Nuclear reaction Analysis, A presentation by Younes Sina, The University of Tennessee, Knoxville

1. Nuclear Reaction Analysis:Particle-Particle Reactions A course of Professor C.J.McHargue Younes Sina

2. The basic principle of Ion Beam Analysis (IBA): Electronic stopping: when an incoming ion beam enters the surface of a thin film it loses some energy while penetrating the film, this stopping process is due to interactions of the beam with the atomic electrons in the energy shells of atoms. Nuclear stopping: The loss of energy can also be caused by the interactions of the beam with nuclei of the target sample.

4. Nuclear reactions are a very useful tool for light elements depth profiling on heavy matrix . The basic principle is when projectiles undergo nuclear reactions with the nuclei of the target material. which leads to the emission of product particles.

5. Nuclear reaction analysis (NRA), like Rutherford backscattering spectrometry (RBS), exploits the interactions of charged particles with atomic nuclei. There is one condition for nuclear reactions to occur, which is the incident particle should possess energy greater than the coulomb barrier potential.

6. Abilities of NRA Different isotopes undergo different nuclear reactions, therefore each of them have unique characteristics such as energy release, excitation states, angular distributions and cross sections . NRA technique has some important features: • High selectivity for the determination of particular light nuclides. • High sensitivity for many nuclides, which are difficult to determine using some other techniques. • It is a non destructive technique • It can be used to analyze more than one light element in near surface layers of materials at once.

7. NRA and RBS techniques are both governed by kinematics equations and their scattering geometry is the same.

8. Coulomb barrier potential is given by: E1 Z1 M1 Z2 M2 M1 Z1 E0

10. Scattering chamber

11. High or ultrahigh vacuum

12. Surface barrier detector

13. Standard nuclear electronics

15. Ein(x),E(Ein(x),Q,θ) E0 x αin θ=θ(αin,αout) αout Absorber foil Eout(x) Detector Eabs(x)

16. The energy of the particles leaving the target is given by: Stopping power

18. Thin detector technique

19. Electrostatic or magnetic deflection technique (also electrostatic detector and magnetic spectrometer)

20. Time -of-flight (TOF) technique

22. Thin detector technique This technique is used when proton and α peaks overlap and αspectrum contains more information than the proton spectrum Nuclear Reaction

23. Electrostatic or magnetic deflection technique (also electrostatic detector and magnetic spectrometer) This techniques are based on the effects of the magnetic and electrostatic fields on energetic charged particles

24. Time -of-flight (TOF) technique This method is used for large number of different particles that are generated simultaneously when they have to be distinguished. The technique is based on the simultaneous measurement of the energy and velocity of the detected particle. From these two values, the mass of the particle can be calculated.

25. Coincidence technique In this method the two reaction products are detected in different detectors in coincidence

26. Measurement Methods Overall near-surface contents Solid angle If (E)=(E0) for E0>E>E-ΔE Cross section Peak area Collected charge

27. Scattered M1 E1 E0 θC θ M1 M2 φ φC Incident Target E2 M2 Recoil

28. Light M3 E3 E0 θC θ M1 M2 φ φC Target Incident E4 M4 Heavy

29. E3=E1αE+βE for θ=165˚

30. Nonresonant depth profiling As the thickness of the sample becomes larger, the peak of the reaction product becomes broader. The peak shape is the convolution of the concentration profile with the cross section and the depth resolution. The depth scale can be calculated from an expression similar to ion backscattering depth scale: Atomic density Nuclear reaction stopping cross-section factor

31. Atomic density Nuclear reaction stopping cross-section factor Fitting parameter in and out can be calculated assuming either a surface-energy or mean-energy approximation

32. A handy approximation by assuming that all energy spread distributions are Gaussian: ΔEd=detector resolution ΔEg=geometrical energy spread ΔEms=energy spread due to multiple scattering ΔEabs=energy straggling in the absorber foil in and out : straggling for incident and reaction product E1 and E2 : energies of the incident ion and the reaction products at depth x

33. Example Consider a SiO2/Si sample where the thickness of the oxide layer is unknown. We want to measure this thickness by measuring the oxygen content of the sample . The best reaction for this measurement is 16O(d,p1)17O. This reaction has a plateau below 900 keV, as shown in the following figure.(between the arrows)

34. The 16O(d,p)17O reaction for both p0 and p1 reaction products. This reaction has a plateau below 900 keV down to 800 keV.

35. Example E0=834 keV θ=135˚ Mylar thickness=12 μm (all of the deuterons with energies below 900 keV are stopped by this foil) Protons: p0(Q=1.919 MeV) and p1(Q=1.048 MeV) The energies of the protons can be calculated from the general kinematic factor: E31/2=B±(B2+c)1/2≈(αEE1+βE)1/2

36. Example

37. Example Energy of proton using 12μm Mylar from appendix 10: E in to Mylar=2.36 MeV for p0 & 1.58 MeV for p1 E out of Mylar= 2.12 MeV for p0 & 1.23 MeV for p1 E taken by Mylar=0.24 MeV for p0 & 0.35 MeV for p1 We use these numbers for our sample

38. Example Eout=2.312 MeV for p0 &1.206 MeV for p1 Detector E3=2.36 MeV for p0 &1.58 MeV for p1 Eout p Light Mylar E3 E0=834 keV θC θ d 16O E0 φ φC Target Incident E4 All of the deuterons with energy below 900 keV are stopped by the foil. 17O Heavy

39. Example E0=834 keV Energy range that we can use ΔE=30 keV S(E)=7.6 eV/Å Average stopping power of deuterons in SiO2

40. The measured spectrum from 834 keV deuterons on a SiO2/Si sample. The protons from the D(d,p)T reaction are not identified here because they have almost the same energy as those from 16O(d,p1)17O reaction.

