Engineering Materials ME Chap 3 10th Ed Part (IV) X-ray ...

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Department of Mechanical EngineeringNational Chung Cheng University

Engineering MaterialsChapter 3 (IV)

The Structure of Crystalline SolidsX-Ray Diffraction

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Prof. J.N. Aoh CCU ME 11.09.2021


Analysis of Crystal StructuresX-RAY Diffraction

Prof. Jong-Ning Aoh2021. 11. 09.

Chap.3 Fundamental of Crystallography

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Fraunhofer Diffraction• Fraunhofer diffraction occurs when the

distances from the object to the source and the object to the image are so large that the incoming and outgoing waves are effectively planar. (_____________optics)

SourceMask

Image

Joseph von Fraunhofer1787-1826


Fraunhofer Diffraction

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d is the distance from the source to mask, L is the distance from mask to image, w is the width of the slit and

λ is the wavelength of the radiation.

If the condition is not met, then curvature of the waves is involved and the resulting effect is known as Fresnel diffraction. (_____________optics)


Intensity pattern for single slit diffraction

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Multiple slits– Grating (______)

S: period of grating

W: width of slit


Diffraction grating

s=12μm s=3μm

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1-D Diffraction patterns


small features of the diffracting object give wide spacings in the diffraction pattern

N=0 N=1

xLs λ=

1-D Diffraction patterns

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Two-dimensional diffraction

• If two diffraction gratings are superimposed perpendicularly, they form a two-dimensional periodic array of apertures.


• The observed diffraction pattern is neither the sum nor the product of the original patterns of the individual gratings, but the separate patterns are repeated to form a two-dimensional array.

Two-dimensional diffraction

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Diffraction occurs when:

• Regularly spaced obstacles – Crystal structures with regular spacing between atoms and

planes• Gratings capable of scattering wave

• ___________ principle• Magnitude of spacing in the similar order of wavelength:• Visible light: λ= ____-____ nm, X-Ray: λ= 0.1-0.5 nm• The typical interatomic spacing ~ 2-3 Å , so the suitable

radiation for the diffraction study of crystals is X-rays.


Huygens principle

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X-Ray Diffraction (XRD) on Crystal StructuresWilliam Henry Bragg and William Lawrence Bragg

-Father and son

1862 1890The Nobel Prize in Physics 1915 was awarded jointly to Sir William Henry Bragg and William Lawrence Bragg "for their services in the analysis of crystal structure by means of X-rays“ --XRD


X-ray discovered by Wilhelm Röntgen

Inte

nsity

Wavelength (λ)0.2 0.6 1.0 1.4

White radiation

Characteristic radiation →due to energy transitionsin the atom

Kβ

KαIntense peak, nearly monochromatic

Target Metal

λ Of Kαradiation (Å)

Mo 0.71Cu 1.54Co 1.79Fe 1.94Cr 2.29

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Bragg’s Law • The Bragg’s law was formulated in 1912 by W. L. Bragg

Incident beamReflected beam

Scattered beam

nλ = 2 dhkl sin θ

For cubic

Wave front

( )222 lkhadhkl

++=

(h k l)

(h k l)


A Bragg condition is such that the scattered rays from twoparallel planes interact with each other in such a way as tocreate ______________ interference.The extra distance that ray B must travel is the distanceX-Y-Z (optical path difference) Thus X-Y-Z = nλ

X-Y = Y-Z = d sin θ

XY

Z

(h k l)

(h k l)

(h k l)

XY+YZ=2 dhkl sinθ = nλ

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In-phase, constructive interference

Out-of-phase, destructive interference


Bragg’s law

θ

θ

A A

BB

X

Y

Zdhkldhkl

XY+YZ=2 dhkl sinθ = nλ

Reflected beam or diffracted beam

transmitted beam

Constructive interference

2θ

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Reflection versus Diffraction

