Download - VIII. Kinematical Theory of Diffraction
VIII. Kinematical Theory of Diffraction8-1. Total Scattering Amplitude
The path differencebetween beamsscattered from thevolume elementapart is
r
rkkrkrk )(2)ˆˆ( ''
The amplitude of the wave scattered from avolume element is proportional to the localelectron concentration ! Why?)(rn
kk
ˆ2
)ˆˆ( ' rkrk
Total scattering amplitude F
dVernrdernF rkki
V
rkki )()( ''
)()(
Define kkk'
dVernF rki )(
)(rn : periodic function in a crystalA periodic function can be expanded byFourier series!Any criteria to expand it?
Crystal translation vector)(rn )( Trn
Physical properties function of a crystal
Crystal: periodic
Periodic function Exponential Fourier Series
k
rkik enrn
2)( k
Tkirkik eenTrn
22)(
12 Tkie
cwbvauT
Translation vectors of the original crystal lattice
1 2 Tkie
for all
u, v, w: integer
*** clbkahk If
Vectors of the reciprocal latticeh, k, l: integer
T
)()( *** cwbvauclbkahTk
)( lwkvhu always integer
1 2 Tkie
Therefore, is not arbitrary! is reciprocallattice, i.e.
k
k
G
G
rGiGenrn
2)(
dVeendVernF rki
G
rGiG
rki
2)(
G
rkGiG
G
rkirGiG dVendVeen
)2(2
G
rkGiG dVenF
)2(
(i) when Gk
2
22 kk VndVnF
*
2222
kk nnVFI
(ii) when Gk
2
G
rkGiG
G
rkGiG dVendVenF
)2()2(
)2( Gk
G
G GknF
)2(
For an infinite crystal
;
Therefore, diffraction occurs at Gk
2
The total scattering amplitude F
dVerndVernF rGirki 2)()(
GrGi NSdVernN
cellunit
2)(
structure factor
For a unit cell, total electron concentration atdue to all atoms in the unit cellr
S
jjj rrnrn
1
)()(
cellunit
2)( dVernS rGiG
dVerrn rGiS
jjj
2
1
)(
S
j
rGijj dVerrn
1
2)(
S
j
rGirrGijj dVeerrn jj
1
2)(2)(
S
j
rrGijj
rGiG dVerrneS jj
1
)(22 )(
jf atomic form factor
S
jj
rGiG feS j
1
2
The scattering amplitude is then expressed as
S
jj
rGiG feNNSF j
1
2
8-2. Atomic form factor calculationThe meaning of form factor is equivalentto the total charge of an atom, which can beobtained by a direct calculation.
With the integral extended over the electronconcentration associated with a singleatom, set the origin at the atom
dVerndVerrnf rGij
rrGijjj
j 2)(2 )()(
Spherical integration dV = dr(rd) (rsind)
http://pleasemakeanote.blogspot.tw/2010/02/9-derivation-of-continuity-equation-in.html
r: 0 - : 0 - : 0 - 2
dVerndVerrnf rGij
rrGijjj
j 2)(2 )()(
0 0
2
0
2 )sin)(()(r
rGijj drrddrernf
rr
ddrrdrrddr
0
2
0
2
0
sin2)sin)((
3
34 r = 2
3
3r
0 0
2
0
cos22 sin)(r
iGrjj ddrderrnf
)()( rnrn jj cosGrrG
assume
rG
0 0
2
0
cos22 sin)(r
iGrjj ddrderrnf
0 0
cos22 sin)(2r
iGrjj dedrrrnf
0
cos2
0
cos2 )cos(sin dede iGriGr
0
cos2 )cos2(2
1 iGrdeiGr
iGr
iGree iGriGr
2
0cos2cos2
GrGr
)2sin(
0
2
2)2sin(2)(2
rjj Gr
Grdrrrnf
0
2
2)2sin()(4
rjj Gr
Grdrrrnf
depends on Gjfsin2* hklG
(chapter 7)
0When 0G 12
)2sin(lim0
Gr
Gr
zdrrrnfr
jj
0
2)(4 total number of
electron in an atom
0 Tabulatedjf Function of element and diffracted angle
Example:
Si, 400 diffraction peak, with Cu K (0.1542 nm)
nm 13577.04/54309.0400 d6.341542.0sin2 400 d
1
368.0sin
sin 0.2
9.67
526.7368.03.04.03.0
22.820.722.8
f
f
0 0.114.0 12.16
0.5 0.66.24 5.31
0.3 0.48.22 7.20
d21sin
or
8-3. Structure Factor Calculation
S
j
cwbvauclbkahij
S
j
rGijG
jjjj efefS1
)()(2
1
2 ***
S
j
lwkvhuijG
jjjefS1
)(2
(a) Primitive cell 1 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; Choose any one will have the same result!
