Download Chapter 12 What is a paramagnetic material?

Chapter 12
What is a paramagnetic material?
Paramagnetic materials are substances that show only miniscule magnetic effects.
They are weakly attracted to a magnetic field.
They can also be characterised by the
extent to which an external magnetic field is able to penetrate into the sample, the magnetic
permeability, , defined by:
 = 0(1 + )
where 0 is a fundamental constant, the permeability of free space.
Paramagnetic
materials have  greater than 0.
Paramagnetic solids are those in which some of the atoms, ions or molecules
making up the solid possess a permanent magnetic dipole moment.
isolated from one another.
magnets.
These dipoles are
The solid, in effect, contains small, non-interacting atomic
In the absence of a magnetic field, these are arranged at random and the solid
shows no net magnetic moment. In a magnetic field, the elementary dipoles will attempt to
orient themselves parallel to the magnetic flux density in the solid, and this will enhance the
internal field within the solid and give rise to the observed paramagnetic effect, (Figure
12.3c, d).
The alignment of dipoles will not usually be complete, because of thermal
effects and interaction with the surrounding atoms in the structure and the dipoles
continually change orientation because of this jostling.
What causes magnetic hysteresis?
Magnetic hysteresis refers to the nature of the curve of the magnetic flux density, B, in
a ferromagnetic solid as the magnetic field, H, is cycled from positive to negative.
The
curve forms a closed loop (provided that the extreme values of H are great enough) called a
hysteresis loop.
Hysteresis is essentially due to the domain structure of a ferromagnetic material.
In
general a ferromagnetic crystal will be composed of an equal number of domains oriented
in all the equivalent directions allowed by the crystal symmetry. The overall magnetization
of the crystal will be zero. If a magnetic field is applied in a nominally positive direction, the
magnetic dipoles will attempt to re-orient themselves in a direction parallel to the applied
field. As the magnetic field increases more dipoles will reorient until ultimately, at high
enough fields, all the dipoles are aligned in the same direction and the crystal will, (in
principle), consist of a single domain.
On reducing and then reversing the applied
magnetic field the converse takes place. This will be opposed by the internal energy of the
solid, as new domain walls have to be created and domain walls must move. The degree
of reorientation of the dipoles will lag behind that encountered at any equal value of H those
on the forward path so that the B - H curve will therefore not follow the original path, but will
trace a new path that lags behind the old one.
Continued cycling of H generates the
hysteresis loop. The definition of hysteresis is “to lag behind”.
What materials are used for magnetic data storage?
The magnetic storage of information relies upon magnetising small volumes of
material, often a thin film deposited on a solid surface or a flexible tape.
Most magnetic
data storage uses the magnetic properties of small individual magnetic particles.
Ideally,
small single domain particles are used, each of which has only two directions of
magnetisation, directed along the + and – directions of a single crystallographic axis. The
direction is switched by the write head and sensed by the read head.
The commonest
magnetic particles in use at present are -Fe2O3, (maghemite), cobalt doped -Fe2O3, and
chromium dioxide, CrO2, all of which have acicular (needle-shaped) crystals. The direction
of magnetisation in all of these compounds corresponds to the needle axis.
For best
recording performance, the crystallites are aligned in a collinear array with the needle axis
parallel to the direction of motion of the tape or disc by applying a magnetic field during the
coating operation.
Quick quiz
1. b; 2. a; 3. a; 4. b; 5. a; 6. c; 7. b; 8. c; 9. b; 10. b; 11. a; 12. c; 13. a; 14.
c; 15. b; 16. c; 17. c; 18. c.
Calculations and questions
1. S, 3/2; L, 6; J, 9/2; m, 3.62 B.
2. S, 7/2; L, 0; J, 7/2; m, 7.94 B.
3. high spin: S, 2; L, 2; J, 0; m, 4.90 B; low spin: S, 1; L, 3; J, 2; m, 2.82 B.
4. high spin: S, 3/2; L, 3; J, 9/2; m, 3.87 B; low spin: S, 1/2; L, 2; J, 5/2; m, 1.73 B.
6. (a) 2.4 x 106 A m-1; (b) 4.92 x 105 A m-1.
