Particle physics experiment
Jiří Dolejší, Olga Kotrbová, Charles University in Prague
We have standard ways of discovering
properties of things around us and to look
“inside” of objects (like the small alarm clock in
the photograph). But these methods may not
be applicable to atom and its components … we
do not have a sufficiently small screwdriver
and we even cannot look at it with sufficient
resolution.
With the best current microscopes
we can only see individual atoms,
like in this picture from a tunneling
microscope, but not the inside of
them.
1
Already in the beginning of the 20th century E. Rutherford and his
collaborators developed a novel method to study the inside of atoms.
They shot a-particles towards a thin gold foil and discovered that
what best corresponds to the experimental results is the idea of atoms
being almost empty with small heavy nuclei and electrons flying around.
Today gold nuclei are collided at RHIC (Relativistic Heavy Ion Collider)
and we expect to learn more about the matter through collisions.
How can the methods based on collisons work? Let‘s try!
A first attempt: I will try to use a collision instead of a screwdriver:
I succeeded in getting
inside!
But not sufficiently deep
into the structure.
Maybe I need higher
energy…
It would probably
be better to start with
a simpler object…
2
Atoms, nuclei, particles … Implicitly we look at them as small balls. Instead of
playing with balls I suggest flicking coins and studying their collisions. Put one
coin on a smooth surface, flick another coin against it and observe the result.
After playing a while with different coins you will get some experience. Look
at the following situations and decide which of the coins is heavier (arrows are
proportional to velocities).
B
A
C
3
I hope your answers are correct: both coins have the same mass in B, the
blue one is lighter in A and heavier in C.
You can repeat the experiment with “coins” and “doublecoins” glued together
with doublesided tape:
What about describing the scattering of coins in the
usual manner in mechanics?
Relevant variables are the masses m1 and m2, the velocities v1 and v2 before
the collision and the velocities v1’ and v2’ after the collision. Important
variables are energy E and momentum p.
v1'
1 2 p2
E
v1
The red coin
is at rest at
the beginning
v2= 0
m1
m2
a
b
v2'
2
mv 
2m
p  mv  2mE
Both energy and momentum in
an isolated system are conserved:
E1  E2  E1  E2
 
 
p1  p2  p1  p2
4
Let‘s work only with momenta:
energy conservation...
… and conservation of both momentum components
p1
p1
p 2


2m1 2m1 2m 2
2
2
p1  p1 cos a  p2 cos b
0  p1 sin a  p2 sin b
2
Let‘s play with momenta: We would like to get rid of p2 and b and it is
appealing to use for that purpose the relation
2
2
cos b  sin b  1
We just rewrite the equations, square each one and sum them:
p1  p1 cos a  p2 cos b
p1 sin a  p2 sin b
 p1  p1 cos a 2
 p1 sin a 2
p2 we can insert into energy
2
equation:

2
 p2 cos 2 b
2
 p2 sin 2 b

m2 2
2
2
2
p1  p1  p1  2 p1 p1 cos a  p1
m1
p12  2 p1 p1 cos a  p1 cos 2 a  p1 sin 2 a  p2 cos 2 b  p2 sin 2 b
2
2
2
p12  2 p1 p1 cos a  p1  p2
2
2
2
5
The last equation


m2 2
2
2
2
p1  p1  p1  2 p1 p1 cos a  p1
m1
is a quadratic equation for p1
2
m2 
m2 
2
2





  0.
p1 1 
 2 p1 p1 cos a  p1 1 

 m1 
 m1 
The discriminant is
2








m
m
m
2
2
2
2
2
2
2
2
1 
  4 p1 cos a  1    
D  4 p1 cos a  4 p1 1 

 m1  m1 
 m1  
and its solutions are
2
2
 m2 
 m2 
2
2
2 p1 cos a  2 p1 cos a  1   
cos a  cos a  1   
m1 
m1 


