Xeds,eels-take Home Exam Answers

Rajasekarakumar Vadapoo, Department of Physics, University of Puerto Rico, PR-00931, USA. More info: http://nanophysics.wordpress.com/

Transmission Electron Microscopy (TEM), Spring 2009. TEST 3- XEDS and EELS, Take home Answers. May, 2009 Part I: XEDS The X-ray emission and absorption edge energies for the most common elements are given in Table 1. Table 1. X-ray emission and absorption edge energies for the most common elements.

Table 1. (continued).

Question 1. ____/2 points Why is it possible to perform XEDS analyses with nanometer scale lateral resolution in a (S)TEM? If you want to obtain an XEDS line scan from a suitable TEM sample, how do you proceed?

In a thin specimen at high energy the electron scattering in the specimen is less and thus there is less interaction between beam and the sample. It gives more X-ray signal and a higher spatial resolution. In TEM we use thin specimen as well as we use very high energy electron beam, which makes it possible to perform XEDS analyses with nanometer scale lateral resolution in TEM. If we want to obtain an XEDS line scan from a sample, we should go for the thinnest region of the sample and go for the highest energy possible in that TEM, so that we can get good resolutions. If we can use any particular filaments which can give a less diverged beam like Field

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emission TEM, it would be an added advantage. We can take the XEDS measurements in spot mode with small spot size and take the measurements for long period which gives the advantage of less interactions of beam and sample and more detectability. To get a better resolution for a particular sample in a TEM, the beam spot size and the time of collecting sample would have a correlative relation. And we can play with tilting the sample, so that the XEDS detector would get maximum signal. Otherwise we can simply go for STEM mode and do the above appropriate conditions to get the line scan.

Question 2. ____/3 points How do you proceed "step by step" to identify the peaks present in an XEDS spectrum?

i.

Select the highest intense peak and try to match with the α-lines (i.e Kα, Lα , Mα ) of the materials possible to fit, then go down with their group of materials.

ii.

If found the α- lines then the other family of lines like β1, β2 , γ1, etc. should be present and could be indexed. All the other lines would be having less intense than α- lines. Sometimes the lines may not be resolvable.

iii.

If Lα line fits there is a possible of Kα/ Kβ pair at higher energy region.

iv.

Look for the next higher intense peak and redo the above procedures.

v.

Careful about the pathological overlap, system peaks, artifact peaks

vi.

If we suspect about any coherent bremsstrahlung peaks then try to redo the spectra with different voltage/ sample rotation and check whether the small peak shift.

vii.

If we suspect any sum up peak at twice the energy of any major peak then reacquire the spectrum with lower dead time and check for is the suspected peak disappear?

Question 3. ____/3 points On what depends the probability that an energetic electron will ionize an atom? XEDS analyses are routinely performed in the SEM; what are the advantages and drawbacks if we perform similar analyses in the TEM?

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The probability that an energetic electron will ionize an atom depends on the energy of the electron which is impinging and its angle of collision, binding energy of the electron which is being knocked out from the atom. Having XEDS in TEM over SEM: Advantages: i.

We can analyze composition for the whole thickness, which may give advantage for a particular case rather than only for surface in SEM.

ii.

We can really go to the atomic size line scan due to the low interaction of electrons with the sample in the case of TEM which have low thickness sample and high energy electrons, which is contrary in SEM.

iii.

Having very high energy of electrons we can get the K-lines of most of the elements. (This would be possible only if the particular SEM can operate at that energy).

Disadvantages: i.

We cannot do elemental analysis for the high thickness specimens.

ii.

Collecting area for the signal are very low due to the design of EM coils in TEM, which gives suffering in efficiency of collecting the signal but contrarily in SEM we have enough room to swing the detector to have more signal collecting area.

Question 4. ____/4 points What is the most probable transition when a K-shell vacancy is created and why? What can you say about the Kb radiation? Also, explain Moseley's law.