41. Example Total number of oxygen atoms Standard : Ta2O5 80 V thick Ta2O5 content 6.69×101716O atoms/cm2 Because the collected charge in both cases are the same: Which corresponds to 1000 Å of SiO2

42. Example Consider the application of the 12C(d,p0)13C reaction in analyzing the amount of carbon in a diamond like carbon (DLC) coating that contains a substantial amount of hydrogen. We will take αin=0° and αout=15°. From Appendix 10: Flat plateau in the region between 0.9-1.0 MeV for the 12C(d,p0)13C reaction.

43. E31/2=B±(B2+c)1/2≈(αEE1+βE)1/2 E31/2≈(0.591×1+2.38)1/2=3 MeV For 1 MeV deuteron in carbon εin=7.19×1015 eV cm2/atom For 1 MeV deuteron in hydrogen: εin=1.99×1015 eV cm2/atom For 3 MeV deuteron in carbon εout=2.15×1015 eV cm2/atom For 3 MeV deuteron in hydrogen: εout=0.47×1015 eV cm2/atom

44. 0.59

45. Assuming that the DLC coating is composed of 40 at.% hydrogen and 60 at.% carbon, the Bragg rule gives:

46. Resonant depth profiling The depth resolution of NRA is generally very poor because of the absorber. The depth resolution can be improved if resonances exist. Resonance energy Ē=(E0-Eresonance)/2

47. If all contributions to the energy spread are Gaussian, the depth resolution is given approximately by: energy spread of the incident beam width of resonance energy spread due to small-angle multiple scattering

48. Example A 18O(p,α)15N 629 keV resonance was used to determine oxygen exchange in high-Tc superconductor material Y1Ba18Cu29O7 (the composition of the sample was measured by RBS) High Tc layer was relatively thin, then we can use an average stopping power of 10.87 eV/Å (calculated using Bragg rule) Eq. shows the depth scale to be 1 keV/10.87 eV/Å=92 Å/keV Γ=2.1 keV (assuming the energy spread is negligible relative to the resonance width) then Δx(depth resolution)=200 Å width of resonance

49. Example The α yield (count per channel) at beam energy E0 is given by: Reaction cross section Scaling factor Concentration of 18O in the jth layer Energy distribution of the incoming beam with initial energy Ei in depth xj, with E(Ei,xj) the beam energy and σj the straggling at depth xj

50. Example Levenberg-Marquardt method to fit the concentration profile: The best fit was achieved by using a 150-Å-thick surface layer enriched to 30% followed by constant volume enrichment to 3.5%

51. The 18O(p,α)15N 629 keV resonance from a Ta2O5 reference sample. This resonance sits on a continues background, and therefore, the spectrum dose not go to zero outside the reaction energy.

52. The 18O(p,α)15N 629 keV resonance from a high-Tc sample following an 18O oxygen-exchange anneal. A peak can be found at the resonance energy that indicates an enriched surface layer.

54. The lateral uniformity over the beam area must be high.

55. The standard should be amorphous to avoid channeling effects.

56. The targets should have long-term stability in air, in vacuum, and under ion bombardment.

58. Example Consider a sample contaminated on the surface by oxygen and carbon. Because it is quite difficult to prepare a carbon standard, a Ta2O5 reference sample will be used. A good oxygen standard is too thick for us to consider the oxygen cross section to be constant at 972 keV, where the ratio σ12C/σ16O is known. Therefore, we first measure the oxygen content of the sample with the 16O(d,p)17O reaction at 850 keV. The cross section is quite constant down to 800 keV.

59. Example We can calculate the oxygen content for 850 keV using: Using cross section ratios σ(12C)/σ(16O)=1.91 from the table and changing the energy to 927 keV, we can measure the oxygen content again and the carbon with the 12C(d,p0)13C reaction. We assume that both σ(12C) and σ(16O) can be considered constant. Counts of carbon Counts of oxygen Absolute counts

61. Number of incident ions

62. Energy width per channelSurface height of element A in the unknown Nuclear reaction stopping cross section for A Surface height of element A in the standard ( known)

63. From the two equations: Were R is given by If unknown and known samples are examined under identical condition (same geometry and detector, integrated charge, and amplifier setting) R=1

64. In the use of x equation we must make a first guess for . As new values for x are calculated, the value of must be changed using Bragg rule.

65. For a binary-element unknown sample of thin film AxBy that x+y=1 Atomic density Thickness

70. Reference: Handbook of Modern Ion Beam Materials Analysis Second Edition Editors: Yongqiang Wang and Michael Nastasi Publisher: Materials Research Society 2009

71. Thank you Chukar (Kaw)-Kurdistan