p1 
 p1
.
m
 m2 
1 2

21 
m1
 m1 
6
Let‘s remember that we are looking for real non-negative solutions.
The simplest case is the case of equal masses m1 = m2:
cos a  cos 2 a
p1  p1
 p1 cos a for a  90 or p1  0
2
So in a collision of two coins with equal masses coin 1 cannot be scattered
backwards. It either continues forward (straight or deflected) or stops.
We can see from the conservation equations that if the coin stops, the target
coin takes over the whole energy and momentum. Try it with billiard balls!!!
Instead of continuing
with the detailed
discussion, we will plot
the dependence of
momentum after the
collision on the
scattering angle:
p1
p1
1,2
1…projectile
2…target
1
m1  12 m2
0,6
0,4
m1  2m2
0,2
The deflection of
the heavier particle
is limited
Only lighter
particle can
be scattered
backwards
0,8
m1  m2
0
0
30
60
90
120
150
180
a
7
Although we have played with coins, we used only the most general mechanical
laws. Our results holds for collisions of any objects, including subatomic
particles. The only “complication” is related to the pleasant fact that
particles can be accelerated to speeds close to the speed of light. The
effects predicted by the special theory of relativity appear - particles have
bigger mass and unstable particles live longer.
We can quite easily modify our calculation:
Instead of counting
1 2 p2
E  mv 
kinetic energy
2
2m
we should deal
with total energy E 
m 2c 4  p 2c 2 .
We will insert p2 into the energy conservation equation (all masses here are
2
rest masses):
m12c 4  p12c 2  m2c 2  m12c 4  p1 c 2  m22c 2  p2 c 2
2
The calculation needs more time, more
paper and more patience than the last.
We will only show you the result of the
simplest case m1 = m2 = m:
p1  p1
2
cos a
or p1  0
2
2
p1 sin a
1
2m( E1  m)
8
The difference between the
non-relativistic and the relativistic
calculation is visible on the graph:
p1
p1
1,2
1
Nonrelativistic
case
0,8
0,6
E1
 10
mc 2
0,4
E1
 1000
2
mc
0,2
Protons are accelerated to energies
more than 1000 times larger
than their rest energies
at Fermilab (USA)
0
0
30
60
90
a
A quantitative understanding of the kinematics
of collisions enables us to compare masses of coins
or particles by just colliding them.
First success! Collisions are at
least good for something, not only
for destroying everything ...
9
The formula for relativistic energy
can be rewritten in the form
E  m 2c 4  p 2c 2
E 2  p 2c 2  m2c 4 .
Energy and momentum have different values in different reference frames –
a bottle in my hand in the train has no kinetic energy with reference to the
train but it may have quite a significant
 energy with reference to the ground.
But the special combination of E and p above always equals the square of the
particle rest mass times c4 (independent of the reference frame), i.e. is
constant. This feature offers a surprisingly simple way to measure the mass
of an unstable particle:
Unstable particle
with unknown mass
E  E1  E2

E1 , p1
M
Measure energies and momenta
of decay
 products 
E1 , p1 and E2 , p2,
then calculate

E, p
  
p  p1  p2

E2 , p2
  2 2
( E1  E2 )  ( p1  p2 ) c  M 2c 4
2
You have got the mass M !
You will find words like energy-momentum four-vector, invariant mass, etc. in advanced
textbooks. They refer to the same things as above. You can learn more ...
10
To measure the masses of particles is clearly not enough. How to get some
deeper insight? How to understand structure, interactions, etc.? Maybe the
way is to look at everything that can happen and how probable it is. We can
start again with our macroworld.
What is the probability that I will
catch a ball shot at me?
The best solution to answer this
question is to do an experiment.
Being exposed to randomly placed
moderately fast shots, I caught all
the yellow shots. After several
repetitions of this experiment
I succeeded catching everything
inside the area with the yellow
boundary.
11
My “ability” to catch balls is characterized by the yellow area - it is about 2 m2
and it means that of the “flow” of shots with a density 10 shots per square
meter I expect to have 20 catches.
My “catching ability” is
characterized by an effective
area which I can cover.
Physicists use a special name for
this quantity - “the cross section”
and typically use the letter s
for it.
If a flux of incoming particles
with a density j hits the target,
then the number of interesting
events N with the cross section s is
N  s  j.
The standard unit for the cross
section is 1 barn = 1 b = 10-28 m2.
12
We can express a lot of information in terms of cross-section. For example
the shots I am catching can vary in velocity (and energy) of the ball. I can
easily catch the slow balls but I will probably try to hide myself from hard
shots.
So for hard shots my cross section
of catching will be zero
and the cross section of a ball
hitting me will be close to the area
of my silhouette (+ the band
around accounting for the diameter
of the ball).
One may consider the special (or
“partial”) cross section for a ball
breaking my glasses etc. All the
possible processes can be
summarized in the total cross
section.
13
The energy dependence of
different cross-sections
relating to the interaction
of me and the ball could
look like this graph (and
betray a lot about me …)
sball-me
2.5
At this energy I try to avoid
getting hit by the ball
[m2] 2.0
total
1.0
0.5
At this energy
I catch best
At this energy I am
frozen from the fear
1.5
0
catch
hit
injury
death
Energy of the ball in apropriate units
The cross-sections for
proton-proton interactions
are displayed on this plot:
Elastic cross section - colliding
particles stay intact and they only
change the direction of their flight
At this energy the colliding protons
have enough energy to create a new
particle – a pion. This is one example
of a process contributing to the
inelastic cross section = stotal - selastic
inelastic
14
Usually bunches of projectiles are scattered of many target particles. The
situation may look like in our picture ...
target
beam
Nbeam
Sbeam
Here are Nbeam incoming particles within
a beam with the area Sbeam. The flux of
incoming particles is then jbeam = Nbeam/ Sbeam
Every target catcher
has a cross section s.
There are Ntarget of them
subjected to beam of
projectiles.
We assume that
the individual catchers
do not collaborate and
do not shield others.
So the number of catches
is Nevents = j .s. Ntarget
t
Reality can be always looked at from different viewpoints ...
N events  jbeam .s .N target
Density
of target
particles
N target t
N beam

