energy loss and energy straggling a presentation by younes sina
DESCRIPTION
Modern Ion Beam Analysis, Energy Loss and Energy StragglingA Presentation by Younes Sina, PhD student of MSE at The University of Tennessee, KnoxvilleTRANSCRIPT
Energy Loss and
Energy Straggling
Younes Sina
The University of Tennessee, Knoxville
Nt
areal density
dE/dx Stopping power, stopping force, specific energy loss[MeV/mm] , [eV/μm]
ΔE/Δx=dE/dxΔx→0
εstopping cross section[eV/1015 atoms/cm2 ],[ KeV/mg/cm2 ]
ε=1/N(dE/dx)
ε=1/ρ(dE/dx)
[atoms/cm2]
Nvolume density [atoms/cm3]
ρ mass density (gr/cm3)
Nt=N .dx
Basic concepts & definitions
Incident particles Transmitted particles
E0
Z1
M1
E0 -ΔE
Δx
Nt
M2
Z1
Z1
M1
Basic concepts and definitions
Uni
t con
vers
ions
Multiply units by For units Example
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
μg/cm2 nm 100 μg/cm2 For C~258 nm
μg/cm2 1015 atoms/cm2 100 μg/cm2
For Au~305x1015 atoms/cm2
eV cm2/1015 atoms MeV/(mg/cm2 ) 100 eV cm2/1015 atoms for Al2O3~2.95 MeV cm2/mg
[M2= (2MAl + 3 MO)/5; MAl=26.98, MO=16.00]
eV cm2/1015 atoms keV/μm 30eV cm2/1015 atoms for Si~150 keV/μm
][1
1
amuM
]3/[
][10661.1 22
cmg
amuM
]/[
103cmg
][661.110
1
3
amuM
1581.0
10389.1 7
][661.11
2 amuM
][661.1
]/[102
32
amuMcmg
Basic Physics
Important factors during interaction of ions and target:
Ion velocityCharge of the ionCharge of target atom
Energy loss of ions:3 regimes for ions:
Low velocityIntermediate velocityHigh velocity
In comparison to the orbital velocity of atomic electron
V (velocity of ions)<<< v0 (velocity of electron at orbital)
V0= Bohr velocity
elastic collision with target nuclei
Nuclear energy loss dominates
nuclear energy loss diminishes as 1/E
Electronic energy loss dominates (inelastic collisions with atomic electrons)
Low velocityIntermediate velocityHigh velocity
ΔE= ΔEn+ΔEe
The ion carries its electrons and tends to neutralize by electron capture
With increasing v
v≈ 0.1v0 to v≈Z12/3v0 ΔEe
v>>v0 : charge state of the ion increases ion becomes fully stripped of its electrons
In the low –ion-velocity range
E
In the high –ion-velocity range
v>>v0 ΔEe chargeion
dE/dx=N .Z2 .(Z1e2)2 f (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/amu contribution of nuclear stopping <1%
For example for Zr (amu=90): at E≥2.22 keV contribution of nuclear stopping <1%
Bethe & Bloch formula for high- velocity regime
EHI= (mHI/mH)EH≈ mHIEH
The scaling rule
γ= fraction effective charge
higher energy of ion γ→1
εHI = εHγHI 2ZHI
2
ε = 1/N(dE/dx)
EH=EHe/mHe→EH =2MeV/4=0.5 MeV
Example
If γHe=1 EHe=2 MeVWhat is EH ?
