Chapter 9 - Spectroscopy methods for material characterization Flashcards

1
Q

What are the basic principles of XPS? What are typical requirements to the equipment and how to perform the analysis of the XPS response?

A

XPS uses the photoelectric effect. We irradiate a sample with x-rays energetic enough to knock out the inner core electrons. We then detect these electrons as a function of their kinetic energy. Since we know the energy of the incoming photons, and we know the work function of the detector, we can calculate the binding energy of the electrons. These energies are characteristic for the element.

The requirements to the equipment is that we have UHV.

We also need advanced detectors that can filter the electrons based on energy, which is the hemi-spherical analyzers.

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2
Q

Analyze advantages and disadvantages of Auger and XPS methods regarding the surface analysis.

A

XPS:
+ Can perform depth profiles non-destructively.
+ Surface sensitive (5-10 nm)
+ Sensitive to chemical shifts
+ Very sensitive, and normally quantitative
+ All elements are detectable, except H and He
+ Chemical/electronic state information is available.
- Requires UHV, and that materials are vacuum

Auger:
+ Surface sensitive (0.5-3 nm)
+ Sensitive to chemical shifts due to bondings
+ Straight-forward depth profiling…
- … but destructive.
- Poorer quantitative results.
- Can get charging effects, which can distort the AES peaks.
- High power electron beams can destroy sample.

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3
Q

What are the principles of ARXPS? How can the depth profile issues be resolved using AES?

A

The principles of Angle Resolved XPS is that we take consecutive XPS measurements at different angles. From this we can get information primarily of the surface (high angle) and more in the bulk (normal incidence). This takes advantage of the escape depth of 5-10 nm, and the fact that a tilt will create an interaction volume closer to the surface.

By using models and iterative algorithms, we can construct the depth profiles. It is a non-destructive method.

We can thus get information about the depth profile of ultrathin layers.

In AES we have to sputter the surface with an ion beam in order to get depth information.

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4
Q

ISS: what are the basic principles and advantages?

A

Ion scattering spectroscopy. Here we get the information from the scattered ions. The spectrum is obtained by recording the number of scattered primary ions collected per second as a function of their energy from zero to the energy of the primary beam. Higher energy scattering peaks are characteristic of heavier elements, and lower energy scattering peaks are characteristic of lighter elements.

This technique is very sensitive to the uppermost layer.

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5
Q

Analyze advantages, disadvantages and challenges of XPS, AES and ISS in characterization of energy materials.

A

XPS:
+ surface sensitive
+ can get depth profiles up to 10 nm without being destructive.
+ nice for quantitative analysis
- must have elements that are vacuum-compatible
- expensive equipment due to fancy detectors and UHV
- can’t visualize
- Insensitive to H and He

Auger:
\+ surface sensitive (more so than XPS)
\+ can get depth profiles
- depth profiling is destructive
- not as good for quantitative measurements
- insensitive to H and He

ISS:
+ sensitive to the top-most layer
+ sensitive to H and He

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6
Q

What can often determine the functionality of energy materials?

A

Surface and subsurface composition of the materials.

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7
Q

What is ESCA?

A

Electron Spectroscopy for Chemical Analysis. The same as XPS.

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8
Q

Why is XPS and AES so surface sensitive?

A

Because of the low energies of the electrons. The escape depth is therefore low, because the electrons from further down will be scattered too much.

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9
Q

Draw schematics of the mechanisms of emission of photoelectron and emission of Auger electron.

A

See slide 9.

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10
Q

What are the two most used x-ray lines? Why?

A

The Mg K alpha-line, and the Al K alpha-line. Because they have a low energy dispersion, and have enough energy to reach many core levels.

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11
Q

Draw schematically the hemi-spherical analyzer.

A

See slide 15 and 16.

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12
Q

What are the sample requirements for an XPS experiment?

A
  • Has to withstand high vacuum (below 10^-7 torr)
  • Has to withstand irraditation by X-rays
  • Sample surface must be clean
  • The sample should be “reasonably” sized.
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13
Q

Give examples of materials that can not be used in XPS chambers.

A
  • Majority of organic compounds.
  • Some steels (only stainless steel is used)
  • Lead (soldering is done with lead-free solder)
  • Indium (low melting point prevents its use in baked systems)
  • Zinc, cadmium (high vapor pressures during system bake-out).
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14
Q

Why can we see F, N and C signals in an ITO-spectrum?

A

F because it is normally a dopant in ITO. N and C are contaminants.

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15
Q

How can one draw conclusions about chemical bonding states at the surface from an XPS spectrum?

A

Chemical bonds cause a small shift of the energy. This means that we can compare this to known data and draw conclusions of the nature of the chemical bonding state.

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16
Q

What is a potential pitfall when interpreting XPS spectra?

A

There can be overlapping binding energies for different species (for example silicides and carbides). This can produce ambiguous spectra.

17
Q

What is an important thing to take into account when constructing depth profiles from ARXPS measurements?

A

That the mean free path in different elements for electrons are different.

18
Q

What is used as the probe in AES?

A

Electrons.

19
Q

Why is the Auger spectrum usually differentiated?

A

The Auger peaks are superimposed on the SE background, and are generally quite weak. The differentiating is used to highlight these peaks.

20
Q

Why are we able to say something about the surface atoms from Ion Scattering Spectroscopy?

A

Because the ratio of the observed energy and the primary energy depends on the masses of the particles colliding (incoming ion and surface atom), and the scattering angle. Since scattering angle is known, we can solve for the mass of the surface atom and thus find out which atom it is.