Transcript Chapter 8

VIII. Kinematical Theory of Diffraction
8-1. Total Scattering Amplitude
The path difference
between beams
scattered from the

volume element r
 ˆ' 
ˆ
apart is ( k  r  k  r )
 



2

'
'
ˆ
ˆ
(k  r  k  r )
 (k  k )  r

2 ˆ 
k k

The amplitude of the wave scattered from a
volume element is proportional to the local

electron concentration n ( r ) ! Why?
Total scattering amplitude F
 i ( k  k ' )  r 
F   n (r )e
dr 
 i ( k  k ' )  r
dV
 n (r )e
V



'
Define k  k   k
  i  k  r
F   n (r )e
dV

n(r ) :
periodic function in a crystal
A periodic function can be expanded by
Fourier series!
Any criteria to expand it?
Crystal: periodic
 

Physical properties function of a crystal n ( r )  n ( r  T )
Crystal translation vector
Periodic function  Exponential Fourier Series

n(r ) 
n

k

k
e
 
2  ik r
 
 n(r  T ) 
e




T  ua  vb  w c
 
2  i k T
n

k

k
e
 
2  ik r
e
 
2  i k T
1
u, v, w: integer
Translation vectors of the original crystal lattice

 e
 1 for all T


*
*
*
If k  h a  k b  l c
 
2  i k T
h, k, l: integer
Vectors of the reciprocal lattice
 


*



*
*
 k  T  ( h a  k b  lc )  (u a  v b  w c )
 ( hu  kv  lw )
 e
 
2  i k T
always integer
1


Therefore, k is not arbitrary! k
lattice, i.e. G
 

2  iG r
n ( r )   n G e

 F   n (r )e

 n

G

G

G
 
 i k r
e
dV 
 
2  iG r
e
 n
 
 i k r

G
dV 

G
e
is reciprocal
 
2  iG r
 n

G
e

G
 
 i k r
e
dV
 

i ( 2  G   k )r
dV
F 

G

G
e
dV


 k  2 G
(i) when
F 
 n
 

i ( 2  G   k )r
n

k 2
dV  Vn

k 2
;
2
I  F
 V n  k
2
*

n
2 k


 k  2 G
(ii) when
For an infinite crystal
F 
F 
n

G


G

G
e
 

i ( 2  G   k )r
dV 

n e

G

G


n G  (  k  2  G )
Therefore, diffraction occurs at
 

i ( 2  G   k )r
dV



 (  k  2 G )


 k  2 G
2
The total scattering amplitude F
  i k r
F   n ( r )e
dV 
 N
  2  iG r
dV
 n ( r )e
  2  iG  r

n
(
r
)
e
dV

NS
G

unit cell
structure factor
For a unit cell, total electron concentration at

r
due to all atoms in the unit cell

n(r ) 
S

 
n j ( r  rj )
j 1
  2  iG  r
dV
 n ( r )e
S G 
unit cell
S


   2  iG  r
n j ( r  r j )e
dV
j 1
S


j 1
S


j 1
   2  iG  r
dV
 n j ( r  rj )e
   2  iG ( r  rj )  2  iG  rj
e
dV
 n j ( r  rj )e
S
S G 
e
 
 2  iG r j
j 1
   2  iG ( r  rj )
dV
 n j ( r  rj )e
 fj
S
 S G 
e
 
 2  iG r j
atomic form factor
fj
j 1
The scattering amplitude is then expressed as
S
F  NS G  N
e
j 1
 
 2  iG r j
fj
8-2. Atomic form factor calculation
The meaning of form factor is equivalent
to the total charge of an atom, which can be
obtained by a direct calculation.
With the integral extended over the electron
concentration associated with a single
atom, set the origin at the atom
   2  iG ( r  rj )
f j   n j ( r  rj )e
dV 
  2  iG r
dV
 n j ( r )e
Spherical integration dV = dr(rd) (rsind)
r: 0 - 
: 0 - 
: 0 - 2
http://pleasemakeanote.blogspot.tw/2010/02/9derivation-of-continuity-equation-in.html

 
2

 dr ( rd  )( r sin  d  )  2
r  0  0

4
 r dr
2
r
r

3
r
3
fj 
 0
3
=2
3
   2  iG ( r  rj )
f j   n j ( r  rj )e
dV 

 sin  d 

 
  2  iG r
dV
 n j ( r )e
2
  2  iG r
dr ( rd  )( r sin  d  )
 n j ( r )e
r0  0  0

fj 

 
2
 n j ( r ) r sin  e
2
 2  iGr cos 
drd  d 
r0  0  0
assume

n j (r )  n j (r)
 
G  r  Gr cos 

G


r


fj 
2
 n j ( r ) r sin  e
 
2
 2  iGr cos 
drd  d 
r0  0  0



f j  2
2
n j ( r ) r dr
 sin  e
 2  iGr cos 
d
 0
r0


 sin  e
 2  iGr cos 
d 
 0


1
2  iGr
e
e
 2  iGr cos 
d (  cos  )
 0

e
 2  iGr cos 
d (  2  iGr cos  )
 0
 2  iGr cos 
e
2 iGr
 2  iGr cos 0

sin( 2  Gr )
 Gr

f j  2

2
n j ( r ) r dr
r0

f j  4

2
n j ( r ) r dr
r0
When 
2  Gr
sin( 2  Gr )
2  Gr
*
sin 
G hkl  2
depends on G
fj
2 sin( 2  Gr )

