Rutherford Backscattering Spectroscopy, Ion Implantation, Modern Ion Beam Materials Analysis, Younes Sina

1. The University of Tennessee, Knoxville
Backscattering Spectrometry
Part 1
Professor C.J.McHargue
Younes Sina

2. Backscattering spectrometry using ion beams with energies in
the megaelectronvolt range has been used extensively for
accurate determination of stoichiometry , elemental areal
density ,and impurity distributions in thin films.

3. Kinematic factor
K E /E
i
i 1 0
2
 (M 2  M 2sin 2 )  M cos  1/ 2
 
2 1 1
K

 M M 
 1 2
Z2
M2
E0 M1 Z1 M1 Z1 E0 -ΔE

5. Areal density
Integrated peak count
Areal density, Nt, as atoms per unit area
Ai cos i
Nt 
i
Qi ( E ,  )
Cross section
Incident ions Detector solid angle

6. Rutherford scattering cross section
e2≈1.44x10-13
2
 Z Z e2  4[(M 2  M 1 sin 2 )  M 2 cos ]
2
2 1/ 2
 ( E , )   
1 2
2
R
 4E  M 2sin 4 (M 2  M 1 sin 2 )
1/ 2
 
2
2
Z2
M2
E0 M1 Z1 M1 Z1 E0 -ΔE

7. The average stoichiometric ratio for the compound film AmBn
nABN 0

AB
N B
MAB
n NB AB A( E , )
  .
m NA AA B ( E , )
Cross section ratio
Ratio of measured
integrated peak count
mABN 0

AB
N A
MAB

8. Physical film thickness AmBn
( Nt ) A ( Nt ) B
t AB
 AB
N A N B
mABN 0 nABN 0
 
AB AB
N A
MAB N B
MAB
MAB  mMA  nMB

9. FUNDAMENTALS
Energy
separation
 dK 
E1  E 0 EM 2
 dM 2 
E Minimum energy separation that can be
M 2  experimentally resolved
 dK 
E 0 
 dM 2 
mass resolution for fixed θ:

11. Elastic scattering cross section
Fraction of incident particles
Average differential cross section
scattered to the detector
 1  dQ( E ) 1
 ( , E )   ( )
 Nt  Q ( )
Detector solid angle
Number of target atoms per unit area

12. Rutherford cross section
Z2
M2
E0 M1 Z1 M1 Z1
2
 Z Z  
 ( E , )  0.02073 
    M 1 
 sin
1 2 4
R    2 
 2   M 2 
 4E  
Unit: barn/ sr
1 b(barn)=10-24 cm2

13. Non-Rutherford cross section
Low energy
L’Ecuyer Eq. for Low energy ion
4/3
  1 0.049Z 1Z 2

R
E CM
Center-of-mass kinetic energy (keV)
Wenzel and Whaling for light ion with
MeV energy ECM≈Elab
7/2
  1 0.0326Z 1Z 2

R
E CM

14. Laboratory projectile kinetic energy
Non-Rutherford
For 1H:
 (0.12  0.01) Z 2  (0.5  0.1)
NR
E
Laboratory projectile kinetic energy(MeV)
Lab
For 4He:
 (0.25  0.01) Z 2  (0.4  0.2)
NR
E Lab
For 7Li:
 (0.330  0.005)Z 2  (1.4  0.1)
NR
E Lab

16. Experimental geometry
θ1
θ2
  π   1  2 cosθ2= cos (π-θ).cosθ1

17. Experimental geometry
For IBM geometry
x x
din  dout 
cos  1 cos  2
For Cornel
x x
din  dout 
cos  1 cos  1, cos  2

18. Effects of energy loss of ions in solids
Atomic density
dE
 N Stopping cross section
dx
stopping power
For a compound of AmBn:
εAB =mεA+nεB
AB
 dE 
  N   N A A  N B B
AB AB AB AB
 dx 
Atomic density
Molecular density