ffeS lkhiG )000(2 for all hkl
(b) Base centered cell 2 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; ½ ½ 0, ½ ½ 1 : equipoints of rank 1; Two points chosen: 000 and ½ ½ 0.
S
jj
lwkvhuiG feS jjj
1
)(2
fSG 2
0GS
If h , k is unmixed or h + k is even
If h , k is mixed or h + k is odd
)021
21(2)000(2 lkhilkhi
G fefeS
]1[ )( khief
The meaning of the reflection condition(reflection rule)
Suppose that we have a square lattice
(a) Primitive unit cell: one atom at [00]
ffeS khiG )00(2
1
1
)(2
jj
kvhuiG feS jj
(b) Unit cell: two atoms at [00] [0]
1
1
)(2
jj
kvhuiG feS jj
fefefeS ihkhikhiG )1(
)021(2)00(2
fSG 2 If h is even0GS If h is odd
The reflection conditions remove the latticepoints [odd, 0] in the reciprocal lattice.
Therefore, the meaning of the reflection rule isto remove the additional lattice points due to theselection of unit cell. After the substraction, the reciprocal latticestructures derived from both cases (primitiveunit cell and unit cell) become the same.
(c) Body centered cell 2 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; ½ ½ ½: equipoints of rank 1; Two points to choose: 000 and ½ ½ ½.
S
j
lwkvhuijG
jjjefS1
)(2
fSG 2
0GS
If h + k + l is even
If h + k + l is odd
)21
21
21(2)000(2 lkhilkhi
G fefeS
]1[ )( lkhief
(d) Face centered cell 4 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½: : equipoints of rank 3; Four points chosen: 000, ½½0, ½0½, 0½½
)21
210(2)
210
21(2
)021
21(2)000(2
lkhilkhi
lkhilkhiG
fefe
fefeS
]1[ )()()( lkilhikhiG eeefS
If h, k, l unmixed
If h, k, l mixed
fSG 4
0 GS
(e) close-packed hexagonal cell 2 atoms/unit cell 8 corner atoms: equipoints of rank 1; 2/3 1/3 1/2: equipoints of rank 1; Choose 000, 2/3 1/3 1/2. (Miller Indices)
)21
31
32(2)000(2 lkhilkhi
G fefeS
]1[)
232(2)
232(2 lkhilkhi
effef
]1][1[)
21
31
32(2)
21
31
32(222 lkhilkhi
G eefSI
]1][1[)
21
31
32(2)
21
31
32(22 lkhilkhi
eefI
]11[)
21
31
32(2)
21
31
32(22
lkhilkhieef
)]}21
31
32(2cos[22{2 lkhf
)]23
2([cos4 22 lkhf
2h + k l
3m3m
3m13m1
oddevenoddeven
)23
2(cos2 lkh
01
0.750.25
04f 2
3f 2
f 2
2
GS
(f) ZnS: four Zinc and four sulfur atoms per unit cell (8 atoms/unit cell)
000 and 100, 010, 001, 110, 101, 011, 111:equipoints of rank 1; (S atom)½½0, ½0½, 0½½, ½½1, ½1½, 1½½:equipoints of rank 3; (S atom)¼¼¼, ¾¾¼, ¾¼¾, ¼¾¾:equipoints of rank 4; (Zn)
Eight atoms chosen: 000, ½½0, ½0½,0½½ (the same as FCC), (Say S atom)¼¼¼, ¾¾¼, ¾¼¾, ¼¾¾ (Say Zn atom)!
