Diffraction - Novel Materials and Ground Statescanfield.physics.iastate.edu/course/Lecture_01_Diffraction_slides.pdf · Vitalij Pecharsky, Peter Zavalij, Fundamentals of Powder Diffraction

$Page 1: Diffraction - Novel Materials and Ground Statescanfield.physics.iastate.edu/course/Lecture_01_Diffraction_slides.pdf · Vitalij Pecharsky, Peter Zavalij, Fundamentals of Powder Diffraction$
Diffraction

2019 Department of Chemistry, Iowa State University. All rights reserved

Diffraction

Sound wavelength ~ 10-0.01 m


A Little Bit of History

The first Nobel prize in Physics?

Wilhelm Conrad Röntgen (1901)“in recognition of the extraordinary services he has rendered by the discovery of the remarkable rays subsequently named after him”

X-rays

3

His wife’s hand


More Nobel PrizesAntoine Henri Becquerel (1903)"in recognition of the extraordinary services he has rendered by his

discovery of spontaneous radioactivity“

Pierre & Marie Curie (1903)"in recognition of the extraordinary services they have rendered by their

joint researches on the radiation phenomena discovered by Professor Henri Becquerel“

Max von Laue (1914)“for his discovery of the diffraction of X-rays by crystals“

Sir William Henry & William Lawrence Braggs (1915)"for their services in the analysis of crystal structure by means of X-rays“

Charles Glover Barkla (1917)"for his discovery of the characteristic Röntgen radiation of the elements"

4 2019 Department of Chemistry, Iowa State University. All rights reserved

DIFFRACTION

Can be observed for any kind of waves

When diffraction occurs from several periodically arranged objects, the waves add up (interference) to produce maxima and minima of intensity

To achieve this effect, the distance between the objects should be comparable to the wavelength

In crystals, interatomic distancesare on the order of 10-10 m = 1 Å

Hence, the X-rays!


X-ray generation

6

2019 Department of Chemistry,Iowa State University. All rights reserved

X-ray generation

7

Sealed tube Microfocus Rotating Anode Liquid Anode1 5-12 15-20 30+

Ga


OUR PLAN

8

Learn the basic theory – how it works- Miller planes and Miller indices- reciprocal lattice- Bragg law- powder diffraction

Additional suggested reading:

-Vitalij Pecharsky, Peter Zavalij, Fundamentals of Powder Diffraction and Structural Characterization of Materials. 2nd Edition. Springer, ISBN: 978-0-387-09579-0.

-Jenkins & Snyder, Introduction to X-ray Powder Diffractometry, 1996

(for those who would like to acquire deeper knowledge of the subject)


MILLER PLANES

Atoms form periodically arranged planes

Any set of planes is characterized by:(1) their orientation in the crystal (hkl) – Miller indices(2) their d-spacing (dhkl) – distance between the planes

h, k, l correspond to the number of segments in which the a, b, c axes, respectively, are cut by the set of planes

On average, the higher (hkl),the closer is the interplanar distance, dhkl

2-D Examples


MILLER PLANES

10

a

b

c


MILLER PLANES

11

a

b

c


MILLER PLANES

12

a

b

c


MILLER PLANES

13

a

b

c


MILLER PLANES

14

a

b

c


MILLER PLANES

15

a

b

c


MILLER PLANES

16

a

b

c


BRAGG VIEW OF DIFFRACTIONX-rays that hit the crystal are elastically scattered by the sets of (hkl) planes

θ θ

θ θdhkl

θ θ

The path difference for rays 1 and 2 equals to the length of two blue lines:

1

2

1′

2′

θsin2)2-1( hkld=∆


BRAGG LAW

The diffraction maxima will be created by reflection from a set of planes at angle θ that results in the integer wavelength difference in the path of the rays:

θdn hkl sin2=λ

Consequence 1:Each set of planes (hkl) is characterized by its own value θhklat which the diffraction maximum is observed

Consequence 2:Each crystalline compound is characterized by a set of reflections at characteristic dhkl , or θhkl(diffraction “fingerprint” of the compound)


BRAGG VIEW OF DIFFRACTION

The deviation of the ray from its initial path equals 2θ

θ θ

dhkl

1 1′

2θ


Diffraction from Polycrystalline Sample (Powder)

The powder sample in the center is rotatinggenerating “2θ cones” or diffracted rays

2θ



Diffraction from Polycrystalline Sample (Powder)

The film arranged cylindrically around the sample records lines in the places where it is intersected by the cones.

film


POWDER DIFFRACTOMETER

This one is a very old model. We hardly use films anymore.But it is good for basic understanding of the diffraction principles.

