Part 8. Diffraction: Intensity(From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1)

Scattering by an electron: An electron which has been set into oscillation by an X-ray beam is continuously accelerating and decelerating during its motion and therefore emits an EM wave being said to scatter x-rays. The emitted X-ray has the same frequency (wavelength) as the excited X-ray beam i.e. coherent. (the phase change on scattering from an electron is /2) * The intensity of the scattered beam is angular dependent. J.J. Thomson: the intensity I of the beam scattered by a single electron of charge e (C) and mass m (kg), at a distance r (meters) from the electron, is given by

r

P

2

202

22

42

00 sinsin

4 r

KI

rm

eII

0: 410-7 mkgC-2

x

y

zIncident beam wavevector alongx-axis, the electric field lies on theyz plane. Electron at O. Diffractedbeam at P on xz plane, i.e. diffractionplane is xz. Every E can be resolvedinto components Ey and Ez.

O

P

Randompolarized

2r

On average, Ey = Ez, since E is perfectly random.

222zy EEE

00022

2

1

2

1IIIEEE zyzy 2

The y component accelerates the electron in thedirection Oy the intensity of the scattered beam atP is

20 r

KII yPy = yOP = /2

The z component accelerates the electron in thedirection Oz the intensity of the scattered beam atP is

2cos220 r

KII zPz = zOP = /2 -2

The total scattered intensity at P is

2

2cos12cos

2

202

2020

r

KI

r

KI

r

KIIII zyPzPyP

Polarization factorr I ; IIf a monochromator is used with the Bragg angle M the polarization factor for powdered crystalline samplebecomes

M

M

2cos1

2cos2cos12

22

It depends on geometry and sample

condition

Another kind of scattering: Compton scattering;when X-ray encounter loosely bound or free electrons;incoherent scattering, wavelengthchanged, no fixed phase relation;belongs to scattering loss(attenuation). X-ray before impacth1 (1), after Compton scatteringh2 (2). 2

12 sin0486.0)(

h1 e

e

h2

2

Atomic scattering (or form) factor: Consider the physics of coherent scattering by atoms. The atomic scattering factor is the ratio of the scattering from an atom to that from a single bound electron.

electron single aby scattered amplitude

atomby scattered amplitudefactor scattering atomic

Assume that each electron has a spherical charge density(r) around the atomic nucleus and that the X-raywavelength is not too close to the absorption edge.The amplitude scattered to the plane P from a smallvolume of the charge density of oneelectron, dV, is given by (r)dV.s0, s: unit vectors presenting the incident and scattered wave;The path different between thewave scattered at O and at dV isR-(x1 + x2).

O

s0

R

r

x1dV

s

x2

2

srsr Rxx 201 ;

The differential atomic scattering factor is defines by

dVerE

df i

e

)]()[/2( 0)(1 ssr Ee: the magnitude of the wave from

a bound electron

Spherical integration, the integrationelement is dV = 2r2sin d dr

dV

S0

S

S-S0 r

S0

S S-S0

2

Evaluate (S - S0)r = | S - S0||r|cos (S - S0)/2 = sin.

cossin2)( 0 r rSS

dVerE

dff i

e

)]()[/2( 0)(1

)( ssr

r

r

ri

e

drdrerE

dff0 0

2/cossin4 sin2)(1

)(

Let /sin4k

200

22 4|cos2sin2 rrdr

cosd

r

r

kri

e

dedrrrE

f0 0

cos2 cos2)(1

)(

kr

kr

ikr

ee

ikr

e ikrikrikr sin2cos

0cos

cos

r

re

drkr

krrr

Ef

0

2 sin)(4

1)(

For = 0, only k = 0 sinkr/kr = 1.

For n electrons in an atom

electronsn 0

2 sin)(4

1)(

r

re

drkr

krrr

Ef

ZdrrrE

fr

re

electronsn

0

2 )(41

)0( Number of electronsin the atom

equal to 1 boundelectrons

Tabulated

If one consider the absorption of the X-ray, the atomicform factor is expressed as a complex number. Thecalculation above is the real part of the factor. Theimaginary part (absorption) is also tabulated for differentmaterials and X-ray source.

