Rutherford Backscattering Spectrometry

The main points to learn are: 1. The scattering is proportional to the squares of both the charge of the incident

particle and the target nucleus. 2. The number of scattered particles varies as the inverse square of the kinetic

energy of the incident particle. 3. The scattering is inversely proportional to the fourth power of sin(θ/2), where θ is

the scattering angle. 4. For thin targets, the scattering is proportional to the target thickness, t.

Elemental analysis technique where alpha particles are scattered off of targets and is used to determine elemental composition and thin-film thickness.

Mtarget

Mprojectile

vo b

Scattering line

Ftangential

Fradial

The Rutherford Scattering Experiment

204

1

r

ZezeFscattering

Incident particle approaches at a separation distance, b, called the impact parameter, from the target nucleus and scatters at an angle with respect to its initial direction due to the Coulomb force. where z = 2 and Z = atomic number of target

p0 = mv0

pf = mvf p

00

cos42

sin2

tangentialtangential

20

2

radial0radial

Ip

dtr

zZeImvp

Momentum triangle and the Impulse equations

Central force means constant angular momentum:

dbv

zZedt

r

zZeI

dbv

rdt

dt

dmrbmvL

cos1

4cos

4 00

2

20

2

radial

0

22

0

Limits of integration and the impact parameter:

2cot

4

2sin2

2cos

4cos

4

22;

200

2

000

22

2

00

2

radial

mv

zZeb

mvpbv

zZed

bv

zZeI radial

fififi

Relation between b and

The Kinematic Factor - K

M

p

m

p

m

pKE

MVvmp

MVvmvmp

M

He

Hef

He

Hei

HefHey

HefHeHeiHex

222:

sinsin0:

coscos:

22,

2,

,

,,

Assuming that the collision is elastic we apply conservation of momentum and kinetic energy to the above situation to obtain:

He

He

M

M

The kinematic factor is the ratio of the scattered alpha particle’s energy to its incident energy gives a convenient method for determining the elemental make up of the target atoms in the sample. To determine K we proceed as follows…

We eliminate the dependence and the momentum of the target nucleus (since it is difficult to measure anyway) in favor of the mass of the target atom, the mass of the helium nucleus and the scattering angle .

Knowing, or calculating the kinematic factor allows you to determine the mass of the target atom and thus chemically identify the atom.

Now what you need is a target and a spectrum of number of detected alpha particles versus the energy of the alpha particles.

To calibrate the energy spectrum (so we know the leading edge energy in order to calculate the kinematic factor) we use a known energy alpha emitter.

Our choice is 241Am whose alpha particle energies are approximately 5.4 MeV.

The Kinematic Factor - K

2

222

0

1 sincos1

HeHeHe

mMmMmE

EK

The RBS Experiment

Start with a beam of 3.3 MeV alpha particles incident on a target of unknown composition.

Determine the elemental composition of the sample based on the energies of the scattered alpha particles.

Chemical identification based on the kinematics factor.

Letting:

M1 = mass of helium ion,

M2 = mass of target species,

= scattering angle.

incident

scattered

E

EMMM

MMK

2

221

221

21

sincos1

Chemical identification of the target elements using the kinematic factor.

E0 = 2.0 MeV HeE1 = 1830 keV

K = 0.9150M2 = 181amuTantalum

This is a spectrum of 2.0 MeV alpha particles incident on a thin film of “unknown composition” scattered at 170o.

Scattered -particle from front edge, KEo

E x

Al peak

Incident -particle, Eo

Incident -particle, Eo

Scattered -particle from back edge, E and emerges with E1

KEoE1

coun

ts

Energy

10 EKEE

Thickness determination from a thin-film.

Inward:

Outward:

If is assumed constant along the inward and outward paths, then

Inward:

Outward:

Here we don’t know the energy at the back edge, rather only the energy that leaves the target having been scattered from the back edge. So we can eliminate this energy E.

E

E

xE

dEx

x

01cos

1

2cos

E

KE

xE

dEx

0

0

EEE

x

xE

dEx

in

E

E

KEEE

x

xE

dEx

out

E

KE

12

1

cos

x

E

Thickness determination from a thin-film.1 -angle between incident beam and horizontal.

2 -angle between exiting beam and horizontal.

Solving for E and equating the two resulting expressions we find:

From our diagram we see that

Which give us a depth scale that is linear in the energy loss of the alpha particle.

S is called the energy loss factor, which involves the stopping power of the alpha particle in the element in question and as the alpha particle travels farther it loses energy in proportion to the penetration depth in the sample.

In general the stopping powers are not constant, but energy dependant. However, if the film is thin then these stopping powers will be roughly constant.

€

E = E0 −∂E

∂x in

Δx =E1

K+

∂E

∂x out

Δx

K cosθ2

KE0 − E1 = ΔE = K∂E

∂x in

+∂E

∂x out

1

cos 180 −θ( )

⎡

⎣ ⎢

⎤

⎦ ⎥Δx = SΔx

Thickness determination from a thin-film.

Surface Energy Approximation for S0 in Thin Films.

If x is small then E will be small and then we can evaluate the stopping powers on the from and back surfaces.

Thus we find:

So, now we have to look up S0 for a particular element and for a given .

After the elements are identified, you can use the tables provided to look up S0 for your element and from E, you can calculate the thickness of the films on the surface of the substrate and the thickness of the substrate.

00

180cos

10

KEE x

EK

x

ES

Now for Ta, we find at 2.0 MeV and = 170o, So = 118.1 eV / A. Thus the thickness of the tantalum foil is:

AeV

S

Ex

Aev

o

3811.118

10045.0 6

For this lab you are given 3 files of .spe type•Open the files with a text editor and delete the header and footer BUT take the calibration data first from the footer.

•The energy calibration data is in the footer and is the only number given that is not zero. This represents the number of keV per channel, of which there are 8192 of them.

•Save the file as a .dat or .txt file.•Import the data in to Excel and make a thin-line curve using the energy calibration that you took.

Determine the elemental composition of the each of the three spectra.

ONLY from the PHY300_RBS_160degree_ManyElements data that you have, determine the thickness of each of the layers that were seen. Be very careful when looking at the graphs you make. There are some very small peaks that need to be determined.

Do not worry about doing an uncertainty analysis for this lab.

The lab report for everyone is due next Thursday, June 6 at noon.