Theory of kinematical diffraction: a practical recap by Carmelo Giacovazzo

Istituto di Cristallografia , CNR Dipartimento Geomineralogico, Bari University

The Thomson scattering model A X-ray plane wave propagates along the x direction, the scatterer is in O. Q is the observation point, in the plane (x,y).

F= e E ; a= F/m

Ieth = Ii sin2 [e4/(m2r2c4)]

is the angle between the acceleration direction of the electron and the observation direction.

If the beam is non-polarised, we can divide the overall incident beam into two components with the same intensity( = 0.5Ii), the first polarized along the y axis , the second along the z axis.

For the first component φ=(π/2-2θ) and Iey= 0.5Ii [e4/(m2r2c4)] cos22.

For the second component φ=π/2 and Iez =0.5 Ii [e4/(m2r2c4)]

Therefore IeTh =Ii [e4/(m2r2c4)] [(1+cos22)/2]

X-Ray versus e-diffraction

-- e-scattering is caused by interaction with the electrostatic field ( sum of nucleus and of electron cloud); --e-scattering is weakly dependent on the atomic number ( light atom localization) -- e- absorption is strong; --e-interaction is thousands stronger than X-ray interaction: nanocrystals, dynamical effects

The Compton effectThe corpuscolar interpretation of the

quantomechanics lead Compton to consider an energy transfer from the photon to the electron.

The radiation is incoherent ( no phase relationship between the incident and the diffuse beam). That induces a wavelength change equal to

=0.024 (1-cos2)

The interference

We will investigate the interference mechanism which governs the diffusion from an extended body.

We will start with two charged particles, one in O and the other in O’, both interacting with a X-ray beam.

4

AO = -r.so; BO= r.s ; AO+BO = r.(s-so)

(AO+BO)/= r.(s-so)/ = r.r* where r*= (s-so)/= 2 r*.r r*=2sin / F(r*) = Ao + Ao’ exp(2 ir*.r)

F(r*) = j Aj exp(2ir*.rj) = j fj exp(2ir*.rj)

where f2 = I/Ieth

If the scatterer distribution is a continuoum then

You can easily recognize that the amplityde F(r*) is the Fourier transform of the electron density.

Viceversa

The above formulas are quite general: they will be applied to the cases in which (r) represents the electron density of an electron, of an atom, of a molecule, of the unit cell, of the full crystal.

r.r*rrr diFS

2exp*

* 2exp**

r.r*rrr diFS

The scattering amplitude of an electron

An example : the atom C (2x1s;2x2s;2x2p) (e)1s = (c1

3/ ) exp(-2c1r);

(e)2s= c25/(96 ) r2 exp(-c2r)

Icoe +Iincoe =IeTh

Iincoe=IeTh- Icoe=IeTh-IeTh fe2

=IeTh(1-fe2)

6

r.r*rrr

rr

difS ee

e

2exp*

2

rr

rrr

factor scatteringAtomic The

aa

Z

jj

Z

jeja

Tf

*

2

11

Icoe=IeTh fa2

=IeTh(Σ j=1,..,Z fei)2

Iincoe=IeTh (Σ j=1,..,Z [1-fej

2])

The atomic thermal factor a(r) is the electron density for an atom isolated, located at the origin. Owing to the thermal agitation, the nucleus moves at the instant t into r’: then the electron density will be a(r-r’). Let p(r’) the probability of that movement: then

)(8

)/sinexp(

)/sin8exp(*2exp*

' 2/'exp2

1'

*.*'2exp'**

'*'

22

22

22222

222/1

'

'

AUBwhere

B

UUrrq

rUwhereUrU

rp

thatnowSuppose

qfdipff

pdp

S aaat

aS aat

rrr.r'*rrrr

rrr'r'r'rr

Scattering amplitude of a molecule Let j(r) be the electron density for an atom isolated, thermically agitated and located at the origin. Then (r-rj) will be the electron density of the same atom when its nucleus is in rj. Let M(r ) be the electron density of a molecule:

.

N

jjj

j

N

jS jjjjM

jj

S

N

jjjM

j

N

jjM

if

diF

TheniablesvarofchangetheroduceintusLet

diF

1

1

1

1

2exp*

.* 2exp*

.