When K-shell vacancy is created then the most probable is an electron from L shell moves to the K shell, thus the atom undergoes the transition from the K state L state. This transition is accompanied by emission of Kα radiation. The next probable one would be Kβ after Kα radiation which happens when the electron from M shell moves to K shell. The Kβ transition needs more energy than Kα transition which would be explained by Moseley’s law. Moseley’s law says that the wavelength of any particular characteristic X-ray lines would be indirectly proportional to the square of Z values. The Moseley’s eqn is as follows: 4

√     Where, C &  are constants. Z- atomic number  - frequency.

Question 5. ____/4 points Is the critical excitation potential always met with the TEM? Give at least one example. Do we need to worry about absorption when we collect XEDS spectra for quantitative measurements in the TEM? How about with the SEM?

It is not necessary that the critical excitation potential always met with the TEM. For an example to excite Kβ line of the heavy element of atomic number 98 (californium) it is only necessary to have only 127.794 eV. Let’s consider a TEM has a maximum operating potential of 200 keV, but it could operate at 100 keV, in this case critical excitation is not possible. Yes, absorption is a concern while doing quantitative measurements with XEDS spectra in TEM even though we could employ the correction. The improper correction leads to a higher order of error which depends on the uniformity of the thickness of the sample and the collection angle (2α). In the case of SEM we can have high collection angle which would reduce the absorption effect but deteriorate if the sample size is large.

Question 6. ____/5 points What do we refer to as being the fluorescence yield? Discuss and illustrate with examples. If you had a lot of money and a good engineering team, would you like to modify your instrument and install an Auger electron spectrometer in your TEM?

The fluorescence yield for the k-shell can be defined as  = no. of atoms that emit K radiation / no. of atoms with a k- shell vacancy. Whether the vacancy is caused by incident X-rays or by electrons.

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Fig.A. the auger electron yield and the fluorescence yield as a function of atomic number.

The fig. A. shows the relation of auger electron yield and the fluorescence (X (X-ray) yield as a function of atomic no. The auger auge electron yield is (1 - florescence yield). By understanding the extraordinarily powerful tool of having Auger spectrometer in a TEM can give us 3-D D chemical analysis of a nanostructure, with the depth profile in a nanometer level, since auger electrons ejected from one to three mono layers. In a normal TEM we can have only 2D map and we cannott have a chemical analysis layer by layer. Let’s try to understand its applicability with the following example.

Fig.B. Metalic structures confined in CNT.

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If we want to have a chemical image of the structure in fig.B then Auger spectroscopy in a TEM would readily provide the 3-D chemical map layer by layer which cannot be possible in conventional TEM combined with the other spectroscopy measurements like EELS & XDS ( only 2D is possible). Having said that about the advantages of having Auger in TEM lets discuss about the discomfort we have in implications. Normally Auger can be done in low energy which we can do it in a SEM itself. In TEM having cylindrical Auger detector would be extremely painful. The sample has to placed at <100 to the beam to get the Auger electrons, which would be challenging in TEM. To perform depth profiling we need to use sputtering. Since sputtering is a destructive testing method, we have to take care of its post effect in the column as well as its effect on samples. We have to think about the difficulties of placing the sputtering set up and place it in an appropriate position inside the column, which would not affect the normal functioning of TEM as well as efficient.

In my opinion, despite all the above challenges, taking in to account the advantage of 3-D chemical analysis in atomic scale, it would helpful in the particular problem that we discussed above (the metal atomic structures confined in CNT). This would lead to the decision of going for its potential scientific implications if I have enough money as well as a good engineering team.

Question 7. ____/4 points Spectral artifacts can arise. Give at least three examples. How do you eliminate or reduce them?

Coherent bremsstrahlung peaks, sum up peaks and Pathological peaks are some of the spectral artifacts can arise during XEDS measurements.

i.

If we suspect about any coherent bremsstrahlung peaks then try to redo the spectra with different voltage/ sample rotation and check whether the small peak shift.

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ii.

If we suspect any sum up peak at twice the energy of any major peak then reacquire the spectrum with lower dead time and check for is the suspected peak disappear?

iii.