7. 2.19 B.
8. (a) 9.07 x 1028 m-3; (b) 0.223 nm.
9. high spin, t2g3eg2.
10. low spin, t2g5, orbital component not completely quenched.
11. [Co(NH3)6]3+: low spin, t2g6; [CoF6]-3: high spin, t2g4 eg2, 4.9 B.
12. [Fe(CN)6]4-: low spin, t2g6; [Fe(NH3)6]2+: high spin, t2g4eg2, 4.9 B.
13. 435 nm.
14. 303 nm.
15. (a) 2.76 x 10-19 J, (b) 1.11 x 10-29
16. 3.71 x 10-24 J, 5.35 cm.
17. 8.69 x 10-24 J, 2.29 cm.
18. 4.64 x 10-24 J, 4.28 cm.
19. 1.11 x 10-23 J, 1.78 cm.
20. 1.53 x 10-7 kg-1.
21. 6.44 x 10-8 kg-1.
22. 1.67 x 10-3 m-3.
23. (a) 1.6792 / T, (b) 168 K.
24. (a) 0.3023 / T; (b) 0.3627 / T; (c) spin-only, 60.5 K; not quenched: 72.5 K.
25. 4.77 x 105 Am-1, 0.60 T.
26. 3.62 x 105 Am-1, 0.455 T.
27. 6.35 x 105 Am-1, 0.8 T.
28. 6.36 x 105 Am-1, 0.8 T.
Solutions
1 Determine the ground state values of S, J, L and m for the f3 ion Nd3+.
Follow the method in Section 1.3.2.
S = ½ + ½ + ½ = 3/2;
L = 3+2+1 = 6
J = L + S … L – S = 6 + 3/2 …6 – 3/2 = 15/2 …9/2
As the shell is less than half-full, take the lower value as the ground state: J = 9/2
m = gJ B [J(J + 1)]½
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
= 1 + (9/2 x 11/2 - 6 x 7 + 3/2 x 5/2) / (2 x 9/2 x 11/2)
= 1 + (-0.2727) = 0.72727
m = 0.72727 x (99/4)½ B = 3.618 B
2 Determine the ground state values of S, J, L and m for the f7 ion Gd3+.
Follow the method in Section 1.3.2.
S = ½ + ½ + ½ … = 7/2;
L = 3+2+1… = 0
J = L + S … L – S = 0 + 7/2 …0 – 7/2 = 7/2 …-7/2
As the shell is half-full, take the higher value as the ground state: J = 7/2
m = gJ B [J(J + 1)]½
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
= 1 + (7/2 x 9/2 - 0 + 7/2 x 9/2) / (2 x 7/2 x 9/2)
= 1 + 1 = 2
m = 2 x (63/4)½ B = 7.937 B
3 Determine the ground state values of S, J, L and m for the d4 ion Mn3+ in both high and
low spin states.
Follow the method in Section 1.3.2.
Mn3+ high spin has all electron spins parallel
S = ½+½+½+½
= 2;
L = 2 + 1 + 0 + -1 = 2
J = L + S … L – S = 2 + 2 …2 – 2 = 4 … 0
As the shell is less than half-full, take the lower value as the ground state: J = 0
For the transition metals, use the spin-only formula:
m = [n(n + 2)]½ B = [4 x 6]½ B = 4.90 B
Mn3+ low spin has 2 electrons paired and 2 electrons with spins parallel
S = ½+½
= 1;
L = 2+1 = 3
J = L+S…L–S = 3+1…3–1 = 4…2
As the shell is less than half-full, take the lower value as the ground state: J = 2
For the transition metals, us the spin-only formula:
m = [n(n + 2)]½ B = [2 x 4]½ B = 2.83 B
4 Determine the ground state values of S, J, L and m for the d7 ion Co2+ in both high and
low spin states.
Follow the method in Section 1.3.2.
