Transcript Solid State Physics

Solid State Physics
2. X-ray Diffraction
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Diffraction
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Diffraction
m
sin  
W
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m  1, 2, 3, ...
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Diffraction
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Diffraction using Light
Diffraction Grating
One Slit
Two Slits
m
sin 
d
http://physics.kenyon.edu/coolphys/FranklinMiller/protected/Diffdouble.html
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Diffraction
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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Diffraction for Crystals
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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Bragg Law
William
Lawrence
Bragg
1980 - 1971
2d sin   n
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Mo
Cu
Co
Cr
0.07 nm
0.15 nm
0.18 nm
0.23 nm
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Monochromatic Radiation
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Diffractometer
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Nuts and Bolts
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
a1  axˆ
a2  ayˆ
2
G1 
xˆ
a
a3  azˆ
2
G2 
yˆ
a
2
G3 
zˆ
a
The reciprocal lattice is itself a simple
cubic lattice with lattice constant 2/a.
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Reciprocal Lattices

BCC Lattice
ˆ
a1  12 a(xˆ  yˆ  z)
ˆ
a3  12 a(xˆ  yˆ  z)
2
G1 
yˆ  zˆ
a
ˆ
a2  12 a(xˆ  yˆ  z)
0  a1 a2  a3  12 a3
2
G2 
xˆ  zˆ
a
2
G3 
xˆ  yˆ
a
The reciprocal lattice is represented by
the primitive vectors of an FCC lattice.
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Reciprocal Lattices

FCC Lattice
2
a1 
yˆ  zˆ
a
2
a2 
xˆ  zˆ
a
2
a3 
xˆ  yˆ
a
0  a1 a2  a3  a3
ˆ
G1  12 a (  xˆ  yˆ  z)
ˆ
G 2  12 a (xˆ  yˆ  z)
ˆ
G 3  12 a (xˆ  yˆ  z)
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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Back to Diffraction
Diffraction is related to the electron density.
Therefore, we have a...
The set of reciprocal lattice vectors
determines the possible x-ray reflections.
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The difference in path length of the of2the
incident
r sin
 wave at the
The
difference
inwave,
phasethe
angle
is difference is  k  r
For
the
diffracted
phase
points
O and r is
k   r
r sin 

So, the total difference in phase angle is (k  k )  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

n (r) e


i (  k)  r
  n (r) e
dV
f 
i k  k  r
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    ns e
i ( s  k)  r
dV
s
s  k
Constructive
Interference
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Diffraction Conditions
k   k  k
s  k
(k  s )  k
2
or
k  s  k
2
2 k s  s
2
2d sin   n
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Diffraction Conditions

For a crystal of N cells, we can write down
F  N  cell n( r ) e

i s  r

dV  NSs
s
n(r )   n j (r  r j )
j 1
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Diffraction Conditions

The structure factor can now be written as
integrals over s atoms of a cell.
Ss  
j
 e
j

cell

n j ( r  rj ) e
 i s  rj

 n j (  )e

i s  r

i s  


dV
dV
i s   
Atomic form
f j  n j (  )e
dV
factor

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Diffraction Conditions

Let r j  x j a1  y j a 2  z j a 3

Then, for an given h k l reflection

1


3
s  rj  (ha  k a 2  l a )  ( x j a1  y j a 2  z j a 3 )
 hx j  ky j  lz j
Ss   f j e

 i 2 hx j  ky j lz j

j
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Diffraction Conditions

For a BCC lattice, the basis has identical atoms
at ( x1 , y1 , z1 )  (0,0,0) and ( x2 , y2 , z2 )  ( 12 , 12 , 12 )

The structure factor for this basis is
SG  f (1  e


i 2 hk l 
)
S is zero when the exponential is i × (odd
integer) and S = 2f 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
1 1 1
1
1 1
000, 0 2 2 , 2 0 2 , and 2 2 0

The structure factor for this basis is
SG  f (1  ei k l   ei hl   ei hk  )


S = 4f 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
h2  k 2  l 2
When combined with the
Bragg law:
 2
2
 sin   2 h  k 2  l 2 
4a
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
(degrees)
11.44
16.28
20.13
23.38
26.33
29.07
34.14
36.53
38.88
X-ray powder pattern

ratiosusing Cuhkl
determined
K
0.0394radiation,
1
100Å
 = 1.542
0.0786
2
110
0.1184
3
111
sin2
0.1575
0.1967
0.2361
4
5
6
200
210
211
0.3151
0.3543
8
9
220
300, 221
0.3940
10
310
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Structure Determination (310)
sin  
2

2
h
2
 k2  l2

4a
2
(1.5420)
10
0.3940
2
4a
 a  3.88 angstroms
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