Transcript Electronic Magnetic Moments

Spin
•Electronic charge in motion
- A current loop behaves as a
magnetic dipole and has a
magnetic moment.
- Note the current direction is
opposite to the electron
velocity, and also the angular
momentum direction is opposite
to the magnetic moment
•Two contributions to the
electronic magnetic moment
-An orbital magnetic moment due to
orbital angular momentum
-A spin magnetic moment due to
electron spin
Electronic Magnetic Moments
-Orbital Contribution
•The orbital motion of an electron around the nucleus
may correspond to a current in a loop of wire having
no resistance where m=(area of loop) (current)
•Note that the angular momentum is continuous (not
quantized), indicating a classical treatment of the
problem
e
evr
m0  r ( )  r (
)

(2r ) / v
2
p0  r  p  rmv
2
e
e
m0  ( ) p 0
2m
2
eh
m0  
 B
4m
Electronic Magnetic Moments
-Spin Contribution
Spin?
- Was postulated in 1925 by Paul Dirac in order to explain
certain features of optical spectra of hot gases subjected to
a magnetic field(Zeeman effect) and later theoretical
confirmation in wave mechanics
- The root cause of magnetism and an intrinsic
property, together with charge and mass, of subatomic
particles of fermions (eg.electrons, protons and neutrons)
and bosons (photons, pions)
It was found, theoretically and experimentally, that
the magnetic moment due to electron spin is equal
to,m0
eh
mS  
 B
4m
Electronic Magnetic Moments
-m vs. p
For a given angular momentum, the spin gives
twice the magnetic moment of orbit
eh
eh p0
e
m0  

 ( ) p 0
4m
4m h
2m
2
eh
eh pS
e
mS  

 ( ) p S
4m
4m sh
m
2
Electronic Magnetic Moments
-Total Moments
The total magnetic moment per electron is the
vector sum of the orbital and spin magnetic
moments
mtot
mtot
e
e
 m0  m S  ( ) p 0  ( 
)2 p S
2m
2m
e
  g ( ) ptot
2m
The term ‘g’ is called the Lande splitting factor
— g=2 for spin only components
— g=1 for orbital only components
Electronic Magnetic Moments
-Lande’Equation
J ( J  1)  S ( S  1)  L( L  1)
g  1
2 J ( J  1)
•Orbital is quenching : L=0, J=S  g=2
•Spin =0
: S=0, J=L  g=1
Schrodinger Equation
Schrodinger Equation
m=hml ml= l(l+1)
Schrodinger Equation
Electronic Magnetic Moments
-Quantum Mechanical
The orbital angular momentum quantum number (l)
h
p0  l ( )
2
The spin angular momentum quantum number (l)
h
p S  s( )
2
The spin angular momentum quantum number (l)
h
h
p J  J ( )  (l  s )( )
2
2
Hund’s Rule
Empirical rules which determine the occupancy
of the Available electronics within an atom
Used to calculate L, S and J for an unfilled shell
1. Maximum total S=max Sz with Sz=imsi
Obeying the Pauli exclusion principle
2. Maximum total L=maxLz with Lz= imli
Minimizing the Coulomb interaction energy
3. Spin-orbit interaction:
L -S 
if less than half-filled
J= L+S 
if more than half-filled
Hund’s Rule - Examples
•Sm3+ ion having 5 electrons in its 4f shell
(n=4, l=3)  S=5/2, L=5, J=L-S=5/2
ml 3 2 1 0 –1 –2 –3
1 1
1 1
1 1 1
ms
2
2
2
2
2
2
2
occupancy s
•Fe2+ ion having 6 electrons in its 3d shell
(n=3, l=2)  S=2, L=2, J=L+S=4.
Actually, however, S=2, L=0 (quenched)
J=S=2
ml 2
1
ms
2
1
1
2
occupancy s
0 –1 –2
1
2
1
2
1
2

2
1
1
2

1
2
0

1
2
-1

1
2
-2

1
2