.s .N target  N beam .s .
.  N beam .s .ntarget t
S beam
S beam t
15
We arrived at an expression which says that the probability of an incoming
particle interacting is
probabilit y of interactio n 
N events
 s .ntarget t.
N beam
In the case when the beam particles are caught (experts speak about absorption) the beam intensity is gradually reduced - Nbeam depends on the depth in the
target.
target
beam
N beam (0)
t
t measures
the depth
in the target
N beam (t )
N beam (t )
t
N beam (t )  N beam (0). exp( s ntargett )
16
Expert pages! It‘s a challenge (but you don´t need to understand them)!
Let‘s try to find the dependence Nbeam(t). We can start from the equation
N events  N beam .s .ntarget t
and reinterpret it.
N beam (0)
N beam (t )
DN beam
• Nevents means the number of absorbed
particles, i.e. the change in Nbeam, which we
will call -DNbeam. The minus sign reminds us
that Nbeam is decreasing.
• t in this equation means the thickness of
the target. Now we should consider the
number of absorbed particles in some layer
of the target with a thickness Dt, where t
describes the depth in the target.
After all these changes we have
DN beam   N beam .s .ntarget Dt
t
Dt
or with infinitesimal Dt
dN beam   N beam .s .ntarget dt
dN beam (t )
  N beam (t ).s .ntarget
dt
where we stressed the
dependence of Nbeam on t.
17
Expert pages! It‘s a challenge (but you don´t need to understand them)!
Instead of being frightened by solving this differential equation remember which function is
almost identical to its derivative - surely you remember the exponential function. It is only
necessary to fix the constant in it:
Assume N beam (t )  C. exp( D.t ) and put it into the differential equation
dN beam (t )
  N beam (t ).s .ntarget
dt
CD. exp( Dt )  C exp( Dt ).s .ntarget .
Clearly we need
D  s .ntarget .
We still have a free constant C in our solution. But this solution should obey the initial condition N beam (0) at the surface of the target at t  0. So we at last arrive at the result
N beam (t )  N beam (0) exp( s .ntargett ).
We will play with this result for a while: The ratio N beam(t ) / N beam(0) tells us the probability of
the incoming particle to survive path t in the target:
probabilit y to survive (t ) 
N beam (t )
 exp( s .ntargett ).
N beam (0)
18
Expert pages! It‘s a challenge (but you don´t need to understand them)!
The probability of survival to depth t and of interaction at this depth is
(probabili ty of interactio n at t ).dt  exp( s .ntargett ).s .ntarget dt.
If the target is thick enough (“infinite”), the beam should die out - the probability to interact
anywhere ( integral from 0 to infinity) should be 1.


