EDX - Science topic

thank you dear Jürgen Weippert . in fact, I have a silica coated wood spect that is good. but on control wood samples there is this Hf peak. there are usually variety of inorganic martials in wood as a sediments in too low amount(less that 0.5 %). but this peak is not normal in related literature

please see attaches file.

The nitorgen peak should appear between the C and O peaks. I wonder if the substrate x-rays are affecting this area of the spectra.

If you system can identify C and O, that means that your specimen has no identifiable quantities of N.

Most FESEM labs can coat a specimen with carbon instead of Pt or other metal. You may want to repeat your analysis with another, carbon coated specimen.

EDS+TEM is pretty difficult technique to implement properly. For example, operator could choose too short dwelling time, and then nothing but main compnents would be found. By the way, how thick are your LiF layer and silver foil?

Alexander Bolshakov Why not ask the manufacturer?

It's a common artifact of EDS. Ignore it.

EDX: refers to the method or EDX analysis.

EDS: refers to the equipment; Energy Dispersive Spectrometer.

Too little information to give really useful advice.

If you are looking for islands of PTFE on Ni, then simple SEM observation could help (in secondary and backscattered electrons). EDS should be used to check whether F is present in islands. For "quantification" you can determine relative area of islands.

Anyway, you can start with observation and checking if F present in different areas of you specimen. May be you'll need to make cross sections of your plating.

Dear Siddig Abuelgasim ,

the penetration depth of the exciting electrons is in the very low µm regime, even below 1µm, and is depending on the acceleration voltage. The penetration depth/ attenuation length of the x-ray fluorescence is also the low µm regime.

Thus, with respect to your 30µm particles, you will not 'see' the core of them.

Unfortunately, low µm regime is not surface or surface near region. Here you have to go the photo electron spectroscopy, as it has been suggested by Jürgen Weippert .

Without analyzing by XRD technique the parallelly oriented clay samples saturated with Ca and K ions and subsequent glycolation of Ca-sample and heating the K-sample , it not possible which clay mineral it is.

From my point of view, EDX is useful just to perform simple and rough elementar chemical analysis. As mentioend before, the % of error is a critical drawback. I reccomend you to search for XPS analysis. Then, you have a characterization technique that will provide elementar composition, molecular information (oxides, nitrides, etc) with a suitable precision for quantification.

Have a good work !!!!!

You may want to try low voltage operations (as was suggested by Adeyinka Aina ). I would choose 3 kV, good enough to excite both F and Al.

Dear Muhammad Aleem Zahid , your question is a bit fuzzy. I believe you made standardless "quantitative" TEM/EDS analysis of Si-O compound and you are not happy with results. You cannot "convert" data you already have. By the way quantification of specimens like yours with TEM/EDS is nearly impossible. Of course if you have suitable (certified?) SiO2 foil to serve as a standard, you may want to try.

To have an idea of the penetration depth in your sample you can use the classical formula of Kanaya Okayama (you can search it on google)

Dear Siddig Abuelgasim ,

thanks for feedback.

You are right; theta is the angle between the incoming primary x-ray beam impinging onto the XRD sample.The angle between the sample and the diffracted x-ray beam leaving the sample and entering the detector also has the value of theta. So the beam geometry of the XRD set up is just a symmetric reflection set up.

I am sure, you are aware of that issue.

This theta is also the parameter of the x-axis of the XRD pattern, but unfortunately displayed as x=2*theta^*) due to historical reasons.

You are right; the penetration depth is not constant for the whole XRD pattern, but varies along the x-axis according to the equation given above.

So the x-ray penetration depth for the 'peaks' (at specific 2*theta positions) is alo different according to that equation.

*) 2*theta is the total deflection angle or scatter angle of the primary beam

Additional remark:

the penetration depth (or better 'information depth') for EDX, as you have asked in your question above, is more complicated due to two facts:

a) different angles with respect to the sample surface (e.g. theta_p for the primary exciting beam and theta_d for the detected beam) and

b) different linear attenuation coefficients for the excitation radiations (e.g. named as µ_e(E_ex)) and the x-ray fluorescence ( e.g. named as µ_f(E_fl)). Both attenuation coefficient depend on the photon energies, which are in use, such as the (mean) excitation energy E_ex and the fluorescence energy E_fl of the detected x-rays

Your query is similar to the attached link. Please find out. It might be helpful to you.

Best regards

Similar to Alexey Fedorov , I recommend not concluding anything from this image. This is not sufficiently resolved, maybe try AFM if your SEM can't get it better?

SEM and EDX cannot be used for nanometer-thick layers. You need at least 100 nm, but reliable results are possible from 1-2 µm.

