XRD allows Crystal Structure Determination

What do we need to know in order to define the crystal structure?

- The size of the unit cell and the lattice type

(this defines the positions of diffraction spots)

- The atom type at each point

(these define the intensity of diffraction spots)

Conclusion: If we measure positions and intensities of many spots, then we should be able to determine the crystal structure.

POSITION OF PEAKS

LATTICE TYPE

WIDTH OF PEAK

PERFECTION OF LATTICE

INTENSITY OF PEAKS

POSITION OF ATOMS IN BASIS

The scattered x-ray amplitude is proportional to:

atomsall

ii

iefA )( Kr

Can break this sum into a sum over all lattices and a sum over all of the atoms

within the basis.

Structure Factor Shkl gives intensity of peaks

lattice basis

ii

hklief )( Kr

Structure factorhklS

To get a diffraction peak, K has to be a reciprocal lattice vector, but even if K is,

f(r)e-irK might still be zero!

ok

'k

K

2

r ○ K

321 lbkbhbK r = n1 a1 + n2 a2 + n3 a3 (real space)

ijji 2ab

Cubic form: hklS

i

Kiihkl

hkliefS )(

r

Where xi, yi and zi are the lattice positions of the atoms in the basis.

In Class: Simple Cubic Lattice

Simplify the structure factor for the simple cubic lattice for a one atom basis. Just let f be a constant.

hklS

F fi exp 2i hxi kyi lzi i

One atom basis: r 0,0,0 F f exp 2i 0h0k0l f exp 2i 0 f exp 0 f

hklS

hklS

Where xi, yi and zi are the lattice positions of the atoms in the basis.

How Do We Determine The Lattice Constant?

For the simple cubic lattice with a one atom basis:

22)0( fSIfefS hklhkli

hkl

Substituting and squaring both sides: 2222

22

4sin lkh

a

Thus, if we know the x-ray wavelength and are given (or can measure) the angles at which each diffraction peak occurs, we can determine a for the lattice! How?

2sin

222 lkh

So the x-ray intensity is nonzero for all values of (hkl), subject to the Bragg condition, which can be expressed . sin2 hkld

2/1222 lkh

adhkl

We know for cubic lattices (a=b=c):

Missing Spots in the Diffraction Pattern

In some lattices, the arrangement and spacing of planes produces diffractions from planes that are always exactly 180º out of phase causing a phenomenon called extinction.

For the BCC lattice the (100) planes are interweaved with an equivalent set at the halfway position, giving a reflection exactly out of phase, which exactly cancel the signal.

Extinction (out of phase) for 100 family of planes in BCC

What about the 101 family of planes? (001)

(-101)

Group: The Structure Factor of BCCWhat values of hkl do not have diffraction peaks?

Analysis of more than one lattice point per conventional unit cellE.g: bcc and fcc lattices

i

iKihkl

hklefS r

bcc lattice has two atoms per unit cell located at r1 = (0,0,0) and r2 = (1/2,1/2,1/2)

ri = xi a1 + yi a2 + zi a3

hklS

zyxa

zyxa

zyxa

a

a

a

21

3

21

2

21

1

yxb

zxb

zyb

a

a

a

23

22

21

Group: Find the structure factor for BCC.

Two atom basis: r 0,0,0 & r 12

,12

,12

F f exp 2i 0h0k0l exp 2i12

h12

k12

l

f 1exp i hk l h, k & l are integers, so hk lN (where N is an integer)

The exponential can then take one of two values:

exp i hk l 1 if Neven

exp i hk l 1 if Nodd

So :

F=2f if Neven

F=0 if Nodd

hklS

hklS

hklS

hklS

Structure Factorbody-centered cubic

• Allowed low order reflections are:– 110, 200, 112, 220, 310, 222,

321, 400, 330, 411, 420 …– Draw lowest on this cube ->

• Forbidden reflections are:– 100, 111, 210– Due to identical plane of atoms

halfway between causes destructive interference

• Real bcc lattice has an fcc reciprocal lattice

002 022

220

020

200

202

000

112

101

011

110

211

121

This kind of argument leads to rules for identifying the lattice symmetry from "missing" reflections.

How to determine lattice parameter this time?

Just as before, if we are given or can measure the angles at which each

diffraction peak occurs, we can graphically determine a for the lattice!

