Beam – Specimen Interactions Signals

Beam – Specimen InteractionsSignals

Backscattered Electrons

Beam electrons scatter, and escape out of specimenprimary signal from elastic scatteringExample: Cu target

70% absorbed30% backscattered

Backscattered Electron Coefficient

η = # BSE / # incident electrons

Backscattered Electrons:Atomic # dependenceMore trajectories intersect surface with higher Z target

1μm η = 14.0% 0.2 μm η = 53.5%

Au – same scale as Al

η = -0.0254 + 0.016Z – 1.86X10-4Z2 + 8.3X10-7Z3

For multi-component material: η = ∑Ciηi

Weak Z contrast

Strong Z contrast

Atomic Number0 20 40 60 80 100

tter c

Si target

00 tilt η=17.5

450 tilt η=30.0

Backscattered electrons:Energy distributionBSE have usually undergone a number of scattering events in target prior to emerging

0 E0 0 E0

Light elements = broad distribution most BSEs E << E0

Heavy elements = distribution skewed toward E0

Backscattered electrons:Spatial distributionElectrons may emerge from an area outside beam incidence area

Light elements = broad distribution Heavy elements = narrow distribution2 μm

Backscattered ElectronsGreater energy loss farther from beam (more inelastic scattering events)Better BSE resolution obtainable if select only highest energy BSEs

Distance from beam →

BSEs within 10% of E0

All BSEs

Beam – Specimen InteractionsSignals From Inelastic Scattering

• Secondary Electrons• X-Rays

ContinuumCharacteristic

• 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 eVPeak emitted ~ 3-4 eVVery shallow sampling depth

Depth Z (nm)0.0 2.5 5.0 7.5

Secondary electron coefficient

δ = # SE / # incident beam electrons

Dependent on atomic numberif λ = mean free pathmaximum emission depth ~ 5λ

For metals: λ ~ 1nmabundant conduction band electronslots of inelastic scattering of SE

Insulators: λ ~ 10 nm

Information depth for SE ~ 1/100 of BSEAbout 1% of primary beam electron range (Kanaya-Okayama range)

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…

primary

BSE5 λ

Secondary electron density (#SE / unit area) defines apparent resolutionGenerate SEI within λ / 2 of beam (~0.5nm for metals)

Primary beam ~ unscattered in 5λ regionDiameter 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

E1 E2E0 incident

Shape due mainly to variation in δ

Some considerations for resolution…

With metal coating, secondary electrons detected primarily from coatingIn some cases, can improve resolution using higher beam energy (remember – higher kV = higher brightness and smaller beam spot size

10kVSi target

20nm Au on Si substrate

30kVSE emission volume

Produced by deceleration of beam electrons in coulomb field of target atoms→ energy lossExpressed 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

Cu Kα

Cu Lα

Energy (keV)

0 2 4 6 8 10

/ e- S

theoretical

Actual – due to sample and window absorption + low detector efficiency

Cu Kβ

X-ray continuumIncrease beam energy

Max continuum energy increases (short λ limit decreases)Intensity at given energy increases

Intensity is a function of Z and photon energyKramers’ 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

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 stateouter 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)

Valence level

Vacuum

LI (2s)

(2p3/2)(2p1/2)

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)

Polar coordinates…

Quantum state of an electron - Quantum numbers

Ψ(r,θ,Φ) = R(r)P(θ)F(Φ)nRadial componentPrincipal quantum number

lColatitudeOrbital quantum number

AzimuthalMagnetic quantum number

Yields three equations for three spatial variables

Space geometry of the solution of the Schrodinger equation for the hydrogen atom…

Rudi WinterAberystwyth University

Quantum state of an electron - Quantum numbersn Principal (shell) 1, 2, 3, …

radial = size K, L, M…

l Orbital – angular momentum (subshell) 0 to n-1s (sharp) l = 0p (principal) l = 1d (diffuse) l = 2f (fundamental) l = 3

shape so if n = 2, l = 1: 2pml Orbital – Orientation (Magnetic, energy shift,

or energy level for each subshell)

orientation l to – lex: for l = 2: ml = -2, -1, 0, 1, 2

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 WinterAberystwyth University

The Orbitron:Mark Winter, Univ. of Sheffield http://winter.group.shef.ac.uk/orbitron

Quantum Numbers for Electrons in Atomic Electron ShellsX-ray

notationModern notation

n l j = l + s (2j + 1)

K 1s 1 0 1/2 2

LI 2s 2 0 1/2 2

LII 2p1/2 2 1 1/2 2

LIII 2p3/2 2 1 3/2 4

MI 3s 3 0 1/2 2

MII 3p1/2 3 1 1/2 2

MIII 3p3/2 3 1 3/2 4

MIV 3d3/2 3 2 3/2 4

MV 3d5/2 3 2 5/2 6

NI 4s 4 0 1/2 2

NII 4p1/2 4 1 1/2 2

NIII 4p3/2 4 1 3/2 4

NIV 4d3/2 4 2 3/2 4

NV 4d5/2 4 2 5/2 6

NVI 4f5/2 4 3 5/2 6

NVII 4f7/2 4 3 7/2 8

Ionization processes - Critical Excitation PotentialWhat voltage is necessary to ionize an atom?

Must overcome the electron binding energy – depends onElectron quantum state (shell, subshell, and angular momentum)Atomic # (Z)

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 30.8112 Mg 1303.0+ 88.7 49.78 49.5013 Al 1559.6 117.8* 72.95 72.5514 Si 1839 149.7*b 99.82 99.4215 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*b82 Pb 88005 15861 15200 13035 3851 3554 306692 U 115606 21757 20948 17166 5548 5182 4303

Electron binding energies (eV)

Electron binding energies (eV) 2kV beam…

ψ1sψp2

Energy (or wavelength) of an X-ray depends onWhich shell ionization took placeWhich shell relaxation electron comes from

K radiationElectron removed from K shellKα electron fills K hole from L shell

Kβ electron fills K hole from M shell

L radiationElectron removed from L shellLα electron fills L hole from M shellLβ 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

Valence level

Vacuum

LI (2s)

(2p3/2)(2p1/2)

Energy level representation of characteristic X-ray emission process

L1 (2s) → K (1s) , why not?

Sufficiently energetic beam electron ionizes K shell…

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

Normal (diagram) levelEnergy level (core or valence) described by removal of single

electron from ground state configuration

Diagram linesOriginate from allowed transitions between diagram levels

Non-diagram (Satellite) linesGenerally originate from multiply-ionized statesTwo vacancies of one shell (e.g. two K ionizations) → hypersatelliteOther effects from: Auger effect,

Coster-Kronig (subshell) transitions, etc.

X-Ray lines and electron transitions

Originally Ionized shell

Filled from…

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 Kα X-Ray

From these principles, Bohr realized he could calculate the energy corresponding to an orbit:

m = mass of electrone = charge of electronħ = 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…

Electron bound to + nucleus

And binding energy…

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 nFrom expression of binding energy, and orbital frequency of electron, and solving for R in terms of physical constants…

m = mass of electrone = chargeε= 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…

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)

Moseley’s LawX-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

Henry Moseley

Produce overall X-ray spectrumCharacteristic peaks superimposed on a continuum background

X-rays can be detected and displayeddiscriminated either by energy (E) or wavelength (λ)

Energy Dispersive Spectrometry (EDS)

Background

Complex spectrum from monazite (Ce, La, Nd, Th) PO4

For heavy elements Complex spectra → peak overlaps

Note low pk / bkg for Th

Wavelength Dispersive Spectrometry (WDS)

Si in garnet (pyrope)TAP monochromator

CaKα (2nd order)

For heavy elements Complex spectra → peak overlaps

Note low pk / bkg for Th

Monazite (LIF monochromator) in wavelength region of NdL

EDS spectrum

Depth of production of X-RaysX-Rays generated over much of the interaction volumeCharacteristic X-Rays produced in electron range where electron energy exceeds critical excitation potentialZ dependent

Recall ionization energies (keV)…K L M

Si 1.55Ca 4.03Fe 7.10Sn 29.1 4.46Pt 13.9 3.3

X-Ray region will be dependant on both Z and density (ρ)

Φ(ρZ)High density = limited depth of productionDeeper production for low energy ionizations

3 g/cm3 10 g/cm320 keV

X-Ray spatial resolution

Run PHIROZ95, Casino, Win X-RayCompare effects of

different beam energiesdifferent materials

Different lines generated in different regions of interaction volumeDepends on electron energy distribution so function of:

