Chapter 8
description
Transcript of Chapter 8
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Chapter 8Chapter 8
Magnetic and optical properties
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Magnetic Properties
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Railgun
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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 + 4 M B : magnetic induction (the field within the body)M : intensity of magnetizationB/Ho = 1 + 4 (M/Ho) = 1 + 4v
M/Ho : dimensionlessv : magnetic susceptibility per volume
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Other definition of magnetic
susceptibility Gram Susceptibility: g = v/d (unit: cm3/g) d = density Molar Susceptibility: m = g
.(MW) (unit: cm3/mol, or emu)MW: molecular weight
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Macroscopic Point of View Magnetic moment, M, can be related to the rate the energy change of a body in the magnetic field, Ho.
M = -EHo
The sign of the induced magnetic moment, M, defines two types of magnetic behaviors:
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Diamagnetism and Paramagnetism
M (and v, g, m) is negative: diamagnetism
Materials with only paired electrons (In fact, all materials exhibit diamagnetism) M (and v, g, m) is positive: paramagnetism
Materials with unpaired electrons (In fact, these materials exhibit both paramagnetism and diamagnetism)
M = -EHo
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Ferromagnetism, antiferromagnetism and
ferrimagnetism
Ferromagnetism, antiferromagnetism and ferrimagnetism can be considered as special forms of paramagnetism.
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Diamagnetism
It arises from the interactions of electron pairs with magnetic field, generating a small magnetic field opposing the applied magnetic field, Ho.
a property of all materials
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Diamagnetism
SN N S
B = Ho + 4 M
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Diamagnetism Diamagnetic materials move to the region of lower field strength
(repelled by Ho)
Ho
M = -EHo
Ho
E> 0
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.
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Superconductors are perfect diamagnetic materials
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How to obtain diamagnetism (dia)?
T = para + dia
To study the paramagnetic behaviors of materials, dia must be subtracted from T
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Pascal’s constants
Calculation of dia from Pascal’s constants:
dia = i iAi + Bi
A: atoms B: bonds
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Table of Selected Pascal’s Constants atom A (x10
-6 emu) atom A (x10
-6 emu) atom A (x10
-6 emu)
H -2.93 F -63 Na+ -6.8
C -6.00 Cl -20.1 K+ -14.9
C(aromatic) -6.24 Br -30.6 bond B (x10-6
emu)
N -5.57 I -44.6 C=C +5.5
N(aromatic) -4.61 Mg2+ -5 C≡C +0.8
N(monamide) -1.54 Zn2+ -15 C=N +8.2
N(diamide, imide) -2.11 Pb2+ -32.0 C≡N +0.8
O -4.61 Ca2+ -10.4 N=N +1.8
O2(carboxylate) -7.95 Fe2+ -12.8 N=O +1.7
S -15.0 Cu2+ -12.8 C=O +6.3
P -26.3 Co2+ -12.8 anions (x10-6
emu)
Hg2+ -40.0 Ni2+ -12.8 C≡N- -13.0
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Example 1:
N
HH
H
H H
pyridine
5 x C (ring) = 5 x (-6.24) = -31.25 x H = 5 x (-2.93) = -14.61 x N (ring) = 1 x (=4.61) = -4.61
iAi = -31.2 + (-14.6) + (-4.61) = -50.3 x 10-6 cm3/mol (or emu)
iBi In this case, is zero, because the “ring values” of C and N are used.
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Example 2: O
CH3H3C
acetone
3 x C = 3 x (-6) = -186 x H = 6 x (-2.93) = -17.61 x O = -4.61
iAi = -18 + (-17.6) + (-4.61) = -40.21 x 10-6 emu
iBi = 1 x (C=O) = +6.3 x 10-6 emu
dia = i iAi + Bi = -33.9 x 10-6 emu
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Example 3: K4Fe(CN)6 Transition metal
complex
4 x K+ = 4 x (-14.9) = -59.61 x Fe2+ = -12.86 x (C≡N-) = 6 x (-13.0) = -78.0
dia = i iAi + Bi = -150.4 x 10-6 emu
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How to obtain dia of Cr(acac)3? Cr(acac)3 is paramagnetic so it is difficult to measure its
diamagnetism directly. Method I.Calculate dia from Pascal’s constants.
Method II. Synthesize Co(acac)3, Co3+: d6 low spin.
