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7/29/2019 Structure Factor Good
1/28
Suggested Reading
ages - n e rae c enry
Pages 59-61 in Engler and Randle
1
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Structure Factor (Fhkl
)
2 ( )i j iN
i hu kv lw
hkl iF f e
Describes how atomic arrangement (uvw)influences the intensity of the scattered beam.
. ., ,expect in a diffraction pattern.
2
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Structure Factor (Fhkl
)
The amplitude of the resultant wave is given by aratio of am litudes:
amplitude of the wave scattered by all atoms of a UCF
amp tu e o t e wave scattere y one e ectron
to |Fhkl|2.
3
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Some Useful Relations
ei = e3i = e5i = = -1
e2
i = e4
i = e6
i = = +1eni = (-1)n, where n is any integer
eni = e-ni, where n is any integer
eix + e-ix =2 cos x
Needed for structure factor calculations
4
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Fhkl
for Simple Cubic
Atom coordinate(s) u,v,w:
2 ( )
1
i j i
Ni hu kv lw
hkl i
i
F f e
hklF fe f
structure factor equation for simple cubic crystals, the solution is alwaysnon-zero.
5
Thus, all reflections are allowed for simple cubic (primitive) structures.
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Fhkl
for Body Centered Cubic
Atom coordinate(s) u,v,w:
2 ( )
1
i j i
Ni hu kv lw
hkl i
i
F f e
, , .h k l
22 (0) 2 2 2
ii
hklF fe fe
1 i h k lF f ehkl
When h+k+l is even Fhkl = non-zero reflection.
6When h+k+l is odd Fhkl = 0
no reflection.
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Fhkl
for Face Centered Cubic
Atom coordinate(s) u,v,w:
2 ( )
1
i j i
Ni hu kv lw
hkl i
i
F f e
,,0; ,0,;
0,,.
2 2 2h k h l k l
i i i
hklF e e e e
7
1
i h k i h l i k l
hkl
F f e e e
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Fhkl
for Face Centered Cubic
1 i h k i h l i k lhklF f e e e
Substitute in a few values ofhkl and you will find
When h,k,l are unmixed (i.e. all even or all odd), thenFhkl = 4f. [NOTE: zero is considered even]
hkl = or mxe n ces .e., a com na on o oand even).
8
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Fhkl
for NaCl Structure
Atom coordinate(s) u,v,w: +
2 ( )
1
i j i
Ni hu kv lw
hkl i
i
F f e
0,0,0; ,,0; This means these
coordinates , , ;
0,,.(u,v,w)
Cl at ,, + FC transl. ,,; ,,
The re-assignment of coordinates is
, , ; , , 1,,1; 0,,0
,1,1. ,0,0
based upon the equipoint concept inthe international tables for
crystallography
9 Substitute these u,v,w values into Fhkl equation.
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Fhkl
for NaCl Structure contd
For Na: 2 ( )1
i j i
Ni hu kv lw
hkl i
i
F f e
2 (0) ( ) ( ) ( )i i h k i h l i k l
( ) ( ) ( )1Na
i h k i h l i k l
Naf e e e
For Cl:
2 ( ) 2 ( ) 2 ( )( ) 2 2 2l k hi h k i h l i k li h k l
Clf e e e e
( ) ( ) ( ) ( )
( ) ( ) ( )
2 2 2 2
( )
2 2i h k l i i ih k h l k lCl
i h k l i i i
Cl
f e e e e
f e e e e
l k h
l k hThese terms are all positive and even.
10
e er e exponen s o or even depends solely on the remaining
h,
k, and
lin each exponent.
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Fhkl
for NaCl Structure contd
2 ( )
1
i j i
Ni hu kv lw
hkl i
i
F f e
Therefore Fhkl:
( ) ( ) ( )
( ) ( ) ( ) ( )
1i h k i h l i k l
hkl Na
i h k l i i i
F f e e e
l k h
which can be simplified to*:
( ) ( ) ( ) ( )1i h k l i h k i h l i k lhkl Na ClF f f e e e e
11
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Fhkl
for NaCl Structure
When hkl are even Fhkl = 4(fNa +fCl)
hkl Na - Cl
Superlattice reflections
When hkl are mixed Fhkl = 0
o re ec ons
12
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(200)100
80
90
NaClCuK radiation
(220)%)
60
70
ntensity(
50
30
40
(222)(420)
(422) (600)(442)
20
13
(311) (331)
(333)(511)
(440)
(531)
20 30 40 50 60 70 80 90 100 110 1202
()0
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Fhkl
for L12 Crystal Structure
Atom coordinate(s) u,v,w: A B
2 ( )
1
i j i
Ni hu kv lw
hkl i
i
F f e
,,0; ,0,;
0,,.
