EXPERIMENT 21
ALPHA PARTICLES
DRAFT
AIM
The aim of this experiment is to investigate the behaviour of alpha particles: their emission, the way
they lose energy when passing through media, and the way they scatter from a thin foil.
Part 2 of the experiment, an investigation of the scattering of alpha particle from a thin foil, runs
overnight and should be set up on your first afternoon in the lab. Leave at least one hour for
setting up.
2
PART 1
ALPHA PARTICLE SPECTRA
Equipment List:(a)
(b)
Vacuum chamber, alpha particle detector, preamp assembly, vacuum gauge and pump
Three alpha particle sources (Ra, AMR 13 and AMR 33) (c) Canberra 816 nuclear
amplifier
Bias supply for alpha detector
Multichannel analyzer card (in the PC and run by the Maestro program)
(d)
(e)
1.
THE DETECTOR AND VACUUM SYSTEM
The detector is a thin slice of n-type silicon on which is deposited a very thin gold surface, forming
a rectifying p-n junction (Knoll 2000) The silicon is doped to form a sufficiently wide depletion
layer that all the energy of the incident alpha particles is absorbed in the depletion layer, and thus
contributes to the resulting voltage pulse. The gold surface also acts as an electrical contact and as a
window for the entry of alpha particles into the depletion layer. It is extremely fragile and must
on no account be touched.
The bias supply for the semiconductor is provided by a 50 volt power supply. . This supply applies
a reverse bias voltage to the p-n junction producing a depletion layer about 100 µm thick. The
electron-hole pairs formed by ionisation when a charged particle traverses the layer are picked up
by the electrodes and input to a preamplifier so that a charged particle results in a pulse being output
by the preamplifier. After the preamplifier, there is an amplifier that makes this pulse even larger.
The reason we need to increase these voltages, which are inherently small, is to be able to acquire
them with data acquisition equipment. The size of the pulses is determined by the gain of the
amplifier, which you can control. However, you cannot control the relative sizes of the pulses,
which are determined by the relative particle energies.
The detector is installed in a vacuum chamber, with the preamplifier mounted above.
•
•
The preamplifier output is connected to the 816 amplifier, the input of which should be set
to receive NEGATIVE UNIPOLAR pulses.
To prevent radiation damage to the surface barrier detector, the bias should be switched off
and the radioactive source removed immediately counting is completed.
The source and detector are placed in an evacuated chamber to avert the interaction of the
alpha particles with air prior to their reaching the detector.
In order to change radioactive sources (and for part of the experiment on RANGE ENERGY
RELATION FOR ALPHA PARTICLES) it is necessary to admit air to the vacuum chamber
containing the source and detector, and evacuate the vacuum system after inserting the required
source.
The following procedures should be adopted to do so:
3
EVACUATING THE CHAMBER
•
•
Close the air inlet valve – the valve with the silver coloured knob - close to the mechanical
vacuum pump, and turn on the pump,
Open the red coloured valve adjacent to the vacuum system to evacuate the chamber.
Vacuum valves operate like taps: turn anticlockwise to open, clockwise to close.
OPENING THE CHAMBER
•
•
•
Air must be admitted into the vacuum chamber before you can open it.
Turn off the vacuum pump.
Now let air into the chamber by unscrewing the air inlet valve. (Open the red-coloured valve
if it is closed.). You should hear air hissing as it rushes into the chamber.
Moments later, the top section of the chamber should be easy to lift.
•
When you have completed the experiment, turned off the pump, and open the air inlet valve.
This is to prevent oil being sucked back, contaminating the vacuum system and damaging the
detector.
2.
CALIBRATION
The voltage from the detector is proportional to the energy lost by the alpha particles in the
depletion layer. The multi-channel analyzer displays the number of counts corresponding to a
particular voltage on the vertical axis versus the amplitude of the voltage pulse on the horizontal
axis. Each position on the horizontal axis is known as a channel. We must now obtain a calibration
that converts the channel number - which is proportional to the pulse amplitude (strictly voltages
between V and ΔV, where the voltage interval ΔV corresponds to the “width” of a channel”) - into
particle energy.
•
Place the AMR33 source (Pu, Am, Cm) on the shelf in the vacuum chamber, close the air
inlet valve, start the pump, and evacuate the chamber until the pressure gauge reads
0 Torr.
We now need to calibrate the analyzer's horizontal scale in keV per channel.