 (probabili ty of interactio n at t ).dt   exp( s .n
target
0

t ).s .ntarget dt  exp(  x) dx 1.
0
0
This was just a consistency check, that our formulas correspond to “common sense”.
What is the mean interaction (absorption) length (let us use a special symbol for it – t)? Is it
really the average distance at which the beam particle is absorbed? We can calculate an average
depth of interaction by weighting t with the probability that a particle interacts at t after
penetrating into this depth t:


0
0
t   t.(probabili ty of interactio n at t ).dt   t. exp( s .ntargett ).s .ntarget
t
1
s .ntarget
With this result we can rewrite the
previous formulas in the nice form

1
dt 
x exp(  x) dx

s .ntarget 0
t
N beam(t )  N beam(0) exp(  )
t
19
Expert pages! It‘s a challenge (but you don´t need to understand them)!
Let‘s continue our games. The exponential function may be unfamiliar to some people - why work
with such a funny number like e? We can replace e in the formula by any other number ...
t
ln 2 t
1
N beam(t )  N beam(0) exp(  )  N beam(0) exp( 
)  N beam(0)  
t
ln 2 t
2
t
t ln 2
.
We may introduce a “half depth” or a “half thickness” t1/2 = t ln2 instead of the mean interaction
length t - after passing t1/2 the intensity of the beam is reduced to one half.
Surely you noticed that the formulas describing the absorption of a particle beam by the target
matter are close to the formulas describing the decay of unstable particles. We can switch
between these two meanings and learn from both:
t is either the depth or time,
t is either the mean interaction length or the mean lifetime,
t1/2 is either the half-thickness or the half-life …
There are more interesting questions, can you answer them?
1. Assume that atoms (considered as black disks) have a cross section for absorbing visible light
of about 10-19 m. How large is the mean absorption length? Does it make sense? How it‘s
possible to look out window?
2. How much is a beam which passed 10 absorption lengths reduced?
3. How many absorption lengths are needed to kill the beam completely?
20
If the ball is not caught it may hit me and continue in flight
in some direction. Even this is important for a goalkeeper (a
ball ends in the net or outside). Let me imagine the (very
unpleasant) situation that the ball hits my head and is
deflected by an angle a from its original direction:
A less painful and more simple model for
the same situation is the elastic scattering
of small light balls (like peas) from a hard
heavy sphere.
We may expect that the number of balls
scattered to any precise angle will be
negligible. We should look at some interval
of angles Da
Db
Unlike to my head, the hard
sphere is completely symmetric
so the scattering from it
will be symmetric around this axis
b
a
Da
aDa
a
I took off
my glasses ...
b-Db
Where is the cross section?
Balls flying further from
the axis are deflected less
21
The cross section was introduced as an effective
area of the flow of balls (or particles) causing
some effect. What is the effective area
corresponding to balls being scattered to the
interval of angles Da? … We will look at it from
the point of view of incoming balls ...
Da
Solid angle
= 2p sina Da
Db
All balls flying through this ring
with the area DS=2pbDb
will be scattered into the angular
b
interval Da rotated around the axis,
i.e. to solid angle DW=2p sina Da
Area = 2pbDb
In this situation we speak about the
differential cross section: ds
2p b db
dW

2p sin a da
The cross section corresponding to
particles being scattered in all directions
we will get by integration:
.
p
ds
2p sin a da .
dW
0
s 
22
To calculate the differential cross section we need to know the relation
between r and a. We will try to derive this relation for the simple case of
a small ball elastically scattering off a heavy hard sphere:
Elastic scattering means that balls
are scattered from the surface
at the same angle they hit it.
For the small green triangle and for
angles a and b we can write
b
 sin b , 2 b  a  p .
R
b
b
b
b
Eliminating b and keeping a
a
R
b  R sin
p a
2
 R cos
a
2
.
We have what we need!
ds

dW
a
a
a
R cos
2 sin cos
2
2p b d b
b db
a
1
R


2 R sin     
2
2 R2   .



2p sin a da sin a da
sin a
2  2
4 sin a
4
23
The minus sign reflects the fact that the angle a decreases with growing r.
We need only the absolute values for calculating the cross section. So we
arrive at the result for the differential cross section for elastic scattering
of a small ball on a big hard heavy sphere
ds R 2 Oh, we are
Hurrah!!!

dW 4 so good!!!
The cross section corresponding to particles being scattered to all directions
we will get by the integration:
p
p
ds
p R2
p R2
 cos a  p0  p R 2 .
s 
2p sin a da 
sin a da 
dW
2
2
0
0
This result corresponds to common sense - the effective area of a hard
sphere is pR2… Unbelievable! It‘s not completely wrong!
The differential cross section does not depend on angle a. This means that
scattered balls fly with equal probability in any direction like light radiates
from an ideal (perfectly isotropic) source.
Calculating the differential cross section in this model is an easy game, isn´t
it? What about more complicated models?
24
An interesting model could be a scattering of charged particles on a heavy
charged object. A calculation a little bit more complicated than what we
have done up to now gives the result
2