Use the dia value of Co(acac)3 as that of Cr(acac)3.
Method III.Measure the dia value of Na(acac), to obtain the dia value of aca
c-.Then include this value in Pascal’s constant calculation.
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Calculated Expt’l Agreement
N
HH
H
H H
-50.4 x 10-6 -49 x 10-6 good
AsMe2
AsMe2
-147 x 10-6 -194 x 10-6 poor
However, since para >> dia , the disagreement usually
does not cause too much problem.
Disagreement of dia
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Paramagnetism
Paramagnetism arises from the interaction of Ho with the magnetic field of the unpaired electron due to the spin and orbital angular momentum.
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Derivation of M and m
from microscopic point of viewFor S = 1/2 system (Spin only, no orbital contribution)
= -gS : magnetic moment
S : spin angular momentum
ge : electron g-factor (ge = 2.0037 for free e-)
( or B) : Bohr magneton of the electron
= 9.3 x 10-21 erg/Gauss
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Interaction energy between magnetic moment and applied
field
The Hamiltonian describing the interaction energy of this moment with the applied magnetic field, Ho, is:
H = -.H = gS.H
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The energies for S = ½ systemThe energies (eigen values) for S = 1/2 (ms = +1/2, -1/2) system are: E = msgHo E1/2 = (1/2)gHo and E-1/2 = –(1/2)gHo
Energy
ms=1/2
ms=1/2
ms=-1/2
ms=-1/2
E = gHo
Ho
Zeeman effect
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Relative populations of ½ and –½ states
When Ho = 25 kG E ~ 2.3 cm-1
At 300 K, kT ~ 200 cm-1
Boltzmann distribution e-E/kT ~ 1P1/2
P-1/2=
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.
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To obtain M ( or m), we need to consider all the energy states that are populated.
∵ H = -.H = gS.H
∴ The magnetic moment, n with the direction // Ho, of an e
lectron in a quantum state n can be obtained by:n = -(∂En)/(∂H) = -msgSo we consider:
(1) The magnetic moment of each energy state. (2) The population of each energy state.
That is, M = Nn.Pn
N : Avogadro’s number Pn : probability in state n.
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n = -(∂En)/(∂H) = -msg
e-En/kT
e-En/kTPn = =Nn
NT
Nn: population of state n
NT: population of all the states
M = Nn.Pn
H = -.H = gS.H
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M = Nn.Pn
Energy
ms=1/2
ms=1/2
ms=-1/2
ms=-1/2
E = gHo
Ho
En1/2gH
-1/2gH
n
-1/2g
1/2g
E = msgHo n = -(∂En)/(∂H) = -msg
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M for S=1/2 system
=N[g/2 -g/2
g/2kT -g/2kTe
+
e ][eg/2kT e-g/2kT]
M=N
msn e-En/kT
ms
e-En/kT
e± x~ 1 x±
when x << 1
Since gH<<kT when Ho ~ 5kG
= [ 1 + g/2kT 1( g/2kT) ]g/2kT+1 + )g/2kT(1
N g2
= Ng24kT M=
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Curie Law of paramagnetic materials
N g2
4kTm = M
=
Curie Law: m = C/T
∴ N g2
4k=C
m
1/T
slope = C
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Curie-Weiss Law
m
T1
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)
m = C/(T-)
is called Weiss constant.If is positive, then the magnets tend to align parallelly.If is negative, then the magnets tend to align antiparallelly.
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For general S values (not only S = 1/2)
En = msgbH ms = -S, -S+1, …. , S-1, S
M=N
ms=-S
ms
S
(-msg)e-msgH/kT
-msgH/kTe= Ng2H
3kTS(S+1)
N g2
3kTm = M
= S(S+1) spin only
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for S = 1/2, 1, 3/2
m
1/T
slope = Curie const.
S=1/2
S=1
S=3/2For S=1/2 N g2
4kTm =
For S=1 m = 2N g2
3kT
For S=3/2 m = 5Ng2
4kT
N g2
3kTm = M
= S(S+1)
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Definition of eff
m S(S+1)N g2
3kT=
eff =( )3kN
1/2
(mT)1/2
~1/2
)mT(2.828
eff = g[S(S+1)]1/2 (BM, Bohr Magneton)
or eff = [n(n+2)]1/2 where n= number of unpaired e-
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Saturation of Magnetization
The Curie-Wiess law does not hold where the system is approaching saturation.
An approximation has been made:
gH << kT so that e± x~ 1 x±
If the applied magnetic field is very large, Curie-Weiss law is not valid.
(M is not proportional to H) N g2
3kTm = M
= S(S+1)
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Saturation of Magnetization
M/mol-1
H/kT
S=1/2
S=1
S=3/2
S=2
1
2
3
4
1 20
follow Curie -Weiss law
Energy
ms=1/2
ms=1/2
ms=-1/2
ms=-1/2
E = gHo
Ho
If H is large enough, the probability of ms= -1/2
populated is close to 100%.
M = -msg M = for 1 magnet.
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Plot of eff vs Temperature
All the molecules are in the S=3/2 state at all temperatures.
temperature
eff
(BM)
3.87
S=3/2 eff = 2[S(S+1)]1/2
eff = 2.828(T)1/2
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Plot of eff vs Temperature
temperature
eff
(BM)
3.87
1.7
S=3/2
S=1/2
The eff value of the system gradually decreases from high-te
mperature value of 3.87 BM (S=3/2) to low-temperature value 1.7 BM (S=1/2)
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Spin equilibrium and Spin Crossover
temperature
eff
(BM)
3.87
1.7
S=3/2
S=1/2temperature
eff
(BM)
3.87
1.7
S=3/2
S=1/2
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Calculation of eff for f-block elements Now, we consider spin-only cases.For f-block elements, spin-orbit coupling is very large
eff = g[J(J+1)]1/2
g = 1+2J(J+1)
S(S+1)-L(L+1)+J(J+1)
g-value for free ions
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Example: calculate the eff of Nd3+ (4f3)
+3 +2 +1 0 -1 -2 -3ml
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 = 9/2Ground state term symbol: 4I9/2
g = 1+2x(9/2)(9/2+1)
3/2(3/2+1)-6(6+1)+(9/2)(9/2+1)= 0.727
eff = g[J(J+1)]1/2 = 0.727[(9/2)(9/2+1)]=3.62 BM
g = 1+2J(J+1)
S(S+1)-L(L+1)+J(J+1)
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eff values of d-block elements
For d-block elements, spin-orbit coupling is less important. In many cases, eff = 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.
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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.
x
y
dx2-y2
x
y
dxy
z
e-
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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.
E dxy
dx2-y2
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Spin-Orbit Coupling
Electrons have to change directions of spins to circulate Little contribution from orbital angular momentum.
E
dxydx2-y2
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Spin-Orbit Coupling
Orbitals are filled. Little contribution from orbital angular momentum.
E
dxydx2-y2
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Spin-Orbit Coupling
Spin-orbit couplings are significant in the above two cases.
E
dxydx2-y2
E
dxydx2-y2
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Other orbital sets that may give spin-orbit coupling.
dx2-y2/dxy rotate about z-axisdxz/dyz rotate about z-axisdxz/dxy rotate about x-axisdyz/dxy rotate about y-axisdz2/dxz rotate about y-axisdz2/dyz rotate about x-axis
There are no spin-orbit coupling contribution for dz2/dx2-y2 and dz2/dxy
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Magic pentagonRelated to “strength” of spin-orbit coupling
dxydx2-y2
dz2
dxz dyz
66
8
2
22 22
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Spin-orbit coupling influences g-value
g = 2.0023 +E1-E2
n
2.0023: g-value for free ion+ sign for <1/2 filled subshell- sign for >1/2 filled subshelln: number of magic pentagon: free ion spin-orbit coupling constant
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For example: CuIIL4 system (Cu2+: d9)
CuII
L
LL
L
D4h point group
(axial symmetry)
dx2-y2
dxy
dz2
dxz dyz = -829 cm-1
E(dxy)-E(dx2-y2) = 15000 cm-1
(obtained by UV spectroscopy)
gz = g// = 2.0023-8(-829)/15000 = 2.44
gx = gy = g⊥ = 2.0023
spin-orbit coupling has little contribution to x and y directions.
g = 2.0023 +E1-E2
n
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Now, we can predict if the angular momentum will be quenched.
Example: check all the electron configurations in an octahedral field.
Which ones of the above electron configuration in Oh field have little spin-orbit contribution (with g ~ 2.0)?
d3, d4(HS), d5(HS), d6(LS), d7(LS), d8, d9, d10
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In low-symmetry field, spin-orbit coupling are quenched
Remember that Oh is a high-symmetry field.If the symmetry is lowered, degeneracy will be destroyed and the orbital contribution will be quenched.
Oh D4h
D4h: all are quenched except d1 and d3
For low-symmetry field, all are quenched and therefore eff = g[S(S+1)]1/2 (spin-only) is valid.
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Van Vleck Equation
In some cases, the paramagnetic behaviors are more complicated due to(1) mixing of low-lying excited state(2) zero field splitting (ZFS)(3) higher order Zeeman effect.
These problems can be treated using Van Vleck Equation.
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Van Vleck Equation En : the energy of state n can be expressed as a power series of H.
En = En(0) + H ・ En
(1) + H2 ・ En(2) + H3 ・ En
(3) + …..
En(0) : En of zero field; can be set to zero by convention.
H : applied magnetic fieldEn
(1) : 1st order Zeeman coefficient
En(2) : 2nd order Zeeman coefficient (related to mixing of
low- lying excited state)En
(3) : 3rd order Zeeman coefficient (very small and can
be neglected)
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Van Vleck EquationEn = En
(0) + H ・ En(1) + H2 ・ En
(2) + H3 ・ En(3) + …..
n =-En
Ho= En-2HEn
(1) (2)
-En/kTe = e
-En-HEn-H2En(0) (1) (2)
kT = e-HEn-H2En
(0) (1) (2)
kT-En
kT x e
= (1- kTHEn
(1)
) x e kT-En
(0)
∵ when x is small, e-x ~ 1-x, and H2En(2) << kT
-
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Van Vleck Equation
M=N n
n e-En/kT
n e-En/kT= Nn
(-En-2HEn)(1)
(1- kTHEn
(1)
) x e kT-En
n (1- kT
HEn(1)
) x e kT-En
(2)
(0)
(0)
For a paramagnetic system, when H = 0, M = 0
and (En(2))2 and (En
(2)˙En(1)) are very small
and can be neglected. m = M/H
m = Nn[(En)
2(1)
kT -2En(2)]e-En
(0)
kT
ne-En
(0)
kT
∴
∴ nEn
(1)e-En(0)
kT = 0
![Page 58: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/58.jpg)
Application of Van Vleck Equation Example 1: S = 1/2, ms=+1/2, -1/2, no excited state En
(2) = 0
En = En(0) + H ・ En
(1) + H2 ・ En(2) + …….
E = msgHo n = -(∂En)/(∂H) = -msg
Energy
ms=1/2
ms=1/2
ms=-1/2
ms=-1/2
E = gHo
Ho
0 1/2g
-1/2g0
En(1)En
(0)Energy
1/2g
-1/2g
![Page 59: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/59.jpg)
Application of Van Vleck Equati
on S=1/2
m = Nn[(En)
2(1)
kT -2En(2)]e-En
(0)
kT
ne-En
(0)
kT
=N
n[(En)
2(1)
kT]e0
n
0em =
[N (g/2)2
kT +(-g/2)
2
kT]
1 + 1
m = Ng2
4kT
this result is the same as what we obtained from simple boltzmann distribution.
N g2
3kTm = M
= S(S+1)
Energy
ms=1/2
ms=1/2
ms=-1/2
ms=-1/2
E = gHo
Ho
0 1/2g
-1/2g0
En(1)En
(0)Energy
1/2g
-1/2g
![Page 60: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/60.jpg)
Example 2: Cr3+, d3, S = 3/2 with zero field splitting (ZFS) = D
ms= 1/2_+
ms= 3/2_+
ms= 3/2_+
ms= 1/2_+
E=ZFS=D
ms=+3/2
ms=-3/2
ms=+1/2
ms=-1/2
0
Energy0 H
D+3/2gH
D-3/2gH
1/2gH
-1/2gH
En(0)
En(1)
D
D
0
0
3/2g
-3/2g
1/2g
-1/2g
Energy
In this case, En(2) = 0.
![Page 61: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/61.jpg)
Example 2
m = Nn[(En)
2(1)
kT -2En(2)]e-En
(0)
kT
ne-En
(0)
kT
m =[(-g/2) +
2(g/2)
2
kTe0 0e (3g/2)+kT kT
2-D/kTe +
(-3g/2)2
kT e-D/kT]e0 0e e-D/kT e-D/kT+ + +
m = Ng2
4kT [1 + 9e-D/kT
e-D/kT1 +
]
N
ms= 1/2_+
ms= 3/2_+
ms= 3/2_+
ms= 1/2_+
E=ZFS=D
ms=+3/2
ms=-3/2
ms=+1/2
ms=-1/2
0
Energy0 H
D+3/2gH
D-3/2gH
1/2gH
-1/2gH
En(0)
En(1)
D
D
0
0
3/2g
-3/2g
1/2g
-1/2g
Energy
![Page 62: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/62.jpg)
Example 2 when D 0 or at high temperature
m = Ng2
4kT [1 + 9e-D/kT
e-D/kT1 +
]
When ZFS is very small (D0) or at high temperature (kT >> D),
e-D/kT 1
m = Ng2
4kTx 10/2 = 5Ng2
4kT
ms= 3/2_+
ms= 1/2_+
ms=+3/2
ms=-3/2
ms=+1/2
ms=-1/2
N g2
3kTm = M
= S(S+1)
with S = 3/2
![Page 63: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/63.jpg)
When D ∞ or at low temperature (kT<<D), e-D/kT 0.
m = Ng2
4kT [1 + 9e-D/kT
e-D/kT1 +
]
m = Ng2
4kT
ms= 3/2_+
ms= 1/2_+
ms= 3/2_+
ms= 1/2_+
E=ZFS=D
If D is very largeonly ms = +1/2, -1/2are populated
N g2
3kTm = M
= S(S+1)
with S = 1/2
The system behaves like:
![Page 64: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/64.jpg)
Value of ZFS can be obtained by curve-fitting
temperature
eff
(BM)
3.87
1.7
S=3/2
S=1/2
m
1/T
S=1/2
S=3/2
HT LT
m = Ng2
4kT [1 + 9e-D/kT
e-D/kT1 +
] ZFS (D) can be obtained by curve-fitting.
![Page 65: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/65.jpg)
Example 3. S =1 system
E=ZFS=D
0
Energy0 H
D+gH
D-gH
En(0)
En(1)
D
D
0
g
-g
0
Energy
ms= 0
ms= 1_+
ms= 0
ms= 1_+
ms=-1
0
ms=+1
ms=0
m = Nn[(En)
2(1)
kT -2En(2)]e-En
(0)
kT
ne-En
(0)
kT
m =[ (-g)
+2
e (g)kT kT
2
e ]e 0
+ +e-D/kT
-D/kT -D/kT
e-D/kT
N
Van Vleck equation
![Page 66: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/66.jpg)
e.g. 3
m =[ (-g)
+2
e (g)kT kT
2
e ]e 0
+ +e-D/kT
-D/kT -D/kT
e-D/kT
N m =1 + 2
[ e-D/kT
e-D/kT
]2N g2
kT
At high temperature, or ZFS is very small (D<<kT) then e-D/kT 1
m = 2Ng2
3kT
At low temperature or very large ZFS (D>>kT), then e-D/kT 0.
m 0The system appears to be diamagnetic because only ms = 0 state is populated.
N g2
3kTm = M
= S(S+1)
The system appears to be like S = 1with no ZFS.
![Page 67: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/67.jpg)
Interactions of micromagnets
No interactions between magnets
Paramagnetic materials
These magnets are oriented randomlyunder zero applied magnetic field.
If there are interactions between these micromagnets, these materials are ferromagnetic, antiferromagnetic or ferrimagnetic.
m
T
paramagnetic
![Page 68: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/68.jpg)
Ferromagnetic, Antiferromagnetic and Ferrimagnetic
m
T
+Curie temperature
paramagnetic
ferromagnetic
m
T
+Neel temperature'
paramagnetic
antiferromagnetic
With interactions among micromagnets
![Page 69: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/69.jpg)
Magnetic domain
H
Average domain size: 20 ~200 nm
![Page 70: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/70.jpg)
Hysteresis curve of M vs H
Hc
MsMr
H
M
Ms : saturation magnetizationMr: remanence magnetizationHc: coercive magnetic field
![Page 71: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/71.jpg)
Magnetic interaction of polynuclear clusters
OCu
O
O
O
H2O Cu
O
O
OH2
O
O
Cu2(OAc)4.(H2O)2
m
T
paramagnetic
m
T
+Neel temperature'
paramagnetic
antiferromagnetic
![Page 72: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/72.jpg)
Antiferromagnetic coupling complexesAntiferromagnetic coupling complexes
OCu
O
O
O
H2O Cu
O
O
OH2
O
O
Some terms to define:
Magnetic orbital: orbital containing an unpaired electron
Exchange interactions: magnetic interactions between metal centers
Exchange parameter: J, coupling constant
![Page 73: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/73.jpg)
Magnetic behavior of d1-d1 dimerO
CuO
O
O
H2O Cu
O
O
OH2
O
O
J: coupling constantJ > 0 : antiferromagnetic couplingJ < 0 : ferromagnetic coupling
S=1
S=0
S=0
S=1J -J
J > 0 J < 0
S=0
S=1
![Page 74: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/74.jpg)
The energy diagram of d1-d1 dimer system
0
Energy 0 H
J+gH
J-gH
En(0) En
(1)
J
g
-g
0
Energy
ms=-1
0
ms=+1
ms=0
J
S = 0
S = 1 ms= 0
J
0
J J
0
En = En(0) + H ・ En
(1) + H2 ・ En(2) + H3 ・ En
(3) +
…..
Antiferromagnetic coupling system
![Page 75: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/75.jpg)
The value of J can be obtained by curve-fitting.
0
Energy 0 H
J+gH
J-gH
En(0) En
(1)
J
g
-g
0
Energy
ms=-1
0
ms=+1
ms=0
J
S = 0
S = 1 ms= 0
J
0
J J
0
m = Nn[(En)
2(1)
kT -2En(2)]e-En
(0)
kT
ne-En
(0)
kT
Van Vleck equation
em =
[ (-g)+
2
e (g)kT kT
2
e ]e0
+
-J/kT -J/kTN-J/kT3 1 e=
[2g2 ekT]
+
-J/kTN
-J/kT3
m 3 e= [
2Ng2
]+J/kT
kT
Bleany-Bowers Equation
m
T
+Neel temperature'
paramagnetic
antiferromagnetic
![Page 76: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/76.jpg)
Two extreme conditions of the d1-d1 system
m 3 e= [
2Ng2
]+J/kT
kT 1 e=[2g2 ekT
]+
-J/kTN
-J/kT3
When J >> kT or T 0, m 0, this system is diamagnetic or a Cu-Cu bond is formed.
When J << kT or at high temperature, m = Ng22/2kT
Ng22/2kT can be considered as two Ng22/4kT, i.e., d1-d1 can be considered as two independent d1 system
OCu
O
O
O
H2O Cu
O
O
OH2
O
O
![Page 77: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/77.jpg)
The relationship between J and TN
m
T
+Neel temperature'
paramagnetic
antiferromagnetic
m 3 e= [
2Ng2
]+J/kT
kT
Solve m)/ (T) = 0
TN ≒ (5/8) J/K
![Page 78: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/78.jpg)
Direct couping through metal-metal bonding or through superexchange via ligands?
OCu
O
O
O
H2O Cu
O
O
OH2
O
O
dxz, dyzdxydz2
dx2-y2
z-axis
Magnetic orbital
Weak-bond is formed between the interaction of magnetic orbitals.
Is the d1-d1 interaction through ) -bonding ?(2) superexchange via ligands ?
![Page 79: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/79.jpg)
Goodgame’s experiment in 1969
OCu
O
O
O
H2O Cu
O
O
OH2
O
O
[Cu2(O2CH)4(NCS)2]2- [Cu2(O2CCH3)4(NCS)2]2-
d(Cu…Cu) = 2.716 Å d(Cu…Cu) = 2.643 Å
S=1
S=0
S=1
S=0
J = 485 cm-1J = 305 cm-1
This results indicate that the d1-d1 interactionis not through metal-metal bonding only.Superexchange mechanism through ligandsmay dominate.
![Page 80: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/80.jpg)
V
L
V
LL
L
L
L
O O
Mo
L
Mo
LL
L
L
L
O O
x
y
Comparison of V4+-V4+ and Mo5+-Mo5+ dimers
z
dxy
dxz,dyzdx2-y2
dz2
S=1
S=0
S=1
S=0
J = 3000 cm-1J = 200 cm-1
d(V…V) = 3.20 Å d(Mo…Mo) = 2.78 Å
Diamagnetic compound
Paramagnetic at HTAntiferromagnetically coupledAt LT
![Page 81: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/81.jpg)
Magnetic phenomena in 1D crystal
![Page 82: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/82.jpg)
Variation of flux density in diamagnetic and paramagnetic substances in a magnetic field
![Page 83: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/83.jpg)
Magnetic susceptibility
![Page 84: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/84.jpg)
Plot of reciprocal susceptibility against temperature
![Page 85: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/85.jpg)
Some properties of ferromagnetic materials
![Page 86: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/86.jpg)
Some Values of magnetic moments
![Page 87: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/87.jpg)
Antiferromagnetic coupling of spins of d electrons on Ni2+ ions through p
electrons of oxide ions
![Page 88: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/88.jpg)
Ferromagnetic ordering in bcc Fe, fcc Ni and hcp Co
![Page 89: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/89.jpg)
Electronic constitution of iron, cobalt and nickel
4.8 + 2.6 = 7.44.8 up + 2.6 down4.8 - 2.6 = 2.2
![Page 90: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/90.jpg)
Occupied energy levels and density of states N(E) for 3d and 4s, 4p
bands
![Page 91: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/91.jpg)
Pauli paramagnetismPauli paramagnetism
H
Pauli paramagnetism is temperature independent but field dependent.
Electrons near Fermi level are paired. temperature independent
Excess unpaired electronsInduced by magnetic field
![Page 92: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/92.jpg)
Curie and Néel temperatures in lanthanides
![Page 93: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/93.jpg)
Magnetic structure of antiferromagnetic and ferromagnetic
spinels
Spinel[M2+]tet[Fe3+]2
octO4
Inverse Spinel[Fe3+]tet[M2+, Fe3+]octO4
Partial inverse Spinel[Fe3+
1-xZn2+x]tet[M2+
1-xFe3+1+x]octO4
![Page 94: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/94.jpg)
Variation in saturation magnetization with composition for ferrite solid
solution
Partial inverse Spinel[Fe3+
1-xZn2+x]tet[M2+
1-xFe3+1+x]octO4
M1-xZnxFe2O4
When x is large, the antiferromagmnetic coupling is destroyed.
OM M
![Page 95: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/95.jpg)
Garnets
![Page 96: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/96.jpg)
Atomic coordinates for Y3Fe5O12 (YIG)
x 3
x 3
x 2
![Page 97: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/97.jpg)
Variation of magnetic moment at 0 K of garnets
![Page 98: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/98.jpg)
Spontaneous Magnetization in Dy3Fe5O12 garnet
The spins on rare earth (Dy3+) sublattice randomize much rapidly than those on Fe3+ sublattice.
Dy3Fe3Fe2O12
![Page 99: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/99.jpg)
Crystallographic data for ilmenite
![Page 100: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/100.jpg)
Schematic representation of the luminescence
![Page 101: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/101.jpg)
Schematic design a fluorescent lamp
![Page 102: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/102.jpg)
Luminescence spectra of activated ZnS phosphors after irradiation with
UV light
![Page 103: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/103.jpg)
Various types of electronic transition in activator ions
![Page 104: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/104.jpg)
Ground state potential energy diagram for a luminescent center in an ionic
host crystal
![Page 105: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/105.jpg)
Ground and excited state potential energy diagrams for a luminescent
center
![Page 106: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/106.jpg)
Non-radiative energy transfer involved in operation of a sensitized
phosphor
![Page 107: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/107.jpg)
Some lamp phosphor materials
![Page 108: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/108.jpg)
Schematic representation of anti-Stokes and normal luminescence
phenomena
![Page 109: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/109.jpg)
Energy levels of the Cr3+ ion in ruby crystal and laser emission
![Page 110: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/110.jpg)
Design of a ruby laser
![Page 111: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/111.jpg)
Ga-As laser
![Page 112: Chapter 8](https://reader035.fdocuments.us/reader035/viewer/2022081420/56815735550346895dc4d60b/html5/thumbnails/112.jpg)
Energy levels of the Nd3+ ion in neodymium lasers