Younes Sina's presentation on Nuclear reaction analysis
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Transcript of Younes Sina's presentation on Nuclear reaction analysis
![Page 1: Younes Sina's presentation on Nuclear reaction analysis](https://reader036.fdocuments.us/reader036/viewer/2022062313/5577de14d8b42a7b7b8b496d/html5/thumbnails/1.jpg)
Nuclear Reaction Analysis:Particle-Particle Reactions
A course of Professor C.J.McHargue
Younes Sina
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The basic principle of Ion Beam Analysis (IBA):
Electronic stopping:
when an incoming ion beam enters the surface of a thin film it loses some energy while penetrating the film, this stopping process is due to interactions of the beam with the atomic electrons in the energy shells of atoms.
Nuclear stopping:
The loss of energy can also be caused by the interactions of the beam with nuclei of the target sample.
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At nuclear level the collisions leading to interactions are divided into two groups :
Nuclear elastic collisions :
Collisions are the ones that leave reaction products at the same state as reactants.
Nuclear inelastic collisions
The reaction products are left at a state different from the one of the reactants (an excited state)
particle
Scattered particle
particle
Newparticle
Energy
Nuclear elastic collisions Nuclear inelastic
collisions
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Nuclear reactions are a very useful tool for light elements depth profiling on heavy matrix .The basic principle is when projectiles undergo nuclear reactions with the nuclei of the target material. which leads to the emission of product particles.
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Nuclear reaction analysis (NRA), like Rutherford backscattering spectrometry (RBS), exploits the interactions of charged particles with atomic nuclei. There is one condition for nuclear reactions to occur, which is the incident particle should possess energy greater than the coulomb barrier potential.
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Abilities of NRADifferent isotopes undergo different nuclear reactions, therefore each of them have unique characteristics such as energy release, excitation states, angular distributions and cross sections .NRA technique has some important features:• High selectivity for the determination of particular light nuclides.• High sensitivity for many nuclides, which are difficult to determine using some other techniques.• It is a non destructive technique• It can be used to analyze more than one light element in near surface layers of materials at once.
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NRA and RBS techniques are both governed by kinematics equations and their scattering geometry is the same.
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Coulomb barrier potential is given by:
3/12
3/11
21
MM
ZZEb
Z1 M1
Z2M2
Z1 M
1
E1
E0
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The equipment required for NRA is essentially the same as that for RBS:
Particle acceleratorScattering chamberHigh or ultrahigh vacuumSurface barrier detectorStandard nuclear electronicsMultichannel analyzer
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To avoid overloading the detector and acquisition electronics with elastically scattered primary beam it is necessary to filter this large flux.
By placing a thin film (often Mylar) in front of the detector, only the energetic particles produced by exothermic nuclear reactions can pass through.
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E0
Ein(x),E(Ein(x),Q,θ)
Eout(x)
Eabs(x)
Absorber foil
Detector
αin
αout
θ=θ(αin,αout)
x
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The energy of the particles leaving the target is given by:
)/(cos
0
)(],),([)(toux
outinout dxESQxEExE
)/(cos
0
0 )()(inx
inin dxESExE Stopping
power
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Method for filtering unwanted particles:
Absorber foil technique Thin detector technique Electrostatic or magnetic deflection technique (also electrostatic detector and magnetic spectrometer) Time -of-flight (TOF) technique Coincidence technique
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Absorber foil technique
The simplest method to stop the scattered beam by placing an absorber foil in front of the detector
Mylar (pinhole-free) and aluminum foil is used as absorbers
absx
absoutabs dxESxExE0
)()()(
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Thin detector technique
This technique is used when proton and α peaks overlap and α spectrum contains more information than the proton spectrum
Nuclear Reaction
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Electrostatic or magnetic deflection technique (also electrostatic detector and magnetic
spectrometer)
This techniques are based on the effects of the magnetic and electrostatic fields on energetic charged particles
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Time -of-flight (TOF) technique
This method is used for large number of different particles that are generated simultaneously when they have to be distinguished.
The technique is based on the simultaneous measurement of the energy and velocity of the detected particle. From these two values, the mass of the particle can be calculated.
