Institutionen för kärn- och partikelfysik
Department of Nuclear and Particle Physics
Laboratory manual
RUTHERFORD BACKSCATTERING
AUTHORS:
A. Lindholm, J. Blomgren and I. Thorslund
REVISED BY:
A. Lundborg, T. Johansson, C. Willman, C. B.-Champagne
DATE:
2009-04-03
TASK:
Determination of the material thickness and composition
using Rutherford backscattering of alpha-particles
LITERATURE:
Satchler, G R: Introduction to Nuclear Reactions
Arya, A P: Fundamentals of Nuclear Physics
Krane, K S: Introductory Nuclear Physics
Nordling, C, and Österman, J: Physics Handbook
INSTRUCTORS: Tord Johansson, tel: 471 3886,
email: [email protected]
Camille Bélanger-Champagne, tel: 471 3828,
email: [email protected]
1
Rutherford backscattering
INTRODUCTION
In 1909 Hans Geiger and Ernest Marsden performed an experiment of utmost importance for our
understanding of the structure of matter. They irradiated a thin foil of gold with alpha particles
and found that almost all of them were unaffected by the foil, in spite of its thickness of several
atom layers. However, a very small fraction of the alpha particles showed dramatically perturbed
trajectories and the kinematics indicated that they had bounced against a very heavy particle.
From the experimental results Rutherford formulated his atom theory where he described the atom
as a very small nucleus, carrying almost all the mass of the atom, surrounded by a cloud of electrons.
This model was powerful enough to reproduce the energy and angular dependence of the cross section
for elastic scattering. The scattering of alpha-particles off a heavy atom nucleus depends, among
other factors, on the mass and charge of the nucleus. In Rutherford backscattering a beam of alpha
particles are scattered off a material and the resulting energy spectrum is used to determine the
material composition.
COULOMB SCATTERING
Figure 1 visualizes how a particle, coming in from the left, is scattered electromagnetically on a
heavy nucleus at the point O. The force on the particle is repulsive. Assume the mass of the particle
to be m, the velocity v and the charge q. Because the nucleus is heavy and there are no external
forces acting, it can be assumed that the momenta far before and after the scattering are P~i and P~f
respectively, and that |P~i | = |P~f |. The impact parameter – i.e. the distance between the direction
of the incoming particle and the nucleus – is b, and the scattering angle is θ. The charge of the
nucleus is Q.
∆P
Pf
F(r)
Pf
∆P
Pi
θ
r
φ
- Pi
π−θ
2
b
Q
O
Figure 1: The hyperbolic trajectory of a scattered particle. The particle is scattered by a conservative
repulsive central force.
The force on the particle, acting from the nucleus, is denoted F~ (r). Conservation of momentum
demands
Z
~
(1)
△P = F~ (r)dt
but △P~ = P~f − P~i (see figure 1) and from the figure it is also clear that
|△P~ | = 2mv sin(θ/2).
(2)
Rutherford backscattering
2
F~ (r) and △P~ form the angle φ. F~ (r) is decomposed into two components; |F~ (r)| cos φ, which is
parallel to △P~ and |F~ (r)| sin φ, which is perpendicular to △P~ . Combining (1) and (2) gives
Z ∞
2mv sin(θ/2) =
|F~ (r)| cos φ dt
(3)
0
0=
Z
∞
|F~ (r)| sin φ dt
(4)
0
Since there are no external forces acting on the system, the angular momentum has to be conserved. When the particle is far from the nucleus the angular momentum is mvb. Close to the nucleus
it can be denoted mr2 ω where the angular velocity can be written as ω = dφ/dt Conservation of
momentum gives
mr2 ω = mvb
(5)
Equation (3) can be rewritten as
2mv sin(θ/2) =
+(π−θ)/2
Z
|F~ (r)| cos φ
dt
dφ
dφ
(6)
−(π−θ)/2
Equation (5) combined with (6) gives
2mv 2 b sin(θ/2) =
Z
r2 |F~ (r)| cos φ dφ
(7)
In equation (7) r is now a function of φ instead of t. If we substitute sin φ = z and dφ = dz/ cos φ,
then equation (7) can be written as
2
2mv b sin(θ/2) =
+ cos(θ/2)
Z
r2 |F~ (r)| dz
(8)
− cos(θ/2)
In order to extract information about the impact parameter b we notice that
|F~ (r)| =
qQ
qQ
→ r2 |F~ (r)| =
4πε0 r2
4πε0
2mv 2 b sin(θ/2) =
b=
2qQ
cos(θ/2)
4πε0
1 qQ
cot(θ/2)
4πε0 mv 2
(9)
(10)
(11)
The formula above describes the connection between the scattering angle θ and the impact
parameter b.
