Department of Physical Electronics
Faculty of Science, Masaryk University, Brno
Physical laboratory 3
Task I Rutherford’s experiment
Tasks
1. Observe the number of detected α particles for a sufficient number of gold foil positions.
Verify formula 18 for Rutherford’s scattering.
2. Verify that the numbers of detected α particles obey Poisson’s distribution (11).
Figure 1: Ernest Rutherford, Hans Geiger a Ernest Marsden.
Introduction
Not much was known about the building blocks of matter in 1909. Joseph John Thomson had
already discovered corpuscular radiation in 1897. The particles of this radiation have been named
electrons. Further measurements of electrons revealed that they are relatively light and negatively
electrically charged.
It still had to be resolved how the electrons and the positive matter are placed inside the atom.
Thomson suggested a model where the positive matter is evenly distributed inside the atom, and
the electrons are placed just like “raisins in pudding”. This model is known as Thomson’s pudding
model.
In 1909 Hans Geiger and Ernest Marsden, under the guidance of Ernest Rutherford, carried
out an experiment where a beam of α particles bombarded a very thin gold foil. A diagram of
this experiment is shown in Figure 2. According to Thomson’s model, the positively charged α
particles should only very slightly scatter from their initial direction by repelling positive charges.
The results of the experiment were surprising. Most of the α particles passed through the foil
without being scattered. On the other hand, some particles scattered to large angles were also
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Figure 2: Historical diagram of the original experiment of Geiger and Marsden. AB source of
α particles, P lead plate, RR thin foil, S scintilating screen, M detector (source: http://www.
chemteam.info/Chem-History/GM-1909.html).
detected. For this to happen, the positively charged matter in the atom had to be concentrated
into a very small volume. This was the foundation of Rutherford’s model of the atom. The model
assumed the existence of a very small and heavy positively charged atomic nucleus with electrons
orbiting the nucleus as planets with lots of free space around them.
This experiment was crucial for understanding the structure of matter. Most parts of modern
physics and chemistry are built upon this model with added quantum theory. And also,
the principle of this scattering has found its use. Nowadays, the method of RBS – Rutherford
backscattering spectrometry is used to analyse the chemical composition of materials. A beam of
high-energy ions is bombarding an unknown sample, and the scattering of these particles on the
sample is measured. And, as the scattering is dependent on the number of protons in the nucleus,
the chemical composition can be determined.
Scattering theory
The principle of Rutherford’s experiment lies in measuring light α particles scattered on heavy
atoms such as gold. Owing to the large difference in the atomic masses, the gold atoms can be
considered stationary. If the incoming α particle gets close enough to the atomic nucleus (inside
its electron shell), its starts to be repulsed by its positive charge. Due to this force, the incoming
α particle is scattered by the angle χ from its initial direction. The magnitude of the angle χ
depends on how close to the nucleus the α particle gets and on its initial velocity.
The fundamental assumption of the experiment is that the incoming particle transfers no
kinetic energy to the nucleus it is scattering on. The magnitude of its initial kinetic energy and
the magnitude of the initial momentum are the same as their final magnitudes after the scattering
process. The incoming particle, however, scatters by the angle χ. The direction of the momentum,
therefore, changes. The magnitude of this change during the whole process is
∆p = 2mv sin
χ
2
, (1)
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+2e
+Ze
b
b
F
Figure 3: Schematic of α particle scattering.
where m is the α particle mass and v is its initial (and final) velocity. The momentum change is
caused by mutual force F between the nuclei
∆p = Fdt. (2)
Figure 3 schematically shows a collision of an α particle with a massive nucleus. The scattering
process can be described by the angle ϕ instead of the time. The magnitude of the momentum
change of the α particle can be described as
∆p =
1
2
(π−χ)
−1
2
(π−χ)
F(ϕ) cos ϕ
dt
dϕ
dϕ, (3)
where dt
dϕ = 1
ω , is the reciprocal of the angular velocity of the particle around the nucleus.
