EXAFS

The 3rd AOFSRR Summer School- Cheiron School 2009 –SPring-8
Nov. 4th, 2009

Extended X-ray Absorption Fine Structure
Iwao Watanabe
Ritsumeikan University

3216800

7000

7200

7400

7600

7800

EXAFS

TheoryQuantum Mechanics
ModelsApproximations
ExperimentLight Source
Monochromator
Higher Harmonics Rejection
Sample Preparation
Detection Methods
Polarization XAFSData Analysis
Limited Usable Range in Experimental
Data
Estimation of Background Curves
Fourier Transform
Multi-Scattering
Curve Fitting Procedure
Phase Problems
Debye-Waller-Like Parameter
Anharmonicity in Potential

X-ray absorption measurement by transmission method
The most reliable and basic method+1000V
double-crystal
monochromatorsynchrotron

white X-ray

monochromatic
X-rayslits

sampleIo

Bragg diffraction

PC and interfaces
Io current

calculation ofabsorbanceA = - log(I)Io

I current+1000VGas
ionizationchamber（He, N2, Ar, Kr, etc）

I

XAS of Fe(III) aqueous solution

I0

µx

I0
lnI

1

intensity

3

I

0.5

21
6800

7000

7200

7400

X-ray photon energy / eV

7600

7800

0

H2O

Fe(III)Fe(III)ion in water

3Fer(Fe-O)=2.01Å

21E006800

7000

7200

7400

X-ray photon energy / eV
0200
400
photoelectron kinetic energy / eV
Focusing on K shell (1selectron) excitation

2 1.3

1photoelectron wave length / Å7600

7800
600

Background
0.8 0.7
0.6 absorption due
0.5 toother atoms and other shellelectrons

XAFS: X-ray Absorption Fine Structure
Fe ions in aqueous solutions

XANES

Fe(III)Fe(II)µ2

EXAFS

106800

7000

7200

7400

X-ray energy / eV7600

7800

XANES: X-ray Absorption Near Edge Structure
Fe ions in aqueous solutions
2µFe(III)
Fe(II)
107100

7120

7140
7160
X-ray energy / eV

7180

7200

1NH 4VO 3 (V)V2O5(V)VO 2(IV)
V2O3(III)0.505460

5470

5480
E / eV

5490

5500

T . Yamamoto (Kyoto Univ.)

Wong et al. Phys.Rev.B 30(1984) 5596

EXAFS: Extended X-ray Absorption Fine Structure
Fe ions in aqueous solutions

XANESµt

2µ0106800Fe(III)Fe(II)

7000

EXAFS

7200

7400

7600

7800

X-ray energy / eVχ (k ) =µ (k ) − µ 0 (k )µ0 (k )2me ( E − E0 )k=h2

EXAFS: χ(k) spectra

Fe(III)Fe(II)

Fe(III)
longer → shorter distance012345678910 k20
40 number,
60
100 wave
140 vector
200: wave

300

10111213k / Å -1

k=2π

λk=

2 me ( E − E 0 )h2400500( E − E0 )kinetic energy ofphotoelectron

600

Simplest model to explain how theEXAFS oscillation occurs

H2O

Fe(III)Fe(III)ion in water

3Fer(Fe-O)=2.01Å

21E006800

7000

7200

X-ray photon energy / eV

7400

0200

photoelectron kinetic energy / eV

2 1.31

photoelectron wave length / Å0.8 0.77600

7800

4006000.60.5

X-ray photon
7213eV

kinetic energy of the photoelectron
photoelectron
1e00

VFeN
M

4s, 4p, 4d3s, 3p, 3d(700〜800eV)

an electron with Ek=100eV behaves asa wave with λ=1.2Å
h: de Broglie
λ=pThis wave length is just the order ofnormal atom-atom bond distance !!!

2s, 2pLenergy of thephoton
absorbed byFe atom

The cause for EXAFS appearance !!!

KE07113eV (1.743Å)

1sphotoelectron
going out(wave)

B
X-ray absorption
at atomA＝black wave:
directly emitted
from the absorbing
atomneighbor atom
scattering thephotoelectron wave
back to theabsorbing atom+

red broken wave:
doubly scattered at
B and Aphotoelectron
going out(wave)＝

black wave:
directly emitted
from the absorbing
atom+

red broken wave:
doubly scattered at B
and A

If they are in phase , (2r)k=2nπ (k=2π/λ) or 2r=nλ,→ larger photoelectron wave
→ larger probability of finding photoelectrons(amplitude squared).

B
X-ray absorption
at atomA

If they are out of phase , (2r)k=(2n+1) π , and equal
amplitudes,→ photoelectron waves disappear→ no photoelectrons found outside.