•
Run the Maestro program to collect the spectrum. The resulting spectrum will consist of 3
peaks corresponding to α-particles of 3 different energies. Note that these peaks are
asymmetric: in addition to each strong peak used for the calibration this radioactive source
emits a-particles of slightly lower energy than the main α-particle groups. These αparticles take no part in the calibration procedures.
•
The energies of the principal α particles, which are given in Table 1, are used to calibrate
the multichannel analyzer.
4
Table 1. Energy of principal α particles emitted by AMR33 source
! energy (MeV)
Isotope
239
241
Pu
5.147
Am
5.428
244
•
•
•
Cm
5.798
Set the gain of the 816 amplifier so that the maximum energy of the multichannel analyzer
is approximately 8 MeV. (The coarse gain switch should initially be set to 16 and the fine
gain set initially to 6.5. It is unlikely that you will need to change the gain during the course
of this experiment.) Check that the 5.798 MeV 244Cm peak is between channels 1350 and
1450.
The maxima of the 3 peaks in the AMR33 spectrum are used to calibrate the multichannel
analyzer. You can click the mouse on the peak, or move the cursor to the peak by means of
the arrow keys, to read the channel number. (At the bottom left-hand corner of the screen.)
Take no notice of the in-built calibration in Maestro. You are about to do your own, which
will be more reliable.
The channel numbers of the peaks in this spectrum and the energies of the corresponding aparticles are used to calibrate the detector/multichannel analyzer system. The response of
the surface barrier detector is linear. We only need 2 of these 3 data points in order to
produce the required linear calibration of the type
E
= a + b Cp
(1)
where E is the energy and C p is the channel number of an alpha particle peak.
•
•
•
C1
To obtain slightly more accurate calibration data, we will use Origin to fit a straight line
through these 3 points.
To do so, open Origin (a new worksheet – blank table – will automatically appear), enter
the α-particle energies in Column A and the corresponding peak channel numbers in
Column B.
Then go to the Tools menu, open Linear Fit and click Fit in the dialogue box that appears
on the screen. Values for a and b will be stored in the the Origin Results Log, which will
appear on screen when you click Results log in the View menu . (This will mostly appear
as a narrow white window immediately below the toolbars, which can be dragged down to
view your results.)
5
3.
•
•
•
•
•
SECULAR EQUILIBRIUM
Remove the AMR33 source and place the Ra source (radium and its decay products) on the
shelf in the vacuum chamber.
There should be a large peak in the lower channels due to beta particles from the radium
and four peaks corresponding to the four alpha particles listed below
Save the spectrum onto disk as a “user.chn” file replacing user with your name and a 2-3
letters/numerals to identify this file. This should be done immediately on completing the
data acquisition.
Immediately use the use the Export… command from the File menu to convert user.chn”
file to ASCII format. When exporting a data file, it is not necessary to change the extension
to your original file name: a file named “user.txt” will be created. You will see this file if
you view the appropriate directory.
Compare the energies of the four peaks with the energies given in (a), (b), (c), and (f) in
the table below. Remember, alpha particles will lose energy as they go through matter e.g.
gold film, air, etc.
When the radium source was prepared it was likely to be almost pure Radium 226. However since
that time, some of its daughter nuclides reached secular equilibrium. This is because their half
lives are short compared with the time since source preparation. Secular equilibrium is defined
for a chain of decays as the situation where all the daughter nuclides have the same activity as the
parent. Note that the daughter gas radon produced in Reaction (a), in the table below, is kept in the
source by a gold covering.
The transitions when radium decays are as follows:
Table 2. Energy of principal α particles emitted by Ra source
Decay process
(a)
226
(b)
222
(c)
218
(d)
214
(e)
214
(f)
214
(g)
210
Pb →
210
(h)
210
Bi →
210
Po →
206
(i)
210
Ra →
Rn →
222
Rn + alpha (4.78 MeV)
218
Half life
1620 year
Po + alpha (5.48MeV)
3.82 day
214
Pb + alpha (6.00MeV)
3.05 minute
Pb →
214
Bi + electron + antineutrino
27 minute
Bi →
214
Po + electron + antineutrino
17.7 minute
Bi →
210
Po →
Pb + alpha (7.68MeV)
164
s
Bi + electron + antineutrino
21 year
Po + electron + antineutrino
?
