Transcript EPR in a nutshell

Magnetochimica
AA 2011-2012
Marco Ruzzi
Marina Brustolon
2. EPR in a nutshell
Electron spin in a magnetic field
e is proportional to S, meaning that e and S are vectors parallel to each other. They
have opposite directions because the proportionality constant is negative. The latter one is
written as the product of two factors:g and B:
μe  gB S 


e   g  B S


g is a number called Landé factor, or simply g factor. For a free electron g = 2.002319.
B  eh 4me  J T   where
me is the electron mass, e is the electron
charge and h = 6.626 10-34 Js is the Planck constant.
B
is the atomic unit of magnetic
moment, called Bohr magneton. The negative value of the electron charge is the reason why
 B  0 
Zeeman effect
Suppose to apply a constant magnetic field B to an electron spin. Since the energy of a
magnetic moment e is given by the scalar product between e and B, the electron spin energy
will depend on the orientation of e with respect to B
E  -μe  B  g | B | S  B
The dot product reduces to a single term if the direction of B coincides with one of the
axes respect to which the B and S are represented. The choice of the reference frame is
arbitrary and it can be chosen in such a way that the z axis is along the direction of B. In this
case the equation for the energy becomes:
Eg | B | B0 S z 
E   ( 1/ 2 )g | B  Β
The splitting of the electron spin energy level into two levels in the presence of
a magnetic field is called Zeeman effect.
The resonance condition
h  E -E  g | B | B0 
In a magnetic field of 3.5 T, which is the standard magnetic field
intensity used in many EPR spectrometers, for g=2.0023
we have = 9.5 GHz. Other regions of higher microwave frequencies
used in commercial EPR spectrometers are Q-band (~ 34 GHz)
and W-band (95 GHz).
E
0
a
Ea=+(1/2)gBB0
a,b
E=gBB0
b
B0=0
Eb=-(1/2)gBB0
B00
Finding the resonance condition
E
E
a
a
hres
0
0
OK
h0
b
b
Easier
experimental
arrangement
Bres
B0 constant
res
g B B0

h
The microwave frequency
should be changed to find
the resonance condition
0 constant
Bres
h 0

g B
The magnetic field should be
changed to find the
resonance condition
Cavità
Una tipica cavità EPR
-30
-20
-10
0
10
20
30
-30
-20
-10
0
 - 0
 - 0
a
b
10
20
30
Note: EPR spectra are recorded usually
by modulating the magnetic field, and the
amplitude of the modulated signal is then
recorded. Therefore the EPR trace
represents the derivative of the original
signal.
I(B)
B
B
segnale EPR
Field
modulation
coils
I(B)
 B
B
Let us consider the simple case of an atom with a closed shell and an extra electron. In
such an atom the electron spins are all coupled in pairs, except one. The electron angular
momentum has two contributions: one arises from the electron spin, and another one arises
from the orbital motion of the electron around the nucleus. The magnetic moment is the sum
of two terms, referring to the two contributions:
e   B l  g  B S
Contrary to atoms which are spherical, molecules are systems of low symmetry. For them the
orbital angular momentum is suppressed (its average is zero), and the electron angular
momentum, in the absence of spin orbit coupling, is only due to spin. The effect of spin orbit
coupling is to restore a little amount of orbit contribution, which results in a deviation
g =g-ge of the g factor
(see the Table in the next slide for examples of the
deviations.).
Spin density on heavy atoms = larger deviation of the g factor from ge
fattore-g
Radicale
e-
2,0023
CH3
fattore-g
Radicale
But
2,0026
N
Bu t
2,0061
O
Bu t