The law of conservation of energy tells thatwhen no photoelectrons are emitted, no photons should be absorbed !!!
if the X-ray photons are absorbed, the photoelectrons must be emitted !!!
Larger probability of finding photoelectrons → STRONG X-ray absorption
Smaller probability of finding photoelectrons → WEAK X-ray absorption

H2OFe(III)Fe(III)ion in water

3Fer(Fe-O)=2.01Å

21E006800

70007200

X-ray photon energy / eV7400

0200

photoelectron kinetic energy / eV

2 1.3

10.8 0.7

7600

7800

400

6000.60.5

photoelectron wave length / ÅBASIC EXAFS equationNi
2 22f(k)exp(−2σk−2r/λ)S( k ) sin(2 kri + φ i ( k ))
ii
0i
2krii

χ (k ) = − ∑BASIC EXAFS equation

Nif ( k )exp( − 2σ i2 k 2 − 2 ri /λ )S02 ( k ) sin(2 kri + φ i ( k ))2 ii k ri

χ (k ) = − ∑oscillation part

amplitude part×kkkχ(k) k 3

Fe(III)Fe(II)

0Fe(III)
longer → shorter distance

1k/Å-1210 203440 60

56100 140

7200

89300

kinetic energy of the photoelectron / eVphotoelectron wave number (vector)
2πk =
λ

10400

1112500

13600

Fourier Transform:
Simplest way to analyze the EXAFS
oscillation

The simplest way of knowing the wave
number (corresponding to the distance) is
Fourier Transformation of wave on k
0.5
Fe(III)Fe(II)

｜FT ｜

0.4
0.3
0.2
0.1
002Fe(III) shoterFeO

Fe(II) longerFeO4r/Å6824-Fe(CN)6N

03k χ(k)CFetiilla

E2k
χ(tacrtxonsco-2)

0510k /Å

-1

Fourier Transform

μx1

｜FT｜

0.8
16800

7000

7200

7400

E / eV

7600

7800

0.6
0.4
0.2
00

123r/Å

456

Fourier Transform (Frequency Filter)
will let you know the frequenciesof the waves
k/Å

-1r/Å

3

1

20
1-1
010sin(kr )

20

0012

34

562
r1=2.01r2=2.50-1-20310or20

r1=2.0+2
101r2=2.5234mixture

Fourier Transform for two-shell model
221

100-1-1-2
031020-2021020

211012340
1234

Fourier Transform for athree-shell model
10.5
1000-10

10k /Å

20-1

12r /Å

3

4

This must be a BACKGROUND
structure, not corresponding to areal atom-atom distance.

k/Å

-1r/Å

3

1

2
0
1-1
0
1

10

20

0
0
3

1

2

3

4

5

6

123

456

1234

56

20
1-1
0

1020

100
3
20
1-10102000k/Å

-1r/Å

3120
1-10
1102000312345612345612345620
1-101020

100
320
1-10

102000Fourier Transform for
フーリエ変換の範囲を変えたもの

1different k ranges

0-10

310 -1k/Å20
2

1If you have aFT
small
1_20.TXT
FT 3.15_15.75.T XT
FT 5.05_12.55.T XT
number of cycles
inXTk,
FT 7.55_10.05.T

melbor
P! 00

123r /Å456BASIC EXAFS equation

Nif ( k )exp( − 2σ i2 k 2 − 2 ri /λ )S02 ( k ) sin(2 kri + φ i ( k ))2 ii k ri

χ (k ) = − ∑amplitude partoscillation part

By comparing the theoretical EXAFS χ(k) andexperimental χ(k), you can determine;
N coordination numberr bond lengthf(k) and φ(k) are element specificatomic type of coordination

You may get in trouble with the data
quality taken by transmission method

Transmission method
Sample: Ge-Si powder
original powdermelboPr

further ground to finerpowder

By Masaharu Nomura,
Photon Factory, KEK!

Fe(III)(III)

χ(k) k3

1melboPr0!

-1
0

5

Fe(III) ions in aqueous
solution
possibly, inhomogeneous
sample thickness

-1k/Å

1015

2

Fe(III)
(III)

10
7100

7120

7140

7160

7180

7200

X-ray beam frommonochromator

X-ray flux is nothomogeneous and thedensity pattern moves andchanges along with themonochromator angle scanInhomogeneous
samplesolid powder

Leads to weaker EXAFS oscillation
amplitude and noisy spectrum

liquid or film

Si(311) double-crystal monochromator
Output pattern

before improvements on the crystal
holding method

after improvements

1mm x 1mm slit

Io0.6mm x 0.6mm hole
I0.6mmX0.6mm hole
Io

1.0

I
0.5

ln(Io/I)ln(Io/I)
Io

00.50I
4500

5000
E / eVTiO2 anatasepowder
1:20 BN
t=1mm
5500
6000
PF/BL9A

Yield methods:
Fluorescence yieldandTotal-conversion-electron-yieldX-ray absorption measurement by yield methods

double-crystal
monochromatorsynchrotron

white X-ray

monochromatic
X-raysampleslit

Io

Bragg diffraction

PC and interfaces
Io current

calculation ofabsorbanceA = - log(I)IoI current

fluorescence X-ray,electron,
luminescence,
heat, etc;
anything proportional to the
X-ray absorption

