I would like to calculate the infinite thickness of a tungsten-containing sample for an x-ray fluorescence experiment. To calculate the infinite thickness, a mass attenuation coefficient is needed. According to the following NIST page, the mass attenuation coefficient changes as a function of photon energy. For instance:
https://physics.nist.gov/PhysRefData/XrayMassCoef/ElemTab/z74.html
What energy value do I use to calculate the MACs for each element in my sample? Is the energy value related to the energy of the primary photon emitted from the x-ray source? I am using a 20 kV high voltage for the x-ray source (not sure if relevant or not). Thanks for reading!
Dear Alex Gupta ,
measurement of the sample thickness by polychromatic radiation is not a trivial thing.
The sample acts as filter, which preferably attenuates the low photon energy part of the x-ray spectrum compared to the high energy part. As a function of sample depth the spectrum gets more and more hardend. So any 'mean photon energy' which could describe the attenuation will be a function of the depth in the sample and therefore the overall attenuation cannot be decribed by a simple exponential law.
However for polychromatic x-rays there are some alternatives.
A) You may use a set of tungsten foils of different thicknesses in order to set up some kind of calibration curve (attenuation versus thickness) and look at which thickness your sample attenuation matches the 'calibration curve'.
B) Another alternative is to go for quasi monochromatic radiation without using a monochromator. Here the Ross-filter or balanced filter approach is the simplest thing. An example of the spectral attenuation of a pair of balanced filter sets is given in the attachment for the case of 5,5µm Co and 5,0µm Ni. This filter set gives access to a quasi monochromatic section of the x-ray spectrum of an x-ray tube around 8keV.
The net x-ray signal from this section is easily calculated by the difference of Ni and Co filtered radiation: I(Ni filtered) - I(Co filtered). For the primary radiation Io you have to do the same measurements.
Now you are able to use the exponential law:
(I(Ni filtered) - I(Co filtered))/( Io(Ni filtered) - Io(Co filtered)) = exp (-(µ/rho) *(rho*d))
with (µ/rho) being the mass attenuation coeffcient, which is well known from the NIST tables you mentioned above. rho is the density and d the thickness.
You mentioned above, that you have got a tungsten containing sample. This means that the tungsten density is not known. So the only parameter, which you can derive from the exponential equation above, is the term (rho*d) which has the dimension of a mass per area; it its some kind of areal density or mass coverage. When setting rho equal to the bulk tungsten density you will get some kind of effective tungsten foil thicknesss of your sample but not the real thickness.
C) You will get improved results when you choose a filter set which surrounds the characteristic lines of your x-ray tube. When using a Cu tube the K-alpha lines are just at about 8keV, so you can use the Co/Ni filter set. Mean photon energy is still 8keV.
D) with respect to your Nafion membrane experiments*) you should do the same for the undoped sample. You will get some transmisson value TNafion for your host matrix.
The above equation equation then has to be rewritten to:
(I(Ni filtered) - I(Co filtered))/( Io(Ni filtered) - Io(Co filtered)) = TNafiion*exp (-(µ/rho) *(rho*d))
*) https://www.researchgate.net/post/How_can_the_dopant_concentration_in_a_Nafion_membrane_be_measured_via_XRF
Good luck and best regards
G.M.
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