Core shell structure can be confirmed by stereo microscope technique

Dear all,

We have just published this work on the development of Datamuncher Gamma, a new software for the analysis of SEM-EDX maps.

Regards

You need to repeat your analysis. There are lines on your spectrum (possible O and Si) that were not taken in account. It looks like Pb alaysis was performed on M lines, which are overlapping with Si line. You need use L lines, but then it's better to repeat X-ray acquisition at 20 kV. Without using L lines it's mostly impossible to say if presence of S is real and not an artefact produced by software. And if you were using standardless analysis (and I am sure you were), then results for C (and O) are mostly useless.

You may wish to look up the work of ML Wolfrom and JR Vercellotti. Also see doi: 10.1002/9780470110447.ch4.. 1977;24:203-312.Methods Biochem Anal

Determination of heparin and related sulfated mucopolysaccharides by L B Jaques

Dear Puja,

many thanks for sharing this very interesting technical question with the RG community. iI fully agree with Alan F Rawle and Alec Ladonis in that key points here are thorough washing of the initially obtained nanoparticles with distilled water (even more often than 6-10 times) and exclusion of moisture from the calcinated end-product .

This leads me to a question I would like to ask all experts in this forum: Is it possible to make ZnO nanoparticles without any annealing / calcination step? In this context please have a look at the two papers cited below in which this has been reported. As you may know, I'm a synthetic organometallic chemist and not a nanoparticle guru. However, to the best of my knowledge these low-temperature routes will only produce Zn(OH)₂, but no ZnO. Please let me know if I'm wright or wrong in this point.

Here are the two articles:

and

Both papers are freely avvailable as public full texts on RG.

With best wishes, Frank Edelmann

Although EDX microanalysis provides some information, it is always better and safer to combine more methods. The corrosion product can be analyzed, for example, by powder X-ray diffraction analysis. Corrosion products are usually mixtures and may contain an admixture of precipitates such as gypsum and calcite or aragonite.

The weight percentage of an element is the weight of that element measured in the sample divided by the weight of all elements in the sample multiplied by 100.

The atomic percentage is the number of atoms of that element, at that weight percentage, divided by the total number of atoms in the sample multiplied by 100.

So the atomic weight percent is calculated from the element weight percentage by dividing each element weight percentage by its atomic weight, do this for all elements in the sample, you will have a list of atomic proportions. Sum these together to obtain a total atomic weight.  Then for each element in the sample divide its atomic proportion by the total and * 100.

Here is an example - using pyrite. EDX will give you

Fe: el wt% 46.55

S: el wt% 53.45

divide each by their atomic weights Fe (55.85) S (32.06) and sum

Fe: 46.55 / 55.85 = 0.833

S: 53.45 / 32.06 = 1.667

divide each by the sum (2.5) and turn into atomic percentage

Fe: 0.833 / 2.5 * 100 = 33 at wt%

S: 1.667 / 2.5 * 100 = 66 at wt%

so twice as much by the number of atoms of S as Fe, FeS2, but similar-ish wt% because Fe atom is much heavier than S atom.

Interesante pregunta, espero las respuestas aportadas por los expertos cumpla sus expectativas como investigadora.

First of all, having a light element matrix might not be detrimental to quantitative EDX (or WDX) : it all depends -as Vladimir Dusevich said- on having a "good" SEM (well calibrated for quanti), being a "good" operator (knowing what to do and more importantly, what NOT to do), and having "good" standards" (or having proved that one of the quant models works on your samples/matrix).

It is easier for example to quantify heavier elements in a light matrix that is suited for quanti, than for instance Carbon in Steel (for which EPMA would be used).

One of the ground rules would be : if your sample is not smooth, dense (non porous), homogeneous, conductive, isotropic and finally thick enough for electrons not to strike through ; then you can probably forget about precise quantitation by SEM-EDS (or WDS, does not matter).

Plus, you should be aware that "standardless" analysis, (whatever the model, ZAF, PhiRhoZ, P&P...) as provided by all manufacturers is always worse than having your personal standards, ideally on the same matrix as that you will be analyzing (most always difficult to do).

If you're not familiar with quant SEM-EDS, you might want to start with : https://www.springer.com/gp/book/9781493966745

D.E.Newbury again...

I could suggest French books as well if you read French (lots of work done in France on quant SEM-EDS models back in the 70s and 80s).

Dear all, if NP's aggregation and/or the dielectric environment are behind such shift, it is good to have a look at the following document. My Regards

Dear Mazhar, many thanks for asking this very interesting technical question. In this context please have a look at the following relevant link which might help you in your analysis:

This paper has been published Open Access and is freely available as pdf file (see attached).