Scattering by a unit cell: Next, we have to consider the diffracted beam from a group of atoms that can make up the crystal. Atoms representing unit cell crystal Diffracted beam from the cell diffracted beam from the crystal Find out the phase difference from a group of atoms representing a unit cell.

(h00)

A

B

C

N M

S R

1 13 3

2 2

a

The path difference between11 and 22 is NCM.

sin2 001122 hdNCM

haACdh / :indicesMillern From 00 The path difference between11 and 33 is SBR.

a

hx

ha

x

AC

ABSBR /1133

the phase difference between 11 and 33 is

a

hx

a

hx 22

1133

If the position of atom B is specified by the fractionalcoordinate (normalized to the magnitude of the basevector), u = x/a.

hua

hx 22

1133

Extending the argument to other two direction, the atomB has actual coordinates x y z or fractional coordinatesx/a y/b z/c (= u v w), respectively. The phase differencebetween the wave scattered by atom B and that scatteredBy atoms A at the origin for the hkl reflection:

)(2 lwkvhu If u v w is integer, no phase difference.

Waves differing in amplitude and phase may also beadded by representing them as vectors.

A1

A2

A3

13

2)2sin( 111 tAE

)2sin( 222 tAE

The analytical expression for a vectorrepresenting a wave A

iiAe 2AI

realAny scattered wave can be expressed inthe complex exponential form:

)(2 lwkvhuii feAe

The resultant wave scattered by all the atoms of the unitcell is called the structure factor (F). F is obtained bysimply adding together all the waves scattered by theindividual atoms. Assume the unit cell contains atoms 1,2, 3, …, N, with fractional coordinates u1v1w1, u2v2w2, …,uNvNwN.

)(2)(22

)(21

222111 NNN lwkvhuiN

lwkvhuilwkvhui efefefF

N

lwkvhuiNhkl

NNNefF1

)(2

F is, in general, a complex number. |F|: amplitude of theresultant wave in terms of the amplitude of the wavescattered by a single electron.

electron single aby scattered amplitude

cellunit a of atoms allby scattered amplitude F

Structure factor calculations:How to choose the groups of atoms to represent a unit cellof a structure? First, determine the number of atoms in theunit cell. Second, choose the representative atoms for acell properly (ranks of equipoints).

Example 1: Simple cubic There are 1 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1. Choose any one will have the same result.

ffeF lkhihkl )000(2

22fFhkl for all hkl

Example 2: Body centered cubic There are 2 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1. Another equipoints of rank 1 is ½ ½ ½. Two points to choose are 000 and ½ ½ ½.

)1( )()2

1

2

1

2

1(2)000(2 lkhilkhilkhi

hkl effefeF

fFhkl 2 when h+k+l is even22

4 fFhkl

0hklF when h+k+l is odd 02 hklF

Example 3: Face centered cubic There are 4 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1. Another equipoints of rank 3 is ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½. Four atoms chosen are 000, ½ ½ 0, ½ 0 ½, 0 ½ ½.

]1[ )()()(

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2)000(2

lhilkikhi

lkhilkhilkhilkhihkl

eeef

fefefefeF

fFhkl 4 when h, k, l is unmixed (all evens or all odds)22

16 fFhkl

0hklF when h, k, l is mixed 02 hklF

Example 4: Diamond Cubic There are 8 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1.

Another equipoints of rank 3 is ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½. The other equipoints of rank 4 is ¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾. Eight atoms chosen are 000, ½ ½ 0, ½ 0 ½, 0 ½ ½ (the same as FCC), ¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾.

