rbs: a lecture by younes sina

7/28/2019 RBS: A lecture by Younes Sina

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Backscattering Spectrometry

Younes Sina

The University of Tennessee, Knoxville


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ion-solid interactions

Interaction with:ElectronsNuclei


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Such interaction will result in:

Ionization the electron is ejected from its atomic orbit

Excitation the electron is raised to an outer orbit

The interaction of an ion with an atomic electron is purely

Coulomb (i.e. interaction governed by the Coulombs law).

An ionized/excited atom will eventually return to its ground state,

accompanied by the emission of one or more x-rays/photons.


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The interaction of an ion with an atomic nucleus can be


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Rutherford scattering is also sometimes referred to asCoulomb scatteringbecause it relies only upon staticelectric (Coulomb) forces, and the minimal distancebetween particles is set only by this potential.

Elastic Backscattering Spectrometry (EBS)(non-Rutherford) is used when the incident particle isgoing so fast that it exceeds the Coulomb barrier" ofthe target nucleus, which therefore cannot be treated

by Rutherfords approximation of a point charge. Inthis case Schrdinger's equation should be solved to

obtain the scattering cross-section.
http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equationhttp://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation


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Ion Beam Analysis Techniques

TECHNIQUE ION BEAM

ENERG

Y

(MeV)

PIXE H+ 1 - 4

RBS 4He+, H+ 2

ERDA 35Cl+, 20Ne+ 2 - 40

NRA H+, D+ 0.4 - 3


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Comparison of RBS,PIXE, and NRA methods

For Best choice Example

Channeling studies of crystal perfection RBS

Surface studies involving some elements

that are difficult for RBS

Combination of

RBS and NRA

Surface coverage of O using

the 16O(d,p)17OreactionLattice location of impurities

(the impurity mass>>host atom mass)RBS Lattice position of

As in Si, Ag in Zr

Light impurity element in a heavy-element host lattice

NRA andcharacteristic X-ray

H to S in a heavier lattice

Intermediatemass impurities in heavyhost lattice

Characteristic X-ray

Channeling studies

With single crystal targets, the effect of channeling also allows the investigation of the

crystalline perfection of the sample


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Scattering

yield

(count

s)

Alignment of a Si (110) crystal for channeling

RBS spectra for a randomly oriented crystal and aligned crystal

random

aligned


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Schematic diagrams for(a) an amorphous layer(b) random polycrystalline layers(c)


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K= (1680/2000)= 0.84


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Mass Resolution

M2=(15/2000)(470) 4

M2=(15/2000)(210) 2


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Energy


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xSE xNE .


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100 200 300 400 500 600 7000

500

1000

1500

2000

Yield

channel

random

Sample 14-15

Zr

Al

O

7.9x1015

Zr+/cm

2E(keV)=2.61 keV/ch * ch + 123 keV

calibration

=

E

N2

2 ~30nm

xNE .


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Rutherford Cross Section

Unit: barn/ sr1 b(barn)=10-24 cm2


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Au

Si

h f h f d b l h l


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The scattering cross section for a given Z2 is Rutherford at energies below the line.

C Si

Non Rutherford

Rutherford

E0.2514+0.4=3.9


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Summary


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Examples


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NdxdE

For a compound of AmBn:AB =mA+nB

B

AB

BA

AB

A

ABAB

AB

NNNdx

dE

Effects of energy loss of ions in solids

Atomic density

Stopping cross section

Molecular densityAtomic density

stopping power

C l l t th t i ti d t i f


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Example:

Al= 44x10-15 eVcm2 O= 35x10-15 eVcm2

Al2O3=(2x44+3x35)x10-15 =193x10-15eVcm2

3

3222

23

30

cm

moleculesOAl1035.2

mol

g102

mol

molecules106

cm

g4

32

M

NN

OAl

A

eV46101931035.20

15223232

32

OAlOAl

OAl

NdxdE

3

OAl

Al cm

atomsAl

N

222232

107.41035.22 3

222232

cm

atomsO101.71035.23

N

OAl

O

Calculate the stopping cross section and stopping power of2 MeV 4He+ in Al2O3 using Bragg rule

