Backscattering Spectrometry

Younes Sina

The University of Tennessee, Knoxville

Rutherford scattering is also sometimes referred to as Coulomb scattering because it relies only upon static electric (Coulomb) forces, and the minimal distance between particles is set only by this potential.

Elastic Backscattering Spectrometry (EBS) (non-Rutherford) is used when the incident particle is going so fast that it exceeds the “Coulomb barrier" of the target nucleus, which therefore cannot be treated by Rutherford’s approximation of a point charge. In this case Schrödinger's equation should be solved to obtain the scattering cross-section.

Chemical identification of the target elements using the kinematic factor.

E0 = 2.0 MeV HeE1 = 1830 keV

K = 0.9150M2 = 181amuTantalum

This is a spectrum of 2.0 MeV alpha particles incident on a thin film of “unknown composition” scattered at 170o.

Scattering Geometry

Mass Resolution

Rutherford Cross Section

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

Before

Before

After

After

Center-of-mass kinetic energy (keV)

ECM≈Elab

Non-Rutherford cross section

L’Ecuyer Eq. for Low energy ion

Wenzel and Whaling for light ion with MeV energy

Energy

x

dxdxdEE0

)/(

E’

0EEdx

dE

dx

dE

constant

xSE

Depth scale

xSE

21 cos

1

cos

1

outin dx

dE

dx

dEKS

xNE .

21 cos

1

cos

1

outinK

xNE ABAB

AA

2,

1 cos

1

cos

1

AB

Aout

AB

inA

AB

AK

x

dxdxdEE0

)/(Energy loss factor

Stopping cross section factor

xSE xNE .

Examples

Ndx

dE

For a compound of AmBn:εAB =mεA+nεB

B

AB

BA

AB

A

ABAB

AB

NNNdxdE

Effects of energy loss of ions in solids

Atomic density

Stopping cross section

Molecular densityAtomic density

stopping power

Example:

εAl= 44x10-15 eVcm2 εO= 35x10-15 eVcm2

εAl2O3=(2x44+3x35)x10-15 =193x10-15eVcm2

3

3222

2330

cm

moleculesOAl1035.2

molg

102

molmolecules

106cm

g4

32

M

NN

OAl

A

eV46101931035.2 0

15223232

32

OAlOAl

OAl

NdxdE

3

OAl

Al cm

atomsAlN 222232

107.41035.22 3222232

cm

atomsO101.71035.23 N

OAl

O

Calculate the stopping cross section and stopping power of 2 MeV 4He+ in Al2O3 using Bragg rule

From appendix 3:

For a compound of AmBn:εAB =mεA+nεB

A

eV46101931035.2 0

15223232

32

OAlOAl

OAl

NdxdE

A

eV46

eV10461035101.71044107.4

0

815221522

cm

x

dxdxdEE0

)/(

Example:

B

AB

BA

AB

A

ABAB

AB

NNNdxdE

Depth scale

xSE

21 cos

1

cos

1

outin dx

dE

dx

dEKS

xNE .

21 cos

1

cos

1

outinK

xNE ABAB

AA

2,

1 cos

1

cos

1

AB

Aout

AB

inA

AB

AK

x

dxdxdEE0

)/(Energy loss factor

Stopping cross section factor

Depth resolution

ExampleCalculate the depth- scattered ion energy differences for 2 MeV 4He+ in Al2O3

θ1=0°and θ2=10°K factor for 4He on Al=0.5525K factor for 4He on O=0.3625

0KEE 1

MeVEAl

105.125525.01

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

eVcmO

Oout

Al

Oout

OAl

Oout

εAB =mεA+nεB ε at E0,surface

ε at E1

2,

10

cos

1

cos

1 323232

][ OAl

Alout

OAl

inAl

OAl

AlK

We can now calculate the stopping cross section factors

21515150

3210350015.110240101935525.0][ eVcm

OAl

Al

2,

10

cos

1

cos

1 323232

][ OAl

Oout

OAl

inO

OAl

OK

21515150

3210326015.110252101933625.0][ eVcm

OAl

O

Using the molecular density N Al

2O

3=2.35x1022 molecules/cm3 we find:

xxNEOAlOAl

AlAl

A

eV03.82

3232

0

xxNEOAlOAl

OO

A

eV06.76

3232

0

Example

Surface spectrum height

10

0 cos][

)(

EQE

H

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

1

0

0,cos0

m)(

AB

A

A

A

QEH

E

2

0

0,cos0

n)(

AB

B

B

B

QEH

E

Energy width per channel

stopping cross section factors

Mean energy in thin films

)()(1

0 NtEESEA

i

r

i

iSEA

in

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

20

)1( ESEA

inEE

)(0

),(

cos NtNtSEA

iEE

i

ii

i EQ

A

Surface Energy Approximation

E0E1E2

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

be calculated using with

E= Ē(1) then ∆Ei(1) and Ē(2)