2exp*

)()(

.r*rr

RRrrRr

Rrr

r.r*rrrr

rrr

Scattering amplitude of an infinite crystal

As we know the electron density of a crystal may be expressed as a convolution:

* * *

1

.*1

*

,,

,,

cbar

rrH

rrrr

rr

*H

*H

*H

*H

*

lkh

where

FV

F

FV

T

thenisamplitudescatteringThe

L

lkhM

lkhM

M

r

Scattering amplitude of a finite crystal

A finite crystal may be represented as the product

where (r) is the function form which is equal to 1 inside the volume of the crystal, equal to 0 outside. Then

rrr .cr

r.rrr

r.rrrrr

rrH

rrr

*

**

*H

*

di

diTD

where

DFV

TTF

S

HM

2exp

2exp)(

1

**

The structure factor

HHHHHH

H

H

HH

*HH

XH

XH

.rr

ABtanawithiFF

fB

fA

where

iBAif

ifF

N

jjj

N

jjj

N

jjj

N

jjj

/ exp

2sin

2cos

2exp

2exp

1

1

1

1

XH

The Ewald sphere

OP=r*H =1/dH = IO sin = 2sin /

The Bragg law

2dH sin = n

It may be written as2(dH/n ) sin = or also 2dH’ sin = where H’= (nh, nk, nl).Saying r*= r*H is equivalent to state the Bragg law:r* = 2sin / =1/dH

The symmetry in the reciprocal space

The Friedel LawFH = AH + iBH; F-H = AH - iBH

-H = - H

IH (AH + iBH )(AH - iBH) = AH2 +BH

2

I-H (AH - iBH )(AH + iBH) = AH2 +BH

2

The intensities show an inversion centre.

The effect of a symmetry operator in the reciprocal spaceThe general expression of the structure factor is

Let C (R,T) be a symmetry operator. Consider the quantity

N

jjj ifF

1

) 2exp( XHH

TH

TH

RXH

THRXHTH

HRH

HRH

HRH

H

RH

2

||||

2exp

)( 2exp

2exp 2exp 2exp

1

1

FF

iFF

FTif

iifiF

j

N

jj

j

N

jj

Find the Laue groupApply the Friedel law and the symmetry induced law | FHR| =| FH|to find the Laue groups.

---------------------------------------------------- Laue groupP1 | Fh k l |= |F-h -k -l| -1P-1 “ “-----------------------------------------------------------P2 |Fhkl| =| F-h k -l |= |F-h-k-l |=| Fh -k l| 2/mPm ‘’ ‘’ ‘’ ‘’P2/m-------------------------------------------------------------P222 | Fhkl| = | Fh -k -l| = |F-h k -l| = |F-h -k l| 2/m 2/m 2/m = |F-h -k -l| = |F-h k l| = | Fh -k l| = |Fh k -l|

Find the systematically absent reflections

Find the set of reflections for which

Once you have found it , check if the product HT is not an integral value. In this case the ( true) equation

is violated, unless |FH| = 0. The reflection is systematically absent.

HRH

THHRH iFF 2exp

Space group determination: the single crystal scheme

a) First step : lattice analysis, which provides the Unit Cell parameters. E.g.

a = 11.10(2), b = 14.27(3), c = 9.72(2) = 90.25(37), = 89.48(30), = 90.32(35)

Is the unit cell orthorhombic?

b) Second step: symmetry analysis In the example above, three two-fold axes should

be present to confirm the orthorhombic nature of the crystal.

The symmetry analysis

1) First step: identification of the Laue group via the recognition of the

“ symmetry equivalent reflections”.

The Laue group generally identifies without any doubt the crystal system ( pseudo-symmetries, twins etc. are not considered).

b) Second step: identification of the systematically absent reflections.

The Laue GroupSymmetry operators : Then

are symmetry equivalent positions in direct space. In the reciprocal space we will observe the symmetry

equivalent reflections

If the space group is n.cs. , the Friedel opposites should be added: the complete set is

m1,...,s ),,( sss TRC

sjsjs TrRr

m1,...,s , s Rh

m1,...,s , s Rh m1,...,s , s Rh

The Laue group

100

010

001

1 R

m

2

m

2

m

2

100

010

001

2 R100

010

001

3 R

100

010

001

4 R

),lkh(l),kh(),lk(h(hkl), 4321 RhRhRhRh

l)k(h),-l(hkkl),-h(),-lkh( 4321 RhRhRhRh

The equivalent reflections are

The systematically absent reflections

Since

sFFS

hThRhhhR 2 ,s,

(1) )2exp( siFFS

hThhR

we have

If hRs =h and hTs=n the relation (1) is violated unless the reflection h is a systematically absent reflection.

The combination of the information on the Laue Group with that on the systematically absent reflections allows the determination of the Extinction Symbol.

