Some of the results of EDX analysis are quite scattered e.g. for Ca K: from about 3% to 27% at. They agree don't with the atomic composition of the compound. Any comment on it? Please answer
Dear Yaseen Muhammad ,
you may share your EDX spectra and the expected (rough) elemental composition.
There might be a superposition of Ca K-radiation with the L radiation of for example of Sb and/or Te:
Ca K-alpha (~3,69keV) Sb L-alpha (~3,6 keV)
or
Ca K-alpha (~3,69keV) Te L-alpha (~3,77 keV)
In order to avoid such superposition you should go for an excitation energy of just above Ca K-excitation (e.g. 4,1keV; Ca K-edge ~ 4,04keV).
The L-edge/binding energies for Sb and Te are above that energy value and their L lines should not show up at all in this case.
However there might be a simple other explanation for that issue; what is about a heterogenous sample ? The Ca might no be homogenously distributed in your sample...
Best regards
G.M.
Atomik özellikleri aynı olmadığı için atomik bileşimleri de aynı değildir. (%3-%27).ancak ayırt edici özellikleri azaldıkça heterojenliği azalır ve homojenliği artar. Bu yüzden band aralığı düzelir ve bağlanma enerjileri birbirine yaklaşır. Bağlanma enerjilerinin birbirine yaklaşması onun homojenliğini arttırır. Birbirine yakın enerji seviyelerine gelerek enerji açığa çıkarıp bileşim oluştururlar.
Specimen roughness, changes in composition in specimen, contaminations, etc. Again, as in another your question, it is impossible to answer with no additional data about your experiment.
Translation from Turkish by Google:
Since their atomic properties are not the same, their atomic compositions are not the same. (3%-27%). However, as its distinctive features decrease, its heterogeneity decreases and its homogeneity increases. Therefore, the band gap improves and the binding energies converge. The convergence of binding energies increases its homogeneity. By coming to energy levels close to each other, they release energy and form a compound.
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