),(

cos

EQ

A

i

ii

iNt

)()1()1(

1

)1( NtE i

r

i

i

inE

2

)1(

0)2( E inEE

Mean energy in thin films

E0E1E2

Example Calculate surface height for 2 MeV 4He+ on Al2O3:Ω=10-3srE=1 keV/channelQ=6.24x1013 incident particles (10μC charge)θ1=0°, θ2=10° (scattering angle=170°)

From appendix 6: σRAl=0.2128x10-24 & σR

O=0.0741x10-24

From previous example:

2150

3210326][ eVcm

OAl

O

2150

3210350][ eVcm

OAl

Al

1

0

0,cos0

)(

AB

A

A

A

EQmEH

cntEQOAl

Al

Al

AlH 7610350

1021024.610102128.0

0

215

33324

320,

cntEQOAl

O

O

OH 4310326

1031024.610100741.0

0

315

33324

320,

For not too thick film

)()(0

)(2

)(

NtEE

NtSEA

i

f

i

f

E=Ē(f)

E0E1E2

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 μC 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 during acquisition of backscattering spectra

Thin-film analysisThe peak integration method

iR

i

R

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 deposited on the sample during the run

solid angle subtended by the detector at the target

an application of the peak integration method of analysis of the two-element thin film

E0=3776 keVθ=170˚θ1=0˚θ2=10˚Ω=0.78 msrCBi=(0.99±0.03)E=(3.742±0.005)keV/channelÉ=(8±3) keVKFe=(170˚)=0.7520KGd=(170˚)=0.90390

E1= nE+ É

Example

Energy intercept

sr

cmE

Fe

R

22424

20 102469.010776.3

521.3)170,(

sr

cmE

Gd

R

22424

20 10510.110776.3

53.21)170,(

998.03776

)26)(2)(049.0(1

3/4

FeR

993.03776

)64)(2)(049.0(1

3/4

GdR

From appendix 6

EZZ

CMR

21

3/4

049.01From

Example

Center-of-mass energy

From Trim 1985:

215104.51)3776( eVcmkeVFe

215102.52)3676( eVcmkeVFe

215103.86)3776( eVcmkeVGd 215105.87)3676( eVcmkeVGd

Example

008.1

)201020(

)1660('

)1757(

01.20'

0,

DTR

cts

nB

nB

CQ

HAB

A

cts

nA

nA

HAB

B)20640(

)1812('

)1910(

0,

Integrated counts in spectral regions of interest (initial and final channel numbers are listed:Channels (789-918)=103978 cts; (920-960)=49 ctsChannels (640-767)=64957 cts; (768-788)=79 cts

Example

From: E1= nE+ É and Ki=Ei1/E0 :

keVEnBB

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

EKA

A

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

Example

Energy intercept

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

Example

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

22436

19

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

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

SEA

Fe

21810)021.0709.0()( cm

atomsNtSEA

Gd

21810)08.068.2(

cmatoms

iR

i

R

Bii

EQ

DTReCAiNt

))(,('

cos)(

1

DTR CBiAi

1

Q’ Ω

0.998

σ

e

Example

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

)()(1

0 NtEESEA

i

r

i

iSEA

in

For the first-order energy loss, ΔESEAin ,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(

Example

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

)()(0

)(2

)(

NEE

NtSEA

i

f

i

f

218

2

)1(/1054.2)(

37763676

)( cmatomsNNtSEA

FeFe

218)1(/10672.0)( cmatomsNt

Gd

Example

Results of an additional iteration of this procedure using the following equations we have: (Note that Fe and Gd are evaluated at Ē(1) )

eVeVE in191)10672.0)(105.87()1054.2)(102.52( 18151815)1(

)()1()1(

1

)1( NtE i

r

i

i

inE

2

)1(

0)2( E inEE

)()(0

)(2

)(

NtEE

NtSEA

i

f

i

f

eVE in3681

2

1923776

)2( 218

2

)2(/10)08.055.2()(

37763681

)( cmatomsNtNtSEA

FeFe

218)2(/10)021.0674.0()( cmatomsNt

Gd

Example

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

),(

),(.

E

E

A

A

N

N

m

n

B

A

A

B

A

B

02.078.3998.0

993.0.

521.3

53.21.

323103823

)26164475(

.)170,(

)170,(

//

0

0

R

RE

E

A

A

N

N

Fe

GdFe

R

Gd

R

Gd

Fe

Gd

Fe

Example

If the molecular formula for the film is written as GdmFen , then:

m=0.209±0.001 and n=0.791±0.001

n+m=1

Example

The value of the physical film thickness:

nmcmtFe 3021044.8

1055.222

18

Elemental bulk density

3220 /1044.8 cmatomsM

NN

Fe

FeFe

3220 /1002.3 cmatomsM

NN

Gd

GdGd

nmcmtGd 2231002.3

10674.022

18

nmtGdFe 525

Example

NNAB

B

B

AB

A

A NtNtt

)()(

),(

cos

EQ

A

i

ii

iNt

Areal density, Nt, as atoms per unit area

Detector solid angle

Integrated peak count

Incident ionsCross section

Areal density

),(

),(.

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 measured integrated peak count

Cross section ratio

AB

AB

M

NmN

AB

A

0

AB

AB

M

NnN

AB

B

0

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

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