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Modern Ion Beam Analysis, Energy Loss and Energy StragglingA Presentation by Younes Sina, PhD student of MSE at The University of Tennessee, Knoxville

### Transcript of Energy loss and energy straggling a presentation by Younes Sina

• 1. The University of Tennessee, Knoxville
Energy Loss and Energy Straggling
Younes Sina

2. Basic concepts & definitions
Nt
areal density
N
volume density
[atoms/cm3]
[atoms/cm2]
dE/dx
Stopping power, stopping force, specific energy loss
[MeV/mm] , [eV/m]

mass density
(gr/cm3)
Nt=N .dx
E/x=dE/dx
x->0

stopping cross section
[eV/1015atoms/cm2],[ KeV/mg/cm2 ]
=1/N(dE/dx)
=1/(dE/dx)
3. Basic concepts and definitions
Nt
x
Incident particles
Transmitted particles
E0 -E
E0
Z1
Z1
Z1
M1
M1
M2
4. Unit conversions
5. Basic Physics
Important factors during interaction of ions and target:
Ion velocity
Charge of the ion
Charge of target atom
6. 3 regimes for ions:
Energy loss of ions:
Low velocity
Intermediate velocity
High velocity
In comparison to the orbital velocity of atomic electron
7. Low velocity
Intermediate velocity
High velocity
V (velocity of ions)v0:
charge state of the ion increases
ion becomes fully stripped of its electrons
v>>v0 Ee
9. Bethe & Bloch formula for high- velocity regime
dE/dx=N .Z2 .(Z1e2)2f (E/M1)
(E/M1) is a function of target
(it is not a function of the projectile)
for most application of ion beam analysis, nuclear stopping is small. Above 200 keV/amucontribution of nuclear stopping EH =2MeV/4=0.5 MeV
12. EHI=mHIEH
Example 2:
EH=ELi/mLi
If Li=1
CalculateLi@2,5,and 10 MeV
EH=ELi/7
EH=2000/7=285 keV
EH=5000/7=714.28 keV
EH=10000/7=1428.57 keV
HI =HHI 2ZHI2
Li =HLi 2ZLi2
H@285 keV= 0.489
H@714 keV= 0.282
H@1428 keV= 0.177
Li=9H
Li=4.40 Mev.cm2/mg
Li=2.54 Mev.cm2/mg
Li=1.60 Mev.cm2/mg
13. If He=1 then:
He 2ZHe2 = (12). (22)=4.00
Effective charge () as a function of Z1 &Z2
If E He=0.5 MeV :
He 2ZHe2 = (He2). (22)=2.88
If E He=1 MeV :
He 2ZHe2 = (He2). (22)=3.46
If E He=1.5 MeV :
He 2ZHe2 = (He2). (22)=3.75
If E He=2 MeV :
He 2ZHe2 = (He2). (22)=3.89
If E He=3 MeV :
He 2ZHe2 = (He2). (22)=3.99
ai:fitting constant value
M1 = 4.0026
E/M1 [keV/amu]
a0=0.2865
a1=0.1266
a2=-0.001429
a3=0.02402
a4=-0.01135
a5=0.001445
14. Effective charge () as a function of Z1 &Z2
Li =A{1-exp[-(B+C)]}
A=1+ (0.007+5x10-5Z2)exp {-[7.6-ln(ELi[keV/amu]2}
B=0.7138+0.002797ELi[keV/amu]
C=1.348x10-6 (ELi[keV/amu]2)
15. Calculation of Li ,stopping in carbon at 2,5 and 10 MeV
285
715
1430
mLi=7
ELi= 2, 5, and 10 Mev
ELi/mLi = ELi/7=
0.8
0.97
1
Li =A{1-exp[-(B+C)]}
Li =
A=1+ (0.007+5x10-5Z2)exp {-[7.6-ln(ELi[keV/amu]2}
B=0.7138+0.002797ELi [keV/amu]
C=1.348x10-6 (ELi[keV/amu]2)
From Example 2:
H@285 keV= 0.489
H@714 keV= 0.282
H@1428 keV= 0.177
Li =HLi 2ZLi2
Li =HLi 2(3)2
Li = 2.81
Li = 2.38
Li = 1.593
Li =9HLi 2
Li =9HLi 2
Li =9HLi 2
Li =9HLi 2
16. Effective charge () for heavy ions : Z > 3
HI =1- exp (-A)[1.034-0.1777 exp(-0.08114 ZHI)]
A=B+0.0378 sin ( B/2)
B=0.1772 (EHI [keV/amu]) 1/2 ZHI-2/3
17. Stopping cross section for compound
Braggs rule
AB =mA+nB
Example :2.0 MeV ion 4He stopping in silicon SiO2
SRIM-2006 gives Si(2.0 MeV)=46.88 eV cm2/1015 atoms and O(2.0 MeV)=38.36 cm2/1015 atoms.
For , SiO2 we then have SiO2 =1Si+2O =41.02 cm2/1015 atoms
18. Stopping cross section and depth scale
E=
dE/dx
Stopping power
X=
=(1/N)(dE/dx)

stopping cross section

Can be evaluated either at E0 or at Eav=E0-E/2
19. Thin targets
E=
E= (dE/dx) (E0)x
E= (dE/dx) (av)x
=(1/N)(dE/dx)
E=(E0)Nt
E=(Eav)Nt
Surface energy approximation
Mean energy approximation
20. Thick targets
Energy loss evaluated at the energy of the ion at the ( i-1) the slab
Stopping cross section evaluated at the energy of the ion at the ( i-1) the slab
Ei= (dE/dx)(Ei-1)xi
Ei=(Ei-1)(Nt)i
E0
21. Example: Proton depth scale in carbon
What is the 2.0 MeV proton energy lost in a carbon target for depth of (a) 1000 nm and (b) 20 m?
From the unit conversion table:
1015 atoms/cm2=
nm
1000 nm= 17.6x 1018 atoms/cm2
20 m= 353x 1018 atoms/cm2
E=(E0)Nt
(E0=2MeV)=2.866 ev cm2/1015 atoms
E=2.866x10-15x17.6x101850 keV
Surface energy approximation
Thin targets
E=(E0)Nt
(E0=2MeV)=2.866 ev cm2/1015 atoms
E=2.866x10-15x353x10181000 keV
22. Example: Proton depth scale in carbon
What is the 2.0 MeV proton energy lost in a carbon target for depth of (a) 1000 nm and (b) 20 m?
From the unit conversion table:
1015 atoms/cm2=
nm
1000 nm= 17.6x 1018 atoms/cm2
20 m= 353x 1018 atoms/cm2
E=(Eav)NtEav=E0-E/2=2000-50/2 keV=1975 keV
(Eav=1975 keV)2.866 ev cm2/1015 atoms
E=2.866x10-15x17.6x101850 keV
Mean energy approximation
Thin targets
E=(Eav)NtEav=E0-E/2=2000-1000/2 keV=1500 keV
(Eav=1500 keV)=3.506 ev cm2/1015 atoms
E= 3.506 x10-15x353x10181235 keV
23. Example: Proton depth scale in carbon
What is the 2.0 MeV proton energy lost in a carbon target for depth of (a) 1000 nm and (b) 20 m?
Thick targets
Ei=(Ei-1)(Nt)i
i=6
(Nt)i =(353/6)x1018 =58.83x1018 atoms/cm2
E1=(E0)(Nt)1= 2.866x10-15x58.83x1018168.5 keV
The energy at the end of the first slab is then E1=E0-E=2000-168.5 keV=1832 keV
Energy loss in the second slab at this energy:
E2=(E1)(Nt)2=3.051x10-15x58.83x1018179.5 keV
E2=E1-E2=1832-179.5keV=1652 keV .
E3= 193.0 , E4= 210.3 ,E5=233.7 , E6= 268.1 keV
E2= Ei(i=1-6)=1253 keV
E0
E2
E1
24. electronic stopping for isotopes
Stopping (medium [ ,Z2]) =stopping (medium [Mav , Z2]).(Mav/ )
25. Straggling
Bohrs theory:
Nt[atoms/cm2]2x1020.
When the energy transferred to target electrons in the individual collisions is small compare to the width of the energy loss distribution, the distribution is close to a Gaussian distribution.
In the limit of high ion velocity, the energy loss is dominated by electronic excitations.
Full width at half- maximum height(FWHM)=2.355
B2[keV2]=0.26Z12Z2Nt[1018 atoms/cm2]
Bohr value for the variance (standard deviation) of the average energy loss fluctuation
26. Example
From the following Equation, we obtain for 4He ions:
B2[keV2]Z2Nt[1018 atoms/cm2]
Helpful for quick estimates of 4He ion Bohr straggling
4% accuracy
B2[keV2]=0.26Z12Z2Nt[1018 atoms/cm2]
27. Corrections to Bohrs theory, other models
Lindhard & Scharff Eq.:
{
0.5 L(x), for E [keV/amu]< 75 Z2
2/B2=
1, for E [keV/amu] 75 Z2
L(x)=1.36 x1/2- 0.16 x3/2
28. Example
Straggling of 5.0 MeV helium ions in gold
From Bohr Eq.:B2[keV2]=0.26Z12Z2Nt[1018 atoms/cm2]
we have:
B2/Nt 82 keV2 cm2/1018 atoms
In a gold layer of 1018 atoms/cm2 (about 170 nm)
B 9 keV
2/B2 0.8 for He ions in gold at 5.0 MeV
=7 keV
29. Straggling in mixtures and compounds
For an compound (mixture) AmBn (m+n=1) with an atom density NAB[atoms/cm3]and the atomic densities NA and NB:
t is the thickness
IfmNAB=NA and nNAB=NBthen:
(AB)2=(A)2+(B)2
30. Example
Bohr straggling of 4He ions in 1018 atoms/cm2 of SiO2
AmBn= SiO2
m=0.33 & n=0.67
Nsi t = 0.33NSiO2t = 0.33x1018 atoms/cm2
NO t = 0.33NSiO2t = 0.67x1018 atoms/cm2
B2[keV2]=0.26Z12Z2Nt[1018 atoms/cm2]
Bohrs Eq.
(BSi)2[keV2]=0.26x 0.33Z12Z2 = 4.80 keV2
(BO)2[keV2]=0.26x 0.67Z12Z2 = 5.57 keV2
(BSiO2)2=(BSi)2+(BO)2
(BSiO2)2=(4.80+5.57)keV2= 3.22 keV
31. Additivity of energy loss fluctuations
(TOT)2=(DET)2+(STR)2 +(BEAM)2
Energy resolution
Energy straggling
Beam energy profile
32. Range
33. Kurdistan, Iran