2sin

222 lkh

For a bcc lattice with a one atom basis, the x-ray intensity is nonzero for all planes (hkl), subject to the Bragg condition, except for the planes where h+k+l is odd. Thus, diffraction peaks will be observed for the following planes:(100) (110) (111) (200) (210) (211) (220) (221) (300) …

A similar analysis can be done for a crystal with the fcc lattice with a one atom basis. For materials with more than one atom type per basis in a cubic lattice, a slightly different rule for the values of (hkl) is generated.

Four atom basis: r 0,0,0 , r 12

,12

,0

, r

12

,0,12

& r 0,

12

,12

F f 1exp i hk

exp i k l

exp i h l

So:

F=4f if h,k,l all even or odd

F=0 if h,k,l are mixed even or odd

Group: Find the structure factor and extinctions for FCC.

hklS

hklS

hklS

hklS

Four atom basis: r 0,0,0 , r 12

,12

,0

, r

12

,0,12

& r 0,

12

,12

F f 1exp i hk

exp i k l

exp i h l

So:

F=4f if h,k,l all even or odd

F=0 if h,k,l are mixed even or odd

Group: Find the structure factor for FCC.

hklS

hklS

hklS

002 022

220

020

200

202

000 111Allowed low order reflections are:111, 200, 220, 311, 222, 400, 331, 310

Forbidden reflections:100, 110, 210, 211

Diamond (Homework due Thursday)Calculate the structure factor and extinctions

for the diamond structure. Lattice = FCC. Basis = (000), (¼ ¼ ¼)

Allowed Diffraction Peaks (Trend?)

The more atoms in basis, the less peaks

Structure FactorNi3Al structure

rAl

0,0,0 , rNi

12

,12

,0

, r

Ni

12

,0,12

& r

Ni 0,

12

,12

F fal

fNi

exp i hk

exp i k l

exp i h l

So:

F=fAl+3f

Ni if h,k,l all even or odd

F=fAl-f

Ni if h,k,l are mixed even or odd

Simple cubic lattice, with a four atom basis

Again, since simple cubic, intensity at all points. But each point is ‘chemically sensitive’.

( )

Common to see an average decrease in intensity of the diffraction peaks despite rules for peak intensities

i

Kiihkl

hkliefS )(

r

Atomic Scattering Factor f (key points)(aka Form Factor)

Only at 2=0 does f=Z

Also, thermal effects increase the effective size of atom

0

10

20

30

40

0 0.5 1.0 1.5

Zr

Zn

Ca

[sin()]/ (Å-1)

Mea

n A

tom

ic S

catte

ring

Fac

tors

Atoms are of a comparable size to the wavelength of the x-rays and so the scattering is not point like. There is a small path difference between

waves scattered at either side of the electron cloud

• This effect increases with angle• For x-rays, scattering strength depends on electron density• All electrons in atom (Z of them) participate, core e- density ~spherical

Na on each fcc site, but with a two atom basis:

rNa 0,0,0 & rCl 12

,12

,12

F fNa fCl exp i hk l

1exp i hk exp i k l exp i h l F 4 fNa fCl if h,k,l all even

F 4 fNa fCl if h,k,l all odd

F 0 if h,k,l mixed

hklS

Structure Factor with Different Atoms NaCl (rock salt) structure

FCC Reminder:

Extra slides

• There is a lot of useful information on diffraction. Following are some related slides that I have used or considered using in the past.

• A whole course could be taught focusing on diffraction so I can’t cover everything here.

XRD: “Rocking” Curve Scan

• Vary ORIENTATION of K relative to sample normal while maintaining its magnitude.How? “Rock” sample over a very small angular range.

• Resulting data of Intensity vs. Omega (w,sample angle) shows detailed structure of diffraction peak being investigated. Can inform about quality of sample.

ik fk

“Rock” Sample

Sample normalK K

XRD: Rocking Curve Example

• Rocking curve of single crystal GaN around (002) diffraction peak showing its detailed structure.

16.995 17.195 17.395 17.595 17.7950

8000

16000

GaN Thin Film(002) Reflection

Inte

nsity

(C

ount

s/s)

Omega (deg)

How do you know if this is good?