2 2 2h k h l k li i i A B B BF f e f e f e f ehkl
14( ) ( ) ( )
A B
i h k i h l i k lF f f e e e
hkl
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Fhkl
for L12 Crystal Structure
( ) ( ) ( )A B i h k i h l i k lF f f e e ehkl
(1 0 0) Fhkl =fA +fB(-1-1+1) =fAfB
(1 1 0) Fhkl =fA +fB(1-1-1) =fAfBA B
(1 1 1) Fhkl =fA +fB(1+1+1) =fA +3fB
(2 0 0) Fhkl =fA +fB(1+1+1) =fA +3fB
= + - + - =
(2 2 0) Fhkl =fA +fB(1+1+1) =fA +3fB
(2 2 1) Fhkl =fA +fB(1-1-1) =fAfB
(3 0 0) Fhkl =fA +fB(-1-1+1) =fAfB
(3 1 0) Fhkl =fA +fB(1-1-1) =fAfB
= =
15
(2 2 2) Fhkl =fA +fB(1+1+1) =fA +3fB
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Intensity (%)
100
(111)
Example of XRD patternfrom a material with an
L12 crystal structure90
70
50(200)
30
(220)(311)
2
10
20(100)
(110)
(210) (211) (221)(300)
310
(222)
320 (321)
16
20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 1150
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Fhkl
for MoSi2
Atom positions: Mo atoms at 0,0,0; ,,
c
a oms a , ,z; , ,z; , , +z; , , -z; z=
MoSi2 is actually body centered tetragonal witha = 3.20 and c = 7.86 z
c c ac
xy
x y
z
xy
z
b
x
y
z
a
b
b b
Viewed down z-axis
17
Viewed down x-axis Viewed down y-axis
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Fhkl
for MoSi2
Substitute in atom positions:
( )2
1
i j iN
i hu kv lw
hkl i
i
F f e
Mo atoms at 0,0,0; ,,
Si atoms at 0,0, ; 0,0,z; ,,+z; ,,-z; z=1/3
2 2 22 ( ) 2 ( )2 (0) 2 2 2 2 2 6 2 2 63 3
i i ii ii
F f e f e f e f e f e f ehkl Mo Mo Si Si Si Si
2 ( ) 2 ( ) 3 33 31
i h i hi ii h k lF f e f e e e e
Mo Si
Now we can plug in different values for h k l to determine the structure factor.
or =
5(0 ) (0)(0) ( 0)3 33 3
1 0 1 02 ( ) 2 ( )1 0 0
21
0
1 1 1 1 1 1 0
i ii ii
hkl Mo Si
hkl
F f e f e e e eF
182You will soon learn that intensity is proportional to ; there is NO REFLECTION!
o
lhkF
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Fhkl
for MoSi2 contd
. For h k l = 0 0 1
5(1) (1)
1 13 33 3
0 0 0 02 ( ) 2 ( )0 0 10
i ii ii
hkl Mo SiF f e e f e e e e
223(1 ) (2 ( ) )
(1 1) ( 1 1) 0
i i
Mo Si
Mo si
f e f COS e
f f
For h k l = 1 1 0
2 0 NO REFLECTION!hkl
F
1 1 0 1 1 0 1 1 00 2 (0) 2 (0)2 (0) (0) 2 2(1 ) ( )
(2) (4)
i i ii i
hkl Mo Si
i i i
Mo Si
Mo si
F f e e f e e e e
f e f e e e e
f f
2 POSITIVE! YOU WILL SEE A REFLECTION
hklF
19
to the rules for diffraction
h + k + l = even
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100(103)
80
90
MoSi2CuK radiation
%)
60
70
(101)
nt
ensity(
50
30
40
(002)(213)
20(112) (200)
(202) (211) (116) (206)
2020 30 40 50 60 70 80 90 100 110 1202
()0
(006)(204) (222) (312)
(314)
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Fhkl
for MoSi2 contd
21
- .PCPDFWIN v. 2.1
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Structure Factor (Fhkl
) for HCP
2N
i hu kv lw
1
hkl i
i
e
influences the intensity of the scattered beam.
i.e.,
It tells us which reflections (i.e., peaks, hkl) toexpect in a diffraction pattern from a given crystal
22
, , .
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, , ,lattice point coordinates are:
0 0 0
1 2 1
3 3 2
Therefore, the structure factor becomes:
2h k l
3 3 21hkl iF f e
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We sim lif this ex ression b lettin :2 13 2
h kg
which reduces the structure factor to:
2 ig
We can simplify this once more using:
from which we find:
cose e x
2 2 2 2h k l 3 2
hkl i
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Selection rules for HCP
0 when 2 3 and oddh k n l 2
2
2when 2 3 1 and eveni
hklf h k n lF
2
4 when 2 3 and even
i
if h k n l
For your HW problem, you will need these things todo the structure factor calculation for Ru.
HINT: It might save you some time if you alreadyhad the ICDD card for Ru.
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List of selection rules for different crystals
Crystal Type Bravais Lattice Reflections Present Reflections Absent
Simple Primitive, P Any h,k,l None
Body-centered Body centered, I h+k+l = even h+k+l = odd
Face-centered Face-centered, F h,k,l unmixed h,k,l mixed
NaCl FCC h,k,l unmixed h,k,l mixed
Zincblende FCC Same as FCC, but if all even
and h+k+l4N then absenth,k,l mixed and if all even
and h+k+l4N then absent
Base-centered Base-centered h,kboth even or both odd h,kmixed
Hexagonal close-packed Hexagonal h+2k=3Nwith l even
h+2k=3N1 with l oddh+2k=3N1 with l even
h+2k=3Nwith l odd
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What about solid solution alloys?
If th ll l k l n r n r r th n
must average the atomic scattering factor.
falloy
=xAf
A+ x
Bf
B
where xn is an atomic fraction for the atomic
consttuent
27
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Exercises
For CaF2 calculate the structure factor anddetermine the selection rules for allowed
reflections.
28