Pb + alpha (5.30MeV)
138 day
6
Determine whether the Ra source is in secular equilibrium. Although you can establish this by
determining if the peak heights are the same, you cannot do so in practice because three of the
peaks overlap and as a result their apparent heights will not be their actual heights. The correct way
to do it is to fit Gaussian functions to the peaks and the area under the fitted Gaussians. using the
Origin data analysis program. .
•
•
•
•
•
•
•
•
•
•
To load the file into Origin use the Import single ASCII… command in the File menu. Two
columns should be loaded into the work sheet. The A(x) column is the x-axis (channel
numbers) and the B(x) is the number of counts from the detector.
Drag the mouse over the A(x) and B(x) in order to select the two columns.
Now plot the data by selecting “Plot” then Line. A plot of the spectrum should appear.
Now select the range on the graph that you want to fit by selecting the following button
from Origin
Arrows will appear on either side of the graph. You can drag these arrows so that they are
on either side of the four peaks of the spectrum.
Now go to the “Analysis” menu, then to Fit Multi-peaks, then Gaussian
Enter “4” for the number of peaks required.
Click “OK” when the Peak half-width estimate is requested.
Now double click the mouse pointer over each peak (the program needs a starting point to
its fitting algorithm). A vertical line should appear on each peak.
Finally, data should appear indicating the area of each peak, with appropriately fitted lines
appearing on the spectrum. If the lines look like they match the spectrum, then you can
regard the fit as being good and take the areas as being correct.
NOTE: You may use QTiPlot to analyse this spectrum if you prefer.
QUESTION: Do you have secular equilibrium? COMMENT.
Tutor checkpoint. Obtain tutor's signature before proceeding.
4.
RANGE ENERGY RELATION FOR ALPHA PARTICLES
Equivalent air thickness of the gold coating of the Ra source
•
•
The radium source has a 3 µm (nominal) gold-alloy coating deposited over the source to
prevent the escape of the radioactive gas radon.
Using the range-energy tables for alpha particles in air (given in the Appendix 4), estimate
the equivalent air thickness of the thin gold film. This can be calculated for each of the four
alpha particles by comparing the range corresponding to the theoretical energy with the
range corresponding to the energy observed in the previous section of this experiment.
Energy lost by α-particles in air
• Let air into the chamber until the pressure is atmospheric and take a spectrum for the AMR
13 americium source. If no peaks are seen the source is too far from the counter: the range of
the alphas in the air is less than the counter-source distance. If this happens reduce the
spacing and repeat the measurements.
• Evacuate the chamber and measure the spectrum of the AMR13 α-particles in a vacuum and
comment on the difference.
7
•
Calculate the effective source-detector separation. Estimate with a ruler the actual separation
given the fact that the face of the detector is 89 mm above the bottom sealing flange. Do not
attempt to measure the position of the detector with a ruler.
Tutor checkpoint. Obtain tutor's signature before proceeding.
5.
HIGH RESOLUTION SPECTROSCOPY
You will now make high-resolution measurements of the energies and abundances of the alphas
emitted from the three isotopes in the AMR33 source used to calibrate the detector. Note that the
alphas emitted from these sources are not monoenergetic, but consist of groups of alpha-particles of
distinct energy as a consequence of a small fraction of the parent nuclei decaying to an excited state
of the daughter nucleus: three for Pu and Am, two for Cm. In order to obtain spectra that enable
both the energy and abundance of the alpha particles to be determined accurately it is necessary to
obtain the highest counts possible. This means locating the source close to the detector to give a
high count rate and counting for as long as possible.
•
•
•
•
•
•
•
Place the AMR33 source (Pu, Am, Cm) in the chamber between 10 and 20 mm from the
detector.
Evacuate the chamber.
Accumulate data until the count in the channels corresponding to the alpha peaks
approaches the maximum count that can be recorded by Maestro (2,048counts).
Calibrate the horizontal scale by taking the energy difference of the main plutonium and
curium peaks at 5.147 and 5.798 MeV, respectively.
Determine the energies and heights of the various α particle peaks by using Origin to fit
gaussian profiles to the peaks. Best results are obtained if you analyse the three groups of
alphas separately.
