P627/Chem 542 Surface/Interface Science Spring 2013 Rutherford Backscattering Lab: Answers to Questions

Transcription

1 Intro Questions: 1) How is a tandem ion accelerator different from a so-called van der Graaf accelerator, and what are some of the advantages (and drawbacks) of the tandem design? A tandem ion accelerator has both the source and the target at (or near) ground potential, and there is a (positive) terminal, at a voltage +V T with an electron stripper midway down the acceleration tube. The source produces negative ions (typically -1 ) that are accelerated towards the (+) terminal in the acceleration tube. Upon reaching the terminal, the stripper will remove (at least) two electrons creating a positive (typically +1) ions which are now accelerated towards ground potential. For an ion that transforms from -1 to +1 will be accelerated to an energy 2eV T. So, for example if V T = 1 MV, then the ion beam will have an energy of 2 MeV. An advantage of the tandem design is that both the source and target can be near ground potential, simplifying operation. In addition, an ion beam can be produced with (at least) twice the energy that a single-ended accelerator could produce with the same terminal potential. Moreover, if the accelerated particle leaves the stripper in a high ionization state (i.e. +2 or greater), then an even higher energy beam can be produced. Disadvantages to the tandem design include the fact that it requires negative ions, which limits possible species for ion beams. Other minor considerations are that the van der Graaf design tends to have better energy control and does not suffer energy spreading that occurs when ions pass through the stripper. Therefore, from a beam perspective, van der Graaf machines are capable of better energy resolution, and thus, in principle, better depth resolution with the appropriate analyzer. 2) Suppose you have two samples comprised of a thin film of Au on a Si substrate. One sample has a carbon film on top of the gold, and the other does not. RBS spectra are taken from each sample under identical scattering conditions. How can I look and the two RBS spectra and, without doing any modeling, tell which sample has the carbon film on the Au? In the spectrum from the sample with C on the Au surface, the peak for the Au would be shifted to lower backscattered energy than the corresponding peak for the clean Au film. Of course the Si feature would also be shifted, and one film would have a C peak, but those features would be obscured by the strong Au signal. 3) Suppose I scatter 2 MeV protons from a thin Au film on top of Si, and measure the backscattered protons. In one case my detector is at a scattering angle of 180, in the other it s at 125. In which spectrum will the Au peak be at higher energy? Justify your answer. The ratio of the incident and backscattered ion energies is given by: [ ( ) ] As shown in plot, this equation is minimum when = 180. So, the energy of a 180 backscattered particle is smaller than the energy of the 125 scattered particle. - 1 of 6 -

2 4) Why is it easy to use RBS to measure the thickness of a gold film on a Si substrate, but hard to measure the thickness of a Si film on a gold substrate? As Au nuclei are much more massive than Si nuclei, backscattering from Au has a high cross section and ions backscattered from Au will have a much higher energy than ions backscattered from Si. So the Au feature will easily overwhelm a Si feature for a Si film on a gold substrate. 5) How can a tandem accelerator with a terminal voltage of 1 MV be used to create a beam of ions that have an energy of 3 MeV when they hit the target? If the source produces singly charged negative ions, but the stripper removes three electrons so that the stripped ion has a charge of 2 +, then the ions will have a kinetic energy of 1 MeV at the terminal, and gain an additional 2 MeV between the terminal and target for a total KE of 3 MeV. 6) Suppose you try to deflect a 1 MeV proton beam using a uniform magnetic field oriented perpendicular to the path of the beam. What must the strength of the field be (in Tesla) so that you deflect the beam by 90 over a distance of 1 m? (Note: Similar fields are used in our apparatus, but only a deflection of about 30 is needed, so the bending magnet is NOT 1 m in length.) A particle of charge q and mass m, moving with a velocity v in a plane perpendicular to a uniform magnetic field B, will experience a Lorentz force of magnitude F = qbv. This force induced a radial acceleration a r such that ma r = mv 2 /r = F r. Thus mv 2 /r = qbv so the particle moves in a circle of radius r = mv/qb. For a circle of radius 1 m, we can solve for the magnetic field as B = mv/q(1 m). The proton mass is 1.67 x kg, proton charge is 1.6 x C. To find the velocity of a 1 MeV proton we can use the Einstein relation E = Mc 2, we can write the rest mass of the proton as M = 938MeV/c 2. So 1 MeV = KE = (1/2)mv 2 = (1/2)(938 Mev/c 2 )v2 or v 2 /c 2 = (2/938) or v = (2/938) 1/2 c. With c = 3 x 10 8 m/s, we have v = 1.39 x 10 7 m/s. So, finally B = (1.67 x kg)(1.39 x 1-7 m/s)/(1.6 x C)(1 m) = T. Note that the earth s magnetic field is about 30 T, so this is about 5000 times the earth s magnetic field. QUESTION: Identify which elements are associated with the peaks and onset near channel 850 in the simulated spectrum. What is the reason the guess got the right edge of the two peaks correct, but the left edge of each is wrong. What is the reason that the simulation shows a large onset at channel ~ 850, while the data has an onset near channel 750? Explain why this is consistent with the way in which it missed the description of the peaks. The rise near 850 in the simulated spectrum is from the Si in the SiN substrate. The right edge of the two peaks is from the surface of the SnS film on top of the SiN substrate. The thickness of the SnS film is underestimated in the simulation. Therefore the falling edges of the Sn and S peaks occur at too high an energy. Similarly, the substrate peaks appear at too high an energy. All of these discrepancies can be attributed to a simulation what assumes an SnS film that is too thin. - 2 of 6 -