From appendix 3:

For a compound of AmBn:AB =mA+nB


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A

eV

46101931035.2 015223232

32

OAlOAl

OAl

Ndx

dE

A

eV46

eV10461035101.71044107.4

0

815221522

cm

x

dxdxdEE0

)/(

Example:

B

AB

BA

AB

A

ABAB

AB

NNNdx

dE


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Depth scale

xSE

21 cos

1

cos

1

outin dx

dE

dx

dEKS

xNE .

21 cos1

cos1

outinK

xNEABAB

AA

2

,

1 cos

1

cos

1

AB

Aout

AB

inA

AB

AK

x

dxdxdEE0

)/(Energy loss factor

Stopping cross section factor

Depth resolution


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Depth resolution

ExampleCalculate the depth- scattered ion energy differencesfor 2 MeV 4He+ in Al2O31=0and 2=10

K factor for 4He on Al=0.5525K factor for 4He on O=0.3625

0KEE 1MeVE

Al105.125525.0

1

MeVEO 725.023625.01 Using the surface-energy approximation

2151515 1019310353104423232

eVcmO

in

Al

in

OAl

in

2151515

,,,10240104631051232

32

eVcmO

Alout

Al

Alout

OAl

Alout

2151515

,,, 10252104831054232

32

eVcm

O

Oout

Al

Oout

OAl

Oout

AB =mA+nB at E0,surface

at E1

Example


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2

,1

0cos

1

cos

1 323232][

OAl

Alout

OAl

inAl

OAl

AlK

We can now calculate the stopping cross section factors

21515150

32

10350015.110240101935525.0][ eVcmOAl

Al

2,1

0

cos

1

cos

1 323232][

OAl

Oout

OAl

inO

OAl

O

K

21515150

32

10326015.110252101933625.0][ eVcmOAl

O

Using the molecular density N Al2O

3=2.35x1022 molecules/cm3 we find:

xxNE OAlOAlAlAl

A

eV

03.82

3232

0

xx

NE

OAlOAl

OO

A

eV

06.76

3232

0

Example

Surface spectrum height


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Surface spectrum height

10

0

cos][

)(

EQEH

Surface height of the two elemental peaks in the compound AmBn are given by

10

0,

cos0

m)(

AB

A

A

A

QEH

E

20

0,

cos

0

n)(

AB

B

B

B

QEH

E

Energy width per channel

stopping cross section factors


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Mean energy in thin films

)()(1

0 NtEESEA

i

r

i

iSEA

in

Mean energy of the ions in the film ,(1)

20

)1( ESEA

inEE

)(0),(

cosNt

Nt

SEA

iEEi

ii

i

EQ

A

Surface Energy Approximation

E0E1E2

h f l


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For the second iteration, the values of (Nt)i(1) should

be calculated using with

E= (1)then Ei(1)and (2)

),(

cos

EQ

A

i

ii

iNt

)()1(

)1(

1

)1( NtE ir

i

iin E

2

)1(

0

)2( EinEE

Mean energy in thin films

E0E1E2

Example Calculate surface height for 2 MeV 4He+ on Al O :


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Example Calculate surface height for 2 MeV 4He+ on Al2O3:=10-3srE=1keV/channelQ=6.24x1013 incident particles (10C charge)1=0, 2=10 (scattering angle=170)

From appendix 6: RAl=0.2128x10-24 & R

O=0.0741x10-24

From previous example:

2150

32 10326][ eVcmOAlO

2150 32 10350][ eVcmOAlAl

10

0,

cos0

)(

AB

A

A

A

EQmEH

cnt

EQOAl

Al

Al

AlH 76103501021024.610102128.0

0

215

33324

320,

cntEQOAl

O

O

OH 43103261031024.610100741.0

0

315

33324

320,


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For not too thick film

)()(

0

)(2

)(

Nt

E

ENt

SEA

i

f

i

f

E=(f)

