My first lecture at UTK

1. Backscattering Spectrometry The University of Tennessee, Knoxville Younes Sina

6. Rutherford scattering is also sometimes referred to as Coulomb scatteringbecause it relies only upon static electric (Coulomb) forces, and the minimal distance between particles is set only by this potential. Elastic Backscattering Spectrometry (EBS) (non-Rutherford) is used when the incident particle is going so fast that it exceeds the “Coulomb barrier" of the target nucleus, which therefore cannot be treated by Rutherford’s approximation of a point charge. In this case Schrödinger's equation should be solved to obtain the scattering cross-section.

12. Chemical identification of the target elements using the kinematic factor. E0 = 2.0 MeV He E1 = 1830 keV K = 0.9150 M2 = 181amu Tantalum This is a spectrum of 2.0 MeV alpha particles incident on a thin film of “unknown composition” scattered at 170o.

14. Scattering Geometry

16. Mass Resolution

22. Rutherford Cross Section Unit: barn/ sr 1 b(barn)=10-24 cm2

23. Before After Before After

26. Non-Rutherford cross section L’Ecuyer Eq. for Low energy ion ECM≈Elab Center-of-mass kinetic energy (keV) Wenzel and Whaling for light ion with MeV energy

31. Energy

37. E’

39. Depth scale Energy loss factor Stopping cross section factor

41. Examples

42. Effects of energy loss of ions in solids Atomic density Stopping cross section stopping power For a compound of AmBn: εAB =mεA+nεB Atomic density Molecular density

43. Calculate the stopping cross section and stopping power of 2 MeV 4He+ in Al2O3 using Bragg rule Example: εAl= 44x10-15 eVcm2 εO= 35x10-15 eVcm2 From appendix 3: For a compound of AmBn: εAB =mεA+nεB εAl2O3=(2x44+3x35)x10-15 =193x10-15eVcm2

44. Example:

45. Depth scale Energy loss factor Stopping cross section factor

46. Depth resolution Calculate the depth- scattered ion energy differences for 2 MeV 4He+ in Al2O3 θ1=0°and θ2=10° K factor for 4He on Al=0.5525 K factor for 4He on O=0.3625 Example ε at E0,surface Using the surface-energy approximation εAB =mεA+nεB ε at E1

47. Example We can now calculate the stopping cross section factors Using the molecular density N Al2O3=2.35x1022 molecules/cm3we find:

48. Surface spectrum height Energy width per channel stopping cross section factors Surface height of the two elemental peaks in the compound AmBn are given by

49. Mean energy in thin films Surface Energy Approximation Mean energy of the ions in the film ,Ē(1) E0 E2 E1

50. Mean energy in thin films For the second iteration, the values of (Nt)i(1) should be calculated using with E= Ē(1) then ∆Ei(1) and Ē(2) E0 E2 E1

51. Example Calculate surface height for 2 MeV 4He+ on Al2O3: Ω=10-3sr E=1keV/channel Q=6.24x1013 incident particles (10μC charge) θ1=0°, θ2=10° (scattering angle=170°) From appendix 6: σRAl=0.2128x10-24 & σRO=0.0741x10-24 From previous example:

52. For not too thick film E=Ē(f) E0 E2 E1

53. Sample analysis Typical experimental operating conditions and parameter ranges used during acquisition of backscattering spectra

54. Thin-film analysis The peak integration method Integrated peak counts That can be accurately determined from the spectrum Correction factor Dead time ratio Integrated charge deposited on the sample during the run Non Rutherford correction factor solid angle subtended by the detector at the target

55. Example an application of the peak integration method of analysis of the two-element thin film E0=3776 keV θ=170˚ θ1=0˚ θ2=10˚ Ω=0.78 msr CBi=(0.99±0.03) E=(3.742±0.005)keV/channel É=(8±3) keV KFe=(170˚)=0.7520 KGd=(170˚)=0.90390 E1= nE+ É Energy intercept

56. Example From appendix 6 From Center-of-mass energy

57. Example From Trim 1985:

58. Example Integrated counts in spectral regions of interest (initial and final channel numbers are listed: Channels (789-918)=103978 cts; (920-960)=49 cts Channels (640-767)=64957 cts; (768-788)=79 cts

59. Example From: E1= nE+ É and Ki=Ei1/E0 : Energy intercept Therefore, element A and B are Gd and Fe, respectively. Note that element A could also be Tb, because KTb=0.9048

60. Example Calculation of elemental areal densities,(Nt) Values of Ai are calculated from the integrated counts in the regions of interests In this case, the background correction is almost negligible

61. Example The areal densities in the surface-energy approximation,(Nt)SEAi using E=E0 1 0.998 DTR CBi Ai e Q’ Ω σ

62. Example The mean energy of the 4He ion in the film, Ē(1), is calculated (to first order) using the following equation For the first-order energy loss, ΔESEAin ,of the ions in the film:

63. Example From the following Eq. we can calculate the areal densities:

64. Results of an additional iteration of this procedure using the following equations we have: (Note that Feand Gdare evaluated at Ē(1)) Example

65. Example The average stoichiometric ratio for this film using the following Eq.:

66. Example If the molecular formula for the film is written as GdmFen , then: m=0.209±0.001 and n=0.791±0.001 n+m=1

67. The value of the physical film thickness: Example Elemental bulk density

68. Areal density Integrated peak count Areal density, Nt, as atoms per unit area Cross section Incident ions Detector solid angle

69. The average stoichiometric ratio for the compound film AmBn Cross section ratio Ratio of measured integrated peak count

70. Physical film thickness

71. My hometown Mahabad, Iran