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Beam – Specimen Interactions Signals Backscattered Electrons Beam electrons scatter, and escape out of specimen primary signal from elastic scattering.

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Presentation on theme: "Beam – Specimen Interactions Signals Backscattered Electrons Beam electrons scatter, and escape out of specimen primary signal from elastic scattering."— Presentation transcript:

1 Beam – Specimen Interactions Signals Backscattered Electrons Beam electrons scatter, and escape out of specimen primary signal from elastic scattering Example: Cu target 70% absorbed 30% backscattered Backscattered Electron Coefficient η = # BSE / # incident electrons

2 Backscattered Electrons: Atomic # dependence More trajectories intersect surface with higher Z target AlAu 1μmη = 14.0% 0.2 μmη = 53.5% Au – same scale as Al

3 η = Z – 1.86X10 -4 Z X10 -7 Z 3 For multi-component material: η = ∑C i η i Weak Z contrast Strong Z contrast Atomic Number Backscatter coefficient η

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5 Si target 0 0 tilt η= tilt η=30.0

6 Backscattered electrons: Energy distribution BSE have usually undergone a number of scattering events in target prior to emerging Al Pb 0 E 0 Light elements =broad distribution most BSEs E << E 0 Heavy elements =distribution skewed toward E 0

7 Backscattered electrons: Spatial distribution Electrons may emerge from an area outside beam incidence area AlPb 0 Light elements =broad distributionHeavy elements =narrow distribution 2 μm

8 Backscattered Electrons Greater energy loss farther from beam (more inelastic scattering events) Better BSE resolution obtainable if select only highest energy BSEs Distance from beam → # BSE BSEs within 10% of E 0 All BSEs

9 Beam – Specimen Interactions Signals From Inelastic Scattering Secondary Electrons X-Rays Continuum Characteristic Auger Electrons Cathodoluminescence Definition: Those electrons emitted with energy less than 50 eV Produced by interaction of beam and weakly-bound conduction band electrons. E transfer = a few eV Peak emitted ~ 3-4 eV Very shallow sampling depth 01.0 III II I E/E 0 dη/d(E/E 0 ) Intensity of SE Depth Z (nm)

10 Secondary electron coefficient δ = # SE / # incident beam electrons Dependent on atomic number if λ = mean free path maximum emission depth ~ 5λ For metals:λ ~ 1nm abundant conduction band electrons lots of inelastic scattering of SE Insulators: λ ~ 10 nm Information depth for SE ~ 1/100 of BSE About 1% of primary beam electron range (Kanaya-Okayama range)

11 Secondary electrons generated by primary beam electrons entering (I), or backscattering (II) Can define two SE coefficients: δ I and δ II SE generation generally more efficient via BSE (δ II / δ I = 3 to 4) Greater path length in region defined by escape depth, plus increased scattering cross sections due to larger energy distribution (extending to lower energies) in BSEs… primary BSE5 λ

12 Secondary electron density (#SE / unit area) defines apparent resolution Generate SE I within λ / 2 of beam (~0.5nm for metals) Primary beam ~ unscattered in 5λ region Diameter of SE I escape = diameter of beam + 2X λ / 2 SE II occur over entire BSE escape area 1 μm or more – but peak sharply in center SE I SE II SE tot distance

13 Influence of beam energy η+δη+δ E1E1 E2E2 E 0 incident Shape due mainly to variation in δ

14 Some considerations for resolution… With metal coating, secondary electrons detected primarily from coating In some cases, can improve resolution using higher beam energy (remember – higher kV = higher brightness and smaller beam spot size 10kV Si target 30kV

15 10kV 20nm Au on Si substrate 30kV SE emission volume

16 Beam – Specimen Interactions Signals From Inelastic Scattering Secondary Electrons X-Rays Continuum Characteristic Auger Electrons Cathodoluminescence Produced by deceleration of beam electrons in coulomb field of target atoms → energy loss Expressed as emission of X-Ray photon Results in continuous spectrum Most energetic = lowest wavelength Short λ limit = Duane-Hunt limit Results in background spectrum

17 X-ray emission from Cu Cu Kα Cu Lα Energy (keV) N x photons / e - Ster theoretical Actual – due to sample and window absorption + low detector efficiency Cu Kβ

18 X-ray continuum Increase beam energy Max continuum energy increases (short λ limit decreases) Intensity at given energy increases Intensity is a function of Z and photon energy Kramers’ Law I (continuum) ~ i p Z (E 0 –E v )/E v i p = probe current Z = atomic # E 0 = beam energy E v = continuum photon energy Background intensity is a determining factor in detection limits

19 Beam – Specimen Interactions Signals From Inelastic Scattering Secondary Electrons X-Rays Continuum Characteristic Auger Electrons Cathodoluminescence INNER SHELL IONIZATION 1)If energy equal or greater than critical excitation potential… Can eject inner shell electron 2)Atom wants to return to ground state outer shell electron fills vacancy – relaxation Outer shell electron = higher energy state relative to inner shell electron some energy surplus in the transition → photon emission (X-ray)

20 The emitted X-ray is characteristic of the target element – Wavelength (or energy) = the transition energy Therefore is a manifestation of the electron configuration. Example:E SiKα =1.740 keV (7.125Å) E FeKα =6.404 keV (1.936Å)

21 Polar coordinates… Quantum state of an electron - Quantum numbers Ψ(r,θ,Φ) = R(r)P(θ)F(Φ) n Radial component Principal quantum number l Colatitude Orbital quantum number M l Azimuthal Magnetic quantum number Yields three equations for three spatial variables Space geometry of the solution of the Schrodinger equation for the hydrogen atom… Z X Y θ Φ

22 Rudi Winter Aberystwyth University Quantum state of an electron - Quantum numbers n Principal(shell)1, 2, 3, … radial = sizeK, L, M… l Orbital – angular momentum (subshell) 0 to n-1 s (sharp) l = 0 p (principal) l = 1 d (diffuse) l = 2 f (fundamental) l = 3 shape so if n = 2, l = 1: 2p m l Orbital – Orientation (Magnetic, energy shift, or energy level for each subshell) orientation l to – l ex: for l = 2: m l = -2, -1, 0, 1, 2 3 spatial coordinates

23 Quantum state of an electron - Quantum numbers m s Spin ½, - ½ Single electron state of motion… n, l, m l, m s or: n, l, j, m l J Total angular momentum (quantum number j ) l +/- ½ = l + m s (except l = 0, where J = ½ only) Rudi Winter Aberystwyth University

24 The Orbitron: Mark Winter, Univ. of Sheffield 1s2p 4f 3d 5f

25 Quantum Numbers for Electrons in Atomic Electron Shells X-ray notation Modern notation nlj = l + s(2j + 1) K1s101/22 LI2s201/22 LII 2p 1/2 211/22 LIII2p3/2213/24 MI3s301/22 MII3p1/2311/22 MIII3p3/2313/24 MIV3d3/2323/24 MV3d5/2325/26 NI4s401/22 NII4p1/2411/22 NIII4p3/2413/24 NIV4d3/2423/24 NV4d5/2425/26 NVI4f5/2435/26 NVII4f7/2437/28

26 Ionization processes - Critical Excitation Potential What voltage is necessary to ionize an atom? Must overcome the electron binding energy – depends on Electron quantum state (shell, subshell, and angular momentum) Atomic # (Z)

27 Element K L-I L-II L-III M-I M-II M-III 1s 2s 2p1/2 2p3/2 3s 3p1/2 3p3/2 6 C 284.2* 7 N 409.9* 37.3* * 41.6* 11 Na Mg Al * Si *b P * 136* 135* 16 S * 162.5* 26 Fe Cu La *b 1209*b 1128*b 82 Pb U Electron binding energies (eV)

28 Element K L-I L-II L-III M-I M-II M-III 1s 2s 2p1/2 2p3/2 3s 3p1/2 3p3/2 6 C 284.2* 7 N 409.9* 37.3* * 41.6* 11 Na Mg Al * Si *b P * 136* 135* 16 S * 162.5* 26 Fe Cu La *b 1209*b 1128*b 82 Pb U Electron binding energies (eV) 2kV beam…

29 Element K L-I L-II L-III M-I M-II M-III 1s 2s 2p1/2 2p3/2 3s 3p1/2 3p3/2 6 C 284.2* 7 N 409.9* 37.3* * 41.6* 11 Na Mg Al * Si *b P * 136* 135* 16 S * 162.5* 26 Fe Cu La *b 1209*b 1128*b 82 Pb U Electron binding energies (eV) 15kV beam…

30 Element K L-I L-II L-III M-I M-II M-III 1s 2s 2p1/2 2p3/2 3s 3p1/2 3p3/2 6 C 284.2* 7 N 409.9* 37.3* * 41.6* 11 Na Mg Al * Si *b P * 136* 135* 16 S * 162.5* 26 Fe Cu La *b 1209*b 1128*b 82 Pb U Electron binding energies (eV) 20kV beam…

31 Element K L-I L-II L-III M-I M-II M-III 1s 2s 2p1/2 2p3/2 3s 3p1/2 3p3/2 6 C 284.2* 7 N 409.9* 37.3* * 41.6* 11 Na Mg Al * Si *b P * 136* 135* 16 S * 162.5* 26 Fe Cu La *b 1209*b 1128*b 82 Pb U Electron binding energies (eV) 50kV beam…

