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

E

ESESN

dEER

ne0 )]()([)(

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