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Chapter 8

Chapter 8. Magnetic and optical properties. Magnetic Properties. Railgun. Magnetization and Magnetic Susceptibility. If a body is placed in a homogeneous magnetic field H o , the magnetic field with the body varies from free-space value . That is, B = H o + 4 p M

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Chapter 8

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  1. Chapter 8 Magnetic and optical properties

  2. Magnetic Properties

  3. Railgun

  4. Magnetization and Magnetic Susceptibility • If a body is placed in a homogeneous magnetic field Ho, the magnetic field with the body varies from free-space value. • That is, • B = Ho + 4p M • B : magnetic induction (the field within the body) • M : intensity of magnetization • B/Ho = 1 + 4p (M/Ho) = 1 + 4p cv • M/Ho : dimensionless • cv : magnetic susceptibility per volume

  5. Other definition of magnetic susceptibility • Gram Susceptibility: cg = cv/d • (unit: cm3/g) • d = density • Molar Susceptibility: cm = cg.(MW) • (unit: cm3/mol, or emu) • MW: molecular weight

  6. Macroscopic Point of View • Magnetic moment, M, can be related to the rate the energy change of a body in the magnetic field, Ho. The sign of the induced magnetic moment, M, defines two types of magnetic behaviors:

  7. Diamagnetism and Paramagnetism M (and cv, cg, cm) is negative: diamagnetism Materials with only paired electrons (In fact, all materials exhibit diamagnetism) M (and cv, cg, cm) is positive: paramagnetism Materials with unpaired electrons (In fact, these materials exhibit both paramagnetism and diamagnetism)

  8. Ferromagnetism, antiferromagnetism and ferrimagnetism • Ferromagnetism, antiferromagnetism and ferrimagnetism can be considered as special forms of paramagnetism.

  9. Diamagnetism a property of all materials • It arises from the interactions of electron pairs with magnetic field, generating a small magnetic field opposing the applied magnetic field, Ho.

  10. Diamagnetism • B = Ho + 4p M

  11. Diamagnetism Diamagnetic materials move to the region of lower field strength (repelled by Ho) For diamagnetic materials, M < 0, ∴ That is, the energy of the system increases with the applied magnetic field, Ho. ∴ the body moves in the direction of lower energy (i.e. lower field). The process is exothermic.

  12. Superconductors are perfect diamagnetic materials

  13. How to obtain diamagnetism (cdia)? • cT = cpara + cdia To study the paramagnetic behaviors of materials, cdia must be subtracted from cT

  14. Pascal’s constants Calculation of cdia from Pascal’s constants: • A: atoms B: bonds

  15. Table of Selected Pascal’s Constants

  16. Example 1: 5 x C (ring) = 5 x (-6.24) = -31.2 5 x H = 5 x (-2.93) = -14.6 1 x N (ring) = 1 x (=4.61) = -4.61 pyridine = -31.2 + (-14.6) + (-4.61) = -50.3 x 10-6 cm3/mol (or emu) In this case, is zero, because the “ring values” of C and N are used.

  17. Example 2: 3 x C = 3 x (-6) = -18 6 x H = 6 x (-2.93) = -17.6 1 x O = -4.61 acetone = -18 + (-17.6) + (-4.61) = -40.21 x 10-6 emu = 1 x (C=O) = +6.3 x 10-6 emu = -33.9 x 10-6 emu

  18. Example 3: K4Fe(CN)6 Transition metal complex 4 x K+ = 4 x (-14.9) = -59.6 1 x Fe2+ = -12.8 6 x (C≡N-) = 6 x (-13.0) = -78.0 = -150.4 x 10-6 emu

  19. How to obtain cdia of Cr(acac)3? Cr(acac)3 is paramagnetic so it is difficult to measure its diamagnetism directly. Method I. Calculate cdia from Pascal’s constants. Method II. Synthesize Co(acac)3, Co3+: d6 low spin. Use the cdia value of Co(acac)3 as that of Cr(acac)3. Method III. Measure the cdia value of Na(acac), to obtain the cdia value of acac-. Then include this value in Pascal’s constant calculation.

  20. Disagreement of cdia Calculated Expt’lAgreement -50.4 x 10-6-49 x 10-6good -147 x 10-6-194 x 10-6poor However, since cpara >> cdia , the disagreement usually does not cause too much problem.

  21. Paramagnetism • Paramagnetism arises from the interaction of Ho with the magnetic field of the unpaired electron due to the spin and orbital angular momentum.

  22. Derivation of M and cmfrom microscopic point of view For S = 1/2 system (Spin only, no orbital contribution) : magnetic moment : spin angular momentum ge : electron g-factor (ge = 2.0037 for free e-) b ( or mB) : Bohr magneton of the electron b = 9.3 x 10-21 erg/Gauss

  23. Interaction energy between magnetic moment and applied field • The Hamiltonian describing the interaction energy of this moment with the applied magnetic field, Ho, is:

  24. The energies for S = ½ system The energies (eigen values) for S = 1/2 (ms = +1/2, -1/2) system are: E = msgbHo E1/2 = (1/2)gbHo and E-1/2 = –(1/2)gbHo Zeeman effect

  25. Relative populations of ½ and –½ states When Ho = 25 kG DE ~ 2.3 cm-1 At 300 K, kT ~ 200 cm-1 Boltzmann distribution The populations of ms = 1/2 and –1/2 states are almost equal with only a very slight excess in the –1/2 state. That means that even under very large applied field, the net magnetic moment is still very small.