By measuring the lines on the film, we can obtain a set of θhkl (or dhkl) corresponding to our powder sample. 23

θdn hkl sin2=λ 2019 Department of Chemistry, Iowa State University. All rights reserved

2θ


24

θdn hkl sin2=λ



X-ray source25

Detector9-sample changer


Applications of Powder Diffractometry

26

Qualitative phase analysis (comparison to the known patterns)

Quantitative phase analysis (intensity measurements)

Unit cell determination (dhkl′s depend on lattice parameters)

Study of phase transitions

Particle size estimation (line width)

Crystal structure refinement (line intensities and profiles)



27








Qualitative Phase Analysis

28

Direct comparison: if you know what phases can be present in your sample, the pattern can be compared against the one calculated from the known crystal structure

- can be done by most of the common structure visualization software (Mercury, CrystalMaker/Diffract, X-Seed, Diamond)

Search/Match comparison: the pattern is compared against the recorded powder patterns stored in the database (PDF from ICDD)

- only compares the position and intensity of peak maxima, but not the peak profiles

- contains both experimental and calculated patterns

- accuracy of the stored patterns is systematically reevaluated


Reciprocal Lattice



30








RECIPROCAL LATTICE

The orientation of a plane is defined by the direction of a normal to the plane:

hklhkl dNd

=


RECIPROCAL LATTICE

32

dhkl vectors in the direct space

d*hkl vectors in the reciprocal space

hklhkl d

Nd

=*

hklhkl dNd

= [Å]

[1/Å]


RECIPROCAL LATTICE

33

The reciprocal lattice is periodic.Therefore, we should be able to define the reciprocal unit cell:

(a*, b*, c*)

such that any vector in the lattice can be described as

d*hkl = ha* + kb* + lc*

Now let’s see how the reciprocal latticeis related to the diffraction pattern


Why Use The Reciprocal Space?

34

Think about the Bragg law:

What do we measure experimentally?

Angles at which diffractionmaxima are observed

θdn hkl sin2=λ

*

21

22sin hkl

hklhkl

dnd

ndnθ ×=

×==

λλλ

×==

hklhkl dn

dnθ 1

22sin λλ


xr1

r2

θ


35

The experimental measurement is directly related to d*hkl OR

The detector is scanning for the reciprocal lattice points d*hkl

xr

dnθ hkl ×=×=2

* 12

sin λ



36Diffraction from a single Xtal

A diffraction pattern is not a direct representation of the crystal lattice The diffraction pattern is a representation

of the reciprocal lattice

In order to find the reciprocal lattice,the diffraction pattern can be indexed


INDEXING PROCEDURE

37

To index a diffraction pattern means:

to find such a basis (a*,b*,c*) that all the diffraction spots (or lines) can be described (indexed) as

d*hkl = ha* + kb* + lc*

with only integer (hkl) values allowed.

All we need to do is to find the value of d* for each spot and then index = let software find the unit cell (in our lazy time).

Basically, we deal with a system of linear equations.

The more reflections we have,the more reliably the unit cell will be determined.


THE RECIPROCAL UNIT CELL

38

The reciprocal UC is related to the direct UC:

By definitiona*·a = 1 b*·a = 0 c*·a = 0a*·b = 0 b*·b = 1 c*·b = 0a*·c = 0 b*·c = 0 c*·c = 1

which means that

a* ⊥ (bc) b* ⊥ (ac) c* ⊥ (ab)

If we can index the diffraction patternand find the reciprocal UC,then we will be able to find the direct UC.

c

ba*

θ


THE RECIPROCAL UNIT CELL

39

The reciprocal UC is related to the direct UC:By definition

a*·a = 1 b*·a = 0 c*·a = 0a*·b = 0 b*·b = 1 c*·b = 0a*·c = 0 b*·c = 0 c*·c = 1

a* ⊥ (bc) b* ⊥ (ac) c* ⊥ (ab)


INDEXING A DIFFRACTION PATTERN



41

The simplest case – cubic symmetry

For a cubic system the expected lines in a powder pattern can be generated from the following relationships

The relationship between the Miller indices of a peak (hkl) and the interplanar d-spacing

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]

Whered = λ/(2sinθ)



42

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]

h k l l2+k2+l2 d (Å) 2θ

1 0 0 1 5.350 16.561 1 0 2 3.783 23.501 1 1 3 3.089 28.882 0 0 4 2.675 33.472 1 0 5 2.393 37.562 1 1 6 2.184 41.302 2 0 8 1.892 48.063 0 0 9 1.783 51.182 2 1 9 1.783 51.183 1 0 10 1.692 54.17



43

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]Manual Indexing

Indexing a powder pattern is the reverse of peak generation.

