QUESTIONS AND ANSWERS

Transcription

1 QUESTIONS AND ANSWERS (1) For a ground - state neutral atom with 13 protons, describe (a) Which element this is (b) The quantum numbers, n, and l of the inner two core electrons (c) The stationary state these inner two core electrons reside in using both spectroscopic and X-ray notation (d) The B.E. XPS of photoelectrons emitted from the 2p 3/2 level of this atom when present in its elemental solid form (refer to Appendix B for tabulated values) (a) Al (b) n = 1, l = 0 (c) 1s, K (d) ev (2) A neutral ground - state atom containing 29 electrons present within its elemental solid with a work function of 4.5 ev experiences X-ray Photoelectron Spectroscopy: An Introduction to Principles and Practices, First Edition. Paul van der Heide John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc. 221

2 222 QUESTIONS AND ANSWERS photoelectron emission upon Al-Kα X-ray irradiation ( ev). The most intense photoelectron emissions have a K.E. XPS energy of ev. Describe (a) Which element the emissions are coming from (b) The B.E. XPS of the respective photoelectrons (c) The stationary state from which these emanated using spectroscopic notation and also X-ray notation (d) The subsequent L 3 V transition (results in Auger emissions or fluorescence) using spectroscopic notation (a) Cu (b) ev (c) 2p 3/2, L 3 (d) 3d 2p 3/2 (3) Al Kα 1 X - rays (Siegbahn notation) arise from KL 3 transitions (X - ray notation). In spectroscopic notation, this fluorescence results from the transition of an electron from the 2p 3/2 level to fill a core hole in the 1s level. The fluorescence energy is thus the energy difference between these states minus some small perturbation energy (final state effects). Assuming no perturbation occurs, what is the energy of the Kα 1 X - rays from (a) Mg and (b) Si? Use values listed in Appendix B. (a) ev (b) ev (4) If you were given a choice between using the Al or Ag anodes in a monochromatic source (from which the Al - Kα or Ag- Lα X-rays are used), which would you use if you were interested in ascertaining the B.E. XPS value of the Si - 1s core electrons from Si? The Ag anode would be used since only the Ag - Lα X-rays have the energy needed to induce the emission of Si - 1s core electrons (see Table 3.3 ). (5) Describe if and how the work function of a sample of interest can be derived using X - ray photoelectron spectroscopy (XPS) or ultraviolet photoelectron spectroscopy (UPS).

3 QUESTIONS AND ANSWERS 223 Assuming sufficiently high - energy resolution conditions are used, the work function of the sample being analyzed can be defined from the low K.E. XPS onset (high B.E. XPS drop - off) of the spectra collected normal to the sample surface, that is, the high B.E.XPS cutoff that appears close to the X - ray energy being used to generate the photoelectrons. The work function should equate to the energy difference between the X - rays used and the B.E. XPS value where the signal drops to 50% of its peak value (this is due to the finite energy spread of the photoelectrons recorded in the respective XPS instruments). Note: UPS using a discharge lamp can provide the work function values to a far superior precision than XPS using a monochromated source due to the narrower line width of discharge lamps relative to monochromated sources. (6) In typical XPS spectra, both photoelectron peaks and Auger electron peaks are present. The questions are (a) What instrumental parameter can be used to distinguish the two? (b) What is the result? (a) Use different X - ray source energies and compare the spectra. (b) The Auger peaks move to different B.E. XPS values as the source energy is changed, whereas the photoelectron peaks remain at the same B.E. XPS. (7) Describe what happens to all the photoelectron peaks observed in the spectra collected from an insulating sample when ineffective charge neutralization conditions are used. All detectable photoelectron peaks from all levels and from all elements present within the analyzed volume move by the same amount to higher B.E. XPS values. In extreme cases, these peaks will become part of the background signal. (8) A compound contains CH-, CO-, and CF-based organics. Using the knowledge that the B.E.XPS shifts for the C - 1s level can be ascribed to typical initial state effects (intra - atomic), list these in the expected B.E. XPS order from lowest to highest.

4 224 QUESTIONS AND ANSWERS CH < CO < CF. This is understood since the electronegativities of these elements scale in the same fashion. (9) What unusual initial state effect causes core - level B.E. XPS values of certain metal atoms/ions to decrease during oxidation? Note: Typically, the oxidation of metals results in an increase in their core - level B.E. XPS values. This unusual effect is, however, observed during the oxidation of select metals, including Sr and Ba. This unusual effect occurs when interatomic initial state effects dominate over intra - atomic effects (typically, this is the other way around). Note: Both have apposing effects on the direction of the ΔB.E.XPS during the oxidation of the photoelectron emitting atom/ion. (10) What could cause the spin orbit splitting value of photoelectrons from a specific level or a specific element to deviate from that typically observed? If the element also suffers multiplet splitting during photoelectron emission, this can cause the apparent spin orbit splitting energy to change. (11) From where do the increased final state effects in the form of rearrangement as well as the requirement to use Russell Saunders coupling arguments ( L-S ) in describing spin orbit splitting for the higher Z elements stem? Increased coupling that occurs between electrons in different stationary states. In the light elements, electrons in different stationary states can be considered as acting independently; that is, l and s (m s ) do not interact. (12) Calculate the theoretical bulk plasmon energy resulting from photoelectron emission from Cs present within its elemental solid (mass density = 1.90 g/cm 3 ). (a) 3.33 ev