E0E1E2

S l l i


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Sample analysis

Experimental Parameter Units Values

Analysis ion energy MeV 1.0-5.0

Beam cross section mm x mm 1.5x1.5

Beam current nA 10-200

Integrated charge C 5-100

Detector energy resolution for 4He ions keV 15

Data acquisition time min 5-10

Vacuum Torr 2x10-6

Pump-down time min 15

Typical experimental operating conditions and parameter ranges used duringacquisition of backscattering spectra

Multiply units by For units Example


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Unitconversions

MeV MeV/amu 4 MeV 4He ~ 1 MeV/amu

v/vo (MeV/amu)1/2 v/vo =1~0.025 MeV/amu

1H

(MeV/amu)1/2 m/s 2 MeV 4He ~ vHE =9.82x 106 m/s

1015

atoms/cm2

nm 1018

Atoms/cm2

For Au~170nm

g/cm2 nm 100 g/cm2 For C~258 nm

g/cm2 1015 atoms/cm2 100 g/cm2

For Au~305x1015 atoms/cm2

eV cm2/1015 atoms MeV/(mg/cm2 ) 100 eV cm2/1015 atoms forAl2O3~2.95 MeV cm

2/mg[M2= (2MAl + 3 MO)/5; MAl=26.98,

MO=16.00]

eV cm2/1015 atoms keV/m 30eV cm2/1015 atoms for Si~150

keV/m

][1

1

amuM

]3/[][10661.1 22

cmgamuM

]/[

103

cmg

][661.1

10

1

3

amuM

1581.0

10389.17

][661.1

1

2 amuM

][661.1

]/[10

2

32

amuM

cmg

Thin film analysis


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Thin-film analysisThe peak integration method

iR

iR

Bii

EQ

DTReCAiNt

))(,('

cos)(

1

Integrated peak countsThat can be accurately determined

from the spectrum

Non Rutherford correction factor

Correction factor

Dead time ratio

Integrated charge depositedon the sample during the run

solid angle subtended by

the detector at the target

Example


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an application of the peak integration method of analysis of the two-elementthin film

E0=3776 keV=170

1=02=10=0.78 msrCBi=(0.990.03)E=(3.7420.005)keV/channel=(83) keVKFe=(170)=0.7520KGd=(170)=0.90390

E1= nE+

Example

Energy intercept

Example


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sr

cmE

Fe

R

22424

20102469.010

776.3

521.3)170,(

sr

cmE

Gd

R

22424

2010510.110

776.3

53.21)170,(

998.03776)26)(2)(049.0(13/4

FeR

993.0

3776

)64)(2)(049.0(1

3/4

GdR

From appendix 6

EZZ

CMR

21

3/4

049.01

From

Example

Center-of-mass energy

Example


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From Trim 1985:

215

104.51)3776( eVcmkeVFe

215102.52)3676( eVcmkeV

Fe

215103.86)3776( eVcmkeV

Gd

215

105.87)3676( eVcmkeVGd

Example

Example


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008.1

)201020(

)1660('

)1757(

01.20'

0,

DTR

cts

nB

nB

CQ

HAB

A

cts

nAnA

HAB

B)20640(

)1812(')1910(

0,

Integrated counts in spectral regions of interest (initialand final channel numbers are listed:Channels (789-918)=103978 cts; (920-960)=49 cts

Channels (640-767)=64957 cts; (768-788)=79 cts

Example


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From: E1= nE+ and Ki=Ei1/E0 :

keVEnB

B

E )62841()38()005.0742.3)(1757('1 E

keVEnAA

E )73413()38()005.0742.3)(1910('1 E

002.0752.0)53776()62841(

0

1

E

EKB

B

002.0904.0)53776(

)73413(

0

1

E

EK

A

A

Therefore, element A and B are Gd and Fe, respectively. Note that element Acould also be Tb, because KTb=0.9048

p

Energy intercept

Example


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Calculation of elemental areal densities,(Nt)

