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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.
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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
Backscattered Electrons: Atomic # dependence More trajectories intersect surface with higher Z target Al Au Au – same scale as Al 1μm η= 14.0% 0.2 μm η= 53.5%
0.6 Weak Z contrast 0.5 0.4 Backscatter coefficient η 0.3 0.2 Strong Z contrast 0.1 0.0 0 20 40 60 80 100 Atomic Number η = -0.0254 + 0.016Z – 1.86X10-4Z2 + 8.3X10-7Z3 For multi-component material: η = ∑Ciηi
Si target 00 tilt η=17.5 450 tilt η=30.0
Backscattered electrons: Energy distribution BSE have usually undergone a number of scattering events in target prior to emerging Al Pb 0 E0 0 E0 Light elements = broad distribution most BSEs E << E0 Heavy elements = distribution skewed toward E0
Backscattered electrons: Spatial distribution Electrons may emerge from an area outside beam incidence area Al Pb 00 2 μm Light elements = broad distribution Heavy elements = narrow distribution
Backscattered Electrons Greater energy loss farther from beam (more inelastic scattering events) Better BSE resolution obtainable if select only highest energy BSEs BSEs within 10% of E0 # BSE All BSEs Distance from beam →
Beam – Specimen Interactions Signals From Inelastic Scattering III • Secondary Electrons • X-Rays • Continuum • Characteristic • Auger Electrons • Cathodoluminescence I dη/d(E/E0) Definition: Those electrons emitted with energy less than 50 eV II 0 1.0 E/E0 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 Intensity of SE 0.0 2.5 5.0 7.5 Depth Z (nm)
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)
primary 5 λ BSE Secondary electrons generated by primary beam electrons entering (I), or backscattering (II) Can define two SE coefficients: δI and δII SE generation generally more efficient via BSE (δII / δI= 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…
Secondary electron density (#SE / unit area) defines apparent resolution Generate SEI within λ / 2 of beam (~0.5nm for metals) Primary beam ~ unscattered in 5λ region Diameter of SEI escape = diameter of beam + 2X λ / 2 SEII occur over entire BSE escape area 1 μm or more – but peak sharply in center SEI SEII SEtot distance
Influence of beam energy η+δ E1 E2 E0 incident Shape due mainly to variation in δ
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
20nm Au on Si substrate 10kV 30kV SE emission volume
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
X-ray emission from Cu 7 Cu Lα 5 Cu Kα theoretical N x 10-6 photons / e- Ster 3 Actual – due to sample and window absorption + low detector efficiency Cu Kβ 1 0 2 4 6 8 10 Energy (keV)
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) ~ ipZ (E0 –Ev)/Ev ip = probe current Z = atomic # E0 = beam energy Ev = continuum photon energy Background intensity is a determining factor in detection limits
Beam – Specimen Interactions Signals From Inelastic Scattering Vacuum Valence level • INNER SHELL IONIZATION • 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) • Secondary Electrons • X-Rays • Continuum • Characteristic • Auger Electrons • Cathodoluminescence M… LIII (2p3/2) LII (2p1/2) LI (2s) K (1s) Kα
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)
Quantum state of an electron - Quantum numbers Z Polar coordinates… θ Space geometry of the solution of the Schrodinger equation for the hydrogen atom… Y Φ Ψ(r,θ,Φ) = R(r)P(θ)F(Φ) X Yields three equations for three spatial variables n Radial component Principal quantum number l Colatitude Orbital quantum number Ml Azimuthal Magnetic quantum number
Quantum state of an electron - Quantum numbers n Principal (shell) 1, 2, 3, … radial = size K, 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 ml Orbital – Orientation (Magnetic, energy shift, or energy level for each subshell) orientation l to – l ex: for l = 2: ml= -2, -1, 0, 1, 2 3 spatial coordinates Rudi Winter Aberystwyth University
Quantum state of an electron - Quantum numbers ms Spin ½, - ½ Single electron state of motion… n, l, ml, ms or: n, l, j, ml J Total angular momentum (quantum number j ) l+/- ½ = l + ms (except l= 0, where J = ½ only) Rudi Winter Aberystwyth University
The Orbitron: Mark Winter, Univ. of Sheffield http://winter.group.shef.ac.uk/orbitron 1s 2p 3d 4f 5f