Co2+ high spin has all electron spins parallel
S = ½+½+½+½+½-½-½
= 3/2;
L = -1 + (-2) = -3 = 3
J = L + S … L – S = 3 + 3/2 … 3 – 3/2 = 9/2 … 3/2
As the shell is more than half-full, take the higher value as the ground state: J = 9/2
For the transition metals, use the spin-only formula:
m = [n(n + 2)]½ B = [3 x 5]½ B = 3.87 B
Co2+ low spin has only 1 unpaired electron
S = ½;
L = 2
J = L+S…L–S = 2+½…2–½
= 5/2 … 3/2
As the shell is more than half-full, take the higher value as the ground state: J = 5/2
For the transition metals, us the spin-only formula:
m = [n(n + 2)]½ B = [1 x 3]½ B = 1.73 B
5 Why don’t the lanthanoid ions possess high spin and low spin states?
In brief: the d orbitals of the transition metals, especially the 3d series, are exposed and
interact strongly with the surroundings.
This gives a large crystal field splitting, allowing a
choice between high and low spin states, depending upon the energy of interaction.
In the lanthanoids the f orbitals are shielded from the surroundings. Thus the crystal
field splitting is small. In this case the lower energy state is with all electrons unpaired and
the ions always show “high spin” characteristics.
6
Estimate the saturation magnetisation, Ms for a sample of ferromagnetic nickel metal,
(a), assuming only the unpaired d electrons contribute to the magnetism and the spins can
be added as if the material were paramagnetic, and (b), using the measured magnetic
moment per nickel atom of 0.58 B.
The metal has an A1 structure with a cubic unit cell
parameter of 0.3524 nm.
(a) In this approximation use the spin-only formula:
m = [n(n + 2)]½ B
The electronic configuration of Ni is 3d8 4s2. The 4s electrons will be paired and 6 of the 8
d electrons will be paired so that n = 2 for Ni
m = [n(n + 2)]½ B = [2 x 4]½ B = 2.828 B
There are 4 Ni atoms in a unit cell so that m(total) = 4 x 2.828 B
The magnetic moment per unit volume Ms = m(total) / a3
= 4 x 2.828 / (0.3524 x 10-9)3
= 2.585 x 1028 B
B = 9.274 x 10-24 J T-1
Ms = 2.585 x 1028 x 9.274 x 10-24 = 2.40 x 106 A m-1
(b) Taking m = 0.58 B
Ms = (4 x 0.58 x 9.274 x 10-24) / (0.3524 x 10-9)3
7
=
4.92 x 105 A m-1
Iron has a saturation magnetisation of 1.72 x 10 6 A m-1.
magnetic moment, in Bohr magnetons, of an iron atom?
cubic unit cell parameter of 0.2867 nm.
What is the measured
Iron has the A2 structure, with a
There are 2 atms of Fe in the unit cell, so:
The magnetic moment per unit volume Ms = m(total) / a3 = 2m / a3
where m is the required magnetic moment.
Ms = (2 x m x 9.274 x 10-24) / (0.2867 x 10-9)3
m = [1.72 x 106 x (0.2867 x 10-9)3 ] / (2 x 9.274 x 10-24) =
2.19 B
8 The saturation magnetisation of cobalt is 1.446 x 10 6 A m-1. (a) Calculate the number of
magnetic dipoles per unit volume in cobalt knowing the effective magnetic moment per
atom is 1.72 B.
(b) If it is assumed that there is one atom per primitive cubic unit cell,
determine the length of the unit cell edge.
(a) Ms = N meff B
1.446 x 106 = N x 1.72 x 9.274 x 10-24
N = 1.446 x 106 / (1.72 x 9.274 x 10-24) = 9.065 x 1028 m-3
(b) If the unit cell contains just 1 atom:
N = 1 / a3 ;
a = (1 / N)1/3 = (1 / 9.065 x 1028)1/3 = .226 x 10-10 m = 0.2226 nm
9
The magnetic moment of Fe3+ ions in the species [Fe(H2O)6]3+ is 5.3 B.
likely electron configuration of the Fe3+ ion?
Use the spin-only formula:
What is the
m = [n(n + 2)]½ B
5.3 = [n(n + 2)]½
28.09 = n(n + 2); n = 4.4
This is closest to the high-spin state, with configuration t2g3eg2 for which m = 5.9 B.
