Hi Gert, regading the comment "the sample (image) changes during exposure": I checked a lot of samples containing a Bi, and under 200kV it will start to move alot. I hope you can see this video:

https://twitter.com/o_CEMM_o/status/875689257180028932

The Bi-compound was partially melted, and you can see the Bi crystals "walking" around under the beam. TEM guys were probably referring to something similar...

Dear Gert,

there is a Bi line (NI,PII.III) at 932 eV which in case of (Bi-)heterogeneity of your sample may contribute to the Cu L alpha1,2 lines at 930eV.

Thanks Gerhard, but can this become that high?

I do not really know. But when there is a significant amount of Bi which in addition blocks (attenuates) the Cu lines it could be.

The neighbouring Bi lines are at 1.354 and 1.330 as well as 1.88 and 1.9, 2.0, 2.24 and 2.4 keV. Do you see them too? (at that position)

I only got the information, that there is some Bi, but in fact I have not seen any Bi-radiation. I only heard from the TEM guys that "the sample (image) changes during exposure" (whatever this means). That in that case the amount of radiation changes is not surprizing, but why the K-to-L ratio in such a considerable way?

If you do not see any further Bi lines the Bi should not be the reason for high Cu L yield.

But another idea:

what is your excitation energy and even excitation energy distribution?

For Cu L lines at bit over 1 keV is necessary but for Cu K you have to have around 9keV and even more.

When exciting with 9keV electrons you will have only 'one' K-excitation per incoming electron.

But electron energy easily goes down during secondary scatter. So some electrons having around 1 keV will come up per one 9keV primary electron. It is a very simple picture*) but may cause unexpected increase of relative L/K ratio.

You should trigger a Geant/ EGS electron excitation simulation of this case of such a thin Cu grid.

*) excitation cross-sections and x-ray escape depths are not considered yet as well as energy dependence of free mean paths of electrons.

This was my idea :-), but the investigations have been performed using 200keV. I have my doubt that a material reduces the energy that much that the Cu grid will "reflect" the electrons with only a few keV above 8. But I do not have any other idea...maybe you? :-)

Thanks!

200keV electron energy is a lot. But we had done a lot of simulations, which show that the kinetic electron energy is easily brought down by the surface near region of a sample. Thus a few electrons above K-edge threshold may remain but much more electrons above L-edge threshold, which are able to excite the Cu grid accordingly.

It is worthwhile to do such simulations and look at the sample thickness dependence of remaining electron energy distribution escaping from the back side of the sample, which impinge onto the Cu grid.

Integrating the geometry of the Cu grid into the simulations and looking for the Cu line output would be the next step.

A nice topic for a master thesis.....

If you are OK with EDS spacial resolution of about 20 nm (it depends mostly on specimen drift) and if you have a spare specimen, you may want to try SEM-EDS. Just mount your specimen on C substrate (you'll lose information about C, of course) or Al substrate (a lot of unwanted radiation). Main problem - visualization, but as I understand you do not have pictures anyway, you just need to find your specimen. I did things like this one with biological specimens stained with Os and for mineral; of course no big problems with them. For non-stained biological specimens I used low voltage (1-3 kV) just to find places whre specimen is present. Not really a great solution, but you will get rid of TEM artifacts (like stray radiation) and you can use low voltages which is much better at low energy end of spectrum.

Dear Vladimir, we have images and we have ED-spectra from TEM. I was only asked for the strange L-to-K ratio they could not explain, and me neither. Since I have not seen anything like this, I asked it here at Researchgate in hope someone is more experienced and has an idea under which conditions the L-radiation of Cu can become bigger than the K-radiation. They also showed me spectra from the same sample where it was OK: Only at certain positions they found this strange behavior.

Dear Gert, what exactly means "certain positions"? Do you mean different positions on grid in X-Y plane, or you tilted the sample against the EDS detector?

I mean at another position on the sample (i.e. also another positions related to the grid). As far as I know, the collection conditions (detector or sample position has not been changed).

Hi Gert, regading the comment "the sample (image) changes during exposure": I checked a lot of samples containing a Bi, and under 200kV it will start to move alot. I hope you can see this video:

https://twitter.com/o_CEMM_o/status/875689257180028932

The Bi-compound was partially melted, and you can see the Bi crystals "walking" around under the beam. TEM guys were probably referring to something similar...

