Transcript Magnetism - Illinois State University

Diamagnetism and
Paramagnetism
Physics 355
Free atoms…
The property of magnetism can have three origins:
1. Intrinsic angular momentum (Spin)
2. Orbital angular momentum about the nucleus
3. Change in the dipole moment due to an applied field
In most atoms, electrons occur in pairs.
Electrons in a pair spin in opposite
directions. So, when electrons are
paired together, their opposite spins
cause their magnetic fields to cancel
each other. Therefore, no net magnetic
field exists. Alternately, materials with
some unpaired electrons will have a net
magnetic field and will react more to an
external field.
Diamagnetism: Classical Approach
Consider a single
closed-shell atom
in a magnetic field.
Spins are all paired and
electrons are distributed
spherically around the
atom. There is no
total angular
momentum.
B
nucleus
r
electron
E
v
Diamagnetism: Larmor Precession
F0  mw r
2
0
B
nucleus
r
electron
0  w0r
e
2
2
eB
w  w0 
2m
v, w0
Lorentz Force
F = -e(v x B)
F = eBrw
Diamagnetism: Quantum Approach
starting
point

e
2m
 L 2 S  
e2 r 2
4m
B
Quantum mechanics makes some useful corrections. The components of L
and S are replaced by their corresponding values for the electron state and
r2 is replaced by the average square of the projection of the electron
position vector on the plane perpendicular to B, which yields
R2 
where R is the new radius of the sphere.
2
3
r2
Diamagnetism: Quantum Approach
If B is in
the z direction
e
e2 2 2
z  
Lz  2S z 
x y B
2m
4m
where
Lz     Lz  d
Sz  
2
2
2
 2
2
x  y    x  y  d


Diamagnetism: Quantum Approach
Consider a single
closed-shell atom
in a magnetic field.
Spins are all paired and
electrons are distributed
spherically around the
atom. There is no
total angular
momentum.
• The atomic orbitals are used to estimate
<x2 + y2>.
• If the probability density *  for a state
is spherically symmetric <x2> = <y2>= <z2>
and <x2 + y2>=2/3<r2>.
• If an atom contains Z electrons in its
closed shells, then
2
Ze
z  
r2 B
6m
• The B is the local field at the atom’s
location. We need an expression that
connects the local field to the applied
field. It can be shown that it is
Blocal  Bapplied  1
3 0 M
Diamagnetism
nZe2 2
M  n z  
r B
6m
0 M
0nZe2 2
 

r
B
6m
Core Electron
Contribution
• Diamagnetic susceptibilities are nearly
independent of temperature. The only
variation arises from changes in atomic
concentration that accompany thermal
expansion.
Diamagnetism: Example
Estimate the susceptibility of solid argon. Argon
has atomic number 18; and at 4 K, its
concentration is 2.66 x 1028 atoms/m3. Take the
root mean square distance of an electron from the
nearest nucleus to be 0.62 Å. Also, calculate the
magnetization of solid argon in a 2.0 T induction
field.
ccp structure
 
0 nZe 2
6m
4  10


r2
7
 1.08  105
T  m/A
 2.66 10
 (18) 1.60 10
6  9.11  10 kg 
28
m
3
31
19
  6.2 10
2
C
11
m

2
Diamagnetism: Example
Estimate the susceptibility of solid argon. Argon
has atomic number 18; and at 4 K, its
concentration is 2.66 x 1028 atoms/m3. Take the
root mean square distance of an electron from the
nearest nucleus to be 0.62 Å. Also, calculate the
magnetization of solid argon in a 2.0 T induction
field.
ccp structure
 B
M 