5 QUESTIONS AND ANSWERS 225 (13) What is the flaw in Koopman s theorem when applied to examining XPS spectra? A frozen orbital approach is used; that is, this assumes no rearrangement/relaxation occurs for the valence electrons of the photoelectron emitting atom/ion. (14) The analysis of an unknown but homogeneous sample is carried out with photoelectrons collected at some off - normal takeoff angle. On rotating the sample around its normal axis (azimuth rotation), reproducible periodic spikes are noted in various core - level photoelectron signals. The questions are (a) What are these variations indicative of? (b) How may they be useful? (a) The variations in photoelectron intensities are due to photoelectron diffraction induced by the crystal the electrons emanate from (the sample) (b) This will only occur on a single - crystal substrate. Photoelectron diffraction can be useful in revealing the element - specific surface crystallographic structure. (15) What is the difference between a CHA and a spherical mirror analyzer (SMA)? Both are types of energy filters and both direct the electrons of interest between two hemispherical plates held at specific potentials. The difference lies in the fact that the average radii of the hemispherical plates used in a CHA are equal to the trajectory of the desired electrons. In an SMA, the radii of the desired electrons are much smaller. Note: SMAs are presently only used in XPS in parallel imaging studies. (16) Can XPS be used to define film thickness? If the film is less than the sampling depth and some spectral difference can be discerned between the film and the underlying

6 226 QUESTIONS AND ANSWERS substrate, then the answer is yes. Indeed, very precise measurements can be carried out on uniform films through a modification of the Beer Lambert Law. (17) Quantification of sputter depth profiles of even the most stable metal oxides (e.g., perovskites) with kev Ar + ions at an angle of 45 relative to the sample surface, using sensitivity factors defined from unsputtered surfaces, tends to reveal less than stroichiometric amounts of O. Explain why this occurs. Preferential removal of light elements relative to heavy elements typically occurs during sputtering of oxides, nitrides, and so on. This occurs as a result of the mass - dependent momentum transfer occurring between the incoming ion (Ar + ) and the element it interacts with, which in effect alters the sputter yields of the respective elements even though sputtered under the same conditions. This effect, termed preferential sputtering, can only be accounted for if the sensitivity factors are adjusted accordingly during quantification. Note: Sputter - induced elemental migration (diffusion, segregation, etc.) can also occur. (18) Scanning electron microscopy (SEM), which records secondary electrons (these peak in energy within the 1- to 10-eV range), is often used in conjunction with Auger electron spectroscopy (AES) and/or XPS to visualize the area of interest (this is done since SEM provides the best spatial resolution of the three, even though it does not provide elemental characterization capabilities). Define which of these three techniques exhibit the greatest surface specificity to the formation of carbon overlayers (this could be in the form of adventitious C, self - assembled monolayers, amorphous carbon layers, etc.) on gold. Assume that in all cases, the data are collected at normal takeoff angles, C - 1s photoelectron emissions resulting from Al - Kα irradiation are examined in XPS, and C KVV Auger emissions are examined in AES. AES will be the most surface specific to the formation of carbon overlayers on gold. This is realized since the inelastic mean free path of the C KVV Auger electrons is the shortest. Indeed, the surface specificity is a function of the electron inelastic mean free path. This exhibits a strong dependence on the electrons kinetic

7 QUESTIONS AND ANSWERS 227 energy ( K.E. ), with a minima in the inelastic mean free path noted at a K.E. of 80 ev for gold. Values then steadily increase with increasing K.E. above this minima and increase more rapidly with decreasing K.E. below this minima (see Fig. 4.6 ). The C KVV (KLL ) energy can be approximated as the difference between the C - 1s and valence levels minus the instruments work function (typically in the 3- to 5-eV range). The K.E. of the C-1s photoelectrons is well in excess of this ( > 1200 ev) and, as stated in the question, the K.E. of the secondary electrons is well below this (these peak at between 1 and 10 ev). Note: The reduced surface specificity of SEM also explains why this technique does not require the stringent vacuum conditions needed in AES as well as XPS; that is, SEM essentially sees through any adventitious C layers formed. The improved surface specificity of AES, however, explains why this technique is commonly used in examining such things as the surface contamination level during epitaxial growth.