ds  Zze
1


.
2


a
dW  8p0 mv  sin 4
2
2
For the derivation see the
following expert pages if
you have enough courage ...
The differential cross section for this model is different from that for the
case of elastic balls! It is not isotropic – it is strongly peaked in the
forward direction!
Model predictions can be compared to experiment
and in such way we can learn which model is a more
appropriate description of the nature.
In this way we can even discover what is inside the atoms, as we wanted from
the begining of our story. Ernest Rutherford and his collaborators Geiger and
Marsden used this method for the first time almost a century ago.
25
Expert pages! It‘s a challenge (but you don´t need to understand them)!
Let‘s try to calculate the differential cross section for the scattering of a particle with mass m,
velocity v and charge ze on an infinitely heavy target with charge Ze (e is the elementary charge).
The scattering is almost the same as the case of hard spheres, only the particles interact, not in
contact, but at a distance according to the Coulomb law:
The first step in the calculation could be to look at
energy and momentum conservation – let‘s try that
and see where it goes. The mass of the target is
infinite, so it will absorb any momentum without
moving and will not carry away any energy (a good
exercise: prove it!). The energy of the incoming
particle is therefore conserved but the momentum
is changed:
The path of the incoming particle
is curved - it is a hyperbola.
In the far past it is close to
the red line at the distance b
from the center of the target.
In the far future the path
is close to the line at the angle a.

Dp
ze
b
a


p  mv
r
Zze 2
F (r ) 
4p0 r 2

a
Dp  2mv sin
2
The momentum change is caused by the Coulomb
force:
Coulomb force
1
a


p   mv 
Ze

Dp 
 
 F ( r ) dt .

26
Expert pages! It‘s a challenge (but you don´t need to understand them)!
There are vectors in this equation,
 but we can avoid them. The easiest way is to project the
equation onto the direction of D p

Dp 


F
 ( r ) dt

Dp 

 F (r ) cos  dt.


a
What about using the angle  to parameterize the
path instead of time t ? To do this we only need to
change the integration variable from t to .

Dp 


 F (r (t )) cos  (t ) dt 


p a
2


r
The angle  changes
from -(p-a)/2 at t=-
to (p-a)/2 at t=+.
p a
2
F (r ( )) cos 
d
d
dt
What is d /dt ? … angular velocity w! And moreover, the angular momentum L=mr2w is conserved
for a particle in the central field! We can also calculate the angular momentum from the initial
condition L=mvb (angular momentum = momentum×radius).
L  mr 2w  mvb
w
vb
r2
Now we are ready to put everything together...
27
Expert pages! It‘s a challenge (but you don´t need to understand them)!

a
Dp  2mv sin
2

Dp 

a
Zze 2
2mv sin 
2 4p0
a


2


p a
2

p a

2
Zze 2
b sin 
2 8p0 mv2


1
d
cos

vb
r2
r2

p a
d
vb
w  2
dt
r
p a
2


1
d
d
dt
F (r (cos( )) cos 
p a
2
Zze 2
F (r ) 
4p0 r 2
p a
p a
cos  d  sin   p 2a  2 sin


p a
p a
2
2
 2 cos
2
Zze 2
a
cos

d


cos

4p0 mv2
2
p a
2
2
b
To calculate the differential cross section we need b(a):
Then as before
2
cotan
a
Zze 2
4p0 mv
2
cotan
2
a
2
.
cotan
a
ds
b d b  Zze  1
1
2   Zze 
2.


 
2 
2 

dW sin a da  4p0 mv  sin a 2 sin 2 a  4p0 mv  2 sin a cos a 2 sin 2 a
2
2
2
2
2
2
2
ds  Zze 2 
1

 
dW  8p0 mv2  sin 4 a
2
28
a
2
Expert pages! It‘s a challenge (but you don´t need to understand them)!
Can one guess the result? Does it correspond to a “common sense“?
Do we understand the formulae qualitatively?
We start with the
relation between the
distance b (called the
impact parameter) and
the scattering angle a.
scattering angle a
p (180 )
A particle flying closer to
the target (with smaller b)
will suffer stronger
Coulombic repulsion and
will be deflected more than
a particle flying farther
away from the target.
The limiting case is b = 0
when the particle clearly
bounces back (a = p).
Thanks to Rupert
Leitner for
stimulating this page
Since the Coulomb interaction
has the infