Optical Characterization in Industry, National Labs, and
Academia
1.
Industrial Materials Characterization using Spectroscopic techiques
2.
Multiuser facilities at National Labs
3.
University Optical Labs
Major suppliers of Spectroscopic Equipment (solutions)
Phys 774:
X-ray and Neutrons
Spectroscopy
JY Horiba
Bruker / Siemens
Sopra
Philips
Accent-Biorad
Fall 2007
Spectra-Physics
Lecture 12
1.
Optical Characterization in Industry:
Optical Characterization in National Labs
Materials and device structure characterization in a
production line
New materials and ideas
Thin films, coatings, multilayers
1.
FT-IR Reflectivity and Transmission (IV)
2.
Composition and strain
2.
Raman Scattering (III-V)
3.
Conductivity / resistance
3.
Spectroscopic ellipsometry (IV)
4.
Nanotechnology and nanoscale
4.
Photoluminescence (III-V)
characterization
5.
In situ characterization (on fly)
5.
Parts: Edmund Optics and CVI
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1.
2.
Emergent materials
1.
Micro FT-IR Reflectivity and
Transmission
New instrumentation / new
techniques
2.
Surface-enhanced Raman Scattering
3.
Bio-optics
3.
Far-IR Spectroscopic ellipsometry (IV)
Micro-photoluminescence (III-V,IV)
4.
Nanomaterials
4.
Near-field Photoluminescence (III-V)
6.
X-ray diffraction (III-V)
5.
In situ characterization
5.
Nano-scale X-ray diffraction (III-V)
7.
X-ray fluorescence
6.
X-ray fluorescence / X-ray standing
waves
Issues:
8.
E-beam spectroscopy
7.
Maintanence
Speed
9.
Interpretation and models
2
Imaging vs. Analythical characterization
3
E-beam bio-imaging
Goals: unique experiments with emergent materials
4
Examples of Optical Characterization in National Labs
Examples of Optical Characterization in National Labs
New materials and ideas
New materials and ideas
5
6
Examples of Optical Characterization in National Labs
New materials and ideas
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8
OUTLINE
How to get an access to multi-user facilities at National Labs
Need ideas
1.
Select Facility (Synchrotron, Observatiry, Neutron source, ...)
2.
Register as a user
X-ray radiation. Sources of X-rays, Detection of X-rays
3.
Write a proposal. Remember that Proposals become „Public Information“
Primary techniques: Fluorescence, Diffraction, Scattering
4.
Schedule time for experiment
Instrumentation: Diffractometers, detectors
5.
Pass safety training
Applications
6.
1.
Arrive to Facility and carry out experiments exactly how it is described in your
proposal
7.
Write a paper
8.
Start this process over
2.
X-ray Spectroscopy
Neutrons Spectroscopy
Sources, Detectors
Techniques: diffraction and Inelastic scattering
Instrumentation and Applications
HW: write a 2-page proposal for any experiment
Follow guidelines for proposal writing at any
Facility, e.g, CHESS
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Discovery of X-rays
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OUTLINE
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12
OTHER USE of X-RAYS
Wavelength and the size of observable objects
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Interaction of Electrons, X-rays, and Neutrons with matter
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OUTLINE
Surface
Neutrons
X-rays
Electrons
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Sources of X-rays
X-ray tubes
X-ray tube
Electronic
transitions
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Synchrotron Radiation
Synchrotron radiation
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HOW CAN WE SELECT A WAVELENGTH
Natural Synchrotron Radiation
Stars and
Galaxies
Accelerating
electron
emits light
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Closer look at the properties of Synchrotron Radiation
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Synchrotrons
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Progress with Brilliance of Synchrotron Radiation
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EXAMPLES of SYNCHROTRON RADIATION HIGHLIGHTS
EXAMPLES of SYNCHROTRON RADIATION HIGHLIGHTS
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SCATTERING
DIFFRACTION vs.
SCATTERING
of X-rays
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Scattering Geometry: General Consideration
Scattering Cross Section
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Scattering Cross Section
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Scattering by individual atoms
(Mutual interference between scattered rays gives diffraction pattern)
Scattering from atom
Consider single electron. Plane wave Ein = E0 e
i ( k ⋅r −ωt )
k=k =
2π
λ
E0 i ( kR −ωt )
e
Scattered field: E ' = f e
fe – scattering length of electron
R
R – radial distance
Two electrons:
E ' = fe
or, more generally
many electrons:
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E0 ikR
e ⎡⎣1 + ei∆k ⋅r ⎤⎦
R
E ' = fe
E0 ikR i∆k ⋅r1 i∆k ⋅r2
⎤⎦
+e
e ⎡⎣e
R
E
E ' = f e 0 eikR ∑ ei∆k ⋅rl
R
l
similar to single electron with
f = f e ∑ e i∆k ⋅32rl
l
intensity:
I~ f
2
= fe
2
∑e
2
Atomic scattering factor (dimensionless) is determined by
electronic distribution.
i∆k ⋅rl
l
this is for coherent scatterers.