10
The magnetic moment of Fe3+ ions in the species [Fe(CN)6]3- is 2.3 B.
likely electron configuration of the Fe3+ ion?
What is the
What can you conclude about the orbital
contribution to the magnetic moment for this species?
(a) Use the spin-only formula:
m = [n(n + 2)]½ B
2.3 = [n(n + 2)]½
5.29 = n(n + 2); n = 1.5
This is closest to the low-spin state, with configuration t2g5.
The orbital angular momentum
contribution is not completely quenched in this case because the observed magnetic
moment, 2.3 B is between that expected for 1 unpaired spin, 1.73 B and two unpaired
spins, 2.83 B.
11 The species [Co(NH3)6]3+ , containing Co3+ ions, gives rise to diamagnetic solids while
the species [CoF6]3-, also containing Co3+ ions, has a strong magnetic moment and gives
rise to paramagnetic solids. (a) What is the likely electron configuration of the Co3+ ions in
these two species? (b) Calculate the expected magnetic moment of Co3+ in [CoF6]3-.
(a)
Co3+ is a d6 ion.
In the low-spin state t2g6, all electrons are paired, there are no
unpaired spins, and the material would be diamagnetic. Hence [Co(NH3)6]3+ contains lowspin Co3+ and is diamagnetic.
In the high-spin state with configuration t2g4eg2 there are 4 unpaired electrons,
resulting in a paramagnetic solid.
Hence [CoF6]3- contains high-spin Co3+ and is
paramagnetic.
(b) Use the spin-only formula:
m = [n(n + 2)]½ B
m = [4(6)]½ = 4.9 B
12 The species [Fe(CN)6]4-, containing Fe2+ ions, gives rise to diamagnetic solids while the
species [Fe(NH3)6]2+, also containing Fe2+ ions, has a strong magnetic moment and gives
rise to paramagnetic solids. (a) What is the likely electron configuration of the Fe 2+ ions in
these two species? (b) Calculate the expected magnetic moment of Fe2+ in [Fe(NH3)6]2+.
(a)
Fe2+ is a d6 ion and everything in the previous question can be repeated. In the low-
spin state t2g6, all electrons are paired, there are no unpaired spins, and the material would
be diamagnetic. Hence [Fe(CN)6]4- contains low-spin Fe2+ and is diamagnetic.
In the high-spin state with configuration t2g4eg2 there are 4 unpaired electrons,
resulting in a paramagnetic solid.
paramagnetic.
(b) Use the spin-only formula:
m = [n(n + 2)]½ B
Hence [Fe(NH3)6]2+, contains high-spin Fe2+ and is
m = [4(6)]½ = 4.9 B
13 The crystal field splitting of diamagnetic [Co(NH3)6]3+ , containing Co3+ ions is 4.57 x 1019
J. What wavelength light would produce photoinduced paramagnetism in this molecule?
The photon energy must be sufficient to promote an electron across the crystal field energy
gap. Hence
E = h = hc /  = 4.57 x 10-19 J
 =
hc / 4.57 x 10-19 = 6.626 x 10-34 x 2.998 x108 / 4.57 x 10-19
= 435 nm
14 The crystal field splitting of diamagnetic [Fe(CN)6]4- , containing Fe2+ ions is 6.56 x 1019
J. What wavelength light would produce photoinduced paramagnetism in this molecule?
The photon energy must be sufficient to promote an electron across the crystal field energy
gap. Hence
E = h = hc /  = 6.56 x 10-19 J
 =
15
hc / 4.57 x 10-19 = 6.626 x 10-34 x 2.998 x108 / 6.56 x 10-19
The light absorbed by the complex ion [FeF 6]3-, peaks at 719 nm.
= 303 nm
This absorption is
due to the promotion of an electron from the lower, (t 2g), to the upper, (eg), state in the Fe3+
ion. (a) Calculate the magnitude of the crystal field splitting of the Fe3+ d-orbitals due to F-.
(b) What is the relative population of the two levels at 300 K?
