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VIII. Kinematical Theory of Diffraction
8-1. 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 crystal
A periodic function can be expanded byFourier series!
Any criteria to expand it?
Crystal translation vector
)(rn
)( Trn
Physical properties function of a crystalCrystal: periodic
Periodic function Exponential Fourier Series
k
rki
kenrn
2)( k
Tkirki
keenTrn
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 lattice
h, 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 kkVndVnF
*
22
22
kknnVFI
(ii) when Gk
2
G
rkGiG
G
rkGiG
dVendVenF
)2()2(
)2( Gk
G
GGknF
)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 at
due 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
rGi
GdVerrneS jj
1
)(22 )(
jf atomic form factor
S
jj
rGi
GfeS j
1
2
The scattering amplitude is then expressed as
S
jj
rGi
GfeNNSF j
1
2
8-2. Atomic form factor calculation
The 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
3
4r
= 2
3
3r
0 0
2
0
cos22 sin)(r
iGrjj ddrderrnf
)()( rnrn jj cosGrrG
assume
r
G
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
1iGrde
iGriGr
iGr
ee iGriGr
2
0cos2cos2
Gr
Gr
)2sin(
0
2
2
)2sin(2)(2
r
jj Gr
Grdrrrnf
0
2
2
)2sin()(4
r
jj Gr
Grdrrrnf
depends on Gjfsin
2* hklG
(chapter 7)
0When 0G 12
)2sin(lim
0
Gr
Gr
zdrrrnfr
jj
0
2)(4 total number of
electron in an atom
0 Tabulated
jf Function of element and diffracted angle
Example:
Si, 400 diffraction peak, with Cu K (0.1542 nm)
nm 13577.04/54309.0400 d
6.341542.0sin2 400 d1
368.0sin
sin
0.29.67
526.7368.03.0
4.03.0
22.8
20.722.8
ff
0 0.114.0 12.16
0.5 0.66.24 5.31
0.3 0.48.22 7.20
d2
1sin
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
lwkvhui
GfeS jjj
1
)(2
fSG
2
0G
S
If h , k is unmixed or h + k is even
If h , k is mixed or h + k is odd
)02
1
2
1(2)000(2 lkhilkhi
GfefeS
]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
kvhui
GfeS jj
(b) Unit cell： two atoms at [00] [0]
1
1
)(2
jj
kvhui
GfeS jj
fefefeS ihkhikhiG
)1()0
2
1(2)00(2
fSG
2 If h is even
0G
S 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
0G
S
If h + k + l is even
If h + k + l is odd
)2
1
2
1
2
1(2)000(2 lkhilkhi
GfefeS
]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
)2
1
2
10(2)
2
10
2
1(2
)02
1
2
1(2)000(2
lkhilkhi
lkhilkhi
G
fefe
fefeS
]1[ )()()( lkilhikhi
GeeefS
If h, k, l unmixed
If h, k, l mixed
fSG
4
0G
S
(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)
)2
1
3
1
3
2(2
)000(2lkhi
lkhi
GfefeS
]1[)
23
2(2)
23
2(2
lkhi
lkhi
effef
]1][1[)
2
1
3
1
3
2(2)
2
1
3
1
3
2(2
22 lkhilkhi
GeefSI
]1][1[)
2
1
3
1
3
2(2)
2
1
3
1
3
2(2
2lkhilkhi
eefI
]11[)
2
1
3
1
3
2(2)
2
1
3
1
3
2(2
2 lkhilkhi
eef
)]}2
1
3
1
3
2(2cos[22{2 lkhf
)]23
2([cos4 22 lkh
f
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)!
)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
)02
1
2
1(2)000(2
lkhi
Zn
lkhi
Zn
lkhi
Zn
lkhi
Zn
lkhi
S
lkhi
S
lkhi
Slkhi
SG
efef
efef
efef
efefS
]1[
][)()()(
)4
1
4
1
4
1(2)000(2
lkiS
lhikhi
lkhi
Znlkhi
SG
efee
efefS
Structure factor of face centered cell
fcc
lkhi
ZnSGSeffS
][
)2
1
2
1
2
1(
h, k, l: mixed 0fccS 0G
S
h, k, l: unmixed 4fccS
][4)
2
1
2
1
2
1( lkhi
ZnSGeffS
when h, k, l are all odd ][4 ZnSGiffS
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 ZnSGffS
]][[)
2
1
2
1
2
1()
2
1
2
1
2
1(22 lkhi
ZnS
lkhi
ZnSfccGeffeffSS
]
[
)2
1
2
1
2
1(
)2
1
2
1
2
1(2222
lkhi
ZnS
lkhi
ZnSZnSfccG
eff
effffSS
)](2
cos2[ 2222lkhffffSS ZnSZnSfccG
)(16 222
ZnSGffS
22)(16 ZnSG
ffS
22)(16 ZnSG
ffS
h+k+l : odd
h+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 e
e
e
e
e
enF
1
1
1
1
1
1
2/2/
2/2/
2/
2/
1
1aDiaDi
aDiMaDiM
aDi
aDiM
aDi
aDiM
ee
ee
e
e
e
e
)2/sin(2
)2/sin(22/
2/
aDi
aDMi
e
eaDi
aDiM
)2/sin(
)2/sin(
)2/sin(
)2/sin(
)2/sin(
)2/sin(
2/)1(2/)1(2/)1(
cD
cDL
bD
bDN
aD
aDM
eeenF cDLibDNiaDMiG
2FI
222
*
)2/sin(
)2/sin(
)2/sin(
)2/sin(
)2/sin(
)2/sin(
cD
cDL
bD
bDN
aD
aDMnn
GG
222
*
)sin(
)sin(
)sin(
)sin(
)sin(
)sin(
LNM
nnGG
2;
2;
2
cDbDaD
where
GkD
2
You should already familiar with the functionof 2
)sin(
)sin(
M
Chapter 4
222
*
)sin(
)sin(
)sin(
)sin(
)sin(
)sin(
LNM
nnIGGConsider
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(
MaG
k 1)
2(
LcG
k 1)
2(
NbG
k 1)
2(
or
Therefore, for a finite crystal, the diffractedintensity is finite based on the condition below.
I 0 if
MaG
k 1)
2(
LcG
k 1)
2(
NbG
k 1)
2(
Example: for a very thin sample
Ewald sphere construction
2;
2
'' k
Sk
S
When G
diffraction occurs dueto “shape effect”.
𝑆 ′ −𝑆
