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Chem 253, UC, Berkeley
What we will see in XRD of simple
cubic, BCC, FCC?
Position
Intensity
Chem 253, UC, Berkeley
Structure Factor:
adds up all scattered X-ray from each lattice points in
crystal













c
z
b
y
a
x
d
lc
kb
ha
K
j
2
)
( k
hkl S
I 





n
j
d
iK
k
j
e
S
1
2
Chem 253, UC, Berkeley
X-ray scattered from each primitive
cell interfere constructively when:

 n
d 
sin
2
For n-atom basis:
sum up the X-ray scattered from the whole basis
1


 R
iK
e
Chem 253, UC, Berkeley
d

R
i
d j
d

k

'
k
)
( j
i d
d
K 

Phase difference:
The amplitude of the two rays differ: )
( j
i d
d
iK
e






 k
k
K
'
3
Chem 253, UC, Berkeley
The amplitude of the rays scattered at d1, d2, d3…. are in
the ratios :
The net ray scattered by the entire cell:
2
)
( k
hkl S
I 
j
d
iK
e







n
j
d
iK
k
j
e
S
1
Chem 253, UC, Berkeley
For simple cubic: (0,0,0)
1
0

 
iK
k e
S
4
Chem 253, UC, Berkeley
For BCC: (0,0,0), (1/2, ½, ½)…. Two point basis
l
k
h 



 )
1
(
1
S=2, when h+k+l even
S=0, when h+k+l odd, systematical absence
)
(
)
(
2
1
0
2
1
1 l
k
h
i
z
y
x
iK
iK
j
d
iK
k
e
e
e
e
S j






















Chem 253, UC, Berkeley
For BCC: (0,0,0), (1/2, ½, ½)…. Two point basis
S=2, when h+k+l even
S=0, when h+k+l odd, systematical absence
(100): destructive
(200): constructive
5
Chem 253, UC, Berkeley
Observable diffraction
peaks
2
2
2
l
k
h 

Ratio
Simple
cubic
SC: 1,2,3,4,5,6,8,9,10,11,12..
BCC: 2,4,6,8,10, 12….
FCC: 3,4,8,11,12,16,19, 24….
2
2
2
l
k
h
a
dhkl




 n
d 
sin
2
Chem 253, UC, Berkeley
S=4 when h+k, k+l, h+l
all even (h,k, l all even/odd)
S=0, otherwise
)
(
)
(
)
(
1 k
l
i
l
h
i
k
h
i
k e
e
e
S 








 


FCC
6
Chem 253, UC, Berkeley
Observable diffraction
peaks
2
2
2
l
k
h 

Ratio
Simple
cubic
SC: 1,2,3,4,5,6,8,9,10,11,12..
BCC: 2,4,6,8,10, 12, 14….
(1,2,3,4,5,6,7…)
FCC: 3,4,8,11,12,16,19, 24….
2
2
2
l
k
h
a
dhkl




 n
d 
sin
2
Chem 253, UC, Berkeley
Observable diffraction peaks
2
2
2
l
k
h 

Ratio
SC: 1,2,3,4,5,6,8,9,10,11,12..
BCC: 2,4,6,8,10, 12, 14….
(1,2,3,4,5,6,7…)
FCC: 3,4,8,11,12,16, 19, 24….
2
2
2
l
k
h
a
dhkl




 n
d 
sin
2
)
(
4
sin 2
2
2
2
2
2
l
k
h
a





7
Chem 253, UC, Berkeley
What we will see in XRD of simple
cubic, BCC, FCC?
)
(
75
.
0
);
(
5
.
0
sin
sin
2
2
FCC
BCC
B
A




2
2
2
l
k
h
a
dhkl



Chem 253, UC, Berkeley
Ex: An element, BCC or FCC, shows diffraction
peaks at 2: 40, 58, 73, 86.8,100.4 and 114.7.
Determine:(a) Crystal structure?(b) Lattice constant?
(c) What is the element?
2theta theta (hkl)
40 20 0.117 1 (110)
58 29 0.235 2 (200)
73 36.5 0.3538 3 (211)
86.8 43.4 0.4721 4 (220)
100.4 50.2 0.5903 5 (310)
114.7 57.35 0.7090 6 (222)

2
sin
2
2
2
l
k
h 

a =3.18 Å, BCC,  W
8
Chem 253, UC, Berkeley





n
j
d
K
i
j
k
j
e
k
f
S
1
)
(
Polyatomic Structures
fj: atomic scattering factor
 No. of electrons













c
z
b
y
a
x
d
lc
kb
ha
K
j
Reading : West Chapter 5

 








n
j
lz
ky
hx
i
j
n
j
d
iK
j
k e
f
e
f
S j
1
)
(
2
1

Chem 253, UC, Berkeley
Simple Cubic Lattice
Caesium Chloride
(CsCl) is primitive
cubic
Cs (0,0,0)
Cl (1/2,1/2,1/2)
What about CsI?
)
( l
k
h
i
Cl
Cs
k e
f
f
S 



 
9
Chem 253, UC, Berkeley
(hkl) CsCl CsI
(100) 
(110)  
(111) 
(200)  
(210) 
(211)  
(220)  
(221) 
(300) 
(310)  
(311) 
Chem 253, UC, Berkeley
FCC Lattices
Sodium Chloride
(NaCl)
Na: (0,0,0)(0,1/2,1/2)(1/2,0,1/2)(1/2,1/2,0)
Cl: (1/2,1/2,1/2) (1/2,0,0)(0,1/2,0)(0,0,1/2)
Add (1/2,1/2,1/2)
10
Chem 253, UC, Berkeley
S=4 (fNa +fCl) when h, k, l, all even;
S=4(fNa-fCl) when h, k, l all odd
S=0, otherwise
]
1
][
[ )
(
)
(
)
(
)
( k
l
i
l
h
i
k
h
i
l
k
h
i
Cl
Na
k e
e
e
e
f
f
S 












 



Chem 253, UC, Berkeley
(hkl) NaCl KCl
(100)
(110)
(111) 
(200)  
(210)
(211)
(220)  
(221)
(300)
(310)
(311) 
11
Chem 253, UC, Berkeley
Chem 253, UC, Berkeley
Diamond Lattice
12
Chem 253, UC, Berkeley
SSi(hkl)=Sfcc(hkl)[1+exp(i/2)(h+k+l)].
SSi(hkl) will be zero
if the sum (h+k+l) is equal to 2 times an odd integer, such
as (200), (222). [h+k+l=2(2n+1)]
SSi(hkl) will be non-zero if
(1)(h,k,l) contains only even numbers and
(2) the sum (h+k+l) is equal to 4 times an integer.
[h+k+l=4n]
Shkl will be non-zero if h,k, l all odd:
Diamond Lattice
)
(
4 si
si if
f 
FCC:
S=4 when h+k, k+l, h+l
all even (h,k, l all
even/odd)
S=0, otherwise
)
(
)
(
)
(
1 k
l
i
l
h
i
k
h
i
fcc e
e
e
S 








 


Chem 253, UC, Berkeley
Silicon Diffraction pattern
13
Chem 253, UC, Berkeley
Chem 253, UC, Berkeley
14
Chem 253, UC, Berkeley
Chem 253, UC, Berkeley
(1/8,1/8,1/8), (3/8,3/8,1/8),(1/8,3/8,3/8) and (3/8,1/8,3/8).
Bond charges form a "crystal" with a fcc lattice with 4
"atoms" per unit cell.
Bond charges in covalent solid
origin : (1/8,1/8,1/8).
Sbond-charge(hkl) =
Sfcc(hkl)[1+exp(i/2)(h+k)+exp(i/2)(k+l)+exp(i/2)(h+l)].
For h=k=l=2:
Sbond-charge(222)=Sfcc(222)[1+3exp(i/2)4]=4Sfcc(222) non-zero!
Bond charge position: (0,0,0), (1/4,1/4,0),(0,1/4,1/4) and (1/4,0,1/4).

