IBeAM Facility for Ion Beam Analysis of Materials

Ion Beam Analysis Robert J. Culbertson

Outline

I. Overview

II. Fundamentals of Ion Scattering

A. Kinematics

B. Scattering Cross Section

C. Penetration into Matter (Energy Loss and Energy

Straggling)

III. Channeling

References

J.W. Mayer and E. Rimini, Ion Beam Handbook for Material Analysis, (Academic Press,

New York, 1977)

W-K. Chu, J.W. Mayer, and M-A. Nicolet, Backscattering Spectrometry, (Academic Press,

New York, 1978)

L.C. Feldman, J.W. Mayer, and S.T. Picraux, Materials Analysis by Ion Channeling,

(Academic Press, New York, 1982)

L.C. Feldman and J.W. Mayer, Fundamentals of Surface and Thin Film Analysis, (North-

Holland, New York, 1986)

S.A.E. Johansson and J.L. Campbell, PIXE: A Novel Technique for Elemental Analysis, (John

Whiley and Sons, Chichester, 1988)

J.R. Bird and J.S. Williams, Ion Beams for Materials Analysis, (Academic Press, Sydney,

1989)

J.R. Tesmer, C.J. Maggiore, M. Nastasi, J.C. Barbour, and J.W. Mayer, Eds., High Energy

and Heavy Ion Beams in Materials Analysis, (Materials Research Society, Pittsburgh, 1990)

J.L. Campbell, K.G. Malmoqvist, and S.A. Johansson, Particle–Induced X–Ray Emission

Spectrometry (PIXE) (John Wiley and Sons, 1995)

J.R. Tesmer and M. Nastasi, Editors, Handbook of Modern Ion Beam Materials Analysis,

(Materials Research Society, 1995).

J. Tirira, Y. Serruys, P. Trocellier, Forward Recoil Spectrometry: Applications of Hydrogen

Determination in Solids (Plenum, 1996)

M. Nastasi, J.W. Mayer, and J.K. Hirvonen, Ion-Solid Interactions: Fundamentals and

Applications, (Cambridge University Press, 1996).

Energy Scale

Egap in HTSC Nuclear Binding Energy

Egap in SC X-Rays Cosmic Rays

10-2

100 10

4 10

2 10

6 10

8 10

10 10

12 10

14 10

16

ION BEAM ANALYSIS

Hel

ium

No

zzle

Beam

Vap

or

So

urc

e

Lar

ge H

adro

n C

oll

ider

Ele

ectr

ost

ati

c A

ccel

erato

rs

Sy

nch

rocy

clo

tro

ns

Sy

nch

rotr

on

s

Lin

ear

Acc

eler

ato

rs

Fer

mi

Lab

ENERGY (eV)

Ion Beam Analysis Techniques

Applications of Ion Beams

Kinematic Factor K

Definition:

0

EK

E

=

!

"m1 m

2

E

E0

Ef

• E0, m1, and ! are usually pre-set experimental parameters

• E is measured during the experiment

2

1

22 2 2sin cos

2 1 1

1 20

m m m

EK

m mE

! !" #$ %& '( +) *& '+ ,

= = & '+& '

& '& '- .

Derivation of K

m1

m2

m1

m2

!

"E

10, v

10E

2f, v

2f

E1f, v

1f

• Conservation of Energy (elastic collision):

2 2 21 1 1

1 10 1 1 2 22 2 2m v m v m v

f f= +

• Conservation of Momentum: cos cos

1 10 1 1 2 2

0 sin sin1 1 2 2

m v m v m vf f

m v m vf f

! "

! "

= +

= +

• After some algebra (eliminate " using sin2"+cos2"=1):

21

22 2 221 sin cos21 2 1 11 10

21

1 2210 1 1

E m m mm vf

Km mE m v

f

! !" #$ %& '( +) *& '+ ,

= = = & '+& '

& '& '- .

Recoil of Target Atom

m1

m2

m1

m2

!

"E

10, v

10E

2f, v

2f

E1f, v1f

Recoil Energy of target atom:

2 212 1 1 10 2 2 102

1 21 2

1 2

4cos

where and ( is the "reduced mass")

f f fE E E E m v EM

m mM m m

m m

µ!