)4

3

4

3

4

1(2)

4

3

4

1

4

3(2)

4

1

4

3

4

3(2)

4

1

4

1

4

1(2

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2)000(2

lkhilkhilkhilkhi

lkhilkhilkhilkhihkl

fefefefe

fefefefeF

)4

1

4

1

4

1(2

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2

1

1

lkhi

lkhilkhilkhi

hkl

e

eeefF

FCC structure factor

Two FCC shifted by ¼ ¼ ¼

)1(4 ifFhkl when h, k, l are all odd22

32 fFhkl fFhkl 8 when h, k, l are all even and h + k + l = 4n

2264 fFhkl

0)11(4 fFhkl when h, k, l are all even and

02 hklFh + k + l 4n

0hklF when h, k, l are mixed 02 hklF

Example 5: Close packed hexagonal cell There are two atoms in a unit cell The eight atoms in the corner are equipoints of rank 1. Choose 000 to represent it. The other point is 1/3 2/3 1/2. (000)

(001)

(100)

(010)

(110)

( 1/3 2/3 1/2)

equipoints

)2

1

3

2

3

1(2

)000(2lkhi

lkhihkl fefeF

Set [h + 2k]/3+ l/2 = g

)1( 2 ighkl efF

gfgfeefF igighkl 2222222

cos4)2cos22()1)(1(

)

23

2(cos4cos4 22222 lkh

fgfFhkl

h + 2k l

3m3m

3m13m1

evenoddevenodd

)

23

2(cos2 lkh 2

hklF

10

0.250.75

4f 2

0f 2

3f 2

Example 6: NaCl (two kinds of atoms), can be considered as 4 Na at 000 FCC structure + 4 Cl at ½ ½ ½ FCC structure

The structure factor:

)2

1

2

1

2

1(2

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2

1

lkhi

ClNa

lkhilkhilkhi

hkl

eff

eeeF

FCC structure factor

)(0 )( lkhiclNahkl effF when h, k, l mixed 0

2 hklF

)(4 clNahkl ffF when h, k, l all even22

)(16 ClNahkl ffF

)(4 clNahkl ffF when h, k, l all odd22

)(16 ClNahkl ffF

You can also think of the structure as simple FCC and the atom is replaced by a pair of atoms (1 Na + 1 Cl) with the new structure factor

)2

1

2

1

2

1(2 lkhi

ClNa eff

Multiplicity Factor: The following factors discussed are related to the diffraction from polycrystalline sample. The multiplicity factor, p, for hkl planes may be defined as the number of permutations of position and sign of h, k, l. It represents the number of equivalent planes to yield the same diffraction cone. E.g. d100, d010, d001, d-100, d0-10, d00-1 form part of the same diffraction cone for cubic crystal p = 6. If the crystal is tetragonal, d100, d010, d-100, d0-10, form part of the same diffraction cone p = 4; d001, d00-1 p = 2 Lorentz factor: Certain trigonometrical factors, including: Angular distribution of the diffraction (finite spreading of the intensity peak, fraction of crystal contributing to a

2B

Inte

nsit

y

Diffraction Angle 2

Imax

Imax/2

2

IntegratedIntensity

B

diffraction peak, and the intensity spreading in a cone. Integrated intensity: (finite spreading) from constructive interference at Bragg angle to complete destructive interference transition region

B2

1

2 B1

B2

1 2 1 2

A Ba

1 2

path difference for 11-22 = AD – CB = acos2 - acos1

= a[cos(B-) - cos (B+)] = 2asin()sinB ~ 2a sinB.

C D

Na

1N

2Na sinB = completely cancellation (1- N/2, 2- (N/2+1) …)

BNa sin2

Maximum angular range of the peak

Imax 1/sinB, Half maximum B 1/cosB (will be shown later) integrated intensity of the reflection ImaxB (1/sinB)(1/cosB) 1/sin2B.