G  0
 0

 f j  4
n
r0
 0
fj
( r ) r dr  z
2
j
lim
 0
(chapter 7)
sin( 2  Gr )
2  Gr
1
total number of
electron in an atom
Tabulated
Function of element and diffracted angle
Example:
Si, 400 diffraction peak, with Cu K (0.1542 nm)
sin 
0
0.1 0.2 0.3 0.4 0.5 0.6
14.0 12.16 9.67 8.22 7.20 6.24 5.31

d 400  0 . 54309 / 4  0 . 13577
nm
2 d 400 sin   0 . 1542    34 . 6
sin 

or
 1

 0 . 368 
8 . 22  7 . 20
8 . 22  f

0 .3  0 .4
0 . 3  0 . 368
 f  7 . 526
sin 


1
2d
8-3. Structure Factor Calculation
S
S G 

f je
 
 2  iG r j
S

j 1
S


f je
j 1
S

G



*
*


*
 2  i ( h a  k b  l c ) ( u j a  v j b  w j c )
f je
 2  i ( hu j  kv j  lw j )
j 1
(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!
S G  fe
2i ( h 0  k 0  l 0 )
 f
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
S

G

e
 2  i ( hu j  kv j  lw j )
fj
j 1
S G  fe
S G  2 f
S G  0
 2i ( h 0  k 0  l 0 )
 fe
 2i ( h
1
2
k
1
2
l0)
 f [1  e
 i ( h  k )
If h , k is unmixed or h + k is even
If h , k is mixed or h + k is odd
]
The meaning of the reflection condition
(reflection rule)
Suppose that we have a square lattice
(a) Primitive unit cell：one atom at [00]
1
S

G

e
 2  i ( hu j  kv j )
fj
j 1
S G  e
2i ( h 0 k 0 )
f  f
(b) Unit cell：two atoms at [00]
1
[ 0]
2
1
S

G

e
 2  i ( hu j  kv j )
fj
j 1
S

G
e
 2i ( h 0 k 0 )
f e
 2i ( h
S G  2 f
If h is even
S G  0
If h is odd
1
2
k 0)
f  (1  e
  ih
)f
The reflection conditions remove the lattice
points [odd, 0] in the reciprocal lattice.
Therefore, the meaning of the reflection rule is
to remove the additional lattice points due to the
selection of unit cell.
After the substraction, the reciprocal lattice
structures derived from both cases (primitive
unit 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
S

G


f je
 2  i ( hu j  kv j  lw j )
j 1
S G  fe
 2i ( h 0  k 0  l 0 )
S G  2 f
S G  0
 fe
 2 i ( h
1
k
2
1
2
If h + k + l is even
If h + k + l is odd
l
1
2
)
 f [1  e
 i ( h  k  l )
]
(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
S G  fe
 2i ( h 0  k 0  l 0 )
 fe
S

G
 2i ( h
 f [1  e
1
 fe
 k 0l
2
 i ( h  k )
1
2
)
 2 i ( h
 fe
e
1
k
2
1
l0)
2
 2i ( h 0  k
 i ( h  l )
1
2
e
l
1
)
2
 i ( k  l )
]
If h, k, l unmixed
 S G  4 f
If h, k, l mixed
 S G  0
(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)
S G  fe
 2i ( h 0  k 0  l 0 )
 f  fe
I  S G
2
 2i (
2hk
3
 f [1  e
2
 fe

l
2
)
 2 i ( h
2
k
3
1
3
 f [1  e
 2i ( h
2
3
k
1
3
l
1
2
)
l
1
)
2
 2i (
2hk
][ 1  e

l
)
3
2
]
2i ( h
2
k
3
1
3
l
1
2
)
]
I  f [1  e
2
 f [1  e
2
 2i ( h
2
k
3
 2i ( h
2
3
1
l
3
k
1
1
)
2
l
3
1
][ 1  e
)
e
2
 f { 2  2 cos[ 2  ( h
2
2
4f
cos [ (
2
2h  k
k

3
k
l
3
l
2
3
1
2
k
3
2i ( h
3
2
2i ( h
1
1
l
3
l
3
1
1
1
2
)
2
)
]
 1]
)]}
2
)]
2
2h + k
l
l 
 2h  k
cos  (
 )