Thanks Janez...I will check, what does "changes" mean :-). Nevertheless, interesting experience.

Hi Gert,

you may consider quick qualitative EELS experiment to cross-check the elements present in your sample without compromising the spacial resolution: you will avoid stray radiation as the EELS signal is highly localized and overcome eventual signal overlapping of Bi and Cu (possible false-positive Cu-L counts). The other reason of higher Cu-L counts might be you actually have Cu in your sample?

Regarding this, there are only two things that come to my mind. 1.) secondary fluorescence caused by stray radiation, 2.) summary peak mentioned by Gerhard above.

I vote for secondary fluorescence and recommend you to perform EELS spectroscopy proposed by Paulinka.

Cu L-edge excitation of via secondary fluorescence from the sample containing C, O and F does not work due to photon energy reasons.

C-, O- and F- K-fluorescence photon energy is much smaller than Cu L-edge threshold(s).

Hi Gerhard,

what else can influence the photons emittance than secondary fluorescence? If you consider that it happend only in certain sample positions, do you have any other idea for explanation? The stray radiation usually comes from inner parts of microscope or some impurities on samples. Don´t forget, Gert had powder sample inside. According to my experiences, together with powder sample the impurities (with very low %) of totally different composition are almost still present.

Dear Vladimir,

please define your term 'secondary fluorescence'.

'Secondary radiation' is known as fluorescence; but the term 'secondary fluorescence' is not known to me, sorry. It may be that we talk about different things.

Nevertheless impurities above atomic number of Na (Z=11) are able to excite Cu L lines via K fluorescence and even Auger electrons (high Auger yield & low fluorescence yield for low atomic numbers). Such impurites may be present in the sample, as you suggested and could explain the only spot-wise appearance of the high Cu L-line yield. I totally agree with you.

However the 4pi emission characteristics of Auger/fluorescence is a very lossy one with respect to subsequent excitation. Here Monte Carlo simulations could help to estimate the integral Cu L line yield from all excitation sources, as I have suggested above.

Dear Gerhard,

The secondary fluorescence mean that fluorescence phenomenon is generated artificialy by another substance, presented around (grid, inner parts of TEM) or in the sample directly (impurities), which underwent itself fluorescence. The photon, emitted from this another substance, can induce fluorescence in the sample.

Yes, but it does not explain why it happens at some positions and at other positions not, does it?

Attached are three spectra acquired at this sample. The upper is the one representing the mysterious Cu-L part. the middle shows the signal which comes from the C foil and the Cu grid. The lower displays the signal of the Bi-O-F particle overlaid by the Cu grid and the C foil.

Thanks Vladimir for clearification.

We both meant the same.

@ Gert:

first of all thanks for sharing the spectra.

The idea behind Vladimirs answer is that impurities randomly spead across the sample may increase Cu L excitation due to its fluorecscence and/or Auger electrons stimulated by the shower of scattered electrons arising from the primary electron beam (or even the primary beam; but the primary electron energy is quite high). Dose enhancement is known for sufficiently high Z material but should in principle also work for low to medium Z material when significant Auger emission also works. My special view here is on 'secondary' Auger emission because Auger electrons will bring up the dose more efficient than fluorescent x-rays.

According to your spectra you have got impurities such as Si, Cl and Bi. Their lines are all able to excite Cu L edge and so the appropriate Auger electrons. Auger yield of Cl and Si is almost one order of magnitude larger than K x-ray fluorescence. So please imagine that the Si and Cl lines of your upper spectrum are about 10 times larger......

Please have a look at Fig. 4 of the link:

https://en.wikipedia.org/wiki/Auger_electron_spectroscopy

Dear Gerhard, this sounds like a reasonable explanation. I will check this.... Thanks a lot!

Hello Gert, I also found an article about this topic:

Article Energy dispersive spectrum (EDS) study of copper grid effect...

I am having the same issue with Bi(0,85)Dy(0,15)FeO3 samples, copper peaks are unusually high. Could this affect the quantification of the elements in the sample? Because the nominal concentration of Dy was supposed to be 15% and it is showing a 55%, I don't know how to explain this. I attach my result. Thank you very much for you help.