µ µ

= " = # = =

= + =+

Example: 2 MeV He incident on Ag

m1 = mHe = 4 amu; E10 = 2 MeV; m2 = mAg = 107 amu

The maximum recoil energy (head-on collision; " = 0):

( )

2 1 22 10 10 10 102 2

1 2

4 4 4 4 107cos (0) 0.14

111f

m mE E E E E

M m m

µ ! !" #= = = =$ %

& '+

Thus, a maximum of 14% of the initial He energy can be transferred to the Ag atom.

Kinematic Factor: Example Consider a graphite substrate with layers of approximately one monolayer each of O, Si, Ni,

Ge, and W. A Rutherford backscattering spectrum using He ions at 2.0 MeV and a scattering

angle of 170 degrees is shown below.

0

50

100

150

200

250

300

350

400

100 200 300 400 500 600

0.5 1 1.5

O, Si, Ni, Ge, and W on Graphite

2 MeV He, 170 degrees

Channel Number

Energy (MeV)

Given the mass of each of the elements we can construct a table of kinematic factors.

Multiplying K by incident energy E0 gives the expected detected energy. Lastly, the “channel

number” for a particular data acquisition system can be determined from calibration constants

(in this example, Energy = 0.003*Channel + 0.08). Element

Mass

(amu)

K

Energy

(MeV)

Channel

Number

C 12.01 0.2529 0.506 142

O 16.00 0.3527 0.725 215

Si 28.09 0.566 1.132 351

Ni 58.71 0.7627 1.525 482

Ge 72.59 0.8033 1.607 509

W 183.85 0.9173 1.835 585

Now C, O, Si, Ni, Ge, and W are readily identified in the spectrum.

Kinematic Factor: Example (cont.)

Plotting the spectrum using a logarithmic vertical scale shows the smaller peaks more clearly.

0.001

0.01

0.1

1

10

100

100 200 300 400 500 600

0.5 1 1.5

O, Si, Ni, Ge, and W on Graphite

2 MeV He, 170 degrees

Channel Number

Energy (MeV)

C

O Si

Ni GeW

Some questions:

What is the peak near channel 240?

Why is the silicon peak not symmetric?

Why do the Ni and Ge peaks overlap?

Why is the carbon data flat, and why does it have a step?

Finding m2

( )( ) ( ) ( ) ( )

1

2 21 2 22 cos 4 cos 4 1 1 2 cos

2 1 1 12 1m Km K Km K K m K K

K! ! !

" #$ %= & & & & + &' () *& + ,- .

Dependence of K on m2

Note that K changes rapidly for small values of m2. This results in high mass resolution.

However, K is rather insensitive to m2 at high values of m2; thus, mass resolution is not as

good at high values of m2.

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

K vs. m2

m1=m

He, !=170

m2

Dependence of K on m1

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

K vs m2 for various m

1

1 H4 He12 C28 Si40 Ar

m2

Conclusion: better mass resolution is obtained with larger m1.

Angular Dependence of the Kinematic Factor K 2

122 2 2

sin cos1 2 1 1, ,

1 21 210

E m m mf

K m mm mE

! !!

" #$ %& '( +) *& '$ % + ,

= = & ') *++ , & '

& '& '- .

Note that this form of K simplifies for !=90° and !=180°.

K m

1

, m

2

, 90°!

"#$

%&=

m

2

'm

1

m

1

+m

2

!

"

###

$

%

&&&

K m

1

, m

2

, 180°!

"#$

%&=

m

2

'm

1

m

1

+m

2

!

"

###

$

%

&&&

2

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200 250

K vs m2 (m

1=4)

90

180

m2

Dependence of K on !

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

KHe

vs. m2 for various !

60

30

90

120

150

180

m2

• K is more sensitive to m2 at high values of m2 for larger !; K is most sensitive at higher

masses for !=180°.

• Many laboratories, including IBeAM, use !=170°.

(Why not use !=180° in the lab?)

Mass Resolution

Example: thin layers of Ag (108 amu), Sn (119), W (184), and Au (197); detector resolution:

15 keV FWHM; beam: 2.0 MeV He; scattering angle: 170°

Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb 41 42 43 44 45 46 47 48 49 50 51

Ta W Re Os Ir Pt Au Hg Tl Pb Bi 73 74 75 76 77 78 79 80 81 82 83

0

10

20

30

40

50

60

70

80

1.65 1.7 1.75 1.8 1.85 1.9

Mass Resolution(2 MeV He, 170 degrees)

Energy (MeV)

Ag Sn

W Au

Note that Ag-Sn are barely resolved; W-Au are not resolved.