Reflection DiffractionOccurs from surface Occurs throughout the bulk

Takes place at any angle Takes place only at Bragg angles

~100 % of the intensity may be reflected

Small fraction of intensity is diffracted

Note: X-rays can ALSO be reflected at very small angles of incidence

scattered ray

diffracted beam (constructive interference)

incident beam

transmitted beam


2θ

2θ

2θ

nλ = 2 dhkl sin θ

Incident angle of X ray

To be determined

Wavelength of X ray

θ

θ

222 lkhadhkl

++=

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θλ sin2 hkldn =

In XRD nth order reflection from (h k l) is considered as 1st

order reflection from (nh nk nl)

θλ sin2n

dhkl=

θλ sin2 n n n lkhd=

1nhnk nl

hkl

dd n

=

300

100

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dd

=

200

100

12

dd

=

Hence, (100) planes are a subset of (200) planes

Important point to note:In a simple cubic crystal, 100, 200, 300… are all allowed ‘reflections’. But, there are no atoms in the planes lying within the unit cell!.


X-ray diffraction on a single crystal of a crystallographic plane

Single crystal X-ray diffraction (XRD), although simple, is often difficult to arrange in practice since it requires a single crystal to be grown. Additionally, there is a very precise requirement on the alignment of the crystal.

(hkl)Reflected beam must satisfy Bragg’s law

λ = 2 dhkl sin θhkl

transmitted beam

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X-ray diffraction on a single crystal of a crystallographic plane

Incident X-ray

X-ray detector

Straight-through X-raydhkl

To satisfy nλ=2 dhkl sinθ


2θ

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X-ray powder diffractionX-ray diffraction for polycrystalline

Powder or polycrystalline sample

Diffraction experiments using either a powder sample, or a polycrystalline sample.

This ideally results in all possible orientationsappearing in the sample.

2 dhkl sinθ = nλ


X-ray powder diffraction

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• If the sample consists of many randomly orientated single crystals, the diffracted beams are seen to lie on the surface of several cones. The cones may emerge in all directions, forwards and backwards.

{h k l}

{h2 k2 l2}


Forward and Back Diffraction

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{h2 k2 l2}

{h k l}

Each cone representing the diffraction of plane {h k l} intersects the film giving diffraction lines. The lines are seen as arcs on the film.

Film and powder Method


• An older method of X-ray diffraction analysis is known as the film method.

ray

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Indexing the film pattern

OR

BaTiO3


Quartz: modern diffractometer output

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Film-counter correlation

(BaTiO3)


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• Bragg's Law: nλ = 2dhklsinθdhkl = nλ/2sinθ

if n=1, dhkl = λ/2sinθ


known

Known from table or derived from a=f (R)

To be determined

222 lkhadhkl

++=


X-Ray Diffraction Pattern

Adapted from Fig. 3.20, Callister 5e.

(110)

(200)

(211)

z

x

ya b

c

Diffraction angle 2θ

Diffraction pattern for polycrystalline α-iron (BCC)

Inte

nsity

(rel

ativ

e)

z

x

ya b

cz

x

ya b

c

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Spacing between planes

+

+++=

+=

++=

222222

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222

222

callkh

ad

lC

khd

lkhad

hkl

hkl

hkl

a

--- Cubic

--- HCP

--- Tetragonal

Miller indices

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Additional diffraction rules

Crystal system

Simple cubic BCC FCC HCP

diffraction every plane

h +k + l=even

h, k, l all even or all odd

No diffraction none

h +k + l=odd

h, k, l mixed

h+2k=?,l=?


cubic {h k l} h2+k2+l2 BCC FCC

{100} 1 - -

{110} 2 {110} -

{111} 3 - {111}

{200} 4 {200} {200}

{210} 5 - -

{211} 6 {211} -

{220} 8 {220} {220}

{221} {300}

9 - -

{310} 10 {310} -

{311} 11 - {311}

{222} 12 {222} {222}

{320} 13

{321} 14 {321}

{400}16 {400}

{400}

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Indexing XRD peaks of unknown element or crystal

