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Lecture 11 Ion Beam Analysis

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Lecture 11 Ion Beam Analysis 1 AP 5301/8301 Instrumental Methods of Analysis and Laboratory Lecture 11 Ion Beam Analysis Prof YU Kin Man E-mail: [email protected] Tel: 3442-7813 Office: P6422 2 Lecture 8: outline . Introduction o Ion Beam Analysis: an overview . Rutherford backscattering spectrometry (RBS) o Introduction-history o Basic concepts of RBS – Kinematic factor (K) – Scattering cross-section – Depth scale o Quantitative thin film analysis o RBS data analysis . Other IBA techniques o Hydrogen Forward Scattering o Particle induced x-ray emission (PIXE) . Ion channeling o Minimum yield and critical angle o Dechanneling by defects o Impurity location 3 Analytical Techniques 100at% 10at% AES XRD IBA 1at% 0.1at% RBS XRF 100ppm TOF- 10ppm SIMS 1ppm TXRF 100ppb Detection Range Detection 10ppb Chemical bonding DynamicS 1ppb Elemental information IMS Imaging information 100ppt Thickness and density information 10ppt Analytical Spot size But IBA is typically quantitative and depth sensitive http://www.eaglabs.com 4 Ion Beam Analysis: an overview Inelastic Elastic Incident Ions Rutherford backscattering (RBS), resonant scattering, Nuclear Reaction Analysis channeling (NRA) p, a, n, g Particle induced x-ray emission (PIXE) Defects generation Elastic recoil detection Ion Beam Absorber foil Modification 5 Ion Beam Analysis 1977 2009 2014 1978 6 Why IBA? . Simple in principle . Fast and direct . Quantitative (without standard for RBS) . Depth profiling without chemical or physical sectioning . Non-destructive . Wide range of elemental coverage . No special specimen preparation required . Can be applied to crystalline or amorphous materials . Simultaneous analysis with various ion beam techniques (RBS, PIXE, NRA, channeling, etc.) . Can obtain a lot of information in one measurement 7 A typical Ion Beam Analysis Facility Experimental chamber bending focusing Accelerator Ion source Beam Fixed particle PIXE collimators detector detector f Moveable particle detector 8 IBA Equipment: accelerator Van de Graaff (single ended) Tandem Pelletron 9 IBA Equipment: tandem pelletron 10 Equipment: particle detector 11 Rutherford Backscattering Spectrometry (RBS) . Relatively simple physics . Requires some experimental skills . Requires some data simulation to obtain the most out of the spectrum . The only quantitative and absolute method to determine ─ atomic concentration with no matrix effect ─ composition variation with depth ─ layer thickness ─ damage distribution and impurity lattice location of multi-elemental, multi-layered films Rutherford Backscattering Spectrometry (RBS) a brief history The original experiment Alpha particles at a thin gold foil experiment by Hans Geiger and Ernest Marsden in 1909. 1) Almost all of the alpha particles went through the gold foil 2) Some of the alpha particles were deflected only slightly, usually 2° or less. 3) A very, very few (1 in 8000 for platinum foil) alpha particles were turned through an angle of 90° or more. "We shall suppose that for distances less that 10-12 cm the central charge and also the charge on the alpha particle may be supposed to be concentrated at a point." (1911) 13 RBS: The Surveyor V experiment Surveyor V, first of its spacecraft family to obtain information about the chemical nature of the Moon's surface, landed in Mare Tranquillitatis on September 11, 1967. "Surveyor V carried an instrument to determine the principal chemical elements of the lunar-surface material," explained ANTHONY TURKEVICH, Enrico Fermi institute and Chemistry Department, University of Chicago. "After landing, upon command from Earth, the instrument was lowered by a nylon cord to the surface of the Moon …” 14 General applications of RBS . Quantitative analysis of thin films . thickness, composition, uniformity in depth . solid state reactions . interdiffusion . Crystalline perfection of homo- and heteroepitaxial thin films . Quantitative measurements of impurities in substrates . Defect distribution in single-crystal samples . Surface atom relaxation in single crystals . Lattice location of impurities in single crystals 15 RBS: basic concepts . Kinematic factor: elastic energy transfer from a projectile to a target atom can be calculated from collision kinematics mass determination . Scattering cross-section: the probability of the elastic collision between the projectile and target atoms can be calculated