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Coincidence technique
In this method the two reaction products are detected in different detectors in coincidence
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Measurement Methods
Overall near-surface contents
If (E)=(E0) for E0>E>E-ΔE
in
c NtEQA
cos
)( 0
standardstandard )/( NtAANt
Cross section
Collected charge
Solid angle
Peak area
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M1 M2
M1
M2
Incident
Target
Recoil
Scattered
E0
E1
E2
θ θC
φ φC
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M1 M2
M3
M4
Target
Heavy
Light
E0
E3
E4
θ θC
φ φCIncident
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Reaction Q(MeV)
αE
(MeV)βE
(MeV)Fitting range(MeV)
14N(d,α)12C 13.579 0.122±0.013 9.615±0.016 0.8-1.4
13C(d,p)14C 5.947 0.571±0.005 5.339±0.006 0.6-2.0
15N(d,α)13C 7.683 0.273±0.010 5.487±0.014 0.6-2.0
9Be(d,p)10Be 4.585 0.456±0.006 3.923±0.008 0.6-2.0
12C(d,p)13C 2.719 0.591±0.003 2.384±0.004 0.6-2.0
16O(d,p)17O 1.919 0.685±0.002 1.722±0.003 0.6-2.0
E3=E1αE+βE for θ=165˚
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Nonresonant depth profiling
As the thickness of the sample becomes larger, the peak of the reaction product becomes broader. The peak shape is the convolution of the concentration profile with the cross section and the depth resolution. The depth scale can be calculated from an expression similar to ion backscattering depth scale:
nrxNE ][Atomic density
Nuclear reaction stopping cross-section factor
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nrxNE ][
Nuclear reaction stopping cross-section factor
Atomic density
out
out
in
inEnr
cos
)cos
(][
in and out can be calculated assuming either a surface-energy or mean-energy approximation
Fitting parameter
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out
out
in
in
in
in
out
outabsmsgd
SEES
EE
xEEEE
x
cos)
cos(
]cos
)(cos
[
2
1
22
2
12
2222
A handy approximation by assuming that all energy spread distributions are Gaussian:
ΔEd=detector resolutionΔEg=geometrical energy spreadΔEms=energy spread due to multiple scatteringΔEabs=energy straggling in the absorber foil
in and out : straggling for incident and reaction productE1 and E2 : energies of the incident ion and the reaction products at depth x
by replaced is factor) ck(kinemati
:RBS from differenceonly The
2
1
E
E
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Example
Consider a SiO2/Si sample where the thickness of the oxide layer is unknown. We want to measure this thickness by measuring the oxygen content of the sample . The best reaction for this measurement is 16O(d,p1)17O. This reaction has a plateau below 900 keV, as shown in the following figure.(between the arrows)
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The 16O(d,p)17O reaction for both p0 and p1 reaction products. This reaction has a plateau below 900 keV down to 800 keV.
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E0=834 keVθ=135˚Mylar thickness=12 μm (all of the deuterons with energies below 900 keV are stopped by this foil)Protons: p0(Q=1.919 MeV) and p1(Q=1.048 MeV)The energies of the protons can be calculated from the general kinematic factor:
E31/2=B±(B2+c)1/2≈(αEE1
+βE)1/2
43
1414
43
2/1131
)(
cos)(
MM
MMEQMC
MM
EMMB
Example
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Reaction Q(MeV)
αE
(MeV)βE
(MeV)Fitting range(MeV)
14N(d,α)12C 13.579 0.122±0.013 9.615±0.016 0.8-1.4
13C(d,p)14C 5.947 0.571±0.005 5.339±0.006 0.6-2.0
15N(d,α)13C 7.683 0.273±0.010 5.487±0.014 0.6-2.0
9Be(d,p)10Be 4.585 0.456±0.006 3.923±0.008 0.6-2.0
12C(d,p)13C 2.719 0.591±0.003 2.384±0.004 0.6-2.0
16O(d,p)17O 1.919 0.685±0.002 1.722±0.003 0.6-2.0
Example
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Energy of proton using 12μm Mylar from appendix 10:
E in to Mylar=2.36 MeV for p0 & 1.58 MeV for p1E out of Mylar= 2.12 MeV for p0 & 1.23 MeV for p1
E taken by Mylar=0.24 MeV for p0 & 0.35 MeV for p1
We use these numbers for our sample
Example
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d 16O
p
17
O
Target
Heavy
LightE0=834 keV
Eout=2.312 MeV for p0 &1.206 MeV for p1
E4
θ θC
φ φCIncident
Detector
Mylar
E3=2.36 MeV for p0 &1.58 MeV for p1
Eout
E3
E0
Example
All of the deuterons with energy below 900 keV are stopped by the foil.