Scattering length of electron:
If random then
[(
I ~ Nf e
) ]
f e = 1 + cos 2 2θ / 2
12
If n(r) is spherically symmetric, then
2
r0
re
f a = ∫ 4πr 2 n(r )
0
e2
classical electron radius re =
≈ 2.8 ×10 −15 m
2
4πε 0 mc
1
In atom,
f e ∑ ei∆k ⋅rl → f e ∫ n(r )ei∆k ⋅rl d 3 r
l
sin (∆k ⋅ r )
dr
∆k ⋅ r
in forward scattering ∆k = 0 so
f a = 4π ∫ r 2 n(r )dr = Z
Z - total number of electrons
n(r) – electron density
f a = ∫ n(r )e i∆k ⋅rl d 3 r - atomic scattering factor (form factor)
Atomic factor for forward scattering is equal to the atomic Z number
(all rays are in phase, hence interfere constructively)
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Scattering Cross Section
34
What we can measure?
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Scattering
Scattering
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Scattering
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Scattering
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Applications of inelastic X-ray scattering:
probing lattice vibrations (similar to Raman scattering)
k
Inelastic X-ray scattering
k
k0
θ
q k = k0 ± q
=ω = =ω0 ± =Ω(q)
q = 2k 0 sin θ = 2n
ω0
c
sin θ
Inelastic X-ray scattering in GaN
assumed Ω(q) << ω0
- true for x-rays:
~104
ħΩ < 100 meV; ħω0
eV
n – index of refraction
k0
θ
q k = k0 ± q
=ω = =ω0 ± =Ω(q)
q = 2k 0 sin θ = 2n
ω0
c
sin θ
measuring ω - ω0 and sinθ one can determine dispersion Ω(q)
main disadvantage – difficult to measure ω - ω0 accurately
This difficulty can be overcome by use of neutron scattering
Energy of "thermal" neutrons is comparable with ħΩ (80 meV for λ≈1Å)
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Scattering from crystal
Since ∆k = G,
the lattice factor becomes
f cr = ∑ e i∆k ⋅rl = ∑ f al e i∆k ⋅R l
crystal scattering factor:
l
l
Rl - position of lth atom, fal - corresponding atomic factor
rewrite
where
Then scattering intensity I ~ |fcr|2
f cr = F ⋅ S
F = ∑ f aj e
S = ∑e
i∆k ⋅R lc
l
Where
c
l
where
l
f cr = F ⋅ N = N ∑ f aj e
iG ⋅s j
j
i∆k ⋅s j
j
and
S = ∑ e iG ⋅R l = ∑ e i 2πm = N
- structure factor of the basis,
summation over the atoms in unit cell
- lattice factor, summation over all
unit cells in the crystal
G = Ghkl = hb1 + kb2 + lb3
Then
F = ∑ f aj e
j
if sj = uja1 + vja2 + wja3
i ( u j a1 + v j a 2 + w j a 3 )( hb1 + kb 2 + lb 3 )
= ∑ f aj e
2πi ( hu j + kv j + lw j )
j
structure factor
R l = R lc + s j
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Example: structure factor of bcc lattice (identical atoms)
structure factor
F = ∑ f aj e
2π i ( hu j + kv j + lw j )
j
Two atoms per unit cell: s1 = (0,0,0); s2 = a(1/2,1/2,1/2)
[
F = f a 1 + eπi ( h + k +l )
]
⇒ F=2fa if h+k+l is even, and F=0 if h+k+l is odd
Diffraction is absent for planes with odd sum of Miller indices
For allowed reflections in fcc lattice h,k,and l are all even or all odd
4 atoms in the basis.
What about simple cubic lattice ?