0

1.08 105
  2.0 T   17.2 A/m
4 107 T  m/A
Paramagnetism
Core Paramagnetism
If <Lz> and <Sz> do not both vanish for an atom, the atom has a
permanent magnetic dipole moment and is paramagnetic.
Some examples are rare earth and transition metal salts, such
as GdCl3 and FeF2. The magnetic ions are far enough apart
that orbitals associated with partially filled shells do not
overlap appreciably. Therefore, each magnetic ion has a
localized magnetic moment.
Suppose an ion has total angular momentum L, total spin
angular momentum S, and total angular momentum J = L + S.
Core Paramagnetism
Landé g factor
g  L 2 S J/ J
μ   g B
2
J
where B is the Bohr magneton
B  e / 2m  9.27 1024 J/T
Hund’s Rules
• For rare earth and transition metal ions, except Eu and Sm,
excited states are separated from the ground state by large
energy differences – and are thus, generally vacant.
• So, we are mostly interested in the ground state.
• Hund’s Rules provide a way to determine J, L, and S.
• Rule #1: Each electron, up to onehalf of the states in the shell,
contributes +½ to S. Electrons
beyond this contribute  ½ to S.
The spin will be the maximum
value consistent with the Pauli
exclusion principle.
Frederick Hund
1896-1997
Hund’s Rules
• Each d shell electron can contribute either 2, 1, 0, +1, or
+2 to L.
• Each f shell electron can contribute either 3, 2, 1, 0, +1,
+2, or +3 to L.
• Two electrons with the same spin cannot make the same
contribution.
• Rule #2: L will have the largest possible value consistent
with rule #1.
Hund’s Rules
• Rule #3:
J  LS
if shell  half full
J  LS
J S
if shell  half full
if shell  half full
Hund’s Rules: Example
Find the Landé g factor for the ground state of a praseodymium
(Pr) ion with two f electrons and for the ground state of an erbium
(Er) ion with 11 f electrons.
Pr
• the electrons are both spin +1/2, per rule #1, so S = 1
• per rule #2, the largest value of L occurs if one electron is
+3 and the other +2, so L = 5
• now, from rule #3, since the shell is less than half full,
J  L  S  5 1  4
J 2  S 2  L2
g  1
2J 2
J ( J  1)  S ( S  1)  L( L  1)
4(5)  1( 2)  5(6)
 1
 1
 0.55
2 J ( J  1)
2( 4)(5)
Hund’s Rules: Example
Find the Landé g factor for the ground state of a praseodymium
(Pr) ion with two f electrons and for the ground state of an erbium
(Er) ion with 11 f electrons.
Er
• per rule #1, we have 7(+1/2) and 4(1/2), so S = +3/2
• per rule #2, we have 2(+3), 2(+2), 2(+1), 2(0), 1(1), 1(2),
and 1(3), so L = 6
• now, from rule #3, since the shell is more than half full,
J = L + S = 15/2
J 2  S 2  L2
g  1
2J 2
15 / 2  (17 / 2)   3 / 2  (5 / 2)  6(7)  1.2
J ( J  1)  S ( S  1)  L( L  1)
 1
 1
2 J ( J  1)
2 15 / 2  (17 / 2)
Paramagnetism
Consider a solid in which all of the magnetic ions are identical, having
the same value of J (appropriate for the ground state).
• Every value of Jz is equally likely, so the average value of the
ionic dipole moment is zero.
• When a field is applied in the positive z direction, states of
differing values of Jz will have differing energies and differing
probabilities of occupation.
• The z component of the moment is given by:
 z   g B
Jz
  g B M J
• and its energy is
E  z B   g B M J B
As a result of these probabilities, the average dipole moment is
given by
J
z 

M J  J
 g B M J e  BM J B / kBT
J

e  g BM J B / kBT
 g B J  J ( g B JB / k BT )
M J  J
  2 J  1 x  1
2J  1
 x 
where  J ( x ) 
coth 
coth  

2J
 2J 
 2J
 2J
M  n z  ngBJ J ( gBJB / kBT )
Brillouin Function
J (x)
x
Paramagnetism
If g B JB  k BT , then nearly all of the ions will be
in the lowest state. All dipoles will be aligned with the
applied field and the magnetization is said to be
saturated. The Brillouin function  1 and the
magnetization M  ng B J .
If g B JB  k BT , the ion has nearly the same probability
of being in any of the states and the magnetization will be
small. In the limit of small x, the Brillouin function  (J  1 )x / 3J
and M  13 ng 2 B2 J ( J  1) B / k BT .
Paramagnetism
The magnetic suspectibility
is
0 M C


B
T
where
C  13 ng 2 0  B2 J ( J  1) / k B
The Curie constant can be rewritten as C  13 np 2 0 B2 / kB
where p is the effective number of Bohr magnetons per ion.