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34 Energy (or wavelength) of an X-ray depends on Which shell ionization took place Which shell relaxation electron comes from K radiation Electron removed from K shell K α electron fills K hole from L shell K β electron fills K hole from M shell L radiation Electron removed from L shell L α electron fills L hole from M shell L β electron fills L hole from M or N shell depends on which  transition – which L level ionized and which M or N level is the source of the de-excitation electron Karl Manne Siegbahn

35 KαKα L1 (2s) → K (1s), why not? Sufficiently energetic beam electron ionizes K shell…

36 Selection rules for allowed transitions involving photon emission (conservation of angular momentum) Change in n (principal) must be ≥ 1 Change in l (subshell) can only be +1 or -1 Change in j (total angular momentum) can only be +1, -1, or 0 The photon, following Bose-Einstein statistics, has an intrinsic angular momentum (spin) of 1. So a K-shell vacancy must be filled by an electron from a p-orbital, but can be 2p (L), 3p (M), or 4p (N) So can’t fill K from L1 (2s) in transitions involving photon emission

37 Normal (diagram) level Energy level (core or valence) described by removal of single electron from ground state configuration Diagram lines Originate from allowed transitions between diagram levels Non-diagram (Satellite) lines Generally originate from multiply-ionized states Two vacancies of one shell (e.g. two K ionizations) → hypersatellite Other effects from: Auger effect, Coster-Kronig (subshell) transitions, etc. X-Ray lines and electron transitions

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39 Originally Ionized shell Filled from…

40 Bohr’s Three Postulates: 1)There are certain orbits in which the electron is stable and does not radiate The energy of an electron in an orbit can be calculated - that energy is directly proportional to the distance from the nucleus Bohr simply forbids electrons from occupying just any orbit around the nucleus such that they can’t lose energy and spiral in… 2)When an electron falls from an outer orbit to an inner orbit, it loses energy …expressed as a quantum of electromagnetic radiation 3)A relationship exists between the mass, velocity and distance from the nucleus of an electron and Planck’s quantum constant… Energy of K α X-Ray

41 From these principles, Bohr realized he could calculate the energy corresponding to an orbit: m = mass of electron e = charge of electron ħ = h / 2π

42 If an electron jumps from orbit n=2 to orbit n, the energy loss is: energy is radiated, and expressing Plank’s relationship in terms of angular frequency (ω), rather than frequency (ν): Bohr theoretically has expressed Balmer’s formula and could calculate the Rydberg constant knowing m, e, c, and ħ

43 Balmer and Paschen series in terms of frequency (n and m are integers)… Multply both sides by Plank’s constant, h …Bohr assumes this is equal to the energy difference between two stationary states…. Single set of energy values to account for E differences… Electron bound to + nucleus And binding energy… n identifies a stationary state

44 Bohr assumes that proton and electron orbit around center of mass to derive orbital frequency of electron, then, arrives at an expression for radiation frequency for electron cascading through stationary states… For large n From expression of binding energy, and orbital frequency of electron, and solving for R in terms of physical constants… m = mass of electron e = charge ε= permittivity constant From Coulomb’s Law

45 Substituting the expression for R into expression for binding energy, gives binding energies of stationary states (Z is atomic #) Now, an electron making K transition moves in field of force – potential energy function: Seeing the charge of the nucleus (Z-1)e, and the other n=1 electron. And from the equation above for binding energy, the transition energy is…

46 Or about (10.2 eV)(Z-1) 2 So: O = 0.5 keV Si = 1.7 keV Ca = 3.7 keV Fe = 6.4 keV An approximate expression for the energy of the Kα X-Ray (Bohr’s early quantum theory)

47 Moseley’s Law X-Ray energy is related to Z… empirical relationship E = A(Z-C) 2 (A and C are constants) Bohr theory prediction for K α …E = (10.2)(Z-1) 2 KβKβ KαKα Niels Bohr Henry Moseley

48 Produce overall X-ray spectrum Characteristic peaks superimposed on a continuum background X-rays can be detected and displayed discriminated either by energy (E) or wavelength (λ) Energy Dispersive Spectrometry (EDS)

49 Background

50 Complex spectrum from monazite (Ce, La, Nd, Th) PO 4

51 For heavy elements Complex spectra → peak overlaps Note low pk / bkg for Th

52 Wavelength Dispersive Spectrometry (WDS) Si in garnet (pyrope) TAP monochromator SiK α CaK α ( 2 nd order) SiK β

53 For heavy elements Complex spectra → peak overlaps Note low pk / bkg for Th

54 Monazite (LIF monochromator) in wavelength region of Nd L EDS spectrum

55 Depth of production of X-Rays X-Rays generated over much of the interaction volume Characteristic X-Rays produced in electron range where electron energy exceeds critical excitation potential Z dependent Recall ionization energies (keV)… KLM Si1.55 Ca4.03 Fe7.10 Sn Pt