Pathological peaks: like the Kα & Kβ of the neighboring transition metals overlap which we can overcome by having the very high resolution XEDS detectors which can resolve the small energy differences between the lines.

Question 8. ____/5 points Because of the large number of elements encountered in materials science analysis, the number of possible interferences is much greater than in biological analysis, and so you must be careful. Particularly insidious is the interference in the first transition metal series, in which the Kb line of an element interferes with the Ka line of the next higher atomic number. Identify the elements concerned and discuss the implications.

These are called pathological overlaps. i.

The Kα & Kβ of the neighboring transition metals overlap particularly a. Ti Kβ ( 4.931 keV) line with V Kα (4.949 keV) b. V Kβ ( 5.426 keV) line with Cr Kα (5.411 keV) c. Mn Kβ ( 6.489 keV) line with Fe Kα (6.398 keV) d. Fe Kβ ( 7.057 keV) line with Co Kα (6.924 keV)

ii.

The Ba Lα line (4.47 keV) and the Ti Kα line (4.51keV)

iii.

The Pb Mα (2.35 keV), Mo Lα (2.29 keV), and S Kα (2.31 keV) line.

iv.

The Ti, V and Cr Lα lines (0.45-0.57 keV) and the K lines of N (0.39 keV) and O (0.52 keV).

Question 9. ____/5 points If we use the titanium specimen holder to analyze quantitatively a sample of SrwBaxTiyOz nanoparticles in the LEO 922 TEM, are there problems encountered? How do you proceed?

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If we want to analyze quantitatively the sample of SrwBaxTiyOz with the titanium specimen holder, then we have to make sure the electron didn’t interact with the holder which is very difficult to verify even then we use the focus beam in a small area of the sample. If we proceed we cannot independently quantify the Titanium. But we can take another path to solve this. Since this material in ABO3 structure B consist of (BaxTiy ), if we find out the stochiometries of Sr, Ba & O then from that we can quantify the Ti. Otherwise simply change the holder and proceed.

Question 10. ____/10 points

Figure 1 shows an XEDS spectrum recorded from an "unknown" sample. Identify each of the 14 peaks and present your results in Table 2. In some cases, there may be several possibilities. Explain your reasoning. Figure 1. XEDS spectrum obtained from an "unknown" sample. 14 peaks are indicated. See Table 2. Table 2. Identification of the 14 XEDS peaks indicated on the spectrum shown in Figure 1. Line Energy (keV) Possibility 1 Possibility 2 Line

Energy (keV)

Possibility 1

Possibility2

1

0.2

C Kα

B K(ab)

2

0.8

Ne Kα

Ne K(ab)

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3

1.4

Al Kα

Mg K(ab)

4

1.8

Si Kβ, α

Si K(ab)

5

2.6

Cl Kα

6

6.4

Fe Kα

Mn Kβ

7

7.1

Fe Kβ

Fe K (ab)

8

7.4

Ni Kα

9

8.0

Cu Kα

10

8.3

Ni Kβ

Ni K (ab)

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9.6

Zn Kβ

Zn K (ab)

12

10.0

Ge Kα

Ga Kβ

13

11.3

Se Kα

14

11.6

As Kβ

The possible X-ray emission peak has been noted. Among that possible absorption edge also

Se Kα As Kβ

Zn Kβ Ge Kα & Ga Kβ

• Cu Kα

Ni Kα Fe Kβ

Fe Kα / Mn Kβ

Cl Kα Al Kα

C Kα

Ne Kα

Ni Kβ +?

Si Kα & Si Kβ

notified to explain the peak no 10.