(a)
E = h = hc / 
=
6.626 x 10-34 x 2.998 x108 / 719 x 10-9
= 2.76 x 10-19 J
(b) Use the Boltzmann equation:
N1 / N0 = exp [-(E1 – E0) / kBT]
N1 / N0 = exp [-2.76 x 10-19 / (1.3807 x 10-23 x 300)] = 1.11 x 10-29
16
Calculate (a) the paramagnetic energy level splitting for a Pr 3+ ion in a magnetic flux
density of 0.5 T, and (b) the corresponding wavelength of radiation for a transition between
these energy levels.
(a) Use
E = gJ B B
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
Pr3+ is an f2 ion. The ground state values of S, L, J are determined as in Section 1.3.2.
S = ½+½
= 1;
L = 3+2
= 5
J = L+S…L–S = 5+1…5–1 = 6…4
As the shell is less than half-full, take the lower value as the ground state: J = 4
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
= 1 + (4 x 5 - 5 x 6 + 1 x 2) / (2 x 4 x 5)
= 1 + (-8 / 40) = 0.80
E = 0.8 x 9.274 x 10-24 x 0.5 = 3.71 x 10-24 J
(b)
E = h = hc /  = 3.71 x 10-24 J
 =
hc / 3.71 x 10-24 = 6.626 x 10-34 x 2.998 x108 / 3.71 x 10-24
= 0.0535 m
= 5.32 cm (microwave)
17
Calculate (a) the paramagnetic energy level splitting for a Ho 3+ ion in a magnetic flux
density of 0.75 T, and (b) the corresponding wavelength of radiation for a transition
between these energy levels.
(a) Use
E = gJ B B
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
Ho3+ is an f10 ion. The ground state values of S, L, J are determined as in Section 1.3.2.
S = ½ + ½…
= 2;
L = 3+2+1
= 6
J = L+S…L–S = 6+2…6–2 = 8…4
As the shell is more than half-full, take the higher value as the ground state: J = 8
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
= 1 + (8 x 9 - 6 x 7 + 2 x 3) / (2 x 8 x 9)
= 1 + (36 / 144) = 1.25
E = 1.25 x 9.274 x 10-24 x 0.75 = 8.69 x 10-24 J
(b)
E = h = hc /  = 3.71 x 10-24 J
 =
hc / 8.69 x 10-24 = 6.626 x 10-34 x 2.998 x108 / 8.69 x 10-24
= 0.02286 m
18
= 2.286 cm (microwave)
Calculate: (a) the paramagnetic energy level splitting for a V 4+ ion in a magnetic flux
density of 0.25 T if the orbital contribution is quenched; (b) the corresponding wavelength of
radiation for a transition between these energy levels.
(a) Use
E = gJ B B
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
V4+ is a d1 ion. If the orbital contribution is quenched L = 0, J = S = ½.
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)] = 2
E = 2 x 9.274 x 10-24 x 0.25 = 4.637 x 10-24 J
(b)
E = h = hc /  = 4.637 x 10-24 J
 =
hc / 4.637 x 10-24 = 6.626 x 10-34 x 2.998 x108 / 4.637 x 10-24
= 0.0428 m
19
= 4.28 cm (microwave)
Calculate: (a) the paramagnetic energy level splitting for a Ni 2+ ion in a magnetic flux
density of 0.6 T if the orbital contribution is quenched; (b) the corresponding wavelength of
radiation for a transition between these energy levels.
(a) Use
E = gJ B B
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]
Ni2+ is a d8 ion. If the orbital contribution is quenched L = 0, J = S = 1.
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)] = 2
= 1 + (1/2 x 3/2 + ½ x 3/2) / (2 x 1/2 x 3/2)
= 2
E = 2 x 9.274 x 10-24 x 0.6 = 1.113 x 10-23 J
(b)
E = h = hc /  = 1.113 x 10-23 J
 =
hc / 1.113 x 10-23 = 6.626 x 10-34 x 2.998 x108 / 1.113 x 10-23
= 0.0178 m
20
= 1.78 cm (microwave)
Calculate the mass susceptibility of NiSO4.7H2O at 20C, (units kg-1), which contains
isolated Ni2+ ions.
The density of the compound is 1980 kg m -3.
Assume that the spin
only formula is adequate.