Rules of Thumb:

• Can identify elements up to mass ~100 amu

• Can resolve isotopes up to elements of mass ~50 amu

Mass Resolution

1

20 2

0

22

E E E Km

E mKEE

mm

! ! !"# $

%& '( = = = & '%# $ # $# $ & '%%& ' & ' ) *& ') *& ' & '%& ' & '%) *

& ') *

Here, E! is the total energy resolution, including the detector resolution and energy

straggling.

22

2 2 2 2 2 22 sin ( ) cos( ) 2 sin ( ) cos( )2 1 1 2 2 1 1

2 32 2 2sin ( )

1 2 2 1 1 2

K

m

m m m m m m m

m m m m m m

! ! ! !

!

"=

"

# $ # $% + % +& ' & '

( ) ( )%# $ # $

+ % +& ' & '( ) ( )

Note: This derivative may be found by hand or from Maple, for example:

> diff(((sqrt(m2^2-m1^2*(sin(q))^2)+m1*cos(q))/(m1+m2))^2,m2);

! 2 ( ) + ! m2 2 m1 2 ( )sin q 2 m1 ( )cos q m2

( ) + m1 m2 2 ! m2 2 m1 2 ( )sin q 2

2 ( ) + ! m2 2 m1 2 ( )sin q 2 m1 ( )cos q2

( ) + m1 m2 3

Mass Resolution

1

2

20

KEm

mE

!"# $

%& '( =& '%& ') *

10

100

1000

104

105

0 50 100 150 200

(dK/dm2)-1 vs. m

2

30

60

90

120

150

180

m2

Mass Resolution: Example

Example: Given 20 keV, 2.00 MeV, 1700

E E! "= = = °

0

5

10

15

20

25

30

0 50 100 150 200 250

Mass Resolution

(2 MeV He, !=170 degrees, "E=20 keV)

m2

1

2

20

KEm

mE

!"# $

%& '( =& '%& ') *

Note that increasing E0 directly improves mass resolution.

Scattering Cross Section

The number of scattered particles detected, nd, depends on:

• The number of incident particles, I

• The “size” of the detector (solid angle), !

• The number of scattering particles (areal density), Nt

• The probability of scattering (cross section), !

n I N td

!= "#" " "

Target

Flux I

Scattered ions

Detector

!

nd

"

Incidentmonoenergeticbeam

Impact Parameter, b

bImpact parameterTarget atom

Incident ions

m1

m2

Notice that:

• Small b, large scattering angle

• Large b, small scattering angle

• Generally, scattering is dominated by small angle scattering, but more information is

obtained from large angle scattering.

Solid Angle

A

R

!

R

s

"

Solid angle: (steradians; dimensionles

24Note

Planar angle: (

that for a comp

radians

lete sp

; dimensio

here,

s)2

nl

4 sr

ess)

2 2

A

A

R

R

R R

s

R!

""#= = =

#=

=

Note: Just as s is the length of the arc and not the length of a straight line, A is the area of a

portion a spherical surface of radius R; it’s not the area of a flat circle, although for small

angles the difference is negligible.

( )

For small and :

22 2

24 4 4 1 ( is the diameter of the "flat" circle)2 2 2 2 4

Rd sA

d

R R R R

!

" !" "

"!

#

#= $ $ = =

Solid Angle of the Sun and Moon

Earth

Moon

Sun

Rs=1.5x1

011 m

Rm=3.38x10

8 m

! = 0.53°

ds =

1.4x1

09

m

dm =

3.5

x1

06

m

( )

( )

20.532 51 1 6.7 10 sr

4 4 180

or

22 91.4 10

54 4 6.8 10 sr 2 2 2

111.5 10

35 10 srASUdetect

Compare: - Almost 100x the size of the full moo !o

nr

sun moon

d

sA

sunR R

!!" !

! !

# $ %& '& ' = = () *+ ,

(

%& = = = =

%& ' (

(

(

Cross Section and Reference Frames

m1

m2

m1

m2

!1

!2

Before After

Lab Frame

m1

m2

m1

m2

!

!

Center of Mass Frame

Before After

Differential Scattering Cross Section

(Rutherford)

In the center of mass reference frame:

2

11 2

4 4sin / 2

Z Zd

R

Ed

cm

!

"

# $% &=% &' # $% &( ) % &

( )

In the laboratory reference frame:

22

11 sin cos

22

11 2

1/ 24 4 2sin / 2

11 sin

2

m

m L LZ Zd

R

Ed

mL

m L

! !"