Dear Lina,

in your case Bi is able to excite Cu K-(K edge at 8,9 keV) and Dy L-(L edges 7,8 to about 9 keV) fluorescence radiation via Bi L-fluorescence and Bi L Auger electrons (energies in the range about 10 to 15 keV). You have got a lot of Bi in your sample and Bi L Auger yield is quite high (~60%). Due to high stopping power of the Auger electrons significant excape from the sample is unlikely.

I do not know whether Auger electron excitation of Cu-K- and Dy-L-edges via Bi L secondary radiation (especially Auger electrons) is taken into account in the ZAF correction of your measurement system.

For fluoresecence and Auger electron yield please have a look at page 11 of the link:

https://public.wsu.edu/~scudiero/documents/571-Auger-Lecture.pdf

Additional remark: Your Fe fluorescence should also be enhanced due to Bi L Auger excitation.

Dear Gerhard,

I am very grateful I can count with your knowledge and expertise through this platform. And I also want to thank you for your soon answer and reference recommendation. I was so troubled with this results and I could finally understand what was happening with my samples, since I asked at University and nobody could help me to explain that, not even the TEM technician who is new and seems that she doesn't know about this issue. Iron concentration appears also high because of Bi excitation.

Thank you very much again.

Best regards,

Lina.

Thanks Lina,

for feed back that Fe also appears higher than expected from chemical composition.

Due to its quite low K-edge energy (~7,1keV) Fe might also be affected a bit by Dy L secondary (Lß) radiation. So you should also expect a bit higher Fe values when, for example, taking a sample without Bi such as DyFeO3.

For Cu the K-edge energy (~8,9keV) is too high and Cu EDX yield cannot be affected by Dy L secondary radiation.

So a check with DyFeO3 (if such a sample is available) is recommended in order to see experimentally that Bi is the reason for Dy and Cu enhancement.

Good luck and best regards,

Gerhard

The Cu K/L ratio strongly depend on absortion effects of the relativly soft Cu-L and hard Cu_K energy. This means that the Cu_L line is absopted much quicker than the Cu_K line. Now comes the critical question where does this absorption coming from.

1. Absorption of the holder. This means that if you have a shadow of the holder towards your detector the hard xrays (Cu_K) are pathing through the holder while the soft Cu_L xrays are absorpt. Tilt towards the detector should change the ratio therefore.

2. Absorption in the window of the detector. This is constant and will not vary with beam positions or tilt.

3.Absorption by a shadow of a rough sample surface towards the detector. this is higlhy sample position dependant

Since in your case the Cu signal is not generated from the material exposed by the electron beam, we have to take other phenomena into acount. We have to understand how the Cu signal , which is obviously steming from the grid is excited via a secondary effect. Here two main effects are to be considered.

1. Spray electrons from the illumnation system.

Since there are apertures in the illumnation system on their edges inelastic scattering effects lead to unfocussed electrons bombarding the entire sample. This leads to a Cu signal as background in your spectrum independnat form the area illumnated. By clever chocie of apertures (C1 and C2 thisd effect can be minimised)

2. Excitation of x-ray by backscattered electrons.

If the primary electron hitting the sample some of them are backscattered. They see the magnitc field of the objective lens and are forced on spiral trajerctories by Lorentz force. If in this trajectory a cu bar of the cu grid is present they excite CuL and CuK xrays. There is a high propability that the CuL xrays are absopted in the thick Cu grid bar and only the CuK xrays are reaching the detector.

This leads to a low CuL/K ratio.(1/10) and can vary dramtically with the postion in respect to the next Cu Grid bar (absortion effect) and the density of the material in the primary electron beam (# of bcksc electrons).

I believe this effect 2. is the dominate one in your case. If you examine thin foils of Cu for example the ratio of CuL to CuK is typically 2 to 5 depending on the thickness of the sample and the detecotr window used not only 1/10. This shows that the self absorption of the CuL xrays of the thick Cu grid leads to this low number you reporting and explains its strong variation.

Dear Gerhard,

Thank you very much. I am going to check if maybe there is DyFeO3 in another lab, so I can provide a proof about this issue.

Thanks again for such complete explanations.

Best regards,

Lina