 = N 0 gJ2 B2 [J(J + 1)] / 3 kB T
In the spin-only situation, J = S and gJ = 2, hence:
 = N 0 22 B2 [S(S + 1)] / 3 kB T
N has units of dipoles / unit volume. There is one dipole (Ni 2+) per formula unit, so that the
molar volume contains NA dipoles. Hence,
N = NA / molar volume = NA x density / molar mass
To obtain the mass susceptibility simply use N = NA x density / molar mass:
 (kg-1) = 0 22 B2 [S(S + 1)] x NA / 3 kB T x molar mass
Ni2+ is a d8 ion with 2 unpaired electrons so that S = ½ + ½ = 1; molar mass = 0.280863 kg
mol-1, T = 293 K
 (kg-1) = [4 x 10-7 x 4 x (9.274 x 10-24)2 x 6.022 x 1023 x 2] /
[(3 x 1.38067 x 1023 x 293 x 0.280863)]
= 1.53 x 10-7 kg-1
21
Calculate the mass susceptibility of CuSO4.5H2O at 20C, (units kg-1), which contains
isolated Cu2+ ions.
The density of the compound is 2284 kg m -3.
Assume that the spin
only formula is adequate.
 = N 0 gJ2 B2 [J(J + 1)] / 3 kB T
In the spin-only situation, J = S and gJ = 2, hence:
 = N 0 22 B2 [S(S + 1)] / 3 kB T
N has units of dipoles / unit volume. There is one dipole (Ni 2+) per formula unit, so that the
molar volume contains NA dipoles. Hence,
N = NA / molar volume = NA x density / molar mass
To obtain the mass susceptibility simply use N = NA x density / molar mass:
 (kg-1) = 0 22 B2 [S(S + 1)] x NA / 3 kB T x molar mass
Cu2+ is a d9 ion with one unpaired electron so that S = ½; molar mass = 0.24968 kg mol-1, T
= 293 K
 (kg-1) = [4 x 10-7 x 4 x (9.274 x 10-24)2 x 6.022 x 1023 x 3/4] /
[(3 x 1.38067 x 1023 x 293 x 0.24968)]
= 6.44 x 10-8 kg-1
22 Calculate the volume susceptibility of MnSO 4.4H2O at 20C, (units m-3), which contains
isolated Mn2+ ions.
The density of the compound is 1980 kg m -3.
Assume that the spin
only formula is adequate.
 = N 0 gJ2 B2 [J(J + 1)] / 3 kB T
In the spin-only situation, J = S and gJ = 2, hence:
 = N 0 22 B2 [S(S + 1)] / 3 kB T
N has units of dipoles / unit volume. There is one dipole (Ni 2+) per formula unit, so that the
molar volume contains NA dipoles. Hence,
N = NA / molar volume = NA x density / molar mass
To obtain the volume susceptibility simply use the preceding equation:
 (m-3) = 0 22 B2 [S(S + 1)] x NA x density / 3 kB T x molar mass
Mn2+ is a d5 ion with 5 unpaired electrons so that S = 5/2, molar mass = 0.223060 kg mol-1,
T = 293 K
 (m-3) = [4 x 10-7 x 4 x (9.274 x 10-24)2 x 6.022 x 1023 x 35/4 x 1980] /
[(3 x 1.38067 x 1023 x 293 x 0.223060)]
= 1.67 x 10-3 m-3
23 (a) Calculate the value of x in the Brillouin function BJ(x), where x = gJ μBJB / kBT for an
Mn2+ ion with five unpaired electrons in an inductance of 0.5 T.
Assume that the orbital
angular momentum is quenched, so that L = 0.
Take S as 5/2, (Table 12.2).
The Curie
law requires that x << 1. (b) Estimate the temperature at which x is 0.01.
(a)
x = gJ μBJB / kBT
Mn2+ is a d5 ion with 5 unpaired electrons so that S = 5/2 and in the spin-only situation, J =
S and gJ = 2, hence:
x = (2 x 9.274 x 10-24 x 5/2 x 0.5) / (1.38067 x 1023 T)
= 1.6792 / T
(b) If x must be 0.01;
T = 1.6792 / 0.01 = 168 K
24 (a) Calculate the value of x in the Brillouin function BJ(x), where x = gJ μBJB / kBT for a
Ti3+ ion with one unpaired electron in an inductance of 0.45 T.