The force acting between the nuclei is an electric force that can be described as FE = 1
4πε0
2Ze2
r2 ,
where r is the instantaneous distance between the nuclei. This relation is valid for one of the nuclei
being an α particle (q=2e) and the other one a nucleus with Z protons. Substituting this into
the relation for the momentum change equation we obtain
2mv sin
χ
2
=
1
2
(π−χ)
−1
2
(π−χ)
1
4πε0
2Ze2
r2
1
ω
cos ϕdϕ. (4)
We can simplify the integral on the right side by realising that the electric force between the
nuclei always acts along the connecting line between the nuclei. Therefore, such force acts with
no torque on the α particle and the momentum of the α particle is conserved throughout the
whole process. The momentum in every point of the α particle’s trajectory must be identical to
the initial momentum. Therefore, we can write mωr2 = mv b, where b is the impact parameter
(see Fig. 3) and v is the initial velocity of the α particle. Substituting this into 4 we obtain
4πε0mv2b
Ze2
sin
χ
2
=
1
2
(π−χ)
−1
2
(π−χ)
cos ϕdϕ = 2 cos
χ
2
. (5)
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This can be rewritten as
b =
Z e2
4πε0Ek
cotg
χ
2
, (6)
where Ek = mv2
2 is the kinetic energy of the incoming α particle. This dependence can be
understood as the relation between the scattering angle χ and the impact parameter b. An
alternative explanation is that all α particles incoming to the area σ = πb2 around the nucleus
are scattered to the angle χ or larger.
In reality, we cannot experiment with a single α particle, and we rather use a beam of α
particles of known kinetic energy and their flight direction. We also do not have a single target
nucleus, but we use a thin foil. Let us assume that the α particle beam impacts an area S of
the foil with the thickness d having (volume) concentration of atomic nuclei n. The incoming α
particles interact with nSd target atomic nuclei. The foil area that the α particles need to aim
for to scatter by the angle χ or more is σnS d. The ratio of the α particles scattered by χ or more
to all the α particles is
f =
σnS d
S
= πb2
nd = πnd
Ze2
4πε0Ek
2
cotg2 χ
2
. (7)
We do not have detectors for counting particles scattered to angles larger than a certain value
in a real experiment. Usually, only detectors capable of counting particles in a small interval of
angles (χ, χ + dχ) are available. The number of α particles in this interval can be obtained by
differentiating the previous formula (7)
df = −πnd
Ze2
4πε0Ek
2
cotg
χ
2
sin−2 χ
2
dχ. (8)
If we place the foil into the centre of an imaginary spherical surface of the radius r, the
particles scattered into the angle interval (χ, χ + dχ) will fly out of this spherical surface through
a circular strip with the area of dSr = 2πr sin χrdχ = 4πr2 sin χ
2 cos χ
2 dχ. The number of α
particles captured by a detector of unit area per one unit of time placed in the distance r from
the foil will be
N =
N0|df|
dSr
= N0nd
Z e2
8πε0rEk
2
1
sin4 χ
2
, (9)
where N0 is the initial number of the α particles hitting the foil. Due to simplicity, we usually do
not use the number of scattered particles onto a unit area in the distance r, but rather onto an
elementary area dS described by a solid angle dS = r2dΩ. The number of α particles scattered
to this solid angle is
dN =
N0|df|
dSr
dS = N0nd
Z e2
8πε0Ek
2
1
sin4 χ
2
dΩ. (10)
Poisson distribution testing
Atomic nuclei are composed of protons and neutrons. Only some combinations of the numbers of
protons and neutrons are stable. Others are unstable, and they undergo one of thy different types
of radioactive decay. An example is Americium 241Am undergoing the α decay
241
95 Am → 4
2He +237
93 Np.
The 4
2He particle is a helium nucleus alternatively called an α particle.
The decay of a nucleus is a random process. It cannot be exactly predicted when it happens.
Also, decays of neighbouring nuclei do not directly affect the time of decay of the nucleus. If
Physical laboratory 3 manuals (version May 10, 2023) 5
the decays do not happen too often, the probability of detecting α particles originating from the
decay in a specific time interval obeys the Poisson distribution
P(n) =
λn
n!
e−λ
. (11)
. Here P(n) is the probability that n decays occur (n α particles are detected), and λ is the
mean (expected) number of decays counted during the measurement interval. The use of the
Poisson distribution primarily predicts the probability of phenomena occurring in some specific
time interval. When we know that a certain phenomenon repeats itself twice per minute on average
and that the Poisson distribution can describe its occurrence, we can determine the probability
of n during two minutes by selecting λ = 2 · 2 = 4. For example, the probability of measuring five
phenomena in two minutes will be
P(5) =
45
5!
e−4 .