 








n
j
lz
ky
hx
i
j
n
j
d
iK
j
k e
f
e
f
S j
1
)
(
2
1

15
Chem 253, UC, Berkeley
Nanocrystal X-ray Diffraction
Chem 253, UC, Berkeley
Finite Size Effect
16
Chem 253, UC, Berkeley
Bragg angle:
in phase, constructive
B

B

 
1
For
Phase lag between two planes:
At j+1 th plane:
Phase lag:
2j planes: net diffraction at :0

 
1
2
3
4
j-1
j
j+1
2j-1
2j
2


 


 j
1

Chem 253, UC, Berkeley
Bragg angle:
in phase, constructive
B

For
Phase lag between two planes:
At j+1 th plane:
Phase lag:
1
2
3
4
5
j-1
j
j+1
2j-1
2j
2


 


 j
B

 
2

 
2j planes: net diffraction at :0
2

17
Chem 253, UC, Berkeley
How particle size influence the peak
width of the diffraction beam.
Full width half maximum (FWHM)
Chem 253, UC, Berkeley
18
Chem 253, UC, Berkeley
The width of the diffraction peak is governed by # of
crystal planes 2j. i.e. crystal thickness/size
Scherrer Formula:
B
B
t


cos
9
.
0
 2
2
2
S
M B
B
B 

BM: Measured peak width at half peak intensity (in radians)
BS: Corresponding width for standard bulk materials (large
grain size >200 nm)
Readily applied for crystal size of 5-50 nm.
Chem 253, UC, Berkeley
• Suppose =1.5 Å, d=1.0 Å, and =49°. Then for a crystal
1 mm in diameter, the breath B, due to the small crystal
effect alone, would be about 2x10-7 radian (10-5 degree),
or too small to be observable. Such a crystal would
contain some 107 parallel lattice planes of the spacing
assumed above.
• However, if the crystal were only 500 Å thick, it would
contain only 500 planes, and the diffraction curve would
be relatively broad, namely about 4x10-3 radian (0.2°),
which is easily measurable.