The second geometrical factor arises from the fact that the integrated intensity depends on the number of crystals orientated at or near the Bragg angle. estimated the fraction of the crystal around the angle. See Fig. 4-16.

crystal plane

rrN B )90sin(2

2

cos

4

)90sin(22

BB

r

rr

N

N

Fraction of crystal:

In accessing relative intensity, one should compare the total diffracted energies in different cone. See, Fig. 4-17. For different cones, the radius of the diffracted energy intersecting the film (Hull/Debye- Scherrer film) is different. Assume the total diffracted energy is equally distributed in the cone (2Rsin2B) the relative intensity per unit length 1/sin2B. Lorentz factor

cossin4

1

2sin

cos

2sin

1cos

2sin

1factor Lorentz

2

2

BBB

B

Lorentz–polarization factor: (omitting constant)

cossin

2cos1factor on polarizati-Lorentz

2

2

Absorption factor: The factor accounts for the X-ray being absorbed during its in and out of the sample. Two geometries are considered. Geometry for Hull/Debye-Scherrer Camera: difficult to calculate, different path different absorption, see Fig. 4-19. In general, the absorption factor is written as A(). A() varies with , qualitatively, A() as . Geometry for Diffractometer:

l

dx

I0

2

A

B

CdID

x

Incident I0, is 1cm2 in cross section, and is incident on the powder plate at an angle . The beam incident on the plate is .

)(0

ABeI

: linear absorption coefficient

Let a: volume fraction of the specimen containing particles having

1cm

the correct orientation for diffraction of the incident beam; b: the fraction of the incident energy which is diffracted by one unit volume. volume = l dx 1cm = ldx. actual diffracted volume = abldx diffracted beam diffracted beam absorbed before escaping from the sample

dxeablI AB)(0

dxeeablI BCAB )()(0

sin;

sin;

sin

1 xBC

xABl

dxeabI

dIx

D

sin

1

sin

1

0

sin

If = = dxeabI

dI xD

sin/20

sin

2sin

2

20

0

sin

20

0

abIxde

abIdII

x

x

xx

x DD

a varies with , being accounted for in the Lorentz factor already. The case discussed is for infinite thickness (defined in the textbook, dID(x = 0)/dID(x = t) = 1000 and = = ). Temperature factor: Also known as Debye Waller factor. Atoms in lattice vibrate (Debye model). Temperature (1) lattice constants 2 ; (2) Intensity of diffracted lines ; (3) Intensity of the background scattering .

d

u

d

u

high B

low B

Lattice vibration is more important at high B (u/d). High B

decrease more! Introducing the temperature factor e-2M.

e-2M

sin /

1

Formally, the factor is included in f as Because F = |f 2| factor e-2M shows up What is M?

Meff 0

2222

2

22 sinsin2

22

BB Bu

d

uM

u

0u

02 u

: Mean square displacement in the direction normalto the diffraction plane, difficult to calculate, Debyehas given the following expression:

2u

2

2

2 sin

4)(

6

Bx

xmk

ThM

h: Plank’s constant; T: absolute temperature; m: mass ofvibrating atom; k: Boltzmann’s constant; : Debyetemperature of the substance; x = /T; (x): tabulatedfunction (appendix 15)

2

4

2

2 1015.16

A

T

mk

ThPut all numbers in and change m to atomic weight (A):

Temperature diffuse scattering: general coherent scattering in all directions by the displacement (u) contributing to the general background of the pattern; as Temperature effect on the diffraction intensity, see Fig. 4-22. The diffraction peak width B slightly as T

Summary of the intensities of diffraction peaks from polycrystalline samples: Hull/Debye-Scherrer Camera:

MeApFI 22

22

)(cossin

2cos1

Diffractometer:

MepFI 22

22

cossin

2cos1

A() independent of .

Effects that make the above intensity equation invalid: (1) Preferred orientation: like films, wire, not true random orientation of the crystal Lorentz factor depends on how the crystals orientated (2) Extinction: for crystals with small sizes (10-4-10-5

cm) kinematical theory: intensity is the sum of all diffraction planes

For large perfect crystal dynamical theory is required; takes into account the multiple reflections between diffracted planes (absorption also increases) Actually to match the intensity calculation with the measured intensity is difficult. But, at least the magnitude (very strong, strong, weak) is qualitatively matched.