Good luck with your work!

When temperature increases, it causes fading in organic matter and impurities

Dear Maria Apriliani Gani,

These may be addition of impurities, high temperature and environmental conditions can reduce the formation of carbonates.

Bone mineral is not pure hydroxyapatite. The small, plate-shaped (20–50 nm long, 15 nm wide, and 2–5 nm thick) apatite crystals contain impurities, most notably carbonate in place of the phosphate groups. The concentration of carbonate (4–6%) makes bone mineral similar to a carbonate apatite known as dahllite. Other substitutions are potassium, magnesium, strontium, and sodium in place of the calcium ions, and chloride and fluoride in place of the hydroxyl groups.These impurities can reduce the crystallinity of the apatite , and, in doing so, may alter certain properties such as solubility. The solubility of bone mineral is critical for mineral homeostasis and bone adaptation.

Carbonate substitution in apatite each control the crystal’s perfection and crystallite domain size, affecting material properties, in different environmental conditions. High temperature substitution methods indicating a higher energy barrier for the formation of carbonate.

Ashish

Thanks Ogechi Nkemdirim

in addtion to what Gerhard Martens has already mentioned, I think that you'll get a much higher background due to Bremsstrahlung radiation in foreward direction. So smaller peaks are added on a larger background ... Best regards, Dirk

Bacterial cells are not stoichiometric. So, I believe you want to know cells composition. EDX is not the best (may be one of the worst) methods to quantify composition of a biological matter. If you are interested in a relative quantities of some heavier element, you can use any grid but the grid containing element of interest. If you want to get "cleaner" spectra (with no stray radiation from a grid) you can use Be grids. For SEM you can use carbon substrate.

Hi Negar,

I'm not sure whether you have collected multiple EDX spectra or just one spectrum. If you collected just one spectrum and saw a significant difference in Ca content compared to other spectroscopic analysis, this could be due to a local aggregation of Calcium in that particular area of the sample. Keep in mind in EDX, with high magnification, the spectrum is collected from a very small area of the sample. Therefore you might want to collect multiple spectra from different locations of the samples before making general conclusions about the sample. Also, EDX is a surface analytical technique. Therefore the surface elemental composition you get from XRD has a possibility to differ from the other spectroscopic methods if you crush / powder the samples for the analysis.

I hope this helps and good luck with your study!

First, you have to move from WT% to AT%. To do that, you have to divide each element WT% value by the corresponding atomic mass, so you get the number of atoms of that kind contained in 100 units of total mass. Then you must sum up all the numbers obtained, to get the total equivalent atoms number in your base 100 unit mass of material. By dividing each number obtained in step 1 by the above said total equivalent number, you get an estimation of the AT% of each component in the probed material.

For example (taken from your table), the WT% of Oxygen you probed is 22.84% (22.84 mass units over 100 mass units). The O atomic mass is 16. You divide 22.84 / 16 to get 1.42, the O equivalent atoms number. Doing the same for all other components you get the Sn, Cu etc equivalent atoms number. Than all such number will sum up to 2.39 (approx). The AT% for Oxygen will be 1.42 / 2.39 = 59.4%.

After you got all the AT% values, you may have a rough idea of the coarse stoichiometry of the oxides (it is clear from your table that a large part will be SnO2 from the FTO substrate).

Quantitative EDX analysis for your specimens is very difficult task because of presence of oxygen. You need highly qualified operator and set of rather expensive standards. If you really need good analysis go for some other method, not EDX. If your specimens do not differ in composition too much, you can compare heights of peaks: higher peak, higher concentration. By the way EDS and EDX are interchangeable abbreviations, EDX being a bit more precise in meaning, and EDS being used a bit more often.

Dear Satyaranjan Bairagi ,

There is not a specific relation between XRD and EDS patterns, because XRD detects the crystalline phases, but EDS shows only the elements. I mean EDS is a chemical analysis but XRD is a phase analysis. In general, EDS cannot detect light elements such as B or C precisely. Meanwhile, low content phases may not be detectable by XRD when their percentage is less than 8. Additionally, you are only able to identify the crystalline phases (compounds or elements) by XRD test. I mean, if a phase is in its amorphous state, it cannot be detected by XRD, whereas its chemical composition is easily detectable by EDS.

In addition to EDS you need to study microstructure to see if you really have a single phase specimen. BSE imaging and/or etching should help, If you want to quantify light elements (O, C) you need standards, standardless analysis is not reliable.