56 X-Ray region will be dependant on both Z and density (ρ) Φ(ρZ) High density = limited depth of production Deeper production for low energy ionizations 3 g/cm 3 10 g/cm 3 20 keV X-Ray spatial resolution

57 Run PHIROZ95, Casino, Win X-Ray Compare effects of different beam energies different materials Different lines generated in different regions of interaction volume Depends on electron energy distribution so function of: Initial voltage Material properties (Z, ρ) Critical excitation potentials for ionization events of interest

58 5% 10% 25% 50% 75% Labradorite [.3-.5 (NaAlSi 3 O 8 ) –.7-.5 (CaAl 2 Si 2 O 8 ), Z = 11] 15 kV Electron energy 100%1% Energy contours

59 Labradorite [.3-.5 (NaAlSi 3 O 8 ) –.7-.5 (CaAl 2 Si 2 O 8 ), Z = 11] 5% 10% 25% 50% 75% 100% 15 kV 10 kV 5 kV 1  m 5% 10% 25% 50% 1 kV (~ Na K ionization energy) (~ Ca K ionization energy)

60 Three main conclusions: For same material:linegeneration volume Mlarge Lmedium Ksmall K line of heavy element is excited from smaller region than K line of light element K line of an element is excited from smaller volumes in denser, or higher average Z materials

61 Putting it together… Pb, Th, and U in monazite Ionization energy for PbM-V level (to generate PbMα) = keV Ionization energy for ThM-V level = keV Ionization energy for UM-IV level (to generate UMβ) = keV will be trace element so ~ double the overvoltage to get reasonable count rates = 8 keV (minimum beam energy)

62 2.484 keV ionization potential… This is the lowest required energy of the three elements (Pb, Th, U) and will, therefore, limit the analytical resolution beam voltage% of beam voltage

63 Monte Carlo simulation Electron paths Energy contours 5 keV(2.484 keV ionization potential for Pb M-V level is ~50% of the beam energy) 50%~ 40nm

64 Monte Carlo simulation Electron paths Energy contours 15 keV(2.484 keV ionization potential for Pb M-V level is ~17% of the beam energy) 17% ~480nm

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66 Remember, voltage limited to minimum of 8 kV (2x ionization energy of UM-IV) Spatial resolution limit is then ~120 nm

67 Analytical spatial resolution: D AR = (D beam 2 + D scattering 2 ) 1/2 Based on depth containing 99.5% of total emitted intensity D beam = beam diameter D scattering = scattering dimension, either depth or radial distribution defined by x- ray emission volumes

68 Based on radius containing 99.5% of intensity

69 D Beam (nm) 300 D Beam (nm) 800

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71 Other signals from inelastic scattering Auger process Core level ionization De-excitation via internal conversion and emission of another electron rather than X-Ray →doubly ionized state Can result in satellite X-ray emission (Characteristic of electron configuration) Very small perturbation on background of emitted electrons - Very low yield Low energy - emitted from surface ~ 0.1nm depth (surface analysis technique) e - (KL I L III ) K2K2 Auger process

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73 Auger spectroscopy Sample upper 20Å or so and evaluate kinetic energy of emitted electrons. Materials Evaluation and Engineering, Inc.

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75 Cathodoluminescence Some insulators and semiconductors emit photons in the visible and UV when exposed to the electron beam ~ empty conduction band ~ full valence band The band gap has characteristic energy 1)Promote electron to conduction band Electron – hole pair 2)Recombination 3)Excess energy = band gap energy Expressed as photon (visible)

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77 Conduction band Almost Empty Valence band Almost Full EgEg bandgap Cathodoluminescence Initial state 1. Inelastic scattering imparts energy to specimen. Electron promoted to conduction band. 2. Recombination of electron-hole pair results in photon emission Electron promoted from impurity donor level Emitted photon energy = full band gap energy ν = E(gap) / h Emitted photon energy = impurity donor level ν = E(gap-d) / h donor level

78 Sandstone, secondary electron image 100  m Panchromatic CL image. Bright = K-fsp, dark = quartz.

79 40x60 micron 560nm CL image of diatoms 16x12 micron 560nm CL image of diatoms Butcher et al. (2003) Photoluminescence and Cathodoluminescence Studies of Diatoms – Nature’s Own Nano- Porous Silica Structures eV. Non-bridging hole centers 2.15eV. Self-trapped excitons related to Si nanoclusters?

80 CL counts In:Ga ratio Peak CL wavelength nm 418nm Edwards et al. (2003) Simultaneous composition mapping and hyperspectral cathodoluminescence imaging of InGaN epilayers Integration of WDS and cathodoluminescence mapping. InGaN epilayers.

81 Cathodoluminescence spectrum Shifts energies and / or intensities due to impurities or crystal dislocations and other defects Spectrum from GaAlAs bulk thin Thin with lattice defects


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