Assume that the orbital
angular momentum is quenched, so that L = 0, and take S as ½, (Table 12.2). (b) Repeat
the calculation assuming that the orbital angular momentum is not quenched and L = 2, J =
3/2. The Curie law requires that x << 1. (c) Estimate the temperature at which x is 0.005.
(a)
x = gJ μBJB / kBT
Ti3+ is a d1 ion with 1 unpaired electron so that S = 1/2 and in the spin-only situation, J = S
and gJ = 2, hence:
x = (2 x 9.274 x 10-24 x 1/2 x 0.45) / (1.38067 x 1023 T)
= 0.3023 / T
(b) In the not quenched case, S = ½, L = 2, J = 3/2
gJ = 1 + {[J(J + 1) – L(L + 1) + S(S + 1)] / [2 J(J + 1)]}
= 1 + {[3/2 x 5/2 - 2 x 3 + ½ x 3 / 2] / (2 x 3/2 x 5/2]} = 1 - 0.2 = 0.8
Hence
x = (0.8 x 9.274 x 10-24 x 3/2 x 0.45) / (1.38067 x 1023 T)
= 0.3627 / T
(c) If x must be 0.005;
spin-only:
T = 0.3023 / 0.005 = 60.45 K
not quenched: T = 0.3627 / 0.005 = 72.5 K
25
Estimate: (a) the saturation magnetisation; (b) the magnetic inductance for the cubic
ferrite CoFe2O4 with the inverse spinel structure. Co2+ is a d7 ion. The cubic unit cell has
a lattice parameter of 0.8443 nm and contains eight formula units. Assume that the orbital
angular momentum is quenched.
CoFe2O4 with the inverse spinel structure (Fe3+)[Fe3+ Co2+]O4 where ( ) represents
tetrahedral sites and [ ] represents octahedral sites.
The spins on the ions in the
octahedral and tetrahedral sites are opposed so that the magnetic moments are arranged
(Fe3+) [Fe3+  Co2+ ]. the Fe3+ moment cancel so we only have to contend with the Co2+
ions.
Co2+ is a d7 ion with 3 unpaired electrons.
(a) Use the spin-only formula:
m = [n(n + 2)]½ B
m = [3(4)]½ = 3.87 B
Ms = N m B
N is the number of Co2+ ions per unit volume. There are 8 Co2+ in the cubic unit cell hence:
N = 8/ a3
= 8 / (0.8443 x 10-9)3
Ms = 8 x 3.87 x 9.274 x 10-24 / (0.8443 x 10-9)3
= 4.77 x 105 A m-1
(b) In a ferrimagnetic material
B = 0H + 0M
approximates to:
Hence:
26
B = 0M
B = 0Ms = 4 x 10-7 x 4.77 x 105 = 0.60 T
Estimate: (a) the saturation magnetisation; (b) the magnetic inductance for the cubic
ferrite NiFe2O4 with the inverse spinel structure. Ni2+ is a d8 ion. The cubic unit cell has a
lattice parameter of 0.8337 nm and contains eight formula units.
Assume that the orbital
angular momentum is quenched.
NiFe2O4 with the inverse spinel structure (Fe3+)[Fe3+ Ni2+]O4 where ( ) represents
tetrahedral sites and [ ] represents octahedral sites.
The spins on the ions in the
octahedral and tetrahedral sites are opposed so that the magnetic moments are arranged
(Fe3+) [Fe3+  Ni2+ ]. The Fe3+ moment cancel so we only have to contend with the Ni2+
ions.
Ni2+ is a d8 ion with 2 unpaired electrons.
(a) Use the spin-only formula:
m = [n(n + 2)]½ B
m = [2(4)]½ = 2.83 B
Ms = N m B
N is the number of Ni2+ ions per unit volume. There are 8 Ni2+ in the cubic unit cell hence:
N = 8/ a3
= 8 / (0.8337 x 10-9)3
Ms = 8 x 2.83 x 9.274 x 10-24 / (0.8443 x 10-9)3
= 3.62 x 105 A m-1
(b) In a ferrimagnetic material
B = 0H + 0M
approximates to:
Hence:
27
B = 0M
B = 0Ms = 4 x 10-7 x 3.62 x 105 = 0.455 T
Estimate: (a) the saturation magnetisation; (b) the magnetic inductance for the
hexagonal ferrite ferroxdur, BaFe12O19.