3 Data-based Lab questions. 1) Use SIMNRA to analyze each spectrum measured in the Lab. What is the composition and thickness of each film? Include a printout of your SIMNRA fit along with your lab report. Please see attached. For the SnS samples, we calculate the thickness (in nm) as follows. The density is 5.08 g/cm 3. Mol weight of SnS is g. So mol volume is weight/density = cm 3. Avogadro s number is x So Atomic density is x )/(29.68)*2 = x atoms/cm 3. From this, a thickness of 1 x atoms/cm 2 = (1 x atom/cm 2 )/(0.406 x atom/cm 3 ) = 2.46 x 10-8 cm = nm. 2) One of your spectra is complicated by the fact that the film is sufficiently thick that backscattered intensity from heavier target nuclei overlap intensity backscattered from lighter ions. Use SIMNRA to estimate the maximum SnS film that would give an RBS spectrum where the signal from Sn and that from S do not overlap. Please see attached. 3) In the initial guess for the example spectrum, the simulation looks like two sharp peaks, rather than the experimental data, which looks more like two plateaus. Why is that? Simulate a SnS spectrum for a film with a smaller thickness than that of the original guess. Explain why the simulated spectrum changes the way it does. The initial guess assumed a target thickness that is less than the depth resolution of the RBS technique. As a result, the signal is confined to a resolution-limited peak whose leading edge is comprised of ions backscattered from the surface of the film. If a thicker target is simulated, the peak widens into a plateau as some ions which scatter from target nuclei deeper in the film loose more energy than the instrument resolution. If a simulation is run for a target thickness smaller than the original guess, the location of the peak does not change, only the integrated intensity. In effect, for thickness below the resolution limit, RBS is not capable of elemental depth profiling. However, since the integrated peak intensity still reflects the absolute concentration of nuclei in the target, quantitative analysis of film composition is still possible. 4) Once you have determined the elemental composition of the third film, and identified the mystery element, simulate spectra using the elements to the right and to the left of the mystery element in the periodic table, and show how neither of these simulations gives a good fit to the data. See attached graphs. 5) For the third film, explain why the area of each peak is approximately the same, even though the concentration of each element may be considerably different. - 3 of 6 -

4 The cross section for Rutherford backscattering goes as the square of the atomic number of the target nucleus. So, for targets of different atomic number, difference in cross section can be balanced by differences in concentration, resulting in peaks of equal area. RBS Spectrum from thin SnS film: SIMNRA fit to thin SnS film. From the concentration shown in the target window, the film is slightly off stoichiometry, with an effective composition of Sn 0.44 S 0.43 O The film thickness is 1320 x atoms/cm 2, or 325 nm. RBS Spectrum from thick SnS film SIMNRA fit to thick SnS film. From the concentration shown in the target window, the film is very close to stoichiometry, with an effective composition of Sn 0.49 S 0.50 O The film thickness is 3800 x atoms/cm 2, or 935 nm - 4 of 6 -

5 Simulation indicating maximum SnS film thickness where Sn and S signals are fully resolved: As can be seen by this SIMNRA fit, for a film thickness corresponding to ~ 3000 x atoms/cm 2 the signal from the Sn and S features begin to overlap. For SnS, this corresponds to a thickness of 738 nm. Spectrum from sample with mystery element. Simulation of mystery spectrum. We see that the composition is approximately La 0.05 Sr 0.12 Fe 0.26 O The thickness corresponds to 150 x atoms/cm 2. Assuming a density similar to LaSrTiO 3 (5 atoms /2.56 x cm 3 ), the thickness is About 80 nm. (Should be thinner. Probably a bad model for density) - 5 of 6 -

6 Spectra assuming neighboring elements for mystery element. Simulation for La 0.05 Y 0.12 Fe 0.26 O Simulation for La 0.05 Rb 0.12 Fe 0.26 O Note that even if we readjust the composition to accommodate the peak intensities, the position of the mystery peak is sufficiently different from the data to invalidate these candidate compositions. The effect would be even more pronounced for the lighter element Fe. - 6 of 6 -