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2
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sin
  m
 m

1 2 3
W
3
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4
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Diffraction Grating
One Slit
Two Slits sin
  m
 d http://physics.kenyon.edu/coolphys/FranklinMiller/protected/Diffdouble.html
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The diffraction pattern formed by an opaque disk consists of a small bright spot in the center of the dark shadow, circular bright fringes within the shadow, and concentric bright and dark fringes surrounding the shadow.
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Photons
Electrons
Neutrons
Diffraction techniques exploit the scattering of radiation from large numbers of sites. We will concentrate on scattering from atoms, groups of atoms and molecules, mainly in crystals.
There are various diffraction techniques currently employed which result in diffraction patterns. These patterns are records of the diffracted beams produced.
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What is This Diffraction?
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William
Lawrence
Bragg
1980 - 1971
2 d sin
  n

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Mo 0.07 nm
Cu 0.15 nm
Co 0.18 nm
Cr 0.23 nm 10
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11
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12
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The Bragg law gives us something easy to use,
To determine the relationship between diffraction
Angle and planar spacing (which we already know
Is related to the Miller indices).
But…
We need a deeper analysis to determine the
Scattering intensity from a basis of atoms.
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Reciprocal Lattices

Simple Cubic Lattice a
1
 a
ˆ
2
 a
ˆ
3
 a
ˆ
G
1

2
 a
G
2

2
 a
G
3

2
 a
The reciprocal lattice is itself a simple cubic lattice with lattice constant 2

/ a.
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Reciprocal Lattices
 a
BCC Lattice
1

1
  
2 a
a
3

1
2 a
 
2

1
2 a (x
 
   
0 a a
1 2 a
3

1
2 a
3
G
1

2
 a
ˆ  ˆ
G
2

2
 a
ˆ  ˆ
G
3

2
 a
The reciprocal lattice is represented by the primitive vectors of an FCC lattice.
ˆ  ˆ
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Reciprocal Lattices

FCC Lattice a
1

2
 a

2
   
0 a a
1 2 a
3
 a
3

2
 a
G
1

1
2 a
   ˆ
G
3

1
2 a
ˆ   ˆ

3

2
 a
ˆ  ˆ
G
2

1
2 a
ˆ   ˆ
The reciprocal lattice is represented by the primitive vectors of an BCC lattice.
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Drawing Brillouin Zones
Wigner –Seitz cell
The BZ is the fundamental unit cell in the space defined by reciprocal lattice vectors.
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Drawing Brillouin Zones
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Diffraction is related to the electron density.
Therefore, we have a...
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r sin

2

So, the total difference in phase angle is r sin

( k

 k


)

 k r
 r
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Diffraction Conditions

Since the amplitude of the wave scattered from a volume element is proportional to the local electron density, the total amplitude in the direction k  is f

 
 n n (
( r ) e i r ) e
 i
 
(

  r k )
 r dV dV k
  k
 k

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Diffraction Conditions

When we introduce the Fourier components for the electron density as before, we get f
  s n e s

(
  dV s
  k
Constructive
Interference
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Diffraction Conditions k
  k
 k
 s
  k
(k
 s )
2  k
2 or 2 k s s
2 k
  k

2 d sin
  n

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Diffraction Conditions

F
For a crystal of N cells, we can write down

N
 cell n ( r ) e

 i s r
 dV

NS s n ( r )
 j s 

1 n j
( r
 r j
)
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Diffraction Conditions

The structure factor can now be written as integrals over s atoms of a cell.
S s
   j cell n j
( r
 r ) j e

 i s r
 dV
  j e

 i s r j

 n j
 e
 i s
 dV
Atomic form factor f j
  n j
 e

  dV
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Diffraction Conditions

Let r j
 x j a
1
 y j a
2
 z j a
3

Then, for an given h k l reflection s
  j h
1

 k a
2

 l

 x j
 hx j
 ky j
 lz j
1
 y j a
2
 z j
S s
  j f e j
 i 2

 hx j
 ky j
 lz j

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Diffraction Conditions

For a BCC lattice, the basis has identical atoms
( x
1
, y
1
, z
1
)

( 0 , 0 , 0 ) ( x
2
, y
2
, z
2
)

( 1
2
, 1
2
, 1
2
)

The structure factor for this basis is
S
G
 f ( 1
 e
 i 2
  h
 k
 l

)


S is zero when the exponential is  i  × (odd integer) and S = 2 f when h + k + l is even.
So, the diffraction pattern will not contain lines for (100), (300), (111), or (221).
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Diffraction Conditions

For an FCC lattice, the basis has identical atoms at
000 , 0 1
2
1
2
, 1
2
0 1
2
, and 1
2
1
2
0

The structure factor for this basis is
S
G
 f ( 1
 e
 i
  k
 l

 e
 i
  h
 l

 e
 i
  h
 k

)


S = 4 f when hkl are all even or all odd.
S = 0 when one of hkl is either even or odd.
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Structure Determination
Simple
Cubic
 d
 a h
2  k
2  l
2
When combined with the
Bragg law:
 sin
2
 

4 a
2
 h
2  k
2  l
2

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
(degrees)
11.44
sin 2
0.0394
X-ray powder pattern
 ratios radiation,

= 1.542 hkl
Å

16.28
0.0786
2 110
20.13
23.38
0.1184
0.1575
3
4
111
200
26.33
29.07
34.14
36.53
38.88
0.1967
0.2361
0.3151
0.3543
0.3940
5
6
8
9
10
210
211
220
300, 221
310
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Structure Determination (310) sin
2
 

4 a
2
 h
2  k
2  l
2

( 1 .
5420 )
2
0 .
3940

 a

4 a
3 .
2
88 angstroms
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