Values of Ai are calculated from the integrated counts in the

regions of interests

ctsAFe

)26164475()128(21

7964957

ctsAGd )323103823()130(41

49103978

In this case, the background correction is almost negligible

p

Example


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The areal densities in the surface-energy approximation,(Nt)SEAiusing E=E0

22436

19

)998.0)(102469.0)(1078.0)(1001.20(

)10602.1)(03.099.0)(008.1)(26164475(

)( cmatoms

Nt

SEA

Fe

21810)021.0709.0()(

cmatomsNt

SEA

Gd

21810)08.068.2(

cmatoms

iR

i

R

Bii

EQDTReCAiNt

))(,('cos)(

1

DTR CBiAi

1

Q

0.998

e

p

Example


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The mean energy of the 4He ion in the film, (1), is calculated (to first order)using the following equation

)()(1

0 NtEESEA

i

r

i

iSEA

in

For the first-order energy loss, ESEAin ,of the ions in the film:

20

)1( ESEA

inEE

keV

eV

EE NtNtESEA

Gd

GdSEA

Fe

FeSEA

in

199

)10709.0)(103.86()1068.2)(104.51(

)()(

18151815

00 )()(

keVE 36762

1993776

)1(

p

Example


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From the following Eq. we can calculate the areal densities:

)()(0

)(2

)(

NE

ENt

SEA

i

f

i

f

218

2

)1(

/1054.2)(

3776

3676)( cmatomsNNt

SEA

FeFe

218)1(

/10672.0)( cmatomsNtGd

Results of an additional iteration of this procedure using theExample


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Results of an additional iteration of this procedure using thefollowing equations we have: (Note that Feand Gdareevaluated at (1))

eVeVEin 191)10672.0)(105.87()1054.2)(102.52(18151815)1(

)()1(

)1(

1

)1( NtE ir

i

iin E 2

)1(

0)2( EinEE

)()(0

)(2

)(

NtE

ENt

SEA

i

f

i

f

eVEin 36812

1923776

)2(

218

2

)2(/10)08.055.2()(

37763681)( cmatomsNtNt

SEA

FeFe

218)2(

/10)021.0674.0()( cmatomsNtGd

Example


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The average stoichiometric ratio for this film using the following Eq.:

),(),(.

EE

AA

NN

mn

B

A

A

B

A

B

02.078.3998.0

993.0

.521.3

53.21

.323103823

)26164475(

.)170,(

)170,(

//

0

0

R

R

E

E

A

A

N

N

Fe

Gd

Fe

R

Gd

R

Gd

Fe

Gd

Fe

Example


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If the molecular formula for the film iswritten as GdmFen , then:

m=0.2090.001 and n=0.7910.001

n+m=1

The value of the physical film thickness:Example


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p y

nmcmtFe 3021044.8

1055.222

18

Elem

entalbulkdensity

3220

/1044.8 cmatomsM

N

N Fe

Fe

Fe

3220 /1002.3 cmatomsM

NN

Gd

GdGd

nmcmtGd 2231002.3

10674.022

18

nmtGdFe 525

NNAB

B

B

AB

A

A NtNt

t

)()(

Areal density


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),(

cos

EQ

A

i

ii

iNt

Areal density, Nt, as atoms per unit area

Detector solid angle

Integrated peak count

Incident ionsCross section


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),(

),(.

E

E

A

A

N

N

m

n

B

A

A

B

A

B

The average stoichiometric ratio for the compound film AmBn

Ratio of measuredintegrated peak count

Cross section ratio

AB

AB

M

NmN

AB

A

0

AB

AB

M

NnN

AB

B

0

BAPhysical film thickness


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AB

AB

M

NmN

AB

A

0

AB

AB

M

NnN

AB

B

0

NNAB

B

B

AB

A

A NtNtt )()(

BAAB nMmMM

nmBAPhysical film thickness

rbs: a lecture by younes sina

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