= 0.156. (12)
This corresponds to the value plotted in Figure 4. It can also be seen that the occurrence of 3 or
4 phenomena is the most probable with the probability of around 20%.
Figure 4: Poisson distribution: probability function and corresponding distribution function.
It is necessary to verify that the Poisson distribution can describe α decay of americium.
This can be done in several ways. The easiest one is to measure for a very long time interval
and determine the mean value of the detected α particles λ0. Then, we can divide this interval
into N equally long time intervals and determine the number of detected α particles in each time
interval. Then the dependence of the number of intervals with n detected particles on n is plotted.
Such dependence can be fitted by the expected shape of the Poisson distribution with the fitting
parameter λ. This value can be compared to its theoretical value λt calculated as λt = λ0T, where
T is the length of one time interval N.
However, such a method is incorrect from the statistics point of view. It does not provide for
a quantitative assessment of the degree of reliability of the obtained result. That’s why statistics
work with probability. First, we need to formulate a hypothesis we want to test. In our case,
the hypothesis is that the detected α particles obey the Poisson distribution. Next, we choose a
parameter a that corresponds to the probability of wrongfully rejecting the hypothesis. The usual
values of a = 0.05 or a = 0.01 are chosen. The number a is called the level of reliability.
Next, we need to choose the correct test that we use to check our hypothesis. In the case of
testing of distribution functions, the so-called “goodness-of-fit tests ” are used. The most used
is the χ2 (chi squared) test, as it can be used to test any discreet distribution. In the χ2 we
compare the measured values to the expected distribution. We work similarly as in the previously
described statically incorrect way. However, the test evaluation is different. We take a very
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long measurement. During the measurement, we count a certain number of α particles. Next,
we separate the whole measurement into N equally long time intervals. The time intervals may
not overlap. We compile the dependence of the number of intervals K(n) with n detected α
particles on n. Next, we will test this experimentally obtained distribution against the theoretical
distribution λ = λ0T, where λ0 is the mean number of the detected particles per unit of time in
the long measurement and T is the length of a single time interval N. Up to now, the method
has been the same as the one described previously.
Before starting the test evaluation, we need to fulfil the conditions necessary for using the χ2
test. The first condition is having at least 5 expected occurrences (NPj(n) ≥ 5) at every point
j where we compare the measured and theoretical values. This is dictated by the continuity of
the χ2 distribution that is not maintained for too low values. If this condition is not met at a
certain point, we merge it with some of the neighbouring points until the condition is fulfilled.
The theoretical probability of this merger is the sum of the probabilities of the individual points.
Instead of the original dependence K(n), we will instead use new dependence Kj(n) having some
of the original points in K(n) merged into one point. The second condition is that the total
probability needs to be one. In other words, we need to compare the whole distribution function
and not only its part. However, the previous condition states that all the points above a certain
n = k need to be merged into a single point jk. The result of this merger is that we have a finite
number of points j. The number of points jk − 1 is called the number of the degrees of freedom
of the χ2 test. The probability of the last point is (1−[probability of all the preceding points]).
In statistical nomenclature, the probability of the last point jk is Pjk
= 1 − F(k − 1),where
F(k) is the so called distribution function. The distribution function F(k) is a formula describing
what is the probability of detecting k and less α particles during the measurement interval. This
can be mathematically written as
F(k) =
n≤k
P(n) = e−λ
k
n=0
λn
n!
. (13)
We then calculate the value of χ2 according to
χ2
=
j
Kj(n) − NPj(n)
2
NPj(n)
, (14)
where N is the chosen number of intervals in which the original long measurement was divided
and Pj(n) is the theoretical value corresponding to the Poisson distribution. The result is a single
number that we compare to the χ2 values summarised in table 1 for the selected level of reliability.
If our calculated number is higher than the tabulated value, we will reject the hypothesis (such
as α particles having Poisson distribution).
Perform the χ2 on the data measured in the laboratory. Moreover, plot the graph of the
measured dependence K(n) together with the theoretical dependence K(n) calculated according
to (11).