19. Example: Calculate the stopping cross section and stopping power of
2 MeV 4He+ in Al2O3 using Bragg rule
From appendix 3: εAl= 44x10-15 eVcm2 εO= 35x10-15 eVcm2
For a compound of AmBn:
εAB =mεA+nεB
εAl2O3=(2x44+3x35)x10-15 =193x10-15eVcm2
g 23 molecules
N 0 4 cm 3  6 10 mol 22 Al2O3molecules
N  M   2.35 10
Al 2 O 3
102
g cm 3
mol
Al 2 O 3
 dE  eV
     2.35 10 22 193 10 15  46
Al 2 O 3 Al 2 O 3
N 0
 dx  A
Al atoms O atoms
 3  2.35 10 22  7.11022
Al 2 O 3
 2  2.35 1022  4.7 1022
Al 2O 3
N Al
cm 3
N O
cm 3

20. Example:
Al 2 O 3
 dE  eV
  N   2.35 10 193 10 15
 46
Al 2 O 3 Al 2 O 3 22
0
 dx  A
15 15 eV
 4.7 10  44 10 22
 7.110  35 10
22
 46 10 8
cm
eV
 46 0
A
x
E   (dE / dx)dx
0

21. Depth scale x
E   (dE / dx)dx
0 Energy loss factor
E  S  x
  dE   dE  1 
S    K   1
  
  dx in cos  1  dx out cos  2 
E   N .x Stopping cross section factor
 1 
    K in 1
  out 
 cos  1 cos  2 
 1 1 
   K A  in   out, A
AB AB AB
E A   A N x 
AB AB
A
 cos  1 cos  2 

22. Depth resolution
Calculate the depth- scattered ion energy differences
Example
for 2 MeV 4He+ in Al2O3
θ1=0°and θ2=10°
K factor for4He on Al=0.5525
K factor for4He on O=03625
 0.5525  2  1.105 MeV
Al
E1  KE0 E 1
 0.3625  2  0.725 MeV
O
E 1
Using the surface-energy approximation ε at E0,surface
εAB =mεA+nεB
  2   in  3   in  2  44 1015  3  35 1015  193 1015 eVcm 2
Al 2 O 3 Al O
in
ε at E1
 out, Al  2   out, Al  3   out, Al  2  511015  3  46 1015  240 1015 eVcm 2
Al 2 O 3 Al O
  2   out,O  3   out,O  2  54 1015  3  48 1015  252 1015 eVcm 2
Al 2 O 3 Al O
out,O

23. Example We can now calculate the stopping cross section factors
 1 
[ ]
1
  K Al  in   out, Al
Al 2 O 3 Al 2 O 3 Al 2 O 3

cos  1 cos  2 
0
Al

[ ]
Al 2O 3
0  0.5525 193 1015  240 1015 1.015  350 1015 eVcm 2
Al
 1 
[ ]
1
  K O  in   out,O
Al 2 O 3 Al 2 O 3 Al 2 O 3

cos  1 cos  2 
0
O

[ ]
Al 2O 3
0  0.3625 193 1015  252 1015 1.015  326 1015 eVcm 2
O
Using the molecular density N Al2O3=2.35x1022 molecules/cm3 we find:
 E Al   0 Al 2 O 3
Al N
Al 2 O 3 
x

eV 
 82.3 0   x

A 
 EO   0 Al 2 O 3
O N
Al 2 O 3 
x

eV 
 76.6 0   x

A 

24. Surface spectrum height
Energy width per channel
 ( E )QE
H 0  [ ] cos  1
0
stopping cross section factors
Surface height of the two elemental peaks in the compound AmBn are given by
A( E 0)QmE

H A, 0
 0 AB
A
cos  1
B ( E 0)QnE

H B ,0
 0 AB
B
cos  2

25. Mean energy in thin films Surface Energy Approximation
Ai cos i SEA
Nt   ( Nt )
i
Qi ( E ,  ) E  E0
i
r
E   ( E ) ( Nt )
SEA i SEA
in 0 i
i 1
 E 0  E in
SEA
(1)
Mean energy of the ions in the film ,Ē(1) E
2
E2 E1 E0

26. Mean energy in thin films
For the second iteration, the values of (Nt)i(1) should
Ai cos i
be calculated using N ti  Qi ( E , ) with
E= Ē(1) then ∆Ei(1) and Ē(2)
r
E  
(1) i (1)
in
E (1)
( Nt ) i
i 1
E
(1)
( 2)
E  E0  in
2
E2 E1 E0

27. Example Calculate surface height for 2 MeV 4He+ on Al2O3:
Ω=10-3sr
E=1 keV/channel
Q=6.24x1013 incident particles (10μC charge)
θ1=0°, θ2=10° (scattering angle=170°)
From appendix 6: σRAl=0.2128x10-24 & σRO=0.0741x10-24
From previous example:
[ ] [ ]
Al 2O 3 Al 2O 3
0  350 1015 eVcm 2 0  326 1015 eVcm 2
Al O
A( E 0)QmE
H A, 0 
 0AB
A
cos  1
AlQ 2 E 0.2128 10 24 10 3  6.24 10 3  2 103
   76cnt
H Al , 0
 0 Al 2 O 3
Al
350 10 15
OQ3E 0.074110 24 10 3  6.24 10 3  3 103
   43cnt
H O,0
 0 Al 2 O 3
O
326 10 15

28. For not too thick film
E=Ē(f)
2
 E( f ) 
( Nt )    ( Nt )
(f) SEA
 E0 
i i
 
E2 E1 E0

29. Sample analysis
Typical experimental operating conditions and parameter ranges used during
acquisition of backscattering spectra
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

30. Thin-film analysis
The peak integration method
Integrated peak counts
That can be accurately determined
from the spectrum
Correction factor
Ai cos  1 CBi e  DTR Dead time ratio
( Nt )i 

Q'  R ( E ,  )( ) i
i
R
Integrated charge deposited
on the sample during the run
Non Rutherford correction factor
solid angle subtended by
the detector at the target

31. Example
an application of the peak integration method of analysis of the two-element
thin film
E0=3776 keV
θ=170˚
θ1=0˚
θ2=10˚
Ω=0.78 msr
CBi=(0.99±0.03)
E=(3.742±0.005)keV/channel
É=(8±3) keV E1= nE+ É
KFe=(170˚)=0.7520 Energy intercept
KGd=(170˚)=0.90390

32. Example
From appendix 6
3.521 cm 2
 ( E0 ,170 )  10  24  0.2469 10 24
Fe
R 2
3.776 sr
21.53 cm 2
 ( E0 ,170 )   24
10  1.510 10  24
Gd 
R 2
3.776 sr
4/3
From   1 0.049Z 1Z 2
R
E CM
 
4/3
(0.049)(2)(26) Center-of-mass energy
   1  0.998
 R  Fe 3776
 
4/3
(0.049)(2)(64)
   1  0.993
 R Gd 3776

33. Example
From Trim 1985:
 (3776 keV )  51.4 10
Fe 15
eVcm 2
 (3676 keV )  52.2 10
Fe 15
eVcm 2
 Gd
(3776 keV )  86.3 10 15
eVcm 2
 Gd
(3676 keV )  87.5 10 15
eVcm 2

34. Example
Q '  20.01 C
nB  (757  1)
nB '  (660  1)
 (1020  20) cts
AB
H A, 0
DTR  1.008
nA  (910  1)
nA'  (812  1)
 (640  20) cts
AB
H B ,0
Integrated counts in spectral regions of interest (initial
and final channel numbers are listed:
Channels (789-918)=103978 cts; (920-960)=49 cts
Channels (640-767)=64957 cts; (768-788)=79 cts

