253A-2016-03.pdf

1
Chem 253, UC, Berkeley
What we will see in XRD of simple
cubic, BCC, FCC?
Position
Intensity
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
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
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
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
For simple cubic: (0,0,0)
1
0

 
iK
k e
S

4
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






















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
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
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
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
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
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



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





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

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
(hkl) CsCl CsI
(100) 
(110)  
(111) 
(200)  
(210) 
(211)  
(220)  
(221) 
(300) 
(310)  
(311) 
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
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 












 



(hkl) NaCl KCl
(100)
(110)
(111) 
(200)  
(210)
(211)
(220)  
(221)
(300)
(310)
(311) 

11
Diamond Lattice

12
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 








 


Silicon Diffraction pattern

13

14
(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
Nanocrystal X-ray Diffraction
Finite Size Effect

16
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

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
How particle size influence the peak
width of the diffraction beam.
Full width half maximum (FWHM)

18
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.
• 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