3
Rutherford backscattering
DIFFERENTIAL CROSS SECTION
Figure 2 visualizes how a particle that hits the area with an impact parameter between b and b + db
is scattered into the solid angle element dΩ between θ and θ + dθ. The part of the cross section
corresponding to this process is
dσ
dΩ (dΩ = 2π sin θdθ).
(12)
dσ = 2πb|db| =
dΩ
The differential cross section is
dσ
2πb|db|
=
dΩ
2π sin θdθ
(13)
Combining (11) and (13) results in
dσ
=
dΩ
1
4πε0
2 qQ
2mv 2
2
1
sin (θ/2)
4
(14)
Expressed in the kinetic energy of the alpha particle, denoted Kα = mv 2 /2, the differential cross
section is
2 2
dσ
qQ
1
1
.
(15)
=
4
dΩ
4πε0
4Kα
sin (θ/2)
db
b
θ
dθ
Nucleus
Figure 2: Schematic drawing illustrating the number of particles between b and
b + db being deflected into an angular region 2π sin θdθ The cross section is, in this case by definition,
the proportionality constant; 2πbdb = −(dσ/d(cos θ))2π sin θdθ.
Eq. (15) gives the cross section for an ideal case of pure Coloumb scattering without recoil and
screening of the nucleus by electrons.
But the cross section is defined as the outcome of an ideal experiment and theoretical formulae
are only models. The cross section is defined as the propotionality constant that appear in the
relation between the number of incident particles, Nincident , impinging on a layer of thickness ∆x
and the number of scattered particles into a given region:
1
Nscattered = Nincident · ∆Ω · σ(E) · ∆x ·
.
(16)
cos α
Here α is the incidence angle (as in fig. 3) and ∆Ω the solid angle covered by the detector (in this
experiment also defined in figure 3).
Rutherford backscattering
4
NON STATIONARY TARGET NUCLEUS
When the mass of the target nucleus is of the same order as the mass of the projectile, the energy
spectrum of scattered alpha particles can only be reproduced if recoil is taken into account. For
scattering into a given angular range a lighter nucleus requires a larger energy transfer than a
heavy nucleus; simply given by conservation of energy and momentum. Apart from the mass of the
projectile m and the mass of the target nucleus M the energy after scattering, E1 , depends on the
scattering angle θ as

#1/2 2
" 2
M
m2
cos θ +
 .
− sin2 θ
E1 = E0
(m + M )2
m
(17)
In Rutherford Backscattering (RBS), the energy spectrum of scattered alpha-particles is used to
determine the composition of an unknown target. Use the internet and your imagination to come
up with ideas on how this could be useful!
ENERGY LOSSES IN MATERIAL
When particles pass through a material they lose energy to the material, mainly by ionization and
excitation of target atoms. The energy loss per unit length in a given material is called stopping
power dE/dx. Measured energy losses can be used to determine the material thickness of an unknown
target. This effect is illustrated in figure 4.
The backscattered alpha-particles will not only be stopped by the material, the energy spectrum
will also be broadened by detector resolution, small-angle-scattering and electronic pile-up (see sec.
4.9 in the SIMNRA users guide for different sources of straggling). These effects are all accounted
for in the simulation program SIMNRA.
5
Rutherford backscattering
EXPERIMENTAL PROCEDURE
The experiment is performed at the tandem accelerator at the Ångström Laboratory, Uppsala. The
accelerator delivers a beam of alpha particles with variable energy. The scattered particles, and other
reaction products, are counted in a solid state detector which can be moved to different scattering
angles.
The laboratory exercise is divided into two parts:
First lesson: Visit to the Tandem Accelerator
• How the accelerator works
• The experimental setup
• Data acquisition at one angle and one energy
Second lesson: Analysis (4 hours) – Nuclear Physics Lab.
• Analysis and evaluation of experimental data. The preparatory exercises below have to be
done in advance!
PREPARATORY EXERCISES
Assume pure Coloumb scattering without recoil.
1. Calculate the differential Coulomb cross section for alpha scattering off 28 Si at 4 MeV at the
angles 25◦ , 30◦ , 35◦ , 40◦ , 45◦ , 50◦ , 55◦ , 75◦ , 80◦ , 90◦ , 100◦ , 110◦, 120◦ , 130◦ , 140◦ .
2. Calculate the differential Coulomb cross section for alpha scattering off
energies 4.0, 5.0, 6.0, 7.0, 7.5, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0 MeV.
28
Si at 100◦ for the
3. Calculate the number of alpha particles scattered off SiO2 into the detector at 60◦ for a 4
MeV beam of 50 µC. Assume the helium ions to have 2+ charge. You also have to take into
account that the target is tilted with respect to the incoming beam. This means that the
effective thickness of the target must be calculated. For information about the experimental
setup and target data see figure 3. Take into account that the detector covers a certain solid
angle seen by the the scattered alpha particles. The differential cross section can be calculated
using knowledge from exercise 1.
ANALYSIS IN SIMNRA
1. Start the program SIMNRA 4.4 from the desktop.
2. Setup the experiment by choosing the SETUP meny and EXPERIMENT. Type the experimental parameters in the window:
• Incident ion
4
He
Energy (keV): 2000
Rutherford backscattering
6
• Geometry
Incident angle (deg):0
Exit angle (deg): 15
Scattering angle (deg): 165.
• Calibration
Calibration offset (keV): 83 Keep this value fixed in the analysis!
Energy per channel (keV/ch): 1 Value to be varied
Quadratic term (keV/ch):0. Keep this value fixed in the analysis!
Particles× steradian: 1.0 E11 Value to be varied
• Energy resolution
Detector resolution (keV): 18
Energy spread of beam (keV): 0
Confirm changes by clicking on OK.
Discuss with your partner about what each parameter mean, how a change would influence the
energy spectrum of scattered alpha particles and what experimental factors that they depend
upon.
3. Calibration and determination of layer thickness and composition of aluminum oxide on a
carbon backing.
• Open the ASCII-file AlO C.txt from the Rutherford Backscattr folder on the desktop, by
choosing File→Read Data→ASCII. This is a Rutherford Backscattering spectrum using
a target with a layer of aluminum oxide on a carbon backing.
• Define the target by choosing Target→ Target. There are two layers. The first layer is
Al and O. The layer thickness and the relative concentrations should be varied until the
simulation matches the experimental spectrum. The sum of the relative concentrations
must of course be kept at 1. The second layer is the backing which is taken to be very
thick and consist of pure carbon.
Layer 1:
Thickness (1E15 atoms/cm2 ):200 Value to be varied.
Number of elements: 2
Element: Al Concentration: 0.5 Value to be varied.
Element: O Concentration: 0.5 Value to be varied.
Click on Add.
Layer 2:
Thickness (1E15 atoms/cm2 ): 100 000 Keep fixed.
Number of elements: 1
Element: C
Concentration:1.
Confirm changes by clicking on OK.
• Make the first calculation by choosing Calculate→ Calculate Spectrum.
• Get the first normalization by varying Setup→ Experiment: Particles×sr to fit the height
of the carbon backing. Fit the calibration constant to the backing edge by varying Setup→
Experiment: Energy per channel. Reiterate if necessary until the normalization and the
backing edge are well matched by simulation.
7
Rutherford backscattering
• Adjust the thickness of layer 1 and the relative Al-O composition to fit the peaks. Note
that you may also have to adjust the calibration during this procedure.
• Print the fitted spectrum and save it using Save As.
• To save the fitted spectrum as an image file to include later in your report, open the
“Edit” menu, choose “Copy”. The spectrum is now in the Clipboard. You can paste it
in another program, for example MS Paint, and save it from there in the format of your
choice (.jpg, .png, etc.)
4. Determination of layer thickness and composition of SiO2 on a silicon backing.
• Open the ASCII-file Si02.txt from the Rutherford Backscattr folder on the desktop by
File→ Read Data→ ASCII. This is a Rutherford Backscattering spectrum on a target of
SiO2 on a silicon backing.
• Define the target as above.