!

!

# $% &' (' () )* +) ), -, -* +. +/ 0, -, -' ( * +, -) ), -' ( , - 1 21 2* +) )3 45 6, - , -=, - ' (7 , -, - % &, -1 2 ' (, - ' (* +1 21 2 , -, -* +., -, -* +, -, -1 21 2* +3 4

Expanding in (m1/m2):

( )

22

11 2 12 44 sin / 2

2

Z Z md

R

E md

!

"

# $% &% & ' () *) *= + + ,,,' () *) *- ' () *) *. /. / ' (0 1

Rutherford Scattering Cross Section: Lab Frame

( )( )

2

24 2for << with in MeV and in b/sr [1b (barn)

2

11 2 1, 0.02073 244 s

= 10 c

in / 22

m ]1 2

Z Z m

ER E

m m ER

m

!

! ""

# $% &% & ' () *) *+ ,' () *) * ' (

,

) *) *- .- . ' (/ 0

This Rutherford cross section is often further approximated by leaving out the mass

correction:

( )( )

The accuracy of this form is given

2

11 2, 0.0207344 sin

in the table below for =4 (He) and =170

/ 2

1

Z

m

Z

ER E

L!

" !!

# $ % &' () * +' ( * +' ( , -. /°

m2 Error 10 -18.71% 20 -4.10% 50 -0.63%

100 -0.16% 200 -0.04%

Deviations from the Rutherford Cross Section

Low Energy Limits

Partial screening by electron shells for both low and high energy ions indicate the need for

screening corrections:

4 /30.0491 21 low energies

Z Z

E

R cm

!

!= "

7 / 20.03261 21 light ions with MeV energies

Z Z

E

R cm

!

!= "

Deviations from the Rutherford Cross Section

High Energy Limits

For each element in the figure, the scattering cross section for a given Z2 is “Rutherford” at

energies below the line.

0

5

10

15

0 10 20 30 40 50

Non-Rutherford Energy ( !Lab

: 160-180 degrees)

Z2

7Li

4He

1H

Deviations from the Rutherford Cross Section

Simple Estimates

Consider the distance of closest approach for a head-on collision: energy conservation

requires that the kinetic energy at large separation E is converted to potential energy U

at closest approach distance d:

2

1 1 2 (SI units)4

0

Z Z e

Ed!"

=

2

21 2 ("IBA units": 14.4 eV A)

Z Z e

E ed

= =

!

Low Energy Limit: Want d<rK (K-shell radius)

• For 2 MeV He in Ag (rK~10-3 nm), d=6.8x10-5 nm; thus d<rK is satisfied

• Some limits: EHe > 10 keV in Si; EHe> 340 keV in Au

High Energy Limit: Want d>rnucleus

rnucleus =R0A1/3, R0~1.4x10-15 m; A is the atomic mass in amu

At d=Rn, Emax=Z1Z2e2/(R0A

1/3)

• For He in Si, Emax~9.6 MeV

Energy Loss

t

N

E0

E0 - !E

0

t dEE dx

dx! = "

Energy

Electronic stopping

Nuclear stopping

dE/dx or !Bethe-Bloch region

Energy Loss

400 nm

Al

1 nm Au1 nm Au

E0

0

500

1000

1500

2000

0.6 0.8 1 1.2 1.4 1.6 1.8 2

Energy (MeV)

Al

Au

(front)

Au

(back)

Normal Incidence 2 MeV He+; 170 degrees Scattering

Energy Loss

Stopping Power: Typical units: eV / AngstromdE

dx

( )2eV cm eV

Stopping Cross Section: Typical units: or 15 210 atoms g / cm

!

µ

21 eV cm 3 where is the atomic density (atoms/cm )1510 atoms

dEN

N dx!

" #$ % & '=( ) & '* + , -

Sometimes used:

( )1 eV 3 where is the mass density ( g/cm )

2g / cm

dE

dx! " µ

" µ

# $% & ' (=) * ' (+ , ' (- .

Rules of thumb:

• N (atomic density) is typically around 22 35 6 10 atoms/cm! "

• is approximately the energy lost per monolayer!