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E0=834 keV
ΔE=30 keVS(E)=7.6 eV/Å
A 4000AeV/ 6.7
keV 30
)(max
ES
Ex
Example
Average stopping power of deuterons in SiO2
Energy range that we can use
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The measured spectrum from 834 keV deuterons on a SiO2/Si sample.
The protons from the D(d,p)T reaction are not identified here because they have almost the same energy as those from 16O(d,p1)17O reaction.
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Total number of oxygen atoms
in
c NtEQA
cos
)( 0 standardstandard )/( NtAANt
Standard : Ta2O5
80 V thick Ta2O5 content 6.69×1017
16O atoms/cm2
17OTaOTa 1069.6
31875
21631)/(
525222 NtYYNt SiOSiO
Because the collected charge in both cases are the same:
Which corresponds to 1000 Å of SiO2
Example
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Example
Consider the application of the 12C(d,p0)13C reaction in analyzing the amount of carbon in a diamond like carbon (DLC) coating that contains a substantial amount of hydrogen. We will take αin=0° and αout=15°.
Flat plateau in the region between 0.9-1.0 MeV for the 12C(d,p0)13C reaction.
From Appendix 10:
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E31/2=B±(B2+c)1/2≈(αEE1
+βE)1/2 E31/2≈(0.591×1+2.38)1/2=3 MeV
εin=7.19×1015 eV cm2/atom
For 1 MeV deuteron in carbon
For 1 MeV deuteron in hydrogen:
εin= 1.99×1015 eV cm2/atom
εout=2.15×1015 eV cm2/atom
For 3 MeV deuteron in carbon
For 3 MeV deuteron in hydrogen:
εout= 0.47×1015 eV cm2/atom
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out
out
in
inE
HC
nr
cos)
cos(][ ,
0.59
atom) /Ccm (eV 2Cnr
1515 1045.610]15cos
15.2)
0cos
19.759.0[(][
atom) /Hcm (eV 2Hnr
1515 1022.110]15cos
47.0)
0cos
99.159.0[(][
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Assuming that the DLC coating is composed of 40 at.% hydrogen and 60 at.% carbon, the Bragg rule gives:
/atom)cm (eV 2
Cnr
Hnr
HCnr
15
,
1036.4
6.04.0
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Resonant depth profilingThe depth resolution of NRA is generally very poor because of the absorber.
The depth resolution can be improved if resonances exist.
out
out
in
in
in
in
out
outabsmsgd
SEES
EE
xEEEE
x
cos)
cos(
]cos
)(cos
[
2
1
22
2
12
2222
in
resonance
ES
EEx
cos/)(0
Resonance energy
Ē=(E0-Eresonance)/2
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If all contributions to the energy spread are Gaussian, the depth resolution is given approximately by:
inbeamin
msbeamstin
ES
EEx
x
cos/)(
cos2 222
width of resonance
energy spread of the incident beam
energy spread due to small-angle multiple scattering
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Example
A 18O(p,α)15N 629 keV resonance was used to determine oxygen exchange in high-Tc superconductor material Y1Ba18Cu29O7 (the composition of the sample was measured by RBS)
High Tc layer was relatively thin, then we can use an average stopping power of 10.87 eV/Å (calculated using Bragg rule)
Eq. shows the depth scale to be 1 keV/10.87 eV/Å=92 Å/keVin
resonance
ES
EEx
cos/)(0
Γ=2.1 keV (assuming the energy spread is negligible relative to the resonance width) then Δx(depth resolution)=200 Å
width of resonance
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The α yield (count per channel) at beam energy E0 is given by:
n
j
Ei
jiji dEEExEfxcKEY1 0
)(),,()(
Scaling factor
Concentration of 18O in the jth layer
2
22
),(exp
22
1),,(
j
ji
j
ji
xEEEExEf
Reaction cross section
Energy distribution of the incoming beam with initial energy Ei in depth xj, with E(Ei,xj) the beam energy and σj the straggling at depth xj
Example
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Levenberg-Marquardt method to fit the concentration profile:
22 )( EYY imes
i
The best fit was achieved by using a 150-Å-thick surface layer enriched to 30% followed by constant volume enrichment to 3.5%
Example
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The 18O(p,α)15N 629 keV resonance from a Ta2O5 reference sample. This resonance sits on a continues background, and therefore, the spectrum dose not go to zero outside the reaction energy.
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The 18O(p,α)15N 629 keV resonance from a high-Tc sample following an 18O oxygen-exchange anneal. A peak can be found at the resonance energy that indicates an enriched surface layer.
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Use of standards
Because of uncertainties in the calculation of the absolute overall surface count of an element using a reference target is suggested.
Requirements for good standards:• The thickness of the layer should be small enough not to cause significant change in the cross section.• The lateral uniformity over the beam area must be high.• The standard should be amorphous to avoid channeling effects.• The targets should have long-term stability in air, in vacuum, and under ion bombardment.• The preparation of the target should be highly reproducible.
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c
refc
refref Q
Q
A
ANtNt
Peak areas in the spectra
Collected charge
Total amount of the measured nuclei
If the thickness of both the reference and the unknown sample are small enough to use an energy-independent cross section , their respective peaks can be fully integrated.If the same amount of charge is
collected in both cases, the term
can be eliminated.
c
refc
Q
Q
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Example
Consider a sample contaminated on the surface by oxygen and carbon. Because it is quite difficult to prepare a carbon standard, a Ta2O5 reference sample will be used. A good oxygen standard is too thick for us to consider the oxygen cross section to be constant at 972 keV, where the ratio σ12C/σ16O is known. Therefore, we first measure the oxygen content of the sample with the 16O(d,p)17O reaction at 850 keV. The cross section is quite constant down to 800 keV.
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Example
We can calculate the oxygen content for 850 keV using:
c
refc
refref Q
Q
A
ANtNt
Using cross section ratios σ(12C)/σ(16O)=1.91 from the table and changing the energy to 927 keV, we can measure the oxygen content again and the carbon with the 12C(d,p0)13C reaction. We assume that both σ(12C) and σ(16O) can be considered constant.
O
cc Y
YNtNt
91.10
Absolute counts
Counts of carbon
Counts of oxygen
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If the standard is a bulk specimen and the unknown sample is either a bulk or a thin film, a surface-energy approximation to the nuclear reaction analysis can be used to obtain the composition at the surface. For a binary-element unknown sample of bulk AxBy that x+y=1If we are interested in the amount of A in our unknown sample, we should use a composition AzCw as our standard. (z+w=1)
inACA
UUUAUA
nr
xQEH
cos][
)( 00
E
inACA
KKKAKA
nr
xQEH
cos][
)( 00
E
Surface height of element A in the unknown
Surface height of element A in the standard ( known)
Nuclear reaction stopping cross section for A
Nuclear reaction cross section for element A
• Variables correspond to the detector solid angle• Number of incident ions• Energy width per channel
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From the two equations:
RzH
Hx
ACA
KA
ABA
UA
nr
nr
][
][
0
0
UUU
KKK
Q
QR
EE
Were R is given by
If unknown and known samples are examined under identical condition (same geometry and
detector, integrated charge, and amplifier setting)
R=1
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In the use of x equation we must make a first guess for .
As new values for x are calculated, the value of must be
changed using Bragg rule.
ABAnr
][
ABAnr
][
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For a binary-element unknown sample of thin film AxBy that x+y=1
in
ABAUUA
A
tNQEA
cos
)()( 0
Atomic density
Thickness
density ArealtN ABA )(
GzH
AtN
ACA
KKA
AABA
nr][
)(0
E
UU
KK
Q
QG
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Reference:
Handbook of Modern Ion Beam Materials AnalysisSecond Edition
Editors:Yongqiang Wang and Michael Nastasi
Publisher:Materials Research Society2009
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Chukar (Kaw)-Kurdistan
Thank you