• Make the first calculation by Calculate→ Calculate Spectrum.
• Get a first normalization by varying Setup→ Experiment: Particles×sr and fit the calibration constant to the backing edge by varying Setup→ cExperiment: Energy per channel.
Reiterate until well fitted.
• Adjust the thickness of layer 1 and the relative Si and O composition to fit the spectrum.
Note that you may have to adjust the calibration during this procedure.
• Print the fitted spectrum and save it by Save As.
5. Determination of the layer thicknesses and ordering of gold, silver and cupper layers on a
silicon backing.
• Open the ASCII-file Cal.txt from the Rutherford Backscattr folder on the desktop by
File→ Read Data→ ASCII. This is a Rutherford Backscattering spectrum with three
metal layers on a silicon backing.
• Define the target as above. Note that four layers must be defined (layer 4 is the backing).
• Make the first calculation by Calculate→ Calculate Spectrum.
• Get a first normalization by varying Setup→ Experiment: Particles×sr and fit the calibration constant to the backing edge by varying Setup→ Experiment: Energy per channel.
Reiterate until well fitted.
• Vary the layer thicknesses and the order of the layers to fit the spectrum. Note that you
may have to adjust the calibration during this procedure.
• Print the fitted spectrum and save it by Save As. Do not close the file.
6. Estimate the accuracy of the extracted thickness by varying the thickness until you see a
discrepancy between data and simulation.
Discuss with your partner why this gives you the accuracy?
Rutherford backscattering
8
REPORT
The laboratory exercise should be documented in a complete report in English. The contents required are an abstract, an introduction, a description of the ion source, the accelerator, the experiment, a theoretical derivation of the Rutherford cross section, applications of RBS, all three
preparatory exercises, the analysis of the data, results and discussion. A suggestion on how to write
this specific report can be found below.
Front page: Title, name, email address of all authors, date and abstract.
1. Introduction
2. Theory and applications (including space physics) of Rutherford Backscattering Spectroscopy
(RBS) including energy loss
3. Experimental setup including the tandem accelerator and the beam transport
3.1 Ion source
3.2 Tandem accelerator
3.3 Analysis magnet and switching magnet
3.4 Quadrupole magnet
3.5 Scattering chamber (target,detector)
4 Analysis procedure
5 Results and conclusions
6 Discuss advantages and disadvantages of the RBS method.
7 References
8 Appendix: Preparatory exercises
9 Appendix: Simulation and dataplots
REFERENCES
When you use information from a reference, be it an article or on the web, you must quote the
source of the information. These are a few suggested resources:
1. http://www.cea.com/techniques/analytical techniques/rbs.php
2. http://nucleus.stanford.edu/∼djconnel/research/Ge graded epi/measurement/rbs/
3. SIMNRA User’s Guide, http://www.rzg.mpg.de/∼mam
4. Journal of Geophysical Research, Vol 74, No. 25, 1969.
5. Journal of Geophysical Research, Vol 75, No. 32, 1970.
6. Energy losses: http://www.mse.arizona.edu/classes/mse480/group3/RBShow.htm
Any other sources you choose to use must also be included in your references section. IMPORTANT: Wikipedia is not an authoritative source and thus is not an appropriate reference for a
scientific report.
9
Rutherford backscattering
Scattering chamber
66o
Beam line
Beam dump
Target
R
Movable
detector
nA
Aperture
h = 8 mm
d = 3 mm
h
r = 1 mm (radii of the corners)
Distance target-aperture:
R = 315 mm (movable detector)
Angle beam-target:
d
66 degrees
Target thickness
32.5 µ g / cm 2
Figure 3: Technical data for the detector and target setup. The incidence angle is the angle between
the beam and the normal to the surface.
Rutherford backscattering
10
Figure 4: An incoming particle with energy E0 impinges on a target with two thin layers, A and B.
When passing layer A, the particle loses energy like dE/dx(E0 )∆x. When scattering on a nucleus
within the layer it bounces back and loses energy according to eq. (17) before traversing and exiting
the layer in the backward direction with even less energy. Without energy losses in the material all
particles scattered against a given type of nucleus would form a peak in the energy spectrum. With
energy losses there is instead a structure with an upper edge which corresponds to the outside surface
of the layer and a lower edge which corresponds to the backside interface of the layer.