theory of kinematical di ffraction: a practical recap by carmelo giacovazzo istituto di...
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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 - -