Namely we are given the peak positions and we use that information to determine:

The hkl index of each peak

The unit cell dimensions

The systematic absences, which gives information about the space group

Manual indexing consists of the following steps:

Determine d-spacing of each peak from its 2θ value (using Braggs Law)

Create a table of 1/d2 values for each peak

Look for a common factor (1/a2) that can be divided into each of the 1/d2 values



44

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]

2θ d (Å) 1000/d2

16.56 5.350 34.923.50 3.783 69.928.88 3.089 104.833.47 2.675 139.837.56 2.393 174.641.30 2.184 209.648.06 1.892 279.451.18 1.783 314.651.18 1.783 314.654.17 1.692 349.3

Lets begin by working the previous example backwards.Starting with the peak positions (recall λ = 1.5406 A).



45

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]

With a little inspection (in this case) we see that all of the1000/d2 values can be divided by 34.9

2θ d (Å) 1000/d2 /34.9 hkl

16.56 5.350 34.9 1 100

23.50 3.783 69.9 2 110

28.88 3.089 104.8 3 111

33.47 2.675 139.8 4 200

37.56 2.393 174.6 5 210

41.30 2.184 209.6 6 211

48.06 1.892 279.4 8 220

51.18 1.783 314.6 9 300

51.18 1.783 314.6 9 221

54.17 1.692 349.3 10 310 2019 Department of Chemistry, Iowa State University. All rights reserved


46

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]The last problem was a particularly easy one because there were no missing peaks. Typically there will be some systematic absences.Consider the following example. Once again λ = 1.5406 A.

2θ d (Å) 1000/d2

28.08 3.175 99.232.53 2.750 132.246.66 1.945 264.355.37 1.658 363.858.03 1.588 396.668.14 1.375 528.975.23 1.262 627.977.55 1.230 661.0



47

1𝑑𝑑2

= 1𝑎𝑎2

[ℎ2 + 𝑘𝑘2 + 𝑙𝑙2]It�s immediately clear that 99.2 is not a common factor here.Though we can see that 132.2-99.2=33 might be a common factor.So we’ll give it a try.

2θ d (Å) 1000/d2 /33 hkl

28.08 3.175 99.2 3 111

32.53 2.750 132.2 4 200

46.66 1.945 264.3 8 220

55.37 1.658 363.8 11 311

58.03 1.588 396.6 12 222

68.14 1.375 528.9 16 400

75.23 1.262 627.9 19 331

77.55 1.230 661.0 20 420





49








DETERMINING PARTICLE SIZE

50

Scherrer equation:

D – the mean crystallite dimensionW – the full width at half maximum of the peak (FWHM)wi – instrumental broadeningθ - diffraction angleK – shape factor (usually ~0.9)

𝐷𝐷 =𝐾𝐾𝐾𝐾

𝑊𝑊 − 𝑤𝑤𝑖𝑖 cos𝜃𝜃


Synchrotrons

APS ANL NSLS Brookhaven

ESRF Grenoble

BLS Beijing

ANSTO Melbourne

CLS SaskatoonDiamond Oxford

NSRRC Taiwan

Alba Barcelona

ALS Berkley

PETRA Hamburg Spring-8 Japan

Applications of Powder Diffraction

53








Split of the peaks upon lowering of symmetry

54

Applying pressure to the primitive cubic perovskite, ABO3, Pm�3m

Lets calculate three diffraction peaks with hkl: 100, 010, and 001

For cubic: a = b = c

2Theta30.0 30.4 30.8 31.2 31.6 32.0 32.4 32.8

0.00

0.20

0.40

0.60

0.80

1.00

Rel

ativ

e In

tens

ity

fe

1,0,

1100,

01

0,

001

hkldθ 2/sin λ=cl

bk

ah

d hkl

2

2

2

2

2

2

21

++=

addd 22

001

2

010

2

100

1111===

2019 Department of Chemistry,

Iowa State University. All rights reserved

2Theta30.0 30.4 30.8 31.2 31.6 32.0 32.4 32.8

0.00

0.20

0.40

0.60

0.80

1.00

Rel

ativ

e In

tens

ity

fe

1,0,

1

1,1,

0

100,

01

000

1


55


For tetragonal: a = b >≈ c

hkldθ 2/sin λ=

Applying pressure to the primitive cubic perovskite, ABO3, Pm-3m

cl

bk

ah

d hkl

2

2

2

2

2

2

21

++=

dd 2

100

2

001

11>

cd 22

001

11=

add 22

010

2

100

111==




56


For orthorhombic: a >≈ b >≈ c

2Theta30.0 30.4 30.8 31.2 31.6 32.0 32.4 32.8

0.00

0.20

0.40

0.60

0.80

1.00

Rel

ativ

e In

tens

ity

fe

0,1,

1

1,0,

1

1,1,

0100

001

010

Applying pressure to the primitive cubic perovskite, ABO3, Pm-3m

hkldθ 2/sin λ=

ddd 2

100

2

010

2

001

111>>

bd 22

010

11=

cd 22

001

11=

ad 22

100

11=

cl

bk

ah

d hkl

2

2

2

2

2

2

21

++=



57

But not all peaks would split:111 in all three cases is a single peak – calculate at home