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The above fig. shows the corresponding peaks indexed. Except the peak 10 all other peaks have been notified with the possible error of +/- 0.5 eV. Even though peak 10 fit with the Ni Kβ , it has higher intensity than its corresponding Kα ( the peak 8). It cannot be. Any Kβ has to have less intensity than its Kα. Taking account of Ni K(ab) edge also coincide with the corresponding Kβ peak, it could be possible to have a background intensity sum up with the Kβ peak intensity. Several absorption peaks are coinciding with the major peaks position, suggesting that there might be a background intensity due to absorption which could explain the high intense of Kβ peak than its Kα intensity position in the above spectrum. Question 11. ____/5 points If you have a bulk geological sample (say a rock) and you want to quantitatively characterize this material by determining its composition, which instrument would you choose and why? If you want to determine the composition of large inclusions (100's of micrometers) in this same material would you still proceed in the same manner? If you want to learn more about the grain boundaries in this material (quantitative elemental analysis), what would you do?

If it’s allowed to do the destructive testing, then I would go for inductively coupled plasma atomic emission spectrometry to know the composition. As we know some of the geological samples we would like to employ NDT (non destructive testing) method to preserve it. In that case, taking account of its huge size and the possibilities of having liquids trapped inside the sample, we cannot go for TEM or SEM which would deteriorate its vacuum systems. As we know X-ray diffraction can give the phase details but if the sample are unknown (mainly in geological sample) we cannot get the conclusion. Contrarily X-ray Fluorescent Spectroscopy (XFS) can give the elemental details but not the phase. So, if we combined the XRD & XFS we can do the quantitative elemental analysis of any unknown sample. We can go for any optical microscopy to see the grain boundary but the sample size and the nature of grain boundary location is important (should be in the surface), still this method would be unsuccessful to find out the stoichiometry. If we want to learn more about grain boundaries, as for as in my knowledge we have to go for a technique which contain microscopy as well as elemental analysis could carried out at near room pressure/ low vacuum would be the

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solution. So, the SEM work at low vacuum combined with WDX would be one of the possible methods to do elemental analysis in the grain boundary areas.

Question 12. ____/5 points It is sometimes possible that automatic recognition programs mistakenly identify elements. Comment on Figures 2 (a) and (b) which are two XEDS spectra recorded on two different samples (FeS2 and KBr).

Figure 2. XEDS spectra recorded on two samples (FeS2 and KBr).

In fig.2 (a), the software selected PbMα instead of S since SKα (2.307 eV) comes in the same range of PbMα (2.342 eV). In fig.2.(b), the software mislead UMα (3.165 eV) instead K Kα (3.312 eV) which are in the same energy region and similarly Sb Lα ( 3.604 eV) instead of K Kβ (3.589 eV) in the same region.

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Part II: EELS Question 1. ____/5 points What are the parameters characterizing an EELS spectrometer?

i.

Dispersion: Dispersion is the distance in the spectrum between the position of electrons which differ by the energy dE, which is a function of strength of the magnetic field and the energy of the incident beam. Typical value of dispersion for SEELS:





 2 µm/eV.

For PEELS:  1.5 µm/eV ii.

Energy resolution (∆E): Full width half maximum (FWHM) of zero loss peak.

iii.

Point spread function: Delocalization of information in the EELS spectrum.

iv.

Spectrometer collection angle (β): The intensity of the spectrum is directly proportional to the collection angle.

v.

TEM: Selection of objective aperture and spectrometer entrance aperture size affects the resolution. Small aperture increases the resolution.

vi.

Spatial resolution: It depends on the TEM mode. For the highest spatial resolution we have to operate in diffraction mode.

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Question 2. ____/5 points What are the origins of the background? How do we subtract it? Multiple scattering and extension of ionization edges with onset at lower energies are the origins of the background in EELS. There is no first principle method exists until now to model the background of EELS to the level of degree possible in XEDS using variations on Kramer’s law. But curve fitting, graphical method and differential spectra are the three methods commonly used to subtract the background.

Question 3. ____/10 points Show schematically the correlation between ELNES and the electron band structure for a NiO thin film material

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A core electron can receive enough energy from the electron beam to be ejected but not enough to escape to the vacuum level. In this case the electron will be in any one of the possible energy level above the Fermi energy. We can find the probability of the electron filling certain electronic states above Fermi energy by correlating with the intensity in the ionization edge. Thus the ELNESS would give local density of states (LDOS).