Lytle detector

Sample: 0.01 mol dm-3 Cu(II) solution
transmittance

By Masaharu Nomura,
Photon Factory, KEK

fluorescencetransmittance
fluorescence
Cu 0.5 mmmelborP

Cu foil 6 µm

fluorescence
fluorescence
Cu foil 6 µm

By Masaharu Nomura,
Photon Factory, KEK

Cu foil 2 µm

Self-absorption effect → incorrect N!

Total-Conversion-Electron-Yield methodcounter electrodesampleX-ray
window
X-ray−1000Vto current
amplifierHe gasX-ray

1.5mm
glass

transmittanceAuger electrons(He gasionization)

This method can be applied to thicksamples owing to the short escapedepth of Auger electron.

total-conversionelectron-yield

2transmission

μx

1µ m
Fe3O4

1〇
7100

07000

7150

7500
8000
E / eV

8500

Can we distinguish oxygen from
nitrogen by EXAFS?

Almost
impossible !

Scattering amplitude

Scattering phase shiftNi
2 22f(k)exp(−2σk−2r/λ)Sii0 ( k ) sin(2 kri + φ i ( k ))2 ii k ri

χ (k ) = − ∑Nif ( k )exp( − 2σ i2 k 2 − 2 ri /λ )S02 ( k ) sin(2 kri + φ i ( k ))2 ii k ri

χ (k ) = − ∑amplitude partf scattering amplitude

oscillation partφ phase shift

Can we distinguish oxygen atoms 15%
distant from others by EXAFS?

James E. Penner-Hahn J. Am. Chem. Soc. 1998,120,8401

Can we distinguish sulfur fromnitrogen (or oxygen) by EXAFS?

Almost
impossible !

Can we detect the end-end atomicinteraction in I-I-I molecule (I3-) byEXAFS?

I K-edge EXAFS FT for I3

Dissolved in organic solvents−I −I −I３Å
３Å

(n-C3H7)4N I3 powder

H. Sakane, T. Mitsui, H. Tanida, I. Watanabe. J. Synchrotron Rad. 8, 674 (2001).

I-I bond is very soft and vibrating: large
Debye-Waller like parameter

Almostimpossible !

Iodine K-edge EXAFS Fourier transform for the compound spectrum made up
from 12 independent spectra for organic solvent solutions.
H. Sakane, T. Mitsui, H. Tanida, I. Watanabe. J. Synchrotron Rad. 8, 674 (2001).

Large symmetrical cluster ofmolybdenum oxide complex

Mo K-edge XAFS

Mo6O192‐Mo

0.5μxO0-0.520000

21000
E / eV

Almost undetectable next
2neighbor oxygens!

1|FT|

Mo-Mo

0-1
010-1k/Å

20

100

123r/Å

456

FEFF Mo-O.SMP

Mo-O 1.6A
Mo-O 1.8A
Mo-O 2.0A
Mo-O 2.2A
Mo-O 2.4A
Mo-O 2.6A

10-1
246-1k/Å

810

EXAFS
Extended X-ray Absorption Fine Structure
Theory; very difficult.
Experiment; looks easy.
Data analysis; looks straight forward.
Thanks to the advanced data
analysis software.

EXAFSIn reality,Theory; becomes even more and more complex anddifficult to understand.
Experiment; to obtain CORRECT spectral data is NOT
an easy task.
Data analysis; no one except for the GOD knows
whether the conclusion from the EXAFS
analysis is CORRECT.

EXAFS is a tricky technique.
Then, what do we have to do ?
Use
★ other analytical methods,★ knowledge of chemistry and physics,
and★ good sense as a scientist
and combine them together withthe EXAFS analysis.

http://cars9.uchicago.edu/~ravel/software/

XAS Analysis Software Using IFEFFIT

freratwfosee

Current release: 0.8.059
Release date: 1 July, 2009ATHENA is an interactive graphical utility for processing EXAFS data. It handles
most of the common data handling chores of interest, including deglitching,aligning, merging, background removal, Fourier transforms, and much more.

Current release: 0.8.013
Release date: 15 December, 2008
ARTEMIS is an interactive graphical utility for fitting EXAFS data using
theoretical standards from FEFF and sophisticated data modelling along withflexible data visualization and statistical analysis. ARTEMIS includes
interfaces to ATOMS and FEFF.