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h+k +l
F (h, k , l )
F (h, k , l ) = f [1 + exp(−iπ (h + k ) + exp(−iπ (h + l ) + exp(−iπ (k + l )]
F
F
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F (h, k , l )
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F
F⋅ F
F⋅ F
F (h, k , l ) = f (exp 0) = 1
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Experimental XRD techniques
Rotating crystal method –
for single crystals, epitaxial films
θ-2θ, rocking curve, ϕ - scan
DIFFRACTION
of X-rays
Powder diffraction
Laue method – white x-ray beam used most often used for mounting
single crystals in a precisely known orientation
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EXPERIMENTAL SETUP for Rotating Crystal Diffraction of
X-rays
X-ray Diffraction Setup
SAMPLE
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High Angular Resolution X-ray Diffraction Setup
The Bragg Law
B11 Tiernan
Conditions for a sharp peak in the
intensity of the scattered radiation:
1) the x-rays should be specularly
reflected by the atoms in one plane
2) the reflected rays from the
successive planes interfere constructively
The path difference between the two x-rays: 2d·sinθ ⇒
the Bragg formula:
2d·sinθ = mλ
The model used to get the Bragg law are greatly oversimplified
(but it works!).
– It says nothing about intensity and width of x-ray diffraction peaks
– neglects differences in scattering from different atoms
– assumes single atom in every lattice point
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– neglects distribution of charge around atoms
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We obtained the diffraction (Laue) condition: ∆k =
Diffraction condition and reciprocal lattice
Vectors G which satisfy this relation form a reciprocal lattice
Von Laue approach:
– crystal is composed of identical atoms placed
at the lattice sites T
A reciprocal lattice is defined with reference to a particular Bravais lattice,
which is determined by a set of lattice vectors T.
– each atom can reradiate the incident radiation
in all directions.
Constricting the reciprocal lattice from the direct lattice:
Let a1, a2, a3 - primitive vectors of the direct lattice; T = n1a1 + n2a2 + n3a3
– Sharp peaks are observed only in the
directions for which the x-rays scattered from
all lattice points interfere constructively.
Consider two scatterers separated by a lattice vector T.
Incident x-rays: wavelength λ, wavevector k; |k| = k = 2π/λ; (k'−k ) ⋅ T = 2πm
Assume elastic scattering: scattered x-rays have same energy (same λ) ⇒
k
k'
wavevector k' has the same magnitude |k'| = k = 2π/λ
k=
k' =
k
k'
Condition of constructive interference: k'−k ⋅ T = mλ or
(
Define
)
∆k = k' - k - scattering wave vector
Then ∆k = G
G where G·T = 2πm
, where G is defined as such a vector for which G·T = 2πm
We got ∆k = k' – k = G ⇒ |k'|2 = |k|2 + |G|2 +2k·G ⇒ G2 +2k·G = 0
Then reciprocal lattice can be generated using the primitive vectors
where V = a1·(a2×a3) is the volume of the unit cell
We have bi·aj = δij
Then vector G = m1b1 + m2b2 + m3b3
Therefore, G·T = (m1b1 + m2b2 + m3b3)·(n1a1 + n2a2 + n3a3) =
2π(m1n1+ m2n2+ m3n3) = 2πm
The set of reciprocal lattice vectors determines the possible scattering wave
vectors for diffraction
2k·G = G2 – another expression for diffraction condition
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We got ∆k =
k' – k = G ⇒ |k'|2 = |k|2 + |G|2 +2k·G ⇒ G2 +2k·G = 0
2k·G = G2 – another expression for diffraction condition
Now, show that the reciprocal lattice vector G = hb1 + kb2 + lb3 is
orthogonal to the plane represented by Miller indices (hkl)
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Now, prove that the distance between two adjacent parallel planes of
the direct lattice is d = 2π/G.
The interplanar distance is given by
the projection of the one of the
vectors xa1, ya2, za3, to the direction
normal to the (hkl) plane, which is
plane (hkl) intercepts axes at points
x, y, and z given in units a1, a2 and a3
the direction of the unit vector G/G
By the definition of the Miller indices:
⇒
The reciprocal vector G(hkl) is associated with the crystal planes (hkl) and
is normal to these planes. The separation between these planes is 2π/G
define plane by two non-collinear vectors u and v lying within this plane:
prove that G is orthogonal to u and v:
analogously show
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2k·G = G2 ⇒ 2|k|Gsinθ = G2
⇒ 2·2πsinθ /λ = 2π/d ⇒ 2dsinθ = λ
2dsinθ = mλ - get Bragg law
k
θ
∆k
k'
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Ewald Construction for Diffraction
Geometric interpretation of Laue condition:
Condition and reciprocal space
2k·G = G2 ⇒
– Diffraction is the strongest (constructive interference) at the
perpendicular bisecting plane (Bragg plane) between two reciprocal
lattice points.
– true for any type of waves inside a crystal, including electrons.
– Note that in the original real lattice, these perpendicular bisecting
planes are the planes we use to construct Wigner-Seitz cell
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Geometric interpretation of Laue condition:
2k·G = G2 ⇒
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