quantitative analysis of atomic composition . Energy Loss: inelastic energy loss of the projectile ions through the target perception of depth These allow RBS analysis to give quantitative depth distribution of targets with different masses 16 Kinematic factor: K Conservation of energy : Conservation of momentum: 12 122 1 m1 v m 1 v 1cosf m 2 v 2 cos m1 v m 1 v 1 m 2 v 2 2 2 2 m1 v 1sinf m 2 v 2 sin 2 2 2 2 E1 (m2 m1 sin ) m1 cos K K,, m21 m m2 E (m m ) o 2 1 Kinematic Factor 2 2 2 2 E (m m sin ) m cos K 1 2 1 1 m2 E (m m ) o 2 1 2 (m m ) K ( 180o ) 2 1 m2 (m2 m1) (m m ) K ( 90o ) 2 1 m2 (m2 m1) When m2>>m1: K ( 180o ) ~ 1 m2 18 Element identification 2.5 MeV He ion with =170o K(197Au) = 0.923 K(109Ag) = 0.864 K(107Ag) = 0.862 K(65Cu) = 0.783 K(63Cu) = 0.777 Mass Resolution, dm2 2 E (m m ) For 180o scattering: K 1 2 1 m2 E0 (m2 m1) 2 d E mm 1 d 21 E 2 0 mm12 For a fixed dE1/E0 (~0.01), heavier projectiles result in better mass resolution However, dE1 for heavier projectiles is higher 20 Mass Resolution: examples With system energy resolution dE = 20keV and E0=2MeV (40 4)3 20 For m =40 dm2 1.48a.m.u. 2 4 4(40 4) 2000 (70 4)3 20 For m = 70 dm 3.84a.m.u. 2 2 4 4(70 4) 2000 Isotopes of Ga (68.9 and 70.9 a.m.u.) cannot be resolved 4 MeV 3 MeV 2 MeV 1 MeV We can also improve mass resolution by increasing Eo 3 mm d E d m 21 1 2 E 4m1 m 2 m 1 0 21 RBS Quantification Backscattering Yield . For a given incident number of particles Q, a greater amount of an element present (N ) should result in Q s a greater number of particles scattered. Thus we need to know how often scattering events should be detected (A) at a characteristic energy (E = KE0) and angle , within our detector’s window of solid angle W. Scattering cross-section If particles are coming in with impact parameters between b and b +db , they will be scattered through angles between and +d. For central forces, we have circular symmetry. 2bdb ( ) 2 sin d Rutherford cross-section: 2 2 2 2 2 1/ 2 2 d Z1Z2e cos (1 A sin ) Z1Z2 m1 ~ A dW 2E 4 2 2 1/ 2 E m o sin (1 A sin ) o 2 2 23 Scattering Yield 1/ 2 2 2 22 Example: 2 cos 1A sin Z12 Z e Yield,Y 2E 4 2 2 1/ 2 0 sin 1 A sin 2 Z1Z1 2 ( ) ~ 1024 cm 2[barn] E0 Mixed Si and W target analzed by a 2MeV He ion beam at 165o scattering angle.angle. (Si)=2.5x10-25cm2/str/str (W)=7.4x10-24cm2/str/str Total incident ions Q=1.5x1014 ions W=1.8mstr Area under Si, A(Si)=200 counts Area under W, A(W)=6000 counts 15 2 (Nt)(Nt)Si=3x10 atoms/cm 15 2 (Nt)(Nt)W=3x10 atoms/cm Energy Loss Electronic stopping dE dx ele Nuclear dE Stopping dx nucl When an He or H ion moves through matter, it loses energy through . interactions with e- by raising them to excited states or even ionizing them. Direct ion-nuclei scattering Since the radii of atomic nuclei are so small, interactions with nuclei may be neglected dE dE dE dE dx total dx ele dx nucl dx ele 25 How to read a typical RBS spectrum A thin film on a light substrate . Most projectile ions experience electronic stopping that results in a gradual reduction of the particle’s kinetic energy (dE/dx). At the same time a small fraction of projectile ions come close enough to the nucleus for large- angle scattering (KE). A detected backscattered particle has lost some energy during initial penetration, then lost a large fraction of its remaining energy during the large-angle scattering event, then lost more energy in leaving the solid. Depth Scale E1 K(E0 Ein ) Eout dE dE t E K(E t) ( ) 1 0 dx dx cos E0 KE0 KEo Total energy loss E=KE0-E1 K dE 1 dE E ( ) t [S ] t cos dx cos dx o 1 E0 2 KE0 E1 KE E 0 [So] is the effective stopping power Stopping cross-section 1 dE eV cm3 eVcm2 : N dx cm atom atom K 1 E N N t N[ ] t E in out o cos1 cos2 Depth scale: thin film 180o- Thickness of film: Eo KEo dE E dE dx KE 1 E (K 0 ) t [S ]t t dx cos o E0 E1 KE E E o t So N[ o ] E obtained from TRIM code or empirical energy loss data fitting E 28 Example: layer thickness consider the Au markers EAu=EAuF - EAuB = [KAu dE/dxEo + 1 / (cos10º) • dE/dx EAuB ] • t dE/dx 3MeV = NAl Al 3MeV = 6.02x1022 • 36.56x10-15 = 2.2x109eVcm-1 dE/dx EAuB = NAl Al 2.57MeV = 6.02x1022 x 39.34x10-15 = 2.37x109 eVcm-1 t = 3945Å EAl = 165keV consider the Al signals EAl = [KAl dE/dxEo + 1 / (cos10º) • dE/dx KEo ] • t EAu = 175keV KAu = 0.9225 t = 3937Å KAl = 0.5525 29 Energy Loss: Bragg’s rule For a target AmBn, the stopping cross- section is the sum of those of the constituent elements weighted by the abundance of the elements.
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