)43
43
41(2)
43
41
43(2
)41
43
43(2)
41
41
41(2
)21
210(2)
210
21(2
)021
21(2)000(2
lkhi
Zn
lkhi
Zn
lkhi
Zn
lkhi
Zn
lkhi
S
lkhi
S
lkhi
Slkhi
SG
efef
efef
efef
efefS
]1[
][)()()(
)41
41
41(2)000(2
lkiS
lhikhi
lkhi
Znlkhi
SG
efee
efefS
Structure factor of face centered cell
fcc
lkhi
ZnSG SeffS
][)
21
21
21(
h, k, l: mixed 0fccS 0 GS
h, k, l: unmixed 4fccS
][4)
21
21
21( lkhi
ZnSG effS
when h, k, l are all odd ][4 ZnSG iffS
when h, k, l are all even and h + k + l = 4n ][4 ZnSG ffS
when h, k, l are all even and h + k + l 4n ][4 ZnSG ffS
]][[)
21
21
21()
21
21
21(22 lkhi
ZnS
lkhi
ZnSfccG effeffSS
]
[
)21
21
21(
)21
21
21(2222
lkhi
ZnS
lkhi
ZnSZnSfccG
eff
effffSS
)](2
cos2[ 2222lkhffffSS ZnSZnSfccG
)(16 222
ZnSG ffS
22)(16 ZnSG ffS
22)(16 ZnSG ffS
h+k+l : oddh+k+l :odd multiple of 2.h+k+l: even multiple of 2
Or
8-4. Shape EffectFor a finite crystal, assuming rectangularvolume
G
rGkiG dVenF
)2(
For a particular G
, if 02 DGk
dVenF rDiG
1
0
1
0
1
0
)(1
0
1
0
1
0
M
x
N
y
L
z
czbyaxDiG
M
x
N
y
L
z
rDiG enenF
1
0
1
0
1
0
M
x
N
y
L
z
czDibyDiaxDiG eeen
cDi
cDiL
bDi
bDiN
aDi
aDiM
G ee
ee
eenF
11
11
11
2/2/
2/2/
2/
2/
11
aDiaDi
aDiMaDiM
aDi
aDiM
aDi
aDiM
eeee
ee
ee
)2/sin(2)2/sin(2
2/
2/
aDiaDMi
ee
aDi
aDiM
)2/sin()2/sin(
)2/sin()2/sin(
)2/sin()2/sin(
2/)1(2/)1(2/)1(
cDcDL
bDbDN
aDaDM
eeenF cDLibDNiaDMiG
2FI 222
*
)2/sin()2/sin(
)2/sin()2/sin(
)2/sin()2/sin(
cDcDL
bDbDN
aDaDMnn GG
222*
)sin()sin(
)sin()sin(
)sin()sin(
LNMnn GG
2;
2;
2cDbDaD
where
GkD
2
You should already familiar with the functionof 2
)sin()sin(
M
Chapter 4
222*
)sin()sin(
)sin()sin(
)sin()sin(
LNMnnI GG
Consider
1st min occurs at0)sin(;0)sin(;0)sin( LNM
LNM ;;i.e.
GkDcDbDaD 2;
2;
2;
2
MaGk 2)2(
LcGk 2)2(
NbGk 2)2(
MaGk 1)
2(
LcGk 1)
2(
NbGk 1)
2(
or
Therefore, for a finite crystal, the diffractedintensity is finite based on the condition below.
I 0 if
MaGk 1)
2(
LcGk 1)
2(
NbGk 1)
2(
Example: for a very thin sample
Ewald sphere construction
2;
2
'' kSkS
When G
diffraction occurs dueto “shape effect”.
𝑆 ′ −𝑆