EHI= (mHI/mH)EH≈ mHIEH
Example 2:
If γLi=1 Calculate εLi@2,5,and 10 MeV
EHI=mHIEHEH=ELi/mLi
EH=ELi/7EH=2000/7=285 keVEH=5000/7=714.28 keVEH=10000/7=1428.57 keV
εH@285 keV= 0.489εH@714 keV= 0.282εH@1428 keV= 0.177
εLi=9εH
εHI =εHγHI 2ZHI
2 εLi =εHγLi 2ZLi
2
εLi=4.40 Mev.cm2/mgεLi=2.54 Mev.cm2/mgεLi=1.60 Mev.cm2/mg
Effective charge (γ) as a function of Z1 &Z2
ai : fitting constant value
E/M1 [keV/amu]
M1 = 4.0026
a0=0.2865a1=0.1266a2=-0.001429a3=0.02402a4=-0.01135a5=0.001445
If E He=0.5 MeV :γHe
2ZHe2 = (γHe
2). (22)=2.88
If E He=1 MeV :γHe
2ZHe2 = (γHe
2). (22)=3.46
If E He=1.5 MeV :γHe
2ZHe2 = (γHe
2). (22)=3.75
If E He=2 MeV :γHe
2ZHe2 = (γHe
2). (22)=3.89
If E He=3 MeV :γHe
2ZHe2 = (γHe
2). (22)=3.99
If γHe=1 then:γHe
2ZHe2 = (12). (22)=4.00
1
5
0
exp1 )][ln(2
M
Ea
i
iiHe
γLi =A1-exp[-(B+C)]
Effective charge (γ) as a function of Z1 &Z2
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)
Calculation of γLi ,stopping in carbon at 2,5 and 10 MeV
mLi=7 ELi= 2, 5, and 10 Mev
ELi/mLi = ELi/7= 2857151430
γLi =A1-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)
γLi =0.80.971
εLi =εHγLi 2ZLi
2
εLi =εHγLi 2(3)2
εLi =9εHγLi 2 εLi = 2.81
εLi = 2.38εLi = 1.593
εLi =9εHγLi 2
εLi =9εHγLi 2
εLi =9εHγLi 2
From Example 2:εH@285 keV= 0.489εH@714 keV= 0.282εH@1428 keV= 0.177
Effective charge (γ) for heavy ions : Z > 3
γHI =1- exp (-A)[1.034-0.1777 exp(-0.08114 ZHI)]
A=B+0.0378 sin (π B/2)
B=0.1772 (EHI [keV/amu]) 1/2 ZHI -2/3
Bragg’s rule
Stopping cross section for compound
εAB =mεA+nεB
Example : 2.0 MeV ion 4He stopping in silicon SiO2
SRIM-2006 gives εSi(2.0 MeV)=46.88 eV cm2/1015 atoms and εO(2.0 MeV)=38.36 cm2/1015 atoms. For , SiO2 we then have εSiO2 =1εSi+2εO =41.02 cm2/1015
atoms
Stopping cross section and depth scale
ΔE=
x
dxdxdE0
)/(
X= 0
)/(
1E
E
dEdxdE
dE/dx Stopping power
εstopping cross section
ε=(1/N)(dE/dx)
εCan be evaluated either at E0 or at Eav=E0-ΔE/2
Thin targets
x
dxdxdE0
)/(ΔE=
ΔE= (dE/dx) (E0)
Δx
ΔE=ε(E0)Nt
ε=(1/N)(dE/dx)
ΔE=ε(Eav)Nt
ΔE= (dE/dx) (av)
Δx
Surface energy approximation Mean energy approximation
Thick targets
ΔEi= (dE/dx) (Ei-1)
Δxi
ΔEi=ε(Ei-1)(Nt)i
n
iiEE
1
E0
Energy loss evaluated at the energy of the ion at the ( i-1)
the slabStopping cross section evaluated at the energy of the ion at the ( i-1) the slab
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 μm?
From the unit conversion table:
1015 atoms/cm2=]3/[
][10661.1 22
cmg
amuM
nm
1000 nm= 17.6x 1018 atoms/cm2
20 μm= 353x 1018 atoms/cm2
ΔE=ε(E0)Nt
ε(E0=2MeV)=2.866 ev cm2/1015 atomsΔE=2.866x10-15x17.6x1018≈50 keV
ΔE=ε(E0)Nt
ε(E0=2MeV)=2.866 ev cm2/1015 atomsΔE=2.866x10-15x353x1018≈1000 keV
Surface energy approximation Thin targets
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 μm?