Some He Stopping Cross Sections (Polynomial Fit)

( ) 2 3 4 5

0 1 2 3 4 5E A A E A E A E A E A E! = + + + + +

(E in MeV, ! in eV/1015 atoms/cm2)

Element Z At. Wt. (amu) A0 A1 A2 A3 A4 A5

N 7 14.007 31.01 64 -76.97 34.76 -7.127 0.5523

O 8 15.999 25.9 73.3 -80.5 35.17 -7.1 0.5462

Al 13 26.982 55.94 0.6773 -4.752 0.3401 0.2662 -0.0405

Si 14 28.086 57.97 56.59 -77.66 36.41 -7.624 0.5995

Ti 22 47.879 64.01 109.8 -125.5 55.67 -11.37 0.8827

Ni 28 58.728 41.59 97.79 -91.19 37.32 -7.482 0.5893

Pd 46 106.441 13.57 285.2 -287.4 124.9 25.24 1.939

W 74 183.842 61.69 156.6 -150.9 62.45 -12.33 0.9421

Au 79 196.967 57.99 193.2 -185.4 77.9 -15.59 1.199

20

40

60

80

100

120

140

0 0.5 1 1.5 2

Stopping Cross Section of Some Elements

Energy (MeV)

OAl

Si

Ti

Ni

W

Au

N

N

Compounds: Bragg’s Rule

For a compound of elements A and B with chemical formula AmBn:

m nA B A B

m n

! ! != +

20

40

60

80

100

120

140

160

180

0 0.5 1 1.5 2

Bragg's Rule

He4 Energy (MeV)

!O

!Si

!SiO2

=!Si

+ 2!O

2eV cm

Note that is in units of 15SiO

10 2

molecules

!" #$ %$ %& '

Energy Loss

0

500

1000

1500

2000

2500

3000

0.2 0.4 0.6 0.8 1 1.2 1.4

2 MeV He+: 30 deg. incidence; 170 deg. scattering

Energy (MeV)

A Note About Scattering Geometry

The scattering angle defines a cone centered about the incident beam direction; a detector is

located somewhere on this cone. Two typical choices involve a detector-beam plane that is

vertical (“Cornell” geometry) or horizontal (“IBM” geometry)

!-"scattering

"scattering

Beam Direction

!2 Cornell

!2 IBM

Vertical p

lane

Horizontal plane

Beam Direction(horizontal)

SAMPLE

SURFACE

ve

rti c

al dir

ect io

n

"-!scattering

"-!scattering

Sample normal

Cross Section

Application: Thin Films

The number of scattered particles detected, n, depends on the number of incident particles, I,

the detector solid angle, ! the number of scattering particles (areal density), Nt, and the

scattering cross section, !

n I N t != "#" " "

Consider two elements in the same spectrum, such as a thin film of Si and Ni shown below

(in this case: 2 MeV He, 170°, total of 1017 cm-2 and equal amounts of Si and Ni). The value

of n for each element is the area of the peak. In this case, nSi=1551 and nNi=6385.

0

200

400

600

800

1000

1200

0.6 0.8 1 1.2 1.4 1.6 1.8 2

Thin Si and Ni Film (1017

cm-2

)

Energy (MeV)

Cross Section

Application: Thin Films (cont.)

If we are interested in the relative amount of Ni to Si, then we must find (Nt)Ni/(Nt)Si.

( )

( )

nNt

I

nNtNi Ni Si

Nt nSi

Si Ni

!

!

!

="

=

The cross section has the form

( )

2

11 2 ( , , and are the same for both Si and Ni cases)4 14 sin / 2

Z Z

Z EE

! ""

# $ % &' () * +' ( * +' ( , -. /

( )

( )

26386 14

1.031551 28

nNtNi Ni Si

Nt nSi

Si Ni

!

!" #" #

= = =$ %$ %& '& '

Thus, the Ni and Si components are nearly equal in this film.

Cross Section

Example: Au on Si and Si on Au

Consider a thin layer of Au on Si and visa versa

mSi=28, ZSi=14, mAu=197, ZAu=79

279Au 3214

Si

!

!" #$ $% &' (

0

1 104

2 104

3 104

4 104

5 104

0.5 1 1.5 2

Au on Si and Si on Au

2 MeV He, Normal Incidence, 170 deg. Scattering

1 ML Au

200 ML Si

Energy (MeV)

Cross Section

Example: Au-Si Alloys

Consider Au-Si alloys

mSi=28, ZSi=14, mAu=197, ZAu=79

279Au 3214

Si

!

