younes sina's presentation on nuclear reaction analysis

Nuclear Reaction Analysis:Particle-Particle Reactions

A course of Professor C.J.McHargue

Younes Sina

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.

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

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.

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.

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.

NRA and RBS techniques are both governed by kinematics equations and their scattering geometry is the same.

Coulomb barrier potential is given by:

3/12

3/11

21

MM

ZZEb

Z1 M1

Z2M2

Z1 M

1

E1

E0

The equipment required for NRA is essentially the same as that for RBS:

Particle acceleratorScattering chamberHigh or ultrahigh vacuumSurface barrier detectorStandard nuclear electronicsMultichannel analyzer

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.

E0

Ein(x),E(Ein(x),Q,θ)

Eout(x)

Eabs(x)

Absorber foil

Detector

αin

αout

θ=θ(αin,αout)

x

The energy of the particles leaving the target is given by:

)/(cos

0

)(],),([)(toux

outinout dxESQxEExE

)/(cos

0

0 )()(inx

inin dxESExE Stopping

power

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

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

)()()(

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

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

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.

Coincidence technique

In this method the two reaction products are detected in different detectors in coincidence

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

M1 M2

M1

M2

Incident

Target

Recoil

Scattered

E0

E1

E2

θ θC

φ φC

M1 M2

M3

M4

Target

Heavy

Light

E0

E3

E4

θ θC

φ φCIncident

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˚

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

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

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

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)

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.

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

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

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

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.

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

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.

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

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:

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

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[(][

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

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

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

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

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

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

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.

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.

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.

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

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.

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

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

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

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

][

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

Reference:

Handbook of Modern Ion Beam Materials AnalysisSecond Edition

Editors:Yongqiang Wang and Michael Nastasi

Publisher:Materials Research Society2009

Chukar (Kaw)-Kurdistan

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