 2+, Ba 2+, La 3+ ) prefer 10-12 coordinate sites A (cf perovskite structure) giving connection between CuO 2 plane and MO layer. (b) Small cations (Ca 2+, Y 3+ ) prefer 8 – coordinate sites B (cf CaF 2 structure) and prevent O intercalation between CuO 2 planes which is detrimental to superconductivity.

19
High T c Copper Oxide Superconductor Structures 3.One or two covalent M oxide (M = Cu, Hg, Tl, Pb, Bi) layers connected via interplanar O 2-. e.g. YBa 2 Cu 3 O 7 has two CuO 2 layers and M = Cu, A = Ba and B = Y; T c = 90 K. HgBa 2 Ca 2 Cu 3 O 9 has three CuO 2 layers, M = Hg, A = Ba and B = Ca; T c = 135 K. 4. Superconductivity is found for Cu oxidation state from 2.06 – 2.24. The maximum T c is found at an oxidation state of 2.16.

20
La 2-x Sr x CuO 4 B = Sr, La

21
Revision Ferromagnet: parallel alignment of spins Antiferromagnet: antiparallel alignment of spins Paramagnet: Random alignment of spins Diamagnet: No unpaired electrons ( is negative) Ferrimagnet: antiparallel alignment of spins but the ions on the different sublattices have different moment values (i.e. alternating Mn and Cu)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google