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253A-2016-03.pdf

  • 1. 1 Chem 253, UC, Berkeley What we will see in XRD of simple cubic, BCC, FCC? Position Intensity Chem 253, UC, Berkeley Structure Factor: adds up all scattered X-ray from each lattice points in crystal              c z b y a x d lc kb ha K j 2 ) ( k hkl S I       n j d iK k j e S 1
  • 2. 2 Chem 253, UC, Berkeley X-ray scattered from each primitive cell interfere constructively when:   n d  sin 2 For n-atom basis: sum up the X-ray scattered from the whole basis 1    R iK e Chem 253, UC, Berkeley d  R i d j d  k  ' k ) ( j i d d K   Phase difference: The amplitude of the two rays differ: ) ( j i d d iK e        k k K '
  • 3. 3 Chem 253, UC, Berkeley The amplitude of the rays scattered at d1, d2, d3…. are in the ratios : The net ray scattered by the entire cell: 2 ) ( k hkl S I  j d iK e        n j d iK k j e S 1 Chem 253, UC, Berkeley For simple cubic: (0,0,0) 1 0    iK k e S
  • 4. 4 Chem 253, UC, Berkeley For BCC: (0,0,0), (1/2, ½, ½)…. Two point basis l k h      ) 1 ( 1 S=2, when h+k+l even S=0, when h+k+l odd, systematical absence ) ( ) ( 2 1 0 2 1 1 l k h i z y x iK iK j d iK k e e e e S j                       Chem 253, UC, Berkeley For BCC: (0,0,0), (1/2, ½, ½)…. Two point basis S=2, when h+k+l even S=0, when h+k+l odd, systematical absence (100): destructive (200): constructive
  • 5. 5 Chem 253, UC, Berkeley Observable diffraction peaks 2 2 2 l k h   Ratio Simple cubic SC: 1,2,3,4,5,6,8,9,10,11,12.. BCC: 2,4,6,8,10, 12…. FCC: 3,4,8,11,12,16,19, 24…. 2 2 2 l k h a dhkl      n d  sin 2 Chem 253, UC, Berkeley S=4 when h+k, k+l, h+l all even (h,k, l all even/odd) S=0, otherwise ) ( ) ( ) ( 1 k l i l h i k h i k e e e S              FCC
  • 6. 6 Chem 253, UC, Berkeley Observable diffraction peaks 2 2 2 l k h   Ratio Simple cubic SC: 1,2,3,4,5,6,8,9,10,11,12.. BCC: 2,4,6,8,10, 12, 14…. (1,2,3,4,5,6,7…) FCC: 3,4,8,11,12,16,19, 24…. 2 2 2 l k h a dhkl      n d  sin 2 Chem 253, UC, Berkeley Observable diffraction peaks 2 2 2 l k h   Ratio SC: 1,2,3,4,5,6,8,9,10,11,12.. BCC: 2,4,6,8,10, 12, 14…. (1,2,3,4,5,6,7…) FCC: 3,4,8,11,12,16, 19, 24…. 2 2 2 l k h a dhkl      n d  sin 2 ) ( 4 sin 2 2 2 2 2 2 l k h a     
  • 7. 7 Chem 253, UC, Berkeley What we will see in XRD of simple cubic, BCC, FCC? ) ( 75 . 0 ); ( 5 . 0 sin sin 2 2 FCC BCC B A     2 2 2 l k h a dhkl    Chem 253, UC, Berkeley Ex: An element, BCC or FCC, shows diffraction peaks at 2: 40, 58, 73, 86.8,100.4 and 114.7. Determine:(a) Crystal structure?(b) Lattice constant? (c) What is the element? 2theta theta (hkl) 40 20 0.117 1 (110) 58 29 0.235 2 (200) 73 36.5 0.3538 3 (211) 86.8 43.4 0.4721 4 (220) 100.4 50.2 0.5903 5 (310) 114.7 57.35 0.7090 6 (222)  2 sin 2 2 2 l k h   a =3.18 Å, BCC,  W
  • 8. 8 Chem 253, UC, Berkeley      n j d K i j k j e k f S 1 ) ( Polyatomic Structures fj: atomic scattering factor  No. of electrons              c z b y a x d lc kb ha K j Reading : West Chapter 5            n j lz ky hx i j n j d iK j k e f e f S j 1 ) ( 2 1  Chem 253, UC, Berkeley Simple Cubic Lattice Caesium Chloride (CsCl) is primitive cubic Cs (0,0,0) Cl (1/2,1/2,1/2) What about CsI? ) ( l k h i Cl Cs k e f f S      