The probe depth of XPS is about 10-nm which is a surface sensitive technique whereas EDX can vary from 1 to 3 microns - "bulk" analysis technique. The thickness and homogeneity of elemental distribution in the sample will thus cause different results by the two techniques.

Quite a strange question. You have a mix of Al2O3 and AlN solid phases and you have no N2 molecules at all, if we say about solid films. So, the answer is: "use the atomic % of N".

Your problem, as I can understand from the data presented is a AlN film on Al2O3 substrate. Is it right?

If so, the EDX method is not good in this case: electrons with energies of more than 5 keV (EDX) penetrate quite deeply into a solid (more than 500 nm) and, apparently, you observe an analytical signal from both the film (AlN) and from the substrate (Al2O3). For an accurate answer, you need a depth profiling analysis using one of the methods: XPS, SIMS, AES, LEIS, RBS ...

Non-conducting specimens for SEM should be coated with conductive material, gold is one of the most common coatings. Looks like operator of your SEM coated specimens with too thick layer of gold and forgot to tell you about it. If possible, coat specimens with carbon. Do not pay attention to amounts of C, O and other light elements in your "quantitative" analysis, they are wrong.

Si is one of the constituents of PEO. Refer

If the Raman modes are collective (i.e. non-localized to a functional group), they are sensitive with respect to the crystallite size. I don't know how MOF modes look like, but you can check that out.

In principle, XRD should also be somewhat sensitive to that; do the reflexes there have the same width for all samples?

Dear Maria Mihalache ,

you can do this. I assume that there is no Uranium in the sample iself.

So the sample will not still emit x-rays or gammas due to Uranium radiation.

Nevertheless you should perform two measurements:

a) perform EDX analysis with electron or x-ray beam OFF ---> spectrum_a

b) perform EDX analysis with electron or x-ray beam ON ---> spectrum_b.

your net EDX spectrum will be netEDXspectrum = spectrum_b - spectrum_a.

Please take care that the measurement times are equal.

Good luck

Proper quantification of light elements (such as N) by EDS is a tough task. It requires use of standards and good understanding of the technique. I believe that your results (as well as results from the paper you mentioned) were obtained by means of standarless analysis, which produces huge errors in light elements analysis. If you need bulk analysis of light elements (not a microanalysis, where EDS excels), use other methods, discard EDS results.

Just because peaks have similar emission energies doesn't mean they have the same cross section. Please perform a stoichiometric analysis of your data (many EDX machines come with recommended parameter sets for that) and check whether there is an actual contradiction between values that you get from analyzing different peak pairs since with the information you gave us so far we can't tell whether there is an actual problem in the first place.

We can't measure it directly from EDX. It's only give us the stoichiometric concentration of the composition.

Of course, the acquisition time is affecting the results:

1. Obviously, there is just a statistical effect: number of counts. The columns in the histogram are "jumping", then become stable. Just the law of large numbers.

2. Beam damages. The beam destroys the sample. This is especially important for thin samples (films, coatings, etc).

3. Carbon deposition under the beam. Even the purest chamber has some carbon contamination (SEM users know this annoying effect very well - dark spots).

4. Charging. Including local degradation of conductive coating.

Thus, in ideal world it would be easy: more counts, higher precision. But the real physical world is more complex. Experienced operators can choose the most optimal parameters of analysis. Sometimes, it is a kind of art.

Good luck!

Dear friends, pleas stop this idiocy about non-detecting light elements. C and O were detected. Try to comment on things you know.

Dear Sachin Jambhale,

the problem you report is quite common and may originate from the material itself, from the environment and other surrounding (even the microscope itself, see ttps://onlinelibrary.wiley.com/doi/pdf/10.1002/sca.1996.4950180402) as well from the sample preparation and storage:

1) carbon contamination- the source of carbon can be from any contact with organic upon preparation or even from the air. Any organic contamination tends to form hydrocarbon under the electron beam on the sample surface, the amount of which can increase within the measurement you perform. It is very common effect in SEM and/or in (S)TEM. The most important is to clean properly the sample (plasma cleaning etc.).

2) oxygen - for many alloys, especially Al alloys it is very difficult to avoid it, since Al will passivate at the first possible contact with oxygen, unless one use a protective gas environment or very high vacuum, which is very difficult to realize (because the sample can not see oxygen at any moment between preparation and investigation) otherwise, just accept it is there and in the case you want to quantify the EDX data, keep it in mind and make corresponding corrections (especially important for TEM samples, for which material surface to volume ration gets significant).

