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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