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

v E

Diamagnetism: Larmor Precession

F

0 

m

w 0 2

r

B

 0 

e

2 w 0

r

2 w  nucleus

r

electron

v,

w

0

w 0 

eB

2

m

Lorentz Force F = -e(v x B)

F = eBr

w

Diamagnetism: Quantum Approach

starting point   

e

2

m



L

 2

S

  2

e r

2 4

m

B

Quantum mechanics makes some useful corrections. The components of

L

and

S

r 2

are replaced by their corresponding values for the electron state and is replaced by the average square of the projection of the electron position vector on the plane perpendicular to

B

, which yields

R

2  2 3

r

2 where

R

is the new radius of the sphere.

Diamagnetism: Quantum Approach

If B is in the z direction 

z

 

e

2

m

where

L z L z



2

S z



L z



d

 

e

2

4

m x

2 

y

2

B S z x

2   

2

y

2  

x

2 

y

2  

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 <

x

2 +

y

2 >.

• If the probability density  *  for a state is spherically symmetric <

x

2 > = <

y

2 >= <

z

2 > and <

x

2 +

y

2 >=2/3<

r

2 >.

• If an atom contains

Z

electrons in its closed shells, then 

z

 

Ze

2 6

m r

2

B

• 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

B

local 

B

applied  1 3  0

M

Diamagnetism

M

 

n



z

     0

M B nZe

2 6

m

 

r

2

B

 0

nZe

2 6

m r

2 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 10 28 atoms/m 3 . 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  4  6

m

  7

r

2 

.

 28 

.

 3

.

 31 kg   19 C

.

 11 m  2  

.

  5

Diamagnetism: Example

Estimate the susceptibility of solid argon. Argon has atomic number 18; and at 4 K, its concentration is 2.66

x 10 28 atoms/m 3 . 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.

M

   0

B

  4   10  5    10  7   ccp structure

Paramagnetism

Core Paramagnetism

If <

L

z > and <

S

z > 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 GdCl 3 and FeF 2 . 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

2

J

2

μ

 

g

 B

J

where

 B

is the Bohr magneton

 B 

e

/

2

m

  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 one half 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 J J

  

L



S

if shell  half full

L S



S

if shell if shell   half full 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 4

g J

2 

S

2 

L

2 2

J

2 1

)

2

)

 1

)

 1

)

  

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

g J

2 

S

2 

L

2 2

J

2 1

)

2

)

 1

)

 1

)

        

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

J

z 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

J

z will have differing energies and differing probabilities of occupation.

• The

z

component of the moment is given by: 

z

 

g

 B

J z

 

g

 B

M J

• and its energy is

E

  

z B

 

g

 B

As a result of these probabilities, the average dipole moment is given by  z 

M

J 



J



J



M

J 



J



g J

 B

e

   B

g

 B

/

where  J 

/

B  2

J

2

J

 1

coth

    2

J

 2

J

1 

g x

   

J

1 2

J

J

(

g

 B

coth

 

x

2

J

 

)

M



n



z 

ng

B

)

 J (

x

)

Brillouin Function

x

Paramagnetism

If

g

 B

JB

 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 magnetization

M



ng

 B J .  1 a nd the If

g

 B

JB

 of being in any of the states and the magnetization will be

,



(J

 1 and

M

 1 3

ng

2  B 2  1

B k T

.

Paramagnetism

The magnetic suspectibility is    0

M B



C T

where C  1 3

ng

2   0 B 2  1

) /

k

B The Curie constant can be rewritten as C  1 3

np

2 2 B where

p

is the effective number of Bohr magnetons per ion.

/

k

B