Recently, we measured EDX/EDS of LaCrO3 crystals synthesized by hydrothermal method, and found that the characteristic energy for La and Cr are nearly the same for all of the noted energy levels. Could any one give some discussion on how the equiped software address this circumstance with a definite quantified analytic results?
Thank you in advance.
Why don't you use Raman or XRD, which can better distinguish the formed energies? You can also check the SEM mapping results. If the percentages are also the same, there might be a calibration issue.
In Your EDS software you do the following steps:
1) peak identification (done)
2) fitting of the identified peaks and the background (flourescence, ...)
3) calculate the concentrations
Dear Long Yuan
please have a look at Fig. 1 of:
Article The optical and high-pressure raman scattering study of La(F...
It shows the XRF pattern of La comprising the La-L_alpha, La-L_beta 1 and La-L_beta2 peaks.
The latter one is at around 5,4keV and its position strongly interferes with the Cr-K_alpha in your sample.
The La-L_alpha(~4,6keV) and the La-L_beta1 (~5,1keV) are not affected by Cr.
La-L_beta2 is quite low relative to La_L_alpha (~4,6keV); so your combined peak at ~5,4keV is mainly due to Cr-K_alpha...
So knowing the 'natural' intensity ratios of the La lines, one should be able to 'separate' the Cr-K_alpha contribution to your EDS peak at around 5,4keV and gather the relative concentrations of La and Cr.
The procedure given by Stefan Baunack will work.
Best regards
G.M.
Concentrations are calculated automatically and implemented by ZFT corrections.
You can manually adjust the energie directly on the spectrum to avoid background.
EDS does not use a single energy to determine the abundance of an element. That is typically done in WDS with better energy resolution and less overlap.
EDS uses known profiles of the peaks to deconvolute the spectrum and determine the intensities due to each element. The profiles of La-L and Cr-K are significantly different so that the intensities can be determined quite well even if Cr-Ka significantly overlaps La-Lb2.
The same is true for other nasty overlaps like Ti-K and Ba-L or Al-K and Br-L.
Now, if both elements are present and one far outweighs the other, the amount of the lesser element will be quite uncertain. This gets worse if there are relatively few counts in the spectrum. More counts translates to better statistics and better certainty.
Many EDS systems can show you an overlay of the fitted spectrum to the original spectrum (or a residual spectrum showing what has not been accounted for). You can evaluated how good the fit is and whether you have accounted for everything. Rare earths are often not pure but could have small amounts of other rare earths that can get lost in that bunch of L lines.
Just to add a little explanation to Gerhard's and Warren's excellent answers:
If you have La, you will have La's L-alpha-1 and L-alpha-2 and L-beta-1 and L-beta-2 lines at about 4.6-4.7, 5.0, and 5.4 keV with well known intensity ratios. If you have La, you will have all of those lines.
If you have Cr, you will have its K-alpha-1 and K-alpha-2 and K-beta lines at about 5.4, 5.9 keV.
So if you have peaks at 4.6 but not 5.9 keV, you have all La and no Cr.
If you have peaks at 5.9 but not 4.6 keV, you have all Cr and no La.
If you have peaks at all of those energies, the software will compute the relative amounts based on the relative intensities of each of the emission lines (and other matrix etc factors).
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