This link could solve your doubt

https://www.globalsino.com/EM/page1738.html

I would expect that you can detect the elements in the sample. Especially because you have high atomic numbers, the X-rays are hardly absorbed by carbon. The other condition is the energy used. To excite the sample volume you need low energies. To excite the elements you need high energies. I assume, you try to detect the M-lines or L-lines. The best chance, I believe is to look for the Pb-lines due to their high absorption you need a lower sample thickness. The M-alfa energy is 2,34 keV transition energy (so you need only 5 keV excitation) or L-alfa (10,5 keV). This is well to excite with 15 keV electrons. With such energies you penetrate the carbon layer and the X-rays do so. Perhaps it helpes.

With Regards

R. Mitdank

Thank you very much, I apprecite it

Just a little hint: MEB is SEM in English (that would attract more people who answer).

When you say you want to analyze the SEM without EDX: What do you want to know? From the image you can see that you have some 90° angles which would be expectable for NaCl and all other kinds of cubic structure salts, but of course the chemical composition is not available without EDX/XPS or maybe also a mass spectrum of your water sample.

please check the attached pdf.

Manisha Bharti Anand Parkash

have a nice day.

Hello Vadim, Suntan and Simone,

many thanks for valuable information. I will check all your suggestions.

Have a nice day

Kim

Cu background is a usual thing.

Mapping is not really good analytical method. Acquire spectra, from a spot at a specimen, and another one from an "empty" space and compare them.

There is only one way to be certain. Solve the structure of the material. Scrape the surface onto a carbon coated grid and use micro diffraction techniques in a tem or backscatter diffraction techniques in an sem. The material may be beam sensitive so good luck but there is only one way to be certain, solve the crystallography.

Please go into the Scherrer equation and roughly calculate the peak width dependence on the crystallite size by your own. In your case of particle size of about 2nm (assuming crystallite size is also about 2nm) you will end up at about 0,15rad peak width in 2 theta; i.e. about 8°.

You know that when peak width is increased the peak height goes down. Thus at your small Ru concentration you will hardly see such a peak.

It might be due to ignorance of the peaks for C while choosing the element peaks for EDX spectra.

By edx you can determine the elements mostly located in the surface of the sample. If you are sure that the sample is homogeneous throughout, then this method will be more reliable.

Mohammed M. Obeid Thank you and good luck to you too for your future work

Dear Prof.Ian which parameter we use in the EDS report for calculation of the composition the wt% or at% and how reliable is this method to know the composition

One of the most common causes of mismatches being images and maps is sample movement. This may be caused by loss of volatiles from the sample (which will cause it to shrink) or, most often, the effects of temperature changes. These may be induced by the electon beam or through environmental changes - try letting the sample equilibriate in the beam for a while.

EDX is not the right technique to measure trace elements and it is really very approximate to measure carbon. The carbon detected here is probably located only on the surface of the sample.

100 to 200 nA? Are you sure? It is huge even for a FIB (typically max ion current is about 100 nA). The spot size will be terrible (micrometer?!). Most common parameters for EDX - current from 0.1 to 1 nA, HT from 10 to 30 kV, almost all normal SEMs meet these conditions.

I second a suggestion to analyze at two different kVs, to check if C is a surface contamination. Then - analysis of a target. It is not unreasonable to see 29% at as a contamination, but your SEM should be quite dirty.

I agree with the Dr. m. Toma, because presputtering time will decide the concentration of contamination on the target surface. The self made targets are rougher than the paid targets.

Dear Rajni Vats,

Formula to convert wt% to mole %

Calculate the molar wt of each oxide you want.

e.g. BaO

Atomic wt of Ba + atomic wt of O = Molar wt of BaO

I hope the information will help you.

With Best Wishes,

Samir G. Pandya

Different peaks for same element in EDAX is due to excitation of atomic energy level of respective element under incident energy of SEM/TEM. This will bring possiblility of XRAYS in energy range of different energy level i.e. k-alpha, L-alpha, k-beta, etc. It is depend on sample stability, incident energy of SEM /TEM to make proper excitation to get Xrays.

EDX measures X-rays that are emitted from an atom after a core electron has been kicked out. The X-ray is emitted when an outer shell electron (with higher energy) relax into the "hole" left by the electron that was kicked out. The energy of the X-ray is then determined by the energy difference between the outer shell electron and the hole. The outer shell electron jumps from a higher energy state to the lower energy state of the hole - the energy difference leads to emission of a photon with that energy - the X-ray. Since the outer shell electron can come from different shells with different energy, X-rays can be emitted with different discrete energies, which gives rise to a number of different peaks in the EDX spectrum. You may be able to see labels on the different peaks that indicate what transition occured to release X-rays at that peak.