From the unit conversion table:
1015 atoms/cm2=]3/[
][10661.1 22
cmg
amuM
nm
1000 nm= 17.6x 1018 atoms/cm2
20 μm= 353x 1018 atoms/cm2
ΔE=ε(Eav)Nt Eav=E0-ΔE/2=2000-50/2 keV=1975 keVε(Eav=1975 keV)≈2.866 ev cm2/1015 atomsΔE=2.866x10-15x17.6x1018≈50 keV
ΔE=ε(Eav)Nt Eav=E0-ΔE/2=2000-1000/2 keV=1500 keVε(Eav=1500 keV)=3.506 ev cm2/1015 atomsΔE= 3.506 x10-15x353x1018≈1235 keV
Mean energy approximation Thin targets
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 μm?
Thick targets
ΔEi=ε(Ei-1)(Nt)i
i=6(Nt)i =(353/6)x1018 =58.83x1018 atoms/cm2
ΔE1= ε(E0)(Nt)1= 2.866x10-15x58.83x1018≈168.5 keVThe energy at the end of the first slab is then E1=E0-ΔE=2000-168.5 keV=1832 keVEnergy loss in the second slab at this energy: ΔE2= ε(E1)(Nt)2=3.051x10-15x58.83x1018≈179.5 keVE2=E1-ΔE2=1832-179.5keV=1652 keV …….E3= 193.0 , E4= 210.3 , E5=233.7 , E6= 268.1 keV
ΔE2=Σ ΔEi(i=1-6)=1253 keV
E0E1E2
electronic stopping for isotopes
Stopping (medium [ ,Z2]) =stopping (medium [Mav , Z2]).(Mav/ )M av
M av
Straggling
Nt[atoms/cm2]≥2x1020. 2Z
1)
Z]/[
(1
2
amuMeVE
Bohr’s theory:
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.
ΩB2[keV2]=0.26Z1
2Z2Nt[1018 atoms/cm2]
Full width at half- maximum height(FWHM)=2.355Ω
Bohr value for the variance (standard deviation) of the average energy loss fluctuation
Example
ΩB2[keV2]=0.26Z1
2Z2Nt[1018 atoms/cm2]
From the following Equation, we obtain for 4He ions:
ΩB2[keV2]≈Z2Nt[1018 atoms/cm2]
Helpful for quick estimates of 4He ion Bohr straggling4% accuracy
Corrections to Bohr’s theory, other models
Ω2/ΩB2= 0.5 L(x), for E [keV/amu]< 75 Z2
1, for E [keV/amu] ≥ 75 Z2
L(x)=1.36 x1/2- 0.16 x3/2
225
]/[
Z
amukeVEx
Lindhard & Scharff Eq.:
Example
Straggling of 5.0 MeV helium ions in gold
From Bohr Eq.: ΩB2[keV2]=0.26Z1
2Z2Nt[1018 atoms/cm2]
we have: ΩB
2/Nt≈ 82 keV2 cm2/1018 atoms In a gold layer of 1018 atoms/cm2 (about 170 nm) ΩB≈ 9 keV Ω2/ΩB
2≈ 0.8 for He ions in gold at 5.0 MeV Ω=7 keV
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:
If mNAB=NA and nNAB=NB then:
(ΩAB)2=(ΩA)2+(ΩB)2
t is the thickness
tB
Bn
tA
Am
t NNNAB
AB
)))(222
((
Example
AmBn = SiO2
Bohr straggling of 4He ions in 1018 atoms/cm2 of SiO2
m=0.33 & n=0.67
Nsi t = 0.33NSiO2t = 0.33x1018 atoms/cm2
NO t = 0.33NSiO2t = 0.67x1018 atoms/cm2
ΩB2[keV2]=0.26Z1
2Z2Nt[1018 atoms/cm2] Bohr’s Eq.
(ΩBSi)2[keV2]=0.26x 0.33Z1
2Z2 = 4.80 keV2
(ΩBO)2[keV2]=0.26x 0.67Z1
2Z2 = 5.57 keV2
(ΩBSiO2)2=(ΩB
Si)2+(ΩBO)2
(ΩBSiO2)2=(4.80+5.57)keV2= 3.22 keV
Additivity of energy loss fluctuations
(ΩTOT)2=(ΩDET)2+(ΩSTR)2 +(ΩBEAM)2
Beam energy profile
Energy resolution
Energy straggling
Range
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ESESN
dEER
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