O
2,0029
S
2,0103
Bu t
O
2,0047
HO O 
Bu t
Bu t
Bu t
O
Bu t
2,0052
2,0140
Hyperfine coupling 1
The energy of nuclear spins is influenced by a magnetic field. This effect
is called nuclear Zeeman effect.
E  g N | N | B0 I z
In the presence of a nuclear spin the electron spin experiences an
additional magnetic field provided by the nuclear magnetic moment, which
affects the resonance conditions. The electron nucleus spin interaction is called
hyperfine interaction. It gives rise to a splitting of the resonance EPR lines into
several components, two components for interaction with a nuclear spin I = 1/2,
three components for a I = 1 nucleus, and in general 2I+1 components for the
interaction with a spin I nucleus.
Hyperfine coupling 2
The hyperfine energy contribution is called contact (or Fermi) contribution and it is:
Ehf  a S  I
where
h.c.c.)
a
is a constant (hyperfine
coupling constant,
which depends on |(0)|2, the square of the wave function which describes the
electron motion calculated in the point where the nucleus is.
| a | g | B | B0
Under these conditions (high field approximation) the energy term En due to the
nuclear spin represents a small perturbation on the electron spin energy, which becomes:
Etot  g | B | B0 S z  g N | N | B0 I z  a S z I z
Hyperfine coupling 3
For a free radical (S = 1/2) containing a single magnetic nucleus with I = 1/2, there are
four possible values of the total energy, corresponding to the electron and nuclear spin
components Sz = ±1/2 and Iz = ±1/2 as illustrated in the next slide.
The corresponding energies are:
E1  1 / 2 g |  B | B0  1 / 2 g N |  N | B0  1 / 4a
E2  1 / 2 g |  B | B0  1 / 2 g N |  N | B0  1 / 4a
E3  1 / 2 g |  B | B0  1 / 2 g N |  N | B0  1 / 4a
E1  1 / 2 g |  B | B0  1 / 2 g N |  N | B0  1 / 4a
EPR consists of transitions between pairs of energy levels characterized by different
values of Sz, but the same Iz value.
Energy levels for S=1/2 coupled to I=1/2
E Nuclear Zeeman
E Electron Zeeman
E
E1
a/2
E2
EPR transitions
0
E3
S z  1 ; I z  0
a/2
E Hyperfine
E4
Scheme of the energy levels for an external magnetic field B0
S=1/2 coupled to I=1/2: the EPR spectrum
E
a
mS
mI
(4)
(3)
1/2
1/2
1/2
-1/2
(2)
-1/2
-1/2
(1)
-1/2
1/2
B
Accoppiamento con più nuclei equivalenti
L’intensità relativa delle righe spettrali dovute
all’accoppiamento con più nuclei equivalenti con I=1/2 è data dai
coefficienti dello sviluppo binomiale (a+b)n. Per calcolare questi
ultimi si può utilizzare il triangolo di Tartaglia.
Nuclei (I=1/2)
Equivalenti
Intensità relativa
0
1
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
9
1
1
8
9
10
35
1
5
15
35
70
126
1
4
20
56
84
3
10
21
1
6
15
28
36
3
5
7
2
4
6
1
6
21
56
126
1
1
7
28
84
1
8
36
1
9
1
H
Radicale Metile
H
C
H
z
23 G
B
Mi +
3
2
+
1
2

1
2

3
2
EPR spectrum of Naphtalene radical anion
1 2 
5 Gauss
A quintet of quintets, each 1:4:6:4:1
Nitroxide radicals
Some stable radicals
O
CH3
H3C
H3C
CH3
N
CH3
O
H3C
H3C
CH3
N
H3C
H3C
Tempone
N
H3C
NO2
N
CH3
CH3
O
Nitronyl Nitroxide
radical
CH3
O
O 2N
N
CH3
di-t-butyl-nitroxide
R
O
CH3
O
CH3
O
Tempo
N
CH3
ONa
N
T
T
T =
NO2
H3C
T
2,2-diphenyl-1picrylhydrazyl DPPH
H3C
Trityl
S
S
S
S
CH3
CH3
H3C
H3C
N
CH3
CH3
O
a
a
10 Gauss
3315 Gauss
14N;
I = 1; a.n. = 99.63
L’intensità
relativa
delle
righe
spettrali
dovute
all’accoppiamento con più nuclei caratterizzati da I>1/2 si
può ricavare dal corrispondente “diagramma a piramide”.
Esempio: 2 nuclei
equivalenti con I=1
B0
Intensità
1
relativa
2
O
O
N
N
3
N
N
O
O
1
8.0 G
Ph
Ph
2
Le bande EPR
Campi di risonanza e l per g = 2 alle frequenze a W
comunemente usate in EPR.
Banda W
Frequenza (GHz)
L
S
X
K
Q
W
1.1
3.0
9.5
24.0
35.0
94.0
B0 (Gauss)
l (mm)
392
1070
3389
8560
12485
33600
330
100
32
12.5
8.6
3.2
N.B. g = h/BB0  alti valori di g danno righe a
campi più bassi!
22