  26. To obtain M ( or cm), we need to consider all the energy states that are populated. ∵ ∴ The magnetic moment, mn with the direction // Ho, of an electron in a quantum state n can be obtained by: mn = -(∂En)/(∂H) = -msgb So we consider: (1)The magnetic moment of each energy state. (2)The population of each energy state. That is, M = NSmn.Pn N : Avogadro’s number Pn : probability in state n.

  27. mn = -(∂En)/(∂H) = -msgb M = NSmn.Pn Nn: population of state n NT: population of all the states

  28. M = NSmn.Pn E = msgbHo mn = -(∂En)/(∂H) = -msgb

  29. M for S=1/2 system Since gbH<<kT when Ho ~ 5kG when x << 1

  30. Curie Law of paramagnetic materials Curie Law: cm = C/T ∴

  31. cm = C/(T-q) Curie-Weiss Law q is called Weiss constant. If q is positive, then the magnets tend to align parallelly. If q is negative, then the magnets tend to align antiparallelly. If the system is not magnetically dilute (pure paramagnetic), the neighboring magnetic moments may exhibit an overall tendency of parallel alignment or antiparallel alignment. (still considered as paramagnetic, not ferromagnetic or antiferromagnetic)

  32. For general S values (not only S = 1/2) En = msgbH ms = -S, -S+1, …. , S-1, S spin only

  33. c for S = 1/2, 1, 3/2 For S=1/2 For S=1 For S=3/2

  34. Definition of meff meff = g[S(S+1)]1/2 (BM, Bohr Magneton) or meff = [n(n+2)]1/2where n= number of unpaired e-

  35. Saturation of Magnetization The Curie-Wiess law does not hold where the system is approaching saturation. An approximation has been made: gbH << kT so that If the applied magnetic field is very large, Curie-Weiss law is not valid. (M is not proportional to H)

  36. Saturation of Magnetization If H is large enough, the probability of ms= -1/2 populated is close to 100%. M = -msgb M = b for 1 magnet.

  37. Plot of meff vs Temperature All the molecules are in the S=3/2 state at all temperatures.

  38. Plot of meff vs Temperature The meff value of the system gradually decreases from high-temperature value of 3.87 BM (S=3/2) to low-temperature value 1.7 BM (S=1/2)

  39. Spin equilibrium and Spin Crossover

  40. Calculation of meff for f-block elements Now, we consider spin-only cases. For f-block elements, spin-orbit coupling is very large meff = g[J(J+1)]1/2 g-value for free ions

  41. Example: calculate the meff of Nd3+ (4f3) Lmax = 3+2+1 = 6 Smax = 3 x 1/2 = 3/2 2S+1= 2 x 3/2 + 1 = 4 Ground state J = L-S = 6-3/2 = 9/2 Ground state term symbol: 4I9/2 meff = g[J(J+1)]1/2 = 0.727[(9/2)(9/2+1)]=3.62 BM

  42. meff values of d-block elements • For d-block elements, spin-orbit coupling is less important. In many cases, meff = g[S(S+1)]1/2 is valid. • The orbital angular momentum is often “quenched” by special electronic configuration, especially when the symmetry is low.

  43. Spin-Orbit Coupling • For example, if an electron can move back and forth between dx2-y2 and dxy orbitals, it can be considered as circulating about the z-axis, giving significant contribution to orbital angular momentum.

  44. Spin-Orbit Coupling • If dx2-y2 and dxy orbitals have different energies in a certain electron configuration, electrons cannot go back and forth between them. • ∴ little contribution from orbital angular momentum.

  45. Spin-Orbit Coupling • Electrons have to change directions of spins to circulate  Little contribution from orbital angular momentum.

  46. Spin-Orbit Coupling • Orbitals are filled.  Little contribution from orbital angular momentum.

  47. Spin-Orbit Coupling • Spin-orbit couplings are significant in the above two cases.

  48. Other orbital sets that may give spin-orbit coupling. dx2-y2/dxy rotate about z-axis dxz/dyz rotate about z-axis dxz/dxy rotate about x-axis dyz/dxy rotate about y-axis dz2/dxz rotate about y-axis dz2/dyz rotate about x-axis There are no spin-orbit coupling contribution fordz2/dx2-y2anddz2/dxy

  49. Magic pentagon • Related to “strength” of spin-orbit coupling

  50. Spin-orbit coupling influences g-value 2.0023: g-value for free ion + sign for <1/2 filled subshell - sign for >1/2 filled subshell n: number of magic pentagon l: free ion spin-orbit coupling constant

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