0.58778 nm, c = 2.1236.
The hexagonal unit cell has parameters a =
There are two formula units in the unit cell.
Assume that the
orbital angular momentum is quenched.
The arrangement of the magnetic moments in BaFe12O19 is:
R block: 1p  3o 2o
S block: 4o 2t
Sum: 1p 7o 2o 2t = 8 - 4 = 4
where p represents a trigonally prismatic coordinated ion, o represents an octahedrally
coordinated ion and t a tetrahedrally coordinated ion.
There are thus 4 Fe 3+ ions to be
counted in a formula unit of BaFe12O19 and as there are 2 formula units per unit cell, there
are 8 Fe3+ to be included.
Assuming the spin-only formula is valid:
m = [n (n + 2)] ½ B
Fe3+ is a d5 ion with 5 unpaired spins, so n = 5 and
m = [5 ( 7 )] ½ B = 5.92 B
(a) Ms = N x m x B
N = 8 / unit cell volume
unit cell volume for a hexagonal crystal
= (3 / 2) a2 c
= (3 / 2) (0.58778 x 10-9)2 (2.31236 x 10-9) = 6.919 x 10-28 m-3
Ms = (8 x 5.92 x 9.274 x 10-24) / (6.919 x 10-28) = 6.35 x 105 Am-1
(b) B = 0 H + 0 Ms
and for a ferrimagnetic material we can approximate B = 0 Ms, hence:
B = 4 x 10-7 x 6.35 x 105 = 0.8 T
28
Estimate: (a) the saturation magnetisation; (b) the magnetic inductance for the
hexagonal ferrite SrFe12O19. The hexagonal unit cell has parameters a = 0.58836 nm, c =
2.30376 nm. There are two formula units in the unit cell. Assume that the orbital angular
momentum is quenched.
The method is exactly the same as in the previous question; the only difference being the
new cell volume.
(a) Ms = N x m x B
N = 8 / unit cell volume
unit cell volume for a hexagonal crystal
= (3 / 2) a2 c
= (3 / 2) (0.58836 x 10-9)2 (2.30376 x 10-9) = 6.906 x 10-28 m-3
Ms = (8 x 5.92 x 9.274 x 10-24) / (6.906 x 10-28) = 6.36 x 105 Am-1
(b) B = 4 x 10-7 x 6.36 x 105 = 0.8 T
29
Derive a formula for the saturation magnetisation of a cubic ferrite A2+Fe2O4, which is
partly inverse.
The fraction of Fe3+ ions on tetrahedral sites is given by , where 0 <  <
0.5 (Section 5.3.9).
Ms = N x m x B
In a mixed spinel A2+B23+O4
 = (B) / [B]
where ( ) represents tetrahedral sites and [ ] represents octahedral sites.
When  = 0 we have a normal spinel (A2+) [B23+]O4
When  = 0.5 we have an inverse spinel (B3+) [A2+B3+]O4
For a mixed ferrite the formula is:
(Fe23+ A2+1-2) [A22+B2 - 23+]O4
The spins on the octahedral and tetrahedral sites are opposed so we can schematically add
the spins to give:
tetrahedral sites: Fe3+ 2
A2+ (1 - 2) 
octahedral sites: Fe3+ (2 - 2) 
A2+ ( 2) 
The total number of Fe3+ spins that count = (2 - 4) per A2+Fe2O4
The total number of A2+ spins that count = (1 - 4) per A2+Fe2O4
The effective magnetic moment per A2+Fe2O4 is:
m = (2 - 4) mFe + (1 - 4) mA
As there are 8 A2+Fe2O4 in a cubic unit cell of side a, the number of magnetic moments that
count, N, = 8 / a3. Hence:
Ms = {8 x [(2 - 4) mFe + (1 - 4) mA] B} / / a3