Compare with the values tabulated values below
Compare the full widths at half maximum (FWHM) of the principal peaks in keV obtained
from the Gaussian fitting procedure with the resolution located in manufacturers or in texts
on nuclear particle detectors.
Table 3. Energies and relative abundances of α particles emitted when the parent nuclei in
the AMR33 source decay to the ground state, and excited states, of the daughter nuclei
Parent nucleus
Daughter
nucleus
Energies and abundances of the
main groups of particles
239Pu
235U
5.147 (72%) 5.134 (17%) 5.096 (11%)
241Am
237Np
5.482 (85%) 5.439 (13%) 5.386 (1.7%)
244Cm
240Pu
5.798 (75%) 5.756 (25%)
Tutor checkpoint. Obtain tutor's signature before proceeding.
8
SUPPLEMENTARY QUESTION (These questions are quite advanced and you may like to
consider them in a report if you have time to do the required research).
Various factors determine the probability of α decay. If you look at the reference book Table of
Isotopes you will be able to observe the following general tendencies.
a)
The higher the energy of the emitted α, the higher the probability of that transition.
b)
The larger the spin change of the parent to daughter nucleus the lower the probability of the
transition.
c)
Transitions with no spin change must be between states of the same parity.
Can you see the theoretical reason for the above? Find out about the "Coulomb Barrier" and the
"Centrifugal Barrier" as applied to alpha decay. Remember the spin of the alpha is zero and its
parity is positive. For further help consult the following references:
Reference for Geiger Nuttall law: RB Leighton, Principles of Modern Physics.
Reference for Centrifugal Barrier: AES Green,
KS Krane, Introductory Nuclear Physics §8.3-§8.5.
See also WE Burcham, Elements of Nuclear Physics
9
PART 2
ALPHA PARTICLE SCATTERING
Equipment List:(a)
(b)
(c)
(d)
(e)
(f)
(g)
Scattering chamber: thin gold foil, and alpha detector in an evacuated chamber.
Preamplifier located at the scattering chamber
NIM bin with Canberra 1431 Single Channel Analyser
Detector bias supply
Surface Barrier Detector
Computer
Interconnection cables
This equipment is located on a bench behind the apparatus for Part 1 of this experiment
1.
INTRODUCTION
Around 1910, Rutherford and co-workers showed that the nucleus of an atom was extremely small
– around 10-4 of the size of an atom – by observation of the scattering of alpha particles off a thin
gold foil. In this experiment Rutherford’s work will be reproduced using modern equipment.
2.
THEORY
Let us assume that the nucleus is a small sphere of radius R with a charge + Ze and we bombard a
thin foil with alpha particles of mass m, charge + Z !e , travelling with initial velocity v. ( Z! = 2 for
alpha particles). Some of these alpha particles will be scattered as they pass through the foil. We
will assume that:
(a)
v = c so that a non-relativistic treatment can be used. For alphas emitted from radioactive
elements, v/c is less than 0.005 so this is reasonable.
(b)
The nucleus is fixed (instead of recoiling). Rutherford et al used a gold foil where the atoms
were 50 times heavier than the alpha particle; so this assumption is reasonable. The
apparatus here also uses a gold foil.
(c)
The alpha particle never scatters more than once when passing through the foil. You will
see that this is justified when you do the experiment: even for small angles of scattering, the
probability of scattering is less than 0.001 so that the probability of an alpha particle of
scattering twice is less than one part in a million.
(d)
The alpha particles do not scatter off electrons. This is not strictly true but the angles of
scattering are small because the ratio of the electron mass to that of the alpha is about
0.00014.
We consider the nucleus to be a small sphere. The force (Coulomb force) between it and the alpha
is repulsive and follows the inverse square law, causing the path of a scattered alpha to be a
hyperbola.
F
=
(Ze )( Z #e)
4! " 0 r 2
(2)
10
The angle between the asymptotes of the hyperbola is the angle of scattering θ. This latter depends
on a quantity called the impact parameter p. This is the perpendicular distance from the nucleus to
the line that the alpha would have followed if it had not been scattered. (See Figure 1 below.)
Asymptote
P
Gold
Nucleus
F
"
#
A
0
"
!'s path
G (focus)
p
Asymptote
Figure 1. Geometry for the scattering of an alpha particle off a nucleus.