Initial voltageMaterial properties (Z, ρ)Critical excitation potentials for ionization events of interest

Labradorite [.3-.5 (NaAlSi3O8) – .7-.5 (CaAl2Si2O8), Z = 11]

Electron energy100% 1%

Energy contours

Labradorite [.3-.5 (NaAlSi3O8) – .7-.5 (CaAl2Si2O8), Z = 11]

5 kV1 mm

1 kV(~ Na K ionization energy)

(~ Ca K ionization energy)

Three main conclusions:

For same material: line generation volumeM largeL mediumK small

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

Putting it together…

Pb, Th, and U in monazite

Ionization energy for PbM-V level (to generate PbMα) = 2.484 keV

Ionization energy for ThM-V level = 3.332 keV

Ionization energy for UM-IV level (to generate UMβ) = 3.728 keVwill be trace element so ~ double the overvoltage to get reasonable count rates

= 8 keV (minimum beam energy)

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 voltage5 49.6810 24.8415 16.5620 12.4225 9.93630 8.28

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

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

0 5 10 15 20 25 30 350

2,500Analysis resolution in monazite

Accelerating voltage (keV)

0 2 4 6 8 10 12 140

800Analysis resolution in monazite

Accelerating voltage (keV)

)Remember, voltage limited to minimum of 8 kV (2x ionization energy of UM-IV)

Spatial resolution limit is then ~120 nm

Analytical spatial resolution:DAR = (Dbeam

2 + Dscattering2)1/2

Based on depth containing 99.5% of total emitted intensity

Dbeam = beam diameterDscattering = scattering dimension, either depth or radial distribution defined by x-ray emission volumes

Based on radius containing 99.5% of intensity

0 5 10 15 20 25 300

E0 keV

φ(ρZ) Analytical Resolu-tionPbMα in Monazite

D Beam (nm)

0 5 10 15 20 25 300

E0 keV

Radial Analytical ResolutionPbMα in Monazite

4003005010

4 5 6 7 8 9 10 11100

E0 keV

Analytical ResolutionPbMα in Monazite

Radialφ(ρZ)

400 nm

300 nm

200 nm

100 nm

10 nm50 nm

Beam diame-ter

The ionization range can be estimated by means of a simple relationship

𝑅 (𝑛𝑚 )=33∗𝐴𝜌∗𝑍 (𝐸0❑1.7−𝐸𝑐❑

1.7)Castaing (1952); Merlet and Llovet (2013)

𝐷𝑟𝑒𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑛𝑚)=√𝑅2+𝐵2

Where R is the ionization range (in nm), and B is the beam diameter (in nm)

where E0 and Ec are expressed in keV, A is the atomic weight (g/mol), Z is atomic number, and ρ is density (g/cm3).

Other signals from inelastic scatteringAuger 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)

M…Valence level

Vacuum

LIILI (2s)

(2p3/2)(2p1/2)

e- (KLILIII)

M…Valence level

Vacuum

LIILI (2s)

(2p3/2)(2p1/2)

X-ray emission

Auger process

Auger spectroscopy

Sample upper 20Å or so and evaluate kinetic energy of emitted electrons.

Materials Evaluation and Engineering, Inc. http://mee-inc.com/

CathodoluminescenceSome insulators and semiconductors emit photons in the visible

and UV when exposed to the electron beam~ empty conduction band~ full valence bandThe band gap has characteristic energy

1) Promote electron to conduction bandElectron – hole pair

2) Recombination3) Excess energy = band gap energy

Expressed as photon (visible)

Conduction band Almost Empty

Valence band Almost Full

Eg bandgap

Cathodoluminescence

Initial state1. 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

Sandstone, secondary electron image

100 mm 100 mm

Panchromatic CL image. Bright = K-fsp, dark = quartz.

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

2.0-1.95 eV. Non-bridging hole centers

2.15eV. Self-trapped excitons related to Si nanoclusters?

CL countsIn:Ga ratio Peak CL wavelength

Edwards et al. (2003) Simultaneous composition mapping and hyperspectral cathodoluminescence imaging of InGaN epilayers

Integration of WDS and cathodoluminescence mapping. InGaN epilayers.

Cathodoluminescence spectrumShifts energies and / or intensities due to impurities or crystal dislocations and other defects

Spectrum from GaAlAs

Thin with lattice defects

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