Difference between revisions of "X-ray absorption & fluorescence"
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For more info, see the X-ray Absorption Information (11BM page) | For more info, see the X-ray Absorption Information (11BM page) | ||
http://11bm.xray.aps.anl.gov/absorption.html | http://11bm.xray.aps.anl.gov/absorption.html | ||
For more practical info about strongly absorbing samples see this wiki page: | |||
https://wiki-ext.aps.anl.gov/ug11bm/index.php/Samples_with_Strong_X-Ray_Absorption | |||
=== Absorption Length === | === Absorption Length === |
Revision as of 00:32, 22 February 2014
X-Ray Absorption
For more info, see the X-ray Absorption Information (11BM page) http://11bm.xray.aps.anl.gov/absorption.html
For more practical info about strongly absorbing samples see this wiki page: https://wiki-ext.aps.anl.gov/ug11bm/index.php/Samples_with_Strong_X-Ray_Absorption
Absorption Length
The X-ray beam intensity I(x) at depth x in a material is a function of the attenuation coefficient mu, and can be calculated by the Beer-Lambert law:
I(x) = Io e^(-mu * x)
The attenuation coefficient mu is typical given in inverse length units of 1/cm, and is a function of the incident wavelength, material chemistry and density. It can be calculated or estimated using resources below.
The Absorption Length (or Attenuation Length) is defined as the distance into a material where the x-ray beam intensity has decreased to a value of 1/e (~ 40%) of the incident beam intensity (Io).
Recall that Euler's number e = 2.72.
This is a convenient description, as absorption length x = 1/mu, as shown below:
(1/e) = e^(-mu * x) ln(1/e) = ln(e^(-mu * x)) 1 = mu * x x = 1/mu
as a simple example, consider a solid Nickel metal sample at room temperature probed by X-rays of energy = 30 KeV (Lambda = 0.41 A).
For Ni with density = 8.908 g*cm-3, we can calculate (using resources below) that mu ~ 85.0 cm-1.
Then absorption length x = 1/85 = 0.011 cm = 110 microns.
Capillary Transmission
Continuing with the above example for capillary transmission X-ray diffraction experiments, we can consider a cylindrically shaped solid Nickel metal sample of radius R = 0.055 mm (or 0.0055 cm). The diameter of this sample is then 2*R = 0.011 cm (see absorption length above)
Since mu = 85 cm-1, then mu*R = 0.935, therefore the % total incident x-rays transmitted through the sample is = e^(-2*muR) = ~ 40%
In general, a mu*R of ~ 1.0 is desired for capillary transmission x-ray samples.
Web Resources
WebAbsorb
a web based calculator to estimate X-ray absorption for powder XRD capillary samples (11BM page) http://11bm.xray.aps.anl.gov/absorb/absorb.php
MuCal
calculate X-ray absorption, fluorescence and more (by C. Segre @ IIT/ANL) http://csrri.iit.edu/mucal.html
Software
The two python GUI programs described below compute approximate x-ray scattering cross sections (f, f' and f") for individual elements using the Cromer & Liberman algorithm.
downloaded both here:
https://subversion.xray.aps.anl.gov/trac/pyFprime/
Fprime
computes and plots elemental scattering factors.
Absorb
computes scattering and absorption for a given composition and makes an attempt to estimate density as well. WebAbsorb provides a web based utility based on this program (see http://11bm.xray.aps.anl.gov/absorb/absorb.php).