Because the Coulomb force between the two particles is central, i.e. along the line joining them, we
can say that the angular momentum of the alpha is conserved in its orbit around the nucleus. With
this and the conservation of energy, we show that p and θ are related by
p =
(Ze )(Z $e )cot (! / 2 )
4"# 0 mv 2
(3)
This relationship between impact parameter and the scattering angle is derived in Appendix 1.
It is of course impossible to aim alphas at a nucleus to get a given value of p - alphas can go
anywhere - but we can mentally draw a ring of radius p and width dp around each nucleus and
handle the problem statistically. The result is that the rate of detection of alphas is given by
R(" ) =
(! A) R0 N t
4r 2
B 2 cosec 4 (" / 2 )
(4)
where R0 is the number of particles bombarding the foil per unit time, is the area of the detector
at a distance r from the foil, N is the number of atoms per unit volume of the foil and t is the foil
thickness.
This expression - the Rutherford scattering law - is derived in Appendix 2.
Thus R(! ) is proportional to:
cosec 4 (! /2 )
(a)
(b)
(c)
R0
t
(d)
1 mv 2 , i.e. to the inverse square of the alpha particle’s energy.
2
( )
The importance of (c) and (d) will become evident later on.
For this part of the experiment equation (3) may be rewritten as
11
R(! ) = cosec 4 (! /2 )
3.
(5)
OBSERVATIONS
These are done with an apparatus described by PJ Ouseph (1976). (You might also like to read
Ramage et al. (1975).) The scattering is done in a vacuum chamber containing a collimated source
of alphas (Americium 241) aimed at a 0.5 µm (0.00965 g/cm2) gold foil. A surface barrier
semiconductor detector registers the scattered alphas at an angle θ which is varied by a stepping
motor under computer control.
The foil is turned with gears so its thickness t decreases with θ as t = t0 sec (! 2 ) so that the
effective thickness of the foil is independent of the scattering angle.
• Evacuate the chamber.
Do not change the settings of the SCA 1431 single channel analyzer.
•
•
•
•
•
The experiment is controlled by a dedicated computer. You will be using the program Alpha
Particle Scattering to take measurements over as long a period as possible (usually over
night).
The program contains various sections, each identified by a "button" on the display screen.
These are:
o Hardware Test. This section of the program checks the operational status of the
counters, the stepping motor and the steering mechanism.
o Manual. This section enables you to control the position of the detector and the
counting time manually.
o Run. In this section the computer controls both the position of the detector and the
counting time to:
Step the counter through the zero position, observing the counts for each
angle, which enables the computer to estimate the true zero position (for
undeflected alphas) of the detector
Record the count rate as a function of the angular position of the detector.
o Enter and plot corrected data. As Origin or QTiPlot are now used to record and
analyse the data, this section is no longer used.
Click on the Hardware Test button to enter this section, and check the operation of the
detector and the stepping motor that controls the position of the detector.
Using Manual control of the system, set the detector to 10° and count for 10 minutes. From
the measured count rate determine the constant C in equation (4)
Use equation (4) to determine the counting time required to obtain an accuracy of 5% in the
count rate when the detector is set at 40°.
Recall that in a counting experiment
Standard deviation of count
Accuracy
=
Count
where N = Total count recorded
=
N
N
=
1
N
Z! = 2
It should be obvious from this calculation that an extremely long counting time would be
required for high accuracy at large scattering angles.
12
•
•
Now click on the Run button and then the Calibrate button. This steps the counter through
the zero position observing the counts for each angle, which enables the computer to more
accurately estimate the true zero position (for undeflected alphas). All subsequent settings
must be corrected for this zero offset. Print out the results of this calibration procedure.
After determining the zero offset click the Continue button to proceed to the computercontrolled measurement of the count rate as a function of scattering angle and a blank table
will appear on screen. Enter the counting times for various scattering angles given in
Table 4. Verify that you have entered the correct values then start these measurements.
Table 4. Recommended counting times for specific scattering angles
Scattering angle
(θ) (degrees)
40
35
30
25
20
15
10
5
0
-5
-10
-15
-20
-25
-30
-35
-40
•
•
•
•
Counting time
(minutes)
180
120
90
60
45
30
20
10
1
10
20
30
45
60
90
120
180
Place a notice adjacent to the experiment to advise that an overnight run is in progress and
notify Marek Dolleiser (the Laboratory Senior Technical Officer) that you have started an
overnight run. (Marek will usually download your results the following morning.)