1. Indexing XRD peaks

2. Determine unknown element or crystal


Relationship between θhkl and h2+k2+l2

nλ=2 dhkl sinθ 2 d Sinλ θ=2 2 2hkl Cubic

adh k l

=+ +

222

222 sin4

lkha

++= θλ θ

λ2

2

2222 sin4)( alkh =++

)(sin4

2222

22 lkha ++=

θλ

θ2222 sin)( ∝++ lkh

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Relationship between θhkl and h2+k2+l2

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1

22 2 2 2 2

1 1 1 12

2 2 2 21 1 1

2 2 2 22 2 2 2

4( ) sin sin

( ) sin( ) sin

ah k l K

h k lh k l

θ θλ

θθ

+ + = =

+ +=

+ +

For two different planes (h1 k1 l1) and (h2 k2 l2)

1, 2, 3, 4, 5, 6, ………………



• First set a table with the following columns in sequence:

• Then calculate the values for each θ.• Find the common K for all sin2 θ to fit (h k l)

• 2θ θ sin θdhkl sin2 θK sin2 θ(h2+k2+l2 )(h k l) a (lattice const) R

2θ θ sinθ sin2θ h2+k2+l2 h k l a R

K

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2 θ θ sin θ sin2 θ Ksin2 θ h2+k2+l2 (h k l) dhkl a R0.235 0.407 0.144

0.2097 0.419 0.148

0.145 0.411 0.145

0.126 0.418 0.148

0.119 0.412 0.146

0.103 0.412 0.146

0.095 0.414 0.146

X-ray λ=0.1296 nm

K=



0.2097 0.419 0.148

0.145 0.411 0.145

0.126 0.418 0.148

0.119 0.412 0.146

0.103 0.412 0.146

0.095 0.414 0.146

X-ray λ=0.1296 nm

K=

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0.2097 0.419 0.148

0.145 0.411 0.145

0.126 0.418 0.148

0.119 0.412 0.146

0.103 0.412 0.146

0.095 0.414 0.146

X-ray λ=0.1296 nm

K=


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(BaTiO3)


peak 2θ θ Sinθ Sin2 θ KSin2 θ h2+k2+l2 Index d a1 38.52 19.26 0.33 0.11 3 111 2.34

2 44.76 22.38 0.38 0.14 4 200 2.03

3 65.14 32.57 0.54 0.29 8 220 1.43

4 78.26 39.13 0.63 0.40 11 311 1.22

5 82.47 41.235 0.66 0.43 12 222 1.17

6 99.11 49.555 0.76 0.58 16 400 1.01

7 112.03 56.015 0.83 0.69 19 331 0.93

8 116.60 58.3 0.85 0.72 20 420 0.91

9 137.47 68.735 0.93 0.87 24 422 0.83

10 163.78 81.89 0.99 0.98 27 333 0.78

Determination of Crystal Structure from 2θ

From the ratios in column 6 we conclude that FCC2 d Sinλ θ= 111 1111.54 2 2 0.33

3ad Sinθ= =

o4.04Aa Al= →

Using

We can get the lattice parameter → which correspond to that for Al

Note that Sinθ cannot be > 1

θ2222 sin)( ∝++ lkhNote

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Figure 3.41 shows the first four peaks of the x-ray diffraction pattern for copper, which has an FCC crystal structure; monochromatic x-radiation having a wavelength of 0.1542 nm was used.(a) Index (i.e., give h, k, and l indices for) each of these peaks.(b) Determine the interplanar spacing for each of the peaks.(c) For each peak, determine the atomic radius for Cu and compare these with the value presented in Table 3.1.


a represents the direction of increasing θ

b represents the direction of increasing 2 θ

X-ray diffractometer

The monochromator is used to ensure a specific wavelength reaches the detector, eliminating fluorescent radiation.

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Trace of intensity vs counter angle

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Diffuse peak from Cu-Zr-Ni-Al-Si Metallic glass

Actual diffraction pattern from an amorphous solid

Sharp peaks are missing. Broad diffuse peak exists → the peak corresponds to the average spacing between atoms

Amorphous solid


Compounds

Engineering Materials ME Chap 3 10th Ed Part (IV) X-ray ...

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