35. Example
From: E1= nE+ É and Ki=Ei1/E0 :
 nB E  E '  (757  1)(3.742  0.005)  (8  3)  (2841  6)keV
B
E 1
 nAE  E '  (910  1)(3.742  0.005)  (8  3)  (3413  7)keV
A
E 1
E1  (2841  6)  0.752  0.002
B Energy intercept
KB 
E0 (3776  5)
E1  (3413  7)  0.904  0.002
A
KA 
E0 (3776  5)
Therefore, element A and B are Gd and Fe, respectively. Note that element A
could also be Tb, because KTb=0.9048

36. Example
Calculation of elemental areal densities,(Nt)
Values of Ai are calculated from the integrated counts in the
regions of interests
79
AFe  64957  (128)  (64475  261)cts
21
49
AGd  103978  (130)  (103823  323)cts
41
In this case, the background correction is almost negligible

37. Example
The areal densities in the surface-energy approximation,(Nt)SEA i
using E=E0 1
Ai cos  1 CBi e  DTR
( Nt )i 

Q'  
( E ,  )(
i
)i
R
R
Ai DTR CBi 0.998
e
SEA (64475  261)(1.008)(0.99  0.03)(1.602 1019 ) atoms
( Nt )  6 3  24
Fe (20.0110 )(0.78 10 )(0.2469 10 )(0.998) cm 2
Q’ Ω σ
 (2.68  0.08) 10 18 atoms
cm 2
SEA
( Nt )  (0.709  0.021) 1018 atoms
Gd cm 2

38. Example
The mean energy of the 4He ion in the film, Ē(1), is calculated (to first order)
using the following equation
r
E  
SEA i SEA
in
( E 0) ( Nt )
i
i 1
For the first-order energy loss, ΔESEAin ,of the ions in the film:
E  
SEA SEA
 
SEA Fe Gd
in
( E 0) ( Nt ) ( E 0) ( Nt )
Fe Gd
 (51.4 10 15 )(2.68 1018 )  (86.3 10 15 )(0.709 1018 ) eV
 199 keV
 E 0  E in
SEA
(1) (1) 199
E  E  3776   3676 keV
2 2

39. Example
From the following Eq. we can calculate the areal densities:
2
E (f) 
  (N )
(f ) S EA
( Nt ) i
 E0  i
 
2
 3676 
( Nt )    (N )
(1) SEA
 2.54 1018 atoms / cm 2
 3776 
Fe Fe
(1)
( Nt )  0.672 1018 atoms / cm 2
Gd

40. 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) )
 E
r (1)
E  
(1) i (1) ( 2)
in
E (1)
( Nt ) i
E  E0
2
in
i 1
2
E ( f ) 
  ( Nt )
( f ) S EA
( Nt ) i
 E0  i
 
E  (52.2 1015 )(2.54 1018 )  (87.5 1015 )(0.672 1018 ) eV  191 eV
(1)
in
192
E  3776   3681 eV
( 2)
in
2
2
 3681 
( Nt )    ( Nt )
( 2) SEA
 (2.55  0.08) 1018 atoms / cm 2
 3776 
Fe Fe
( 2)
( Nt )  (0.674  0.021) 1018 atoms / cm 2
Gd

41. Example
The average stoichiometric ratio for this film using the following Eq.:
n NB AB A( E , )
  .
m NA AA B ( E , )
N Fe

AFe
 
Gd
( E0 ,170 ) /  R 
( E ,170 )  /  R 
R Gd
.
N Gd AGd 
Fe
R 0 
Fe
(64475  261) 21.53 0.993
 . .  3.78  0.02
103823  323 3.521 0.998

42. 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

43. Example The value of the physical film thickness:
( Nt ) A ( Nt ) B
t AB
 AB
N A N B
 Fe N 0
Elemental bulk density
N Fe   8.44 1022 atoms / cm 3
M Fe
Gd N 0
N Gd   3.02 1022 atoms / cm 3
M Gd
2.55 1018 0.674 1018
t Fe  cm  302 nm tGd  cm  223 nm
8.44 10 22
3.02 10 22
tGdFe  525 nm

44. Thank you
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