After your run, obtain a print-out of your data and use Origin to plot the final data in log-log
format to determine the slope. From the slope, determine whether your observations of
count rate as a function of scattering angle are consistent with equation (4)above equations
are followed.
If you have time, you may wish to correct your raw data for:
o The finite width of the (imperfectly) collimated alpha beam and the finite solid angle
subtended by the detector
o The finite width of the (imperfectly) collimated alpha beam and the finite solid angle
subtended by the detector at the gold foil
The procedures for adjusting your raw data to take account of these errors are discussed in
Appendix 3.
All the computer instructions are described in greater detail in accompanying bench file.
Tutor checkpoint.
13
References
Knoll Glen F (2000) Radiation Detection and Measurement, (New York: John Wiley)
Ouseph PJ (1976) Am J Phys 44, 1012
Ramage JC et al (1975), Am. J. Phys. 43, 51
14
Appendix 1
Derivation of the relation between impact parameter p and the scattering angle θ.
Because the force between the two particles is inverse square repulsive we can say that the path of
the alpha will be a hyperbola with the gold nucleus at one focus.
Asymptote
P
Gold
Nucleus
F
"
#
0
"
!'s path
G (focus)
A
p
Asymptote
Figure 1. Geometry for the scattering of an alpha particle off a nucleus.
(Repeated for your convenience.)
.
Figure scattering of n a particle off a thin gold foil.
In the figure we have:
!
FP - PG = k
FA - AG = k
Similarly cos ! = k a
= 180o # 2"
(a constant)
(the same constant)
}
a property of a hyperbola
where a = FG. (We have just sent P off to infinity along an asymptote!)
Finally we have p = FO sin ! = a /2 sin " . By elimination we obtain FA (which we'll call r),
which is given by
r =
p (tan (! 2 )+ sec (! 2 ))
(6)
If the velocity of an α particle at ∞ is v and the velocity at A (where the a particle is closest to the
nucleus) is w, then the conservation of angular momentum gives
mvp = mwr
(7)
15
From conservation of energy
1 2 mv 2
= 1 2 mw2 + Z Z #e2
(4! " 0 r )
(8)
Equations (7) and (8), together with equation (6), which links r and p, provides the required result
by elimination of w and r.
Z Z $e2 cot (! 2 )
(9)
p =
4" # 0 mv 2
Appendix 2
Derivation of the Rutherford scattering law
The analysis presented in this Appendix is applicable to any situation where one particle
(molecular, atomic or nuclear) is scattered from another.
If we mentally drawn a ring of radius p and width dp around each nucleus in the scattering foil, the
area of this ring is the differential cross-section for the scattering, d! . Differentiating p we obtain
Z Z $e2 cosec 2 (! 2 )
dp =
4" # 0 mv 2
d!
(10)
The negative sign can be dropped because cross-sections cannot be < 0.
Such an encounter will result in the alpha being scattered through an angle between θ and dθ. The
scattering foil can be pictured as having N′ points per unit area representing the nuclei in it. Around
each one there is a "ring" of radius p and width dp. Thus the probability of a scattering through an
angle θ to θ + dθ can be written
dP =
N #d ! =
Nt d! =
Nt (2 " p dp )
where N is the number of nuclei (= number of atoms) per unit volume in the foil and t is the foil
thickness. To detect the scattered alphas we can use a detector of sensitive area δA at a distance r
such that ! A = 4" r 2 . Considering δA to be the small annulus lying between θ and θ + dθ we
find the following relation between δA and dθ
! A = 2" r 2 sin # d#
= 4" r 2 sin (# 2 ) cos (# 2 ) d#
The rate of detection of alphas will be R(θ) = R dP
where R is the number of alphas bombarding the foil per unit time and, from equation (10),
# ! A R0 Nt $ 2
R (" ) = %
& B F (" ) .
