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What is X-ray Absorption Spectroscopy (XAS)?
This is an inner shell spectroscopy technique
Inner shell means that an X-ray interacts primarily with a deep-core electron rather than with a valence electron.
Spectroscopy means that some aspect of the interaction changes as a function of photon energy
What are the steps of diamond XAS?
Electrons fired into straight accelerator, or linac, and accelerated to ~100 MeV.
Boosted in small synchrotron and injected into storage ring at 3 GeV.
Magnets in ~560 m large ring bend and focus electrons, which are accelerated to close to speed of light.
Energy lost emerges down beamlines as highly focused light at X-ray wavelengths
What equation is used for electron speed?
Eo = mc2
Only true for an electron (e-) at rest (E0).
At rest, the energy of an electron is 0.5 MeV.
If I accelerate that e-, of course its energy increases.
If I define: then E = γmc2 (E is total energy – rest + kinetic)
Look at this, this is why reaching the speed of light for a particle with mass is, shall we say, difficult.
As , the energy required approaches (+) infinity.
Because electrons have very low mass, we can get them close to the speed of light (c).
• m(electron) = 9.11×10-31 kg
• m(proton) = 1.67×10-27 kg
How is radiation produced in synchrotrons?
When charged particles are accelerated in a magnetic field, they emit radiation.
At low velocity, e- radiation pattern is shaped like a pumpkin.
At relativistic speeds, e- radiation pattern is shaped like a tiny needle pointing tangent to the radius of trajectory.
Radiation from e- at relativistic speeds has a strong component of X-ray photons. Very tightly collimated beam of X-rays.
Define critical energy εc – divides the power of the beam in half (50% higher energy, 50% lower energy).
ε_c (keV)=(2.218 E^3 (GeV))/R(m)
X-ray energies produced span orders of magnitudes of eV, which is needed for XAS.
Most element characteristic X-rays in ~2-20 keV region.
So this all makes sense, except that critical energy εc is inversely dependent on radius of the ring… so why do we build big synchrotrons?
Because energy loss per circuit of an individual electron depends strongly on radius:
∆E (per circuit)=(K'γ^4)/R
What would be the keV per circuit for a 3.5 GeV ring?
with R = 1 m → –15.0 keV per circuit; with R = 100 m → –0.15 keV per circuit.
With larger R, less energy needed to bend e-, less energy needed to keep them at relativistic speeds.
→ bending high velocity electrons around corners in a magnetic field creates synchrotron radiation, doing this in a large circle makes it easier to sustain the process.
Inside the ring, because they are charged particles, can alter e- path to change the nature of the emitted synchrotron radiation.
Done by “insertion devices”, increase flux of photons and shift energy distribution to higher energies. These devices are called wigglers and undulators.
Why is a broad spectrum of wavelengths useful?
Synchrotron provides us with a very bright, collimated, X-ray beam.
User can select wavelength required for experiment from a broad spectrum that covers from microwaves to hard X-rays.
It’s perfect for a wide range of X-ray experiments.
In Earth and Environmental Sciences, the most useful are fluorescence, X-ray absorption spectroscopy, Raman scattering, diffraction, and tomography.
How is synchrotron and X-ray absorption spectroscopy done?
For both fluorescence and absorption experiments, a single energy incident beam is selected using Bragg’s law.
OF CRITICAL IMPORTANCE is that this energy IS NOT FIXED as it is in the X-ray tube in a conventional XRD instrument.
You can rotate the crystals and change the incident energy!
nλ = 2d.sin(θ) and E = h c / λ,
→ E = h c / 2d.sin(θ)
Change θ → change E to produce a monochromatic beam at your energy of choice!
What is the XAS experimental setup?
Beam of monochromatic X-rays travels through matter, and lose some intensity via interaction with the material.
Loss is proportional to original intensity I0, thickness of sample x, and linear absorption coefficient μ (in cm-1), which is dependant on physical/chemical state of the absorber and X-ray energy.
Beer’s law: It = I0 exp(-μx)
→ If x is constant, changing E changes It by changing μ.
What is the relationship between absorption edge and x-ray energy?
XAS spectra plotted as linear absorption coefficient μ vs energy.
· Pre Edge ~2-10 eV before main edge
· XANES (X-ray absorption near edge structure) up to ~50 eV above edge (can include pre edge)
o XANES – energy close to that of the edge → multiple scattering of photoelectrons from surrounding atoms
· EXAFS (Extended X-ray Absorption Fine Structure) 50-1000 eV above edge
EXAFS – energy >> that of the edge → single scattering of photoelectrons from surrounding atoms
XANES is sensitive to oxidation state, site geometry, and bond angles.
· Useful for C, N, Fe, Mn, Cu, S, As, Cr, etc.
How is EXAF affected by absorption edge?
At E ≥ absorption edge binding energy Eo, incident X-ray knocks out e-
Importantly, emitted core electron propagates as a spherical wave with a given wavelength λ.
The electron wave is scattered by surrounding atoms, and the scattered waves can interfere with the outgoing wave. This interference is the origin of the EXAFS oscillations.
As energy of incident beam increases, energy of the photoelectron (Ew) increases, so corresponding λw decreases.
Ability to change incident energy at a synchrotron by rotating the monochromator allows you to change the wavelength λw of outgoing electron wave, thus finding absorption coefficient maxima that correspond to nearest neighbour atomic distances.
• The frequency of EXAFS oscillations depends on distance to surrounding atoms (all are convolved together...)
• Amplitude depends on the number and type of backscatters, heavy elements are better scatterers.
• All “shells” contribute; makes data fitting complicated.
• Summary of the information EXAFS provides:
• Z: ±3 – differentiate O&S, not O&N
• N: ±20%
• R: ±0.02 Angstrom
What equations are used for EXAF spectrum data processing?
• Absorbance is converted to χ:
χ(E)= (μ(E)-μ_0 (E))/(〖∆μ〗_0 (E)) , with μ0 = background and Δμ0 = edge jump
• Energy is converted to k, the wave-vector:
k= √(2m/h^2 (E-E_0)), with m = mass e- and E0 = energy photoe- at k = 0
• Finally, signal is weighted, here by k3, to enhance features at high energy.