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Calculating the transmission of a 1mm thick silicon wafer at various wavelengths |
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| Nov24-12, 05:47 AM | #1 |
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Calculating the transmission of a 1mm thick silicon wafer at various wavelengths
I am intending to perform extended absorption fine structure (EXAFS) experiments on a thin film in order to investigate the near order around Europium ions. The thin film is going to be deposited on a silicon substrate, and the question is in reality related to how thick the substrate should be in order to be suitable for doing EXAFS in transmission.
My supervisor asked me to calculate the transmission of a 1 mm thick silicon substrate at the wavelengths used for investigating Eu. This would be at 48.5 keV (i.e. wavelength of 2.56E-11 m), or possibly 8.05 keV (1.54E-10 m). Now I would assume I can just use Beer's law, but I don't know the absorption coefficient of silicon at these wavelengths, and can't seem to find it anywhere. Don't know if they have to be determined experimentally or if I'm just missing something, or if there is some other way to go about it? If anyone think they can somehow help me with this problem it would be greatly appreciated. |
| Nov24-12, 01:31 PM | #2 |
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Assume that all scattering occurs through Compton scattering process. Compton scattering is when a photon scatters off of an electron with complete conservation of kinetic energy and momentum. Every electron in the silicon will scatter as though the nucleus wasn't there. Assume that there is no absorption of xray energy, no inelastic scattering, and no phonon scattering. The extinction cross section is going to be dominated by Compton scattering. The extinction coefficient in this case is not an absorption coefficient, but Beer's Law remains the same. The differential cross section for Compton scattering is known. You can calculate the total cross section from the differential cross section. On the other hand, you may be able to look up the expression for total cross section. In any case, each electron has a scattering cross section equal to the total cross section for Compton scattering. The density of atoms in silicon is known. The number of electrons in an atom of silicon are known. Expressions for Compton scattering are known. These are all you need to calculate the extinction coefficient. |
| Nov24-12, 05:34 PM | #3 |
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Thanks for the help. I'm not completely sure whether the terminology is exactly the same, but I did a quick calculation where I use the total scattering cross section of Si at the given energy. This seems to be sort of the same as the extinction coefficient from what I gather. For a density, ρ, of 2.33 g/cm3 and a total scattering cross section,σ(E) = 65.6 cm2/g, at 48.5 keV, I get the following from Beer's law.
T = e(-ρσ(E)x) ≈ 89 % If this is correct, it's quite good news I suppose :) A similar calculation for energy of 8.05 gave T = 10-7, so even if the substrate thickness is decreased to 0.01 mm it would still absorb everything. Figure it limits what kind of synchrotron I may go to in order to do the experiment, but that shouldn't be a problem. |
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