2Theta15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0

0.00

0.20

0.40

0.60

0.80

1.00

Rel

ativ

e In

tens

ity

Fe

1,0,

00,

0,1

1,1,

01,

0,1

1,1,

1

2,0,

0

0,0,

2

2,1,

0

1,0,

2

2,1,

11,

1,2

tetragonal

2Theta15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0

0.00

0.20

0.40

0.60

0.80

1.00R

elat

ive

Inte

nsity

Fe

1,0,

0

1,1,

0

1,1,

1

2,0,

0

2,1,

0

2,1,

1

cubic


hkldθ 2/sin λ=



58

Effect of the wave-length

sinθ = 𝐾𝐾/2𝑑𝑑ℎ𝑘𝑘𝑘𝑘Cu (1.54 Å)

Cr (2.29 Å)

Co (1.79 Å)

Mo (0.71 Å)

Ag (0.56 Å)

Synchrotron (0.1-3 Å)


59



Cr (2.29 Å)

Co (1.79 Å)

Mo (0.71 Å)

Ag (0.56 Å)



60



Cr (2.29 Å)

Co (1.79 Å)

Mo (0.71 Å)

Ag (0.56 Å)



61

X-ray Fluorescence

sinθ = 𝐾𝐾/2𝑑𝑑ℎ𝑘𝑘𝑘𝑘

Emitting of characteristic X-rays - a principle of XRF method But high background in XRD


CoSe2 sample measured on Cu radiation


CoSe2 sample measured on Cu radiationwith energy discrimination ON








67

X-ray Fluorescence


Cr (2.29 Å)

Co (1.79 Å)

Mo (0.71 Å)

Ag (0.56 Å)


λ =


68

X-ray Absorption

sinθ = 𝐾𝐾/2𝑑𝑑ℎ𝑘𝑘𝑘𝑘

Radiation C P Se Cs Bi

Cu (1.54 Å) 3.33 75.5 80.0 317 244

Mo (0.71 Å) 0.58 7.97 69.5 40.7 126

https://11bm.xray.aps.anl.gov/absorb/absorb.php


https://11bm.xray.aps.anl.gov/absorb/absorb.php

69

Synchrotron X-ray: 11-BM at APS ANL

sinθ = 𝐾𝐾/2𝑑𝑑ℎ𝑘𝑘𝑘𝑘 https://11bm.xray.aps.anl.gov/description.html


https://11bm.xray.aps.anl.gov/description.html

70

Synchrotron X-ray: 11-BM at APS ANL

sinθ = 𝐾𝐾/2𝑑𝑑ℎ𝑘𝑘𝑘𝑘 https://11bm.xray.aps.anl.gov/description.html


https://11bm.xray.aps.anl.gov/description.html

Lax Ba8−x Cu16P30: lab XRD

La = 1.0

La = 1.4

La = 1.6

La = 1.8

La = 2.0

La = 0.0


11-BM data on the same samples

Synchrotron XRD11-BM APS ANL


Lax Ba8−x Cu16P30: lab XRD

La = 1.0

La = 1.4

La = 1.6

La = 1.8

La = 2.0

La = 0.0


11-BM data on the same samples

Synchrotron XRD11-BM APS ANL


Structure: Lax Ba8−x Cu16P30 and Cex Ba8−x Cu16P30

Chem, 2018, 4, 1465. 2019 Department of Chemistry, Iowa State University. All rights reserved

Structure: Lax Ba8−x Cu16P30 and Cex Ba8−x Cu16P30

Chem, 2018, 4, 1465. 2019 Department of Chemistry, Iowa State University. All rights reserved

Cu/P ordering: La1.6Ba6.4Cu16P30


Cu/P ordering: La1.6Ba6.4Cu16P30

Cu is all the time surrounded by 4 P

Minimizing Cu-Cu interactions Maximizing Cu-P interactions


Documents

Diffraction - Novel Materials and Ground Statescanfield.physics.iastate.edu/course/Lecture_01_Diffraction_slides.pdf · Vitalij Pecharsky, Peter Zavalij, Fundamentals of Powder Diffraction