3
2 

3m
3m
3m1
3m1
odd
even
odd
even
0
1
0.75
0.25
2
2
S G
0
4f 2
3f 2
f2
(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)!
S G  f S e
 2i ( h 0  k 0  l 0 )
 fSe
 2i ( h
 k 0l
2
 f Zn e
 f Zn e
S G  [ f S e
1
 2i ( h
)
2
1
k
4
 2i ( h
1
 fSe
3
1
l
4
k
4
1
4
1
3
4
 2i ( h 0  k 0  l 0 )
 [1  e
i ( h  k )
 fSe
)
4
l
e
 2i ( h
)
k
2
1
l0)
2
 2i ( h 0  k
1
l
2
 f Zn e
 f Zn e
 f Zn e
i ( h  l )
1
 2i ( h
3
 2i ( h
1
k
4
 fSe
3
l
4
k
4
1
)
2
4
 2i ( h
1
k
3
4
)
4
l
4
1
1
l
3
)
4
1
4
i ( k  l )
)
]
]
Structure factor of face centered cell
S G  [ f S  f Zn e
i ( h
k
2
h, k, l: mixed 
1
l
2
1
2
)
]  S fcc
S fcc  0
h, k, l: unmixed 
S G  4 [ f S  f Zn e
1
 S G  0
S fcc  4
 i ( h
1
2
when h, k, l are all odd
k
1
2
l
1
2
)
]
S G  4 [ f S  if Zn ]
when h, k, l are all even and h + k + l = 4n
S G  4 [ f S  f Zn ]
when h, k, l are all even and h + k + l  4n
S G  4 [ f S  f Zn ]
Or
2
S G
2
S G
2
 S fcc [ f S  f Zn e
2
 S fcc [ f
2
S
 f S f Zn e
2
S

G
2
S

G
 S fcc [ f
 16 ( f
2
S
1
1
2
2
i ( h  k
2
S
 f
 f
2
Zn
2
Zn
)
1
2
k
1
l
2
 f S f Zn e
l
1
2
1
2
)
][ f S  f Zn e
i ( h
1
k
2
1
2
l
1
i ( h
1
k
2
1
2
)
2
)
]
 2 f S f Zn cos

( h  k  l )]
2
h+k+l : odd
2
 16 ( f S  f Zn )
2
h+k+l :odd multiple of 2.
2
 16 ( f S  f Zn )
2
h+k+l: even multiple of 2
S G
S G
2
 f
2
Zn
i ( h
l
1
2
)
]
8-4. Shape Effect
For a finite crystal, assuming rectangular
volume
F 
 n

G

G
e

 
 i (  k  2  G )r
For a particular
F 
n

G
e

 iD r
, if
n

G
e

 iD r
x0 y0 z0
 n G



 k  2 G  D  0
dV
M 1 N 1 L 1
F 

G
dV
M 1 N 1 L 1

n
x0 y0 z0
M 1
 N 1
 iD  x a
 L 1
 iD  y b
x0
y 0
z0
e
e
e

 iD  z c

G
e



 iD ( x a  y b  z c )
F  n G
1 e
 
 iM D  a
1 e
1 e
 
 iD a
 
 iM D  a
1 e
 
 iD a

1 e
1 e
e
e
F  n G e
e
 
 iD a / 2
 
 iM D  a / 2
e
 
 i ( M 1 ) D a / 2
 
 i D b
 
 iM D  a / 2
e

 
 iN D b
 
 iD a / 2
 
 i ( N  1 ) D b / 2
1 e
 
 iL D  c
1 e
e
 
 iD c
 
iM D  a / 2
 
iD a / 2
e
 
 iM D  a / 2
 
 iD a / 2
e
 
2 i sin( M D  a / 2 )
 
2 i sin( D  a / 2 )
e
 
 i ( L 1 ) D c / 2
e
 
 
 
sin( M D  a / 2 ) sin( N D  b / 2 ) sin( L D  c / 2 )
 
 
 

sin( D  a / 2 ) sin( D  b / 2 ) sin( D  c / 2 )
I  F
2
 
* sin( M D  a / 2 )

 
 n G n G
sin( D  a / 2 )
 n n

G
*

G
sin( M  )
where
2
2
 
sin( N D  b / 2 )
 
sin( D  b / 2 )
sin( N  )
2
sin( L  )
2
 
sin( L D  c / 2 )
 
sin( D  c / 2 )
2
sin(  )
sin(  )
sin(  )
 
 
 
D a
D b
D c
 
; 
; 
2
2
2



D   k  2 G
You should already familiar with the function
2
of
sin( M  )
Chapter 4
sin(  )
2
Consider
I  n G n
*

G
sin( M  )
sin(  )
2
sin( N  )
sin(  )
2
sin( L  )
sin(  )
1st min occurs at
sin( M  )  0 ; sin( N  )  0 ; sin( L  )  0
M    ; N   ; L  
 
 
 


D a
D b
D c 
 
; 
; 
; D   k  2 G
2
2
2
i.e.

  2
 (  k  2 G )  a 
M

  2
(  k  2 G )  c 
L

  2
(  k  2 G )  b 
N
2

k
 
1
(
 G)a 
2
M

 
k
1
(
 G)c 
2
L
or

k
 
1
(
 G)b 
2
N
Therefore, for a finite crystal, the diffracted
intensity is finite based on the condition below.
I  0 if

k
 
1
(
 G)a 
2
M

 
k
1
(
 G)c 
2
L

k
 
1
(
 G)b 
2
N
Example: for a very thin sample
Ewald sphere construction


'


k
k
'
S 
;S 
2
2

When G  𝑆 ′ − 𝑆
diffraction occurs due
to “shape effect”.