' 4r 2 (
16
where
F (! ) =
and
B =
cot (! 2 ) cosec (! 2 )
sin (! 2 ) cos (! 2 )
= cosec4 (! 2 )
Z Z $ e2
4"# mv 2
B has the dimensions of length. It is in fact half the distance of approach of the alpha particle to the
nucleus when θ = 0 ie the distance of closest approach. For 10 MeV alphas on gold, this distance
is greater than the nuclear radius so that the alphas do not see the nuclear force. If this was not so
then there would be a deviation from the Rutherford formula at large θ.
Appendix 3
Corrections to Data
a)
Correction to Angles
The finite width of the (imperfectly) collimated alpha beam and the finite solid angle subtended by
the detector mean that the scattering angle as set is not the same as the "effective" scattering angle.
If we assume that the distribution function describing the incident beam and detector widths is
triangular in form, Tri(θ), then the effective angle is then the mean of the distribution
Tri(θ)cosec4 (θ/2). We have this computed using Simpson's rule. The result, in the form of a
correction to the set angle, is given in the following table.
Table 5. Correction (in degrees) to be subtracted from set angle to obtain effective angle.
4 deg
Half Width of Distribution
5 deg
6 deg
7 deg
Set Angle
5
10
15
20
25
30
35
40
45
50
55
60
8 deg
9 deg
5.1
3.0
2.1
1.6
1.3
1.1
0.89
0.74
0.62
0.52
0.43
6.8
3.9
2.7
2.1
1.7
1.4
1.2
0.95
0.79
0.66
0.54
Corrections to be applied
2.6
1.1
0.70
0.51
0.40
0.32
0.26
1.8
1.1
0.80
0.62
0.50
0.41
0.34
0.28
2.6
1.6
1.2
0.90
0.72
0.59
0.50
0.41
0.38
0.29
3.7
2.2
1.6
1.2
1.0
0.81
0.68
0.57
0.47
0.40
0.32
The computer’s printout of the graph drawn when θ = 0 was found may be used to correct for the
imperfect collimation of the alpha particle beam and the finite acceptance angle of the detector. Fit
visually – i.e. by trial and error - a triangular distribution to this graph. The half width of this
distribution, which is required to apply the correction discussed above is half the base of this
triangle.
17
b)
Correction to Rates
A correction has also to be made to the rates. Since the scattering foil has a finite thickness the
alphas have less energy at the point of scattering than they have when they enter the foil. This is
because of the interaction of the alphas with electrons. Ordinarily this would cause no complication
but as mentioned above the foil is turned by gearing so that its thickness t increases with θ.
The average energy of scattering thus decreases with t. Recall that the rate of detection of alphas
is proportional to the inverse of the this energy. In addition we have, in effect, a thicker foil at
larger angles (which will conversely increase the rate).
Both these affects can be allowed for with a single correction term applied to the rates of
ft
"! #
cos % & $ 0
'2(
E02
where E0 is the incident energy of the alpha and f is a constant which we now define.
The energy loss of the alphas can be approximated by the formula !
dE
dx
=
f
E
where f is a
constant for each absorber and incident particle.
This integrates to E 2 = E02 ! 2 ft and using the dependence t
= t0 sec (! 2 ) this becomes
E2
= E02 " 2 ft0 sec (! 2 ) for scattering occurs on the far side of the foil and
E2
= E02 " ft0 sec (! 2 )
when the scattering occurs halfway through the foil, which we will
use as an “average” scattering event.
Since, from the Rutherford formula
Robs
Thus
"
t0 sec (! 2 )
E02 % ft0 sec (! 2 )
Robs
=
=
R ! t E2
we have
t0
E02 cos (! 2 ) % ft0
=
t0
#
ft $
E02 &cos (! 2 ) % 02 '
E0 ')
(&
Rtrue
"%cos (! 2 ) $ 0.045#&
ft
The value of f is approximately 1500 MeV2 cm2 g-1 and thus the factor 0 is approximately
E02
0.045.
so that
Rtrue
= Robs "%cos (! 2 ) $ 0.045#&
This means one should obtain a line a slope -4 on a log log plot of Rtrue against sin (! 2 ) .
where θ is the corrected scattering angle as discussed earlier in this Appendix.
18
Appendix 4
Table 6. Alpha particle energy vs. mean range in air at 15ºC and 760 mm Hg
Mean range (cm)
Alpha energy (MeV)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
0.00
0.99
1.99
2.77
3.42
4.00
4.50
4.99
5.44
5.87
6.28
6.67
7.04
7.40
7.75