  • 9. 9 Chem 253, UC, Berkeley (hkl) CsCl CsI (100)  (110)   (111)  (200)   (210)  (211)   (220)   (221)  (300)  (310)   (311)  Chem 253, UC, Berkeley FCC Lattices Sodium Chloride (NaCl) Na: (0,0,0)(0,1/2,1/2)(1/2,0,1/2)(1/2,1/2,0) Cl: (1/2,1/2,1/2) (1/2,0,0)(0,1/2,0)(0,0,1/2) Add (1/2,1/2,1/2)
  • 10. 10 Chem 253, UC, Berkeley S=4 (fNa +fCl) when h, k, l, all even; S=4(fNa-fCl) when h, k, l all odd S=0, otherwise ] 1 ][ [ ) ( ) ( ) ( ) ( k l i l h i k h i l k h i Cl Na k e e e e f f S                   Chem 253, UC, Berkeley (hkl) NaCl KCl (100) (110) (111)  (200)   (210) (211) (220)   (221) (300) (310) (311) 
  • 11. 11 Chem 253, UC, Berkeley Chem 253, UC, Berkeley Diamond Lattice
  • 12. 12 Chem 253, UC, Berkeley SSi(hkl)=Sfcc(hkl)[1+exp(i/2)(h+k+l)]. SSi(hkl) will be zero if the sum (h+k+l) is equal to 2 times an odd integer, such as (200), (222). [h+k+l=2(2n+1)] SSi(hkl) will be non-zero if (1)(h,k,l) contains only even numbers and (2) the sum (h+k+l) is equal to 4 times an integer. [h+k+l=4n] Shkl will be non-zero if h,k, l all odd: Diamond Lattice ) ( 4 si si if f  FCC: S=4 when h+k, k+l, h+l all even (h,k, l all even/odd) S=0, otherwise ) ( ) ( ) ( 1 k l i l h i k h i fcc e e e S              Chem 253, UC, Berkeley Silicon Diffraction pattern
  • 13. 13 Chem 253, UC, Berkeley Chem 253, UC, Berkeley
  • 14. 14 Chem 253, UC, Berkeley Chem 253, UC, Berkeley (1/8,1/8,1/8), (3/8,3/8,1/8),(1/8,3/8,3/8) and (3/8,1/8,3/8). Bond charges form a "crystal" with a fcc lattice with 4 "atoms" per unit cell. Bond charges in covalent solid origin : (1/8,1/8,1/8). Sbond-charge(hkl) = Sfcc(hkl)[1+exp(i/2)(h+k)+exp(i/2)(k+l)+exp(i/2)(h+l)]. For h=k=l=2: Sbond-charge(222)=Sfcc(222)[1+3exp(i/2)4]=4Sfcc(222) non-zero! Bond charge position: (0,0,0), (1/4,1/4,0),(0,1/4,1/4) and (1/4,0,1/4).            n j lz ky hx i j n j d iK j k e f e f S j 1 ) ( 2 1 
  • 15. 15 Chem 253, UC, Berkeley Nanocrystal X-ray Diffraction Chem 253, UC, Berkeley Finite Size Effect
  • 16. 16 Chem 253, UC, Berkeley Bragg angle: in phase, constructive B  B    1 For Phase lag between two planes: At j+1 th plane: Phase lag: 2j planes: net diffraction at :0    1 2 3 4 j-1 j j+1 2j-1 2j 2        j 1  Chem 253, UC, Berkeley Bragg angle: in phase, constructive B  For Phase lag between two planes: At j+1 th plane: Phase lag: 1 2 3 4 5 j-1 j j+1 2j-1 2j 2        j B    2    2j planes: net diffraction at :0 2 
  • 17. 17 Chem 253, UC, Berkeley How particle size influence the peak width of the diffraction beam. Full width half maximum (FWHM) Chem 253, UC, Berkeley
  • 18. 18 Chem 253, UC, Berkeley The width of the diffraction peak is governed by # of crystal planes 2j. i.e. crystal thickness/size Scherrer Formula: B B t   cos 9 . 0  2 2 2 S M B B B   BM: Measured peak width at half peak intensity (in radians) BS: Corresponding width for standard bulk materials (large grain size >200 nm) Readily applied for crystal size of 5-50 nm. Chem 253, UC, Berkeley • Suppose =1.5 Å, d=1.0 Å, and =49°. Then for a crystal 1 mm in diameter, the breath B, due to the small crystal effect alone, would be about 2x10-7 radian (10-5 degree), or too small to be observable. Such a crystal would contain some 107 parallel lattice planes of the spacing assumed above. • However, if the crystal were only 500 Å thick, it would contain only 500 planes, and the diffraction curve would be relatively broad, namely about 4x10-3 radian (0.2°), which is easily measurable.