!" #$ $% &' (

0

5000

1 104

1.5 104

2 104

2.5 104

3 104

3.5 104

4 104

0.6 0.8 1 1.2 1.4 1.6 1.8 2

Au-Si Alloys

2 MeV He, Normal Incidence, 170 deg. Scattering

99% Si 1% Au

50-50 Si-Au

Energy (MeV)

Energy Loss

Surface Energy Approximation:

Stopping Power Factor [ ] and Energy Loss Factor [ ]S!

!1

!2

t

E0

kE0

E

Scattering from a surface atom:

• Energy before scattering: E0

• Energy after scattering: kE0

• Detected energy: kE0

At depth t: the ingoing and outgoing paths are

cos

1

tdin != and ("IBM" geometry)

cos2

tdout !

=

or ("Cornell" geometry)cos cos

1 2

tdout ! !

=

(continued)

Energy Loss

Surface Energy Approximation:

Stopping Power Factor [ ] and Energy Loss Factor [ ]S!

Scattering from an atom at depth t:

Energy at surface: E0

Energy at depth t, just before scattering:

cos

1

cos0 00 1

0

t

dE t dEE E dx E

dxdx E E

!

!

" #$ %& '(= ) * )+ , -. /& ' =0 1

Energy at depth t, just after scattering: kE!

Energy after leaving surface:

cos

2

cos0 2

0

cos cos0

1 20 0

t

dE t dEE kE dx kE

dxdx E kE

t dE t dEk E

dx dxE E E kE

!

!

! !

" #$ %& '( (= ) * )+ , -. /& ' =0 1

$ %" # " #, -& '$ % $ %& '* ) ), - , -. /& ' . /& '= =, -0 1 0 1. /

(continued)

Energy Loss

Surface Energy Approximation:

Stopping Power Factor [ ] and Energy Loss Factor [ ]S!

The difference in energy between an ion that scattered from a surface atom and an ion that

scattered from an atom at depth t:

1 1

cos co

cos cos0 0

s

1 20 0

01 20 0

dE d

t dE t dEE kE E kE k

EE k t S t

dx dxE E E kE

Edx dxE E E kE!

! !

!

" #$ %& ' & '( () *+ , + ,$ % $ %- = . / . . .0 1

$ %& ' & ') *& ' & '+ , + ,- = + = $ %+ , + ,) * 2 3+ , + ,

) *2 3 2 3+ ,

4 5 4 5= =+ , + ,) *4 5 4 52

+ ,= =( () *4 5 4 52 36 7

3

(continued)

Energy Loss

Surface Energy Approximation:

Stopping Power Factor [ ] and Energy Loss Factor [ ]S!

The quantity [S] is called the energy loss factor. Instead of using stopping powers (dE/dx) to

calculate [S], we can use stopping cross sections to compute the stopping cross section factor

[ ]! , where

( ) ( )1 1

[ ]0 cos cos

0 01 2

E kE E k Nt NtE E E kE

! ! !" "

# $% & % &' () * ) *+ = , - + =' (= =) * ) *) * ) *' (. / . /0 1

Beware of this odd notation! Note that [ ]! is not simply “! ” with brackets around it – it really

means the entire expression inside the brackets, with k, 1! ,

2! , and ! evaluated at two

energies.

Accuracy of the Surface Energy Approximation

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 100 200 300 400 500

2 MeV He, Normal Incidence, 170 deg Scattering

(Platinum)

Full Numerical Integration (RUMP)

Surface Energy Approximation

Depth (nm)

For this case (Pt, 2 MeV He, normal incidence, 170° scattering):

-15 2207 10 eV-cm

1370 eV/nmS

! =" #$ %

=" #$ %

Depth Resolution

Angular Dependence

Consider a 20 nm layer of Au on Si with 2 MeV He in normal incidence with “IBM”

geometry.

!1

!2

t E0

kE0E

0

5000

1 104

1.5 104

2 104

1 1.2 1.4 1.6 1.8 2

Depth Resolution: Angular Dependence

20 nm Au on Si

100 deg

140

170

Energy (MeV)

Energy Loss

Energy Straggling

42.35 2.35 2 (fwhm)1 2

E Z e Z Nts B

! "# $

= % = & '( )

Energy

!E

"!E

!x

N

Incident BeamTransmitted Beam

Depth Resolution

Limits

Limiters

• Detector Resolution Ed

!

• Energy Straggling Es

!

Depth resolution is improved by increasing path length, but it is ultimately limited by energy

straggling.

[ ] [ ]

2 22E E E

s d

E Et

S N

! ! !

! !!"

# $ # $= +% & % &' ( ' (

= =