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)
Electron binding energies (eV) 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* 8 0 543.1* 41.6* 11 Na 1070.8+ 63.5+ 30.65 30.81 12 Mg 1303.0+ 88.7 49.78 49.50 13 Al 1559.6 117.8* 72.95 72.55 14 Si 1839 149.7*b 99.82 99.42 15 P 2145.5 189* 136* 135* 16 S 2472 230.9 163.6* 162.5* 26 Fe 7112 844.6+ 719.9+ 706.8+ 91.3+ 52.7+ 52.7+ 29 Cu 8979 1096.7+ 952.3+ 932.7 122.5+ 77.3+ 75.1+ 57 La 38925 6266 5891 5483 1362*b 1209*b 1128*b 82 Pb 88005 15861 15200 13035 3851 3554 3066 92 U 115606 21757 20948 17166 5548 5182 4303
Electron binding energies (eV) 2kV beam… 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 C284.2* 7 N409.9* 37.3* 8 0543.1* 41.6* 11 Na1070.8+ 63.5+ 30.65 30.81 12 Mg1303.0+ 88.7 49.78 49.50 13 Al1559.6 117.8* 72.95 72.55 14 Si1839 149.7*b 99.82 99.42 15 P2145.5 189* 136* 135* 16 S 2472 230.9 163.6* 162.5* 26 Fe 7112 844.6+ 719.9+ 706.8+ 91.3+ 52.7+ 52.7+ 29 Cu8979 1096.7+ 952.3+ 932.7 122.5+ 77.3+ 75.1+ 57 La389256266 5891 5483 1362*b 1209*b 1128*b 82 Pb88005 15861 1520013035 3851 3554 3066 92 U 115606 21757 20948 171665548 5182 4303
Electron binding energies (eV) 15kV beam… 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 C284.2* 7 N409.9* 37.3* 8 0543.1* 41.6* 11 Na1070.8+ 63.5+ 30.65 30.81 12 Mg1303.0+ 88.7 49.78 49.50 13 Al1559.6 117.8* 72.95 72.55 14 Si1839 149.7*b 99.82 99.42 15 P2145.5 189* 136* 135* 16 S 2472 230.9 163.6* 162.5* 26 Fe 7112 844.6+ 719.9+ 706.8+ 91.3+ 52.7+ 52.7+ 29 Cu8979 1096.7+ 952.3+ 932.7 122.5+ 77.3+ 75.1+ 57 La389256266 5891 5483 1362*b 1209*b 1128*b 82 Pb88005 15861 1520013035 3851 3554 3066 92 U 115606 21757 20948 171665548 5182 4303
Electron binding energies (eV) 20kV beam… 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 C284.2* 7 N409.9* 37.3* 8 0543.1* 41.6* 11 Na1070.8+ 63.5+ 30.65 30.81 12 Mg1303.0+ 88.7 49.78 49.50 13 Al1559.6 117.8* 72.95 72.55 14 Si1839 149.7*b 99.82 99.42 15 P2145.5 189* 136* 135* 16 S 2472 230.9 163.6* 162.5* 26 Fe 7112 844.6+ 719.9+ 706.8+ 91.3+ 52.7+ 52.7+ 29 Cu8979 1096.7+ 952.3+ 932.7 122.5+ 77.3+ 75.1+ 57 La389256266 5891 5483 1362*b 1209*b 1128*b 82 Pb88005 15861 1520013035 3851 3554 3066 92 U 115606 21757 20948 171665548 5182 4303
Electron binding energies (eV) 50kV beam… 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 C284.2* 7 N409.9* 37.3* 8 0543.1* 41.6* 11 Na1070.8+ 63.5+ 30.65 30.81 12 Mg1303.0+ 88.7 49.78 49.50 13 Al1559.6 117.8* 72.95 72.55 14 Si1839 149.7*b 99.82 99.42 15 P2145.5 189* 136* 135* 16 S 2472 230.9 163.6* 162.5* 26 Fe 7112 844.6+ 719.9+ 706.8+ 91.3+ 52.7+ 52.7+ 29 Cu8979 1096.7+ 952.3+ 932.7 122.5+ 77.3+ 75.1+ 57 La38925 6266 5891 5483 1362*b 1209*b 1128*b 82 Pb88005 15861 1520013035 3851 3554 3066 92 U 115606 21757 20948 171665548 5182 4303
ψp1 ψp Kα ψp2 ψ1s
ψp1 ψp Kβ ψp2 ψ1s
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 Kα electron fills K hole from L shell Kβ electron fills K hole from M shell L radiation Electron removed from L shell Lα electron fills L hole from M shell Lβ electron fills L hole from M or N shell depends on which b transition – which L level ionized and which M or N level is the source of the de-excitation electron Karl Manne Siegbahn
Vacuum Valence level M… Energy level representation of characteristic X-ray emission process LIII (2p3/2) LII (2p1/2) LI (2s) Kα K (1s) Sufficiently energetic beam electron ionizes K shell… L1 (2s) → K (1s) , why not?
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
X-Ray lines and electron transitions 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.
Originally Ionized shell Filled from…
Energy of Kα X-Ray 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…
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
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
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… And binding energy… Electron bound to + nucleus n identifies a stationary state
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
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…
An approximate expression for the energy of the Kα X-Ray (Bohr’s early quantum theory) 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
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 Kα … E = (10.2)(Z-1)2 Niels Bohr Kα Kβ Henry Moseley
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)