I was wondering since we have few choices as an XRD source material. And all of them have different atomic mass/number, producing different characteristic wavelengths. Since, during the production of X-Ray, we also must take care of the heating effect, I was wondering which is a better choice.
Dear Ashish Kumar Patel ,
you are right, some aspects such as wavelength (related to atomic numbers) and heating (melting point, thermal conductivity) have to be considered for the choice of the 'optimal' target material.
The most used target material is Cu because of its thermal properties (high intensity output) and machinability (price issue).
Target materials with lower*) atomic number (Fe, Co, Cr) are used to prevent/reduce fluorescence background due to Fe and similar kinds of elements in the sample.
Target material with higher atomic number**) (such as for example Mo) is used to extend the XRD pattern with respect to the magnitude of the momentum transfer vector q ( = 4pi/lambda * sin(theta) ); especially for small angle scattering applications and dealing with amorphous samples (integrals over q have to be used in the data evaluation here).
*) lower than Cu (Z=29)
**) shorter wavelength compared to Cu
Best regards
G.M.
Dear Gerhard Martens ,
May I ask you a couple of follow-up questions?
1. What effect the fluorescence have in regards to XRD?
2. Can you elaborate on the final paragraph about extending the XRD pattern a bit more?
Dear Ashish Kumar Patel ,
(1) with respect to the fluorescence issue, you know, that for example the Cu K-alpha radiation (~8keV) is able to excite the K-shells of Co (binding energy ~ 7,7keV), Fe (binding energy ~ 7,1keV), Mn (binding energy ~ 6,5keV), etc. The fluorescence is showing up, when the core hole is refilled by an electron; it is spreading out uniformly in the whole 4pi angular space; thus also in the direction of the XRD detector, irrespective on its actual angular position relative to the sample surface. So in consequence the fluorescence photons arising from the sample will give rise to a background onto the diffraction pattern, unless there is no monochromator in front of the detector or an energy resolving detector in use. The background will 'modify' the amorphous contributions in the XRD pattern and will increase the noise. Thus it is advised to avoid or at least 'reduce' possible fluorescence via a 'clever' choice of the x-ray tube target material.
So when you have some Co or Fe in your sample, you should go for a Fe target (Co K-alpha ~ 6,9keV), a Fe target (Fe k-alpha ~6,4keV) or a Cr target (Cr K-alpha ~5,4keV).
(2) When dealing with amorphous sample materials one has to evaluate the pattern*) by some kind of integrals over the whole pattern. The results will be more reliable the larger the theta space is. But due to geometrical issues the theta range is limited. A way out of this dilemma is to shift the peak positions (even in the case of amorphous peaks or 'humps') to the lower theta range. This is done by lowering the wavelengths in use.
From the Bragg equation (lambda = 2*d*sin(theta)) you see, that for smaller lambda you will have smaller thetas; thus resulting in a shift of the pattern to lower thetas. The quantity, which is involved here, is the magnitude of the momentum transfer vector q, which is given by: q = 4*pi/lambda * sin(theta). When taking the Bragg equation you will see that in q-space an XRD pattern showing up crystalline peaks will have them at special values of q = 2*pi/d.
Smaller wavelengths are associated with larger photon energies; thus here one will go for higher atomic numbers of the x-ray target compared to Cu (Z=29). Mo is a commonly used candidate for this application; ZMo = 42, Mo K-alpha ~ 17,4keV.
*) here we deal with small angle x-ray scattering (SAXS); but you know that XRD is some kind of x-ray scatter process in a regular lattice, which includes constructive interference...
Best regards
G.M.
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