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  4. 021: XPS For Beginners – Fundamentals 1

021: XPS For Beginners – Fundamentals 1

  • Created February 23, 2021
  • Author Mark Isaacs
  • Category HarwellXPS Beginners Course

Photoionisation cross-section calculator may be found at the Elettra sincotrone website.

Questions

Figure 1: Ag metal XPS spectrum
Figure 2: Si foil XPS spectrum

1: Above are two XPS spectra (Figures 1&2), one from Ag metal, one from a Si foil. Identify which spectra is which based on how many core orbitals you expect to observe.

Answer

Since we are looking only at single elements, we need only consider how many core orbitals exist. For Si, we have 3p valence orbitals, followed by 3s, 2p, 2s and 1s core orbitals – however due to orbital hybridisation, this 3s orbital will be used in sp3-type valence bonding and as such, will not be present as a core-orbital.

This leaves us with 2p, 2s and 1s. Ag, on the other hand, has many core-orbitals (4p, 4s, 3d, 3p, 3s, 2p, 2s, 1s). Given the spectrum in figure 1 has more than 3 peaks, we can rule out silicon and so it must be silver.

2: Based on the electronic structure of Ag, determine the orbital identity of the peaks labelled A, B, C, D and E

Answer

This is perhaps best approached working back from the right hand side to the left, since we are going from our core-orbitals closest to the valence electrons and working down.

This leaves us with:

  • E = 4p
  • D = 4s
  • C = 3d
  • B = 3p
  • A = 3s

Some of you may have assigned E as either 4d or 5s, as silver tends to complete it’s d-shell with a 5s electron – this is a good guess! The 4d and 5s orbitals are so close in energy they tend to all sit within the valence structure at the fermi energy. As we move on with this course you will learn about the different peak structures resulting from different orbital symmetries which will allow you to more confidently distinguish between f, d, p and s orbitals.

3: Which orbital would you expect to find in the region marked F?

Answer

As mentioned in the above answer, the valence region actually contains both 5s and 4d electrons. These are not readily distinguished using standard XPS, however some high resolutions scans, or the use of UV radiation as an exciting source, permits the deconvolution of the Ag 5s and 4d orbitals.

4: Assuming all instrumental factors are constant, and sample is homogeneous – which peak in an XPS spectrum would you expect to be larger given equimolar concentrations of lithium and chlorine and an excitation energy of 1000 eV? Li 1s or Cl 2p? Hint: use the photoionisation cross-section calculator from Elettra sincotrone.

Answer

So if we use the calculator linked above, we can determine the photoionisation cross-sections for both Li 1s and Cl 2p at 1000 eV exciting radiation.

  • Li 1s = 0.002628 at 1000 eV
  • Cl 2p = 0.1028 at 1000 eV

As we can see, the cross-section for Cl 2p is nearly 40 times larger! This means we will much more readily observe the Cl 2p peak.

5: Copper metal was determined to produce a Cu 3s peak of highly Lorentzian character and a FWHM of 2.27 eV (Figure 3). What do you think is the dominant cause of the peak broadening in this example?

Answer

Intrinsic broadening due to shorter core-hole lifetime. In fact, if we look at another peak, that of the Cu 2p (strongest emission and therefore most widely studied) we can see that this peak reports a much lower FWHM (0.87 eV) and still produces a highly Lorentzian peak – suggesting that instrument broadening factors in this case are relatively negligible to to resultant peak profile (samples were recorded using a monochromated X-ray source, which reduces the X-ray spread).

Figure 3: Cu 3s peak from copper metal

We will continue onto our next session, which will look at the factors affecting peak binding energy and how we can use these to determine chemical identities. We will also learn about ‘Auger’ peaks.

Continue to section 2

Further Reading

Electronic structure in solids – definitions (by HarwellXPS)

X-ray Photoelectron Spectroscopy (by Thermo Scientific)

Transcript

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