Question 4. ____/5 points

400 eV (K(ab))

186 eV (K(ab))

What are the advantages of EELS versus XEDS? Comment on Figure 3.

Figure 3. EELS spectrum of BN. Advantages of EELS vs XEDS: i.

EELS gives more information.

ii.

Enable identification of elements and quantifying it

iii.

Give electronic structure

iv.

More efficient than XEDS (scattering in forward direction also enable this)

Disadvantage of EELS: i.

Interpretation is difficult because of complex absorption edges.

ii.

Rapid change in background and the overlap of absorption edges.

Comment on fig: The adsorption edge of the spectrum exactly matches with the boron and nitrogen as shown in fig. So, the given sample contains Boron and Nitrogen. C- content also present, it could be from the grid. The adsorption edges show some fine structures which corresponds to the Density of states (DOS).

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Question 5. ____/5 points One advantage of the LEO 922 is the easiness to apply a zero loss filter. What are the significances, applications and benefits?

Significances: i.

Due to the omega filter in Leo922, electron following particular path through the spectrometer can be selected by the post spectrometer slit, which gives better leverage in operations.

ii.

We can select the intensity in any part of the EELS spectrum and form the image.

Application & Benefits: iii.

With Parallel EELS, it is possible to acquire the spectrum really fast and we can do the images by post processing whenever we want with pixel by pixel.

iv.

We can analyze the spectrum afterwards without going back to microscope once again.

v.

We can use the spectrum for post processing in several ways and give the possibility of unexpected correlation needed

vi.

All the information in the EELS spectrum can be mapped discretely like valence state images, thickness images, elemental images, dielectric constant images, etc.

Question 6. ____/5 points Comment on Figure 4.

Figure 4. Obtained from BN. 16

The above figs are CBED pattern of BN. Since BN exhibits two different crystal structure diamond-cubic and hexagonal, the viewing direction of the above patterns could be [111] for the case of cubic and [0001] or [-12-13] for the case of hexagonal. The first one is the unfiltered pattern and the second one is the energy filtered one which gives superior resolution. The corresponding densitometer traces are over the non central pattern. The second spectrum clearly reveals the peaks with high resolution corresponding to its pattern.

Question 7. ____/10 points Carefully describe and comment Figures 5 and 6.

Figure 5. Series of electron spectroscopic diffraction patterns of a [111] oriented Si TEM sample (about 50 nm thick) with excess Kikuchi bands at higher energy losses: (a) unfiltered, (b) E = 0 eV, (c) 16 eV, (d) 100 eV, (e) 1,800 eV below and (f) 2,000 eV beyond the Si K edge.

The absorption edges for Si exist in the following energies LIII @102 eV, LII @103 eV & K(ab) @1842 eV. We expect the above figures shows how the increase in electron loss energy (inelastic scattered e-) would change the DP. As we have seen (b) shows the sharp pattern along with the Kikuchi line which would happen when we filtered the electrons except E=0 compare to the unfiltered one. The fig. (c) is at 16 eV, which would be around the Plasmon energy. It might be due to the convolution in that region the DP is diffused. The fig. (d) where we have an absorption edge of Si which naturally diffuse the bragg reflection due to the fluctuation in the intensity. Fig. (e) which is below the Si K edge, means that the energy is almost near to remove 17

an electron from K-shell, which gives the Kikuchi lines. Above the Si K edge fig.(f) which means the energy focus near to the nucleus of the Si atom, where the Kikuchi line intensity is further increased.

Figure 6. Series of electron spectroscopic diffraction patterns of a [111] oriented Si TEM sample (about 800 nm thick) with defect Kikuchi bands: (a) E = 100 eV, (b) 500 eV and (c) 1,300 eV. Here because of increase in thickness we get the defect Kikuchi lines in contrary to the high energy low thickness DP fig 5(e). So, this effect is something due to the thickness. We know that by increasing the thickness of the sample, we increase the chance of multiple scattering. So, it could be due to that. Further increase in energy the central bright diffused disk is increasing in its diameter.

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