χ2
test evaluation example
We can illustrate the χ2 evaluation procedure of a random process on the example of classical
six-sided dice. We want to determine if the dice is evenly weighted on all sides. We will throw
the dice 60 times (N = 60) and mark the individual results’ frequency. If the dice is even, the
probability of every possible result should be P(n) = 1/6. From the measurements, we get the
following table:
n 1 2 3 4 5 6
K(n) 12 3 9 15 7 14
P(n) 1
6
1
6
1
6
1
6
1
6
1
6
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Degrees of freedom 0.05 0.01
1 3.841 6.635
2 5.991 9.210
3 7.815 11.341
4 9.483 13.277
5 11.070 15.086
6 12.592 16.812
7 14.067 18.475
8 15.507 20.090
9 16.919 21.666
10 18.307 23.209
11 19.675 24.725
12 21.026 26.217
13 22.362 27.688
14 23.685 29.141
15 24.996 30.578
16 26.296 32.000
17 27.587 33.409
18 28.868 34.805
19 30.144 36.191
20 31.410 37.566
Table 1: χ2 test critical values.
First, we will check the conditions for the use of the χ2 test. The first condition states that we
should expect at least 5 scores for every possible value on the dice. This condition is fulfilled. We
threw the dice 60 times and for the probability P(n) = 1/6 we expect NP(n) = 10 scores for
every dice value. However, if we threw the dice only 20 times, we would get NP(n) = 3,33 < 5.
In such a case, we would need to merge several points, and the table could look something like
this:
n 1 + 2 3 + 4 5 + 6
Kj(n) 5 9 6
Pj(n) 1
3
1
3
1
3
The second condition states that we need to compare the whole distribution function. This
condition is fulfilled as no other result apart from the 6 different numbers is possible.
Now we can calculate the value of χ2 according to (14) as
χ2
=
5
j=1
Kj(n) − NPj(n)
2
NPj(n)
=
(12 − 10)2
10
+
(3 − 10)2
10
+
(9 − 10)2
10
+
(15 − 10)2
10
+
+
(7 − 10)2
10
+
(14 − 10)2
10
= 10,4
Now we compare this value to the values shown in table 1. For the level of reliability of 0.05 and
5 degrees of freedom we have the critical value of χ2 = 11.07. As we obtained a lower number we
cannot reject the hypothesis that the dice is even.
Physical laboratory 3 manuals (version May 10, 2023) 8
source
foil
detector
Figure 5: Experimental setup.
Experiment description
The first part of the manual has shown us that the number of α particles scattered under the
angle χ to the element of the solid angle dΩ per a unit of time can be written as
dn = N
K1
sin4 χ
2
dΩ, (15)
where N is the number of particles hitting the foil, and K1 is a constant determined by the
experimental parameters. A diagram of the experimental setup is shown in Figure 5. Let us
assume that the source of the α particles has a relatively small area Sz and that the α particles
fly out into all directions equally at the rate of N0 per unit of time. Let us assume that the gold
foil thickness is negligibly small. Then, the α particles hitting the foil are those incoming from
the solid angle under which the source is in the line of sight from any point on the foil. This can
be written as
N = N0
Sz cos α
r2
1
. (16)
Particles scattered on thy foil fly in different directions. Only those scattered to the solid angle
Ωd =
Sd cos β
r2
2
(17)
can get to the detector with the area Sd.
To determine the number of particles counted by the detector per unit of time, we should
integrate equation 15 within limits for Ωd. However, for simplification, we can assume that the
source and detector areas are small compared to the dimensions of the whole apparatus. We then
can consider the scattering angle χ independent of the point on the point of the detector that
was hit by the α particle. We can obtain the total number of detected α particles per unit of
time simply by multiplying equation 15 by the solid angle Ωd. If Rutherford’s description of the
scattering process is valid, the number of detected α particles per unit of time n in our experiment
can be calculated according to
n = K
cos α cos β
r2
1 r2
2 sin4 χ
2
, (18)
where K = N0SzSdK1 is a constant given by the experimental setup.
Physical laboratory 3 manuals (version May 10, 2023) 9
Figure 6: Influence of the gold foil position on the number of detected α particles.
Measurement planning
Before the laboratory, think over what gold foil positions you need to use to verify the formula describing
Rutherford’s scattering. Before making this choice, it is convenient to plot the dependence
of the solid angle on the gold foil position.