Compare to literature to see how good (some

materials naturally easier than others)

Generally limited by quality of substrate

X-ray reflectivity (XRR) measurement

Si

Mo

Mo

Mo

r t [Å] s[Å]0.68 19.6 5.8

0.93 236.5 34.0

1.09 14.1 2.71.00 5.0 2.7

1.00 2.8

Calculation of the density, composition, thickness and interface roughness for each particular layer

W

The surface must be smooth (mirror-like)

0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,010

0

101

102

103

104

105

106

Inte

nsity

(a.

u.)

Diffraction angle (o2)

Kiessig oscillations (fringes)

A glancing, but varying, incident angle, combined with a matching detector angle collects the X rays reflected from the samples surface

The X-ray Shutter is the most important safety device on a diffractometer

• X-rays exit the tube through X-ray transparent Be windows.

• X-Ray safety shutters contain the beam so that you may work in the diffractometer without being exposed to the X-rays.

• Being aware of the status of the shutters is the most important factor in working safely with X rays.

Cu

H2O In H2O Out

e-

Be

XRAYS

windowBe

XRAYS

FILAMENT

ANODE

(cathode)

AC CURRENT

window

metal

glass

(vacuum) (vacuum)

Primary

Shutter

Secondary

Shutter

Solenoid

SAFETY SHUTTERS

XRD: Reciprocal-Space Map

• Vary Orientation and Magnitude of k.• Diffraction-Space map of GaN film on AlN buffer

shows peaks of each film.

/2

GaN(002) AlN

If the wavelength of the incident x-rays and the scattering angle are known, then one can deduce the distance (already done) between the planes, dhkl, responsible for each scattering peak. The following twelve lines were obtained from a crystalline powder, known to belong to a cubic system.

Line d(Å) relative intensity1 3.157 942 1.931 1003 1.647 354 1.366 125 1.253 106 1.1150 167 1.0512 78 0.9657 59 0.9233 710 0.9105 111 0.8634 912 0.8330 3

Index the lines in terms of their Miller indices (hkl) and calculate the lattice constant of the cubic lattice. Establish the type of cubic lattice.

Conditions for Peak

SC All points

BCC Sum = even

FCC All Even/odd

Diam All Odd or sum4n

NaCl All Even/odd FCC

Ni3Al All Points (SC)

Similar to HW, but turned theta dependence to d. How?

Neutron

λ = 1A°

E ~ 0.08 eV

interact with nucleiHighly Penetrating

Electron

λ = 2A°

E ~ 150 eV

interact with electronLess Penetrating

Non-xray Diffraction Methods(more in later chapters)

• Any particle will scatter and create diffraction pattern

• Beams are selected by experimentalists depending on sensitivity– X-rays not sensitive to low Z elements, but neutrons are– Electrons sensitive to surface structure if energy is low– Atoms (e.g., helium) sensitive to surface only

• For inelastic scattering, momentum conservation is important

X-Ray

λ = 1A°

E ~ 104 eV

interact with electronPenetrating

Group: Consider Neutron Diffraction

• Qualitatively discuss the atomic scattering factor (e.g., as a function of scattering angle) for neutron diffraction (compared to x-ray) by a crystalline solid.

• For x-rays, we saw that f is related to Z and has a strong angular component. For neutrons?

• The same equation applies, but since the neutron scatters off a tiny nucleus, scattering is more point-like, and f is ~ independent of .

Systematic Extinction• Systematic extinction is

a consequence of lattice type

• At right is table of systematic extinctions for symmetry elements

• Other extinctions can occur as a consequence of screw axis and glide plane translations (Dove, Ch.6 Structure and Dynamics)

• Accidental Extinctions may occur resulting from mutual interference of other scattering vectors

Symmetry Extinction Conditions

simple none

C hkl; h + k = odd

B hkl; h + l = odd

A hkl; k + l = odd

body hkl; h + k + l = odd

All faces hkl; h, k, l mixed even and odd

Key: C, B, A = side-centered on c-, b-, a-face; I = body centered; F = face centered (001)

Preferred Orientation (texture)

• Preferred orientation of crystallites can create a systematic variation in diffraction peak intensities– can qualitatively analyze using a 1D diffraction pattern– a pole figure maps the intensity of a single peak as a

function of tilt and rotation of the sample• this can be used to quantify the texture

(111)

(311)(200)

(220)

(222)(400)

40 50 60 70 80 90 100Two-Theta (deg)

x103

2.0

4.0

6.0

8.0

10.0

Inte

nsity

(Cou

nts)

00-004-0784> Gold - Au

Diffracting crystallites