The extinction symbols (ES)

In the first position: cell Centric type ( e.g., P - - a , in orth.)

Then the reflection conditions for each symmetry direction are given.

Symmetry directions not having conditions are represented by a dash.

A symmetry direction with conditions is represented by the symbol of the screw axis or glide plane.

Table of ES are given in IT per crystal system. There are 14 ES for monoclinic, 111 for orthorhombic, 31 for tetragonal, 12 for trigonal-hexagonal, 18 for cubic system

Extinction symbol and compatible space groups

• ES

P - - - P222, Pmmm, Pm2m, P2mm, Pmm2(in orth.)

P - - a Pm2a, P21ma, P mma(in orth.)

P - - - P4, P-4, P4/m,(in tetrag.) P422, P4mm, P-42m, P-4m2 P4/mmm

P61-- P61 P65

P6122 P6522

Find the reflections with symmetry restricted phase values

Look for the set of reflections for which

Since HR = H - 2 HT

for that set of reflections it will be -H = H -2 HT, or 2H=2 HTor H = HT+ n

HRH

Some exercises

Find the rotation and translation matrices for the space group P41, and P31;

Find the equivalent reflections for the same space groups;

Find the systematically absent reflections for the space group P21/c, P41,P31, P212121

The diffraction intensities

IH = K1 K2 I0 L P T E |FH|2

where

I0 is the intensity of the direct beam, IH is the integrated intensity of the reflection H,K1 = e4/(m2c4)K2 = 3 /VP polarisation factor (e.g., [1+cos22]/2L Lorentz factorT Transmission fctor ( e.g., I/Ioss)

Why integrated intensity?The diffraction conditions are verified on a domain owing to:a) the finitness of the crystalb)non vanishing divergence of the incident radiation;c) non-monochromatic incident radiation;d) crystal defects

Electron density calculation

*

** 2expS

diF r.rrrr *

lkhhkl iF

V ,,

2exp1

XH

) 2(cos1

) 2(exp

) 2(exp1

0

0

XH

XH

H

HH

HH

Hlkh

lkh

FV

iF

iFV

XH

Electron density map and finitness of the diffraction data

The limiting sphere (and/or other factors) limits the number of measurable diffraction intensities. Let (r*) be the form function of the measured reciprocal space( equal to 1 in the volume containing the measured reflections, equal to 0 elsewere).

***

*

*

*

.

rrrr

rr

rr*

TFT

ThenavailableisFOnly

FT

r

Code SG ASU COLL RES/NRES COMPRES Rint

aker a P -4 21 m

O14 Mg2 Al6 Si4 Ca4 P 0.40/339 23 8.8

anatase b

I 41/a m d O8 Ti4 P 0.25/267 22 13.8

barite c P n m a O16 S4 Ba4 P-A 0.77/355 82 15.3charoite d

P 21/m Ca24 K14 Na10 Si72 O186 P-A 1.18/2878

97 13.3

cnba e P 21/c C29 N H17 A 0.76/3975

83 21.9

gann f P 63 m c Ga2 N2 P 0.25/170 32 4.5mayenite g I -4 3 d O88 Al28 Ca24 P 0.54/228 44 21.5mfu41 h F m -3 m

Zn2 Cl N2 C3 O P-A 1.09/655 96 29.2

mullite i P b a m Al4.56 Si1.44 O9.72 P-A 0.76/213 86 30.9natio j C 2/m O3 Ti1.5 Na0.5 P-A 0.76/517 73 14.2natrolite k F d d 2 Na2(Al2 Si3O10) (H2O)2 P-A 0.75/743 99 20.6srtio3 l P m -3

m O3 Ti1 Sr1 P 0.26/180 43 18.8

srtio3_s m P b n m

O12 Ca4 Ti4 P 0.26/464 6 13.8

zn8sb7 n P -1 Zn32 Sb28 P-A 0.77/3943

54 17.0

Code ORDL EXT ORD7 ORD8aker 1 P -4 21 m 3 3anatase 1 I 41/a m d 1 1barite 1 P n m a 1 1charoite 1 P 21/m 1 1cnba 1 P 21/c 1 1gann 1 P 63 m c 1 1mayenite 1 I -4 3 d 1 2mfu41 1 F m -3 m 1 >100mullite 1 P b a m 3 3natio 1 C 2/m 1 1natrolite 1 F d d 2 1 1srtio3 1 P m -3 m 3 1srtio3_s 1 P b n m 60 36zn8sb7 4 P -1 - -