Also, think how long you will measure each point – how many α particles you should detect
without too high data noise and error. For this, you should understand the properties of the
Poisson distribution: the mean value (µ) of a quantity with the Poisson distribution equals the
parameter λ. Next, data scattering (s2) is also equal to λ and the standard deviation (s) is
calculated as s =
√
λ. Most of the measured points will fall into the interval < µ−s; µ+s >.
(This majority of points roughly equals 68 %, although, in the case of the Poisson distribution,
this value is strongly dependent also on µ.) The measure of the relative uncertainty of the number
of particles is the ratio of s/µ, which, in the case of Poisson distribution, can be calculated as
1/
√
λ. These facts can help us assess the relative uncertainty of the measured data we can expect
for a certain number of detected α particles.
It is moreover necessary to judge the time available for your measurements. When the gold
foil is in the centre between the α particle source and the detector, you can expect approximately
30 detected particles per minute. This information, together with the formula (18), is enough to
help you judge how time-consuming your measurement plan will be and if it is realistic.
Measurement setup description
The whole measurement setup is shown in Figure 7. It consists of a glass vacuum tube having an
α particle source (Americium 241Am, half-life 432.2 years) inside. The α particle detector is on
the other side of the tube. A movable foil holder is placed between the source and the detector.
A strong magnet on the outside of the tube is used to move the foil holder. CAUTION! Do not
try to remove the magnet from the glass – the glass may break. The gold foil is in the shape of
an annulus with the mean radius of v = 2 cm, the distance between the source and the detector
is 22.7 cm. A scale on the outer wall of the glass chamber is used to read the position of the foil.
The mean free path of the α particles in the air under ambient pressure is very small. This
negatively influences or even completely prohibits the measurement as the α particles scatter not
only on the foil but on the gas molecules during their whole path. To avoid such a problem, we
need to decrease the pressure in the tube as much as possible. This is done using a membrane
vacuum pump (2). The pressure in the tube is measured by a manometer (4). As neither the
pump nor the manometer is leak-free, a throttling valve (3) is used to seal the tube. A venting
valve is used to vent the tube after the measurement.
α particles arriving on the detector cause rise of electric signals. These are first amplified in a
preamplifier (1) and amplifier (5). These impulses are then shown and counted by an oscilloscope
(7).
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21 3 4 5 6 7
Figure 7: Measurement setup: (1) preamplifier, (2) vacuum pump, (3) throttle valve, (4) manometer,
(5) amplifier and discriminator, (6) vacuum tube (containing (a) detector, (b) gold foil and
(c) α particle source), (7) oscilloscope.
Experimental procedure
Decrease the pressure in the tube to about 1 kPa using the membrane pump. Seal the tube by
the throttling valve and turn off the pump. Turn on the oscilloscope and connect a USB drive
(bring your own!). Set a correct measurement time on the oscilloscope and move the x-axis to
have the trigger on the left side of the screen. Make sure that after pressing the Acquire button,
the acquisition option in the menu is set to Peak Detect. The menu can be closed by pressing
the Menu On/Off button. Set an appropriate position of the gold foil. Start the measurement
on the oscilloscope using the SINGLE button. Save the measurement to your USB disc using the
Save/Recall button.
Measure the dependence of the detected α particles on the scattering angle. Plot this dependence
in a graph and verify the formula (18) for the Rutherford scattering.
Before verifying the Poisson distribution of the detected α particles, move the gold foil into
the position that corresponds to the maximum number of the detected α particles. Set the longest
possible measurement time on the oscilloscope. Carry out the measurement, save the data and
repeat the measurement at least three times to have a long combined measurement time interval.
When processing the data, you will divide this interval into shorter ones. You will find out if the
number of the α particles in the short intervals corresponds to the Poisson distribution.
Move the foil holder close to the source or the detector after the measurement.
References
[1] Jiří Anděl, Statistické metody, Matfyzpress, Praha 2003
[2] H. Geiger, E. Marsden, On a Diffuse Reflection of the α-Particles, Proc. Roy. Soc. vol. 82, pp.
495-500, 1909
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[3] E. Rutherford, The Scattering of α and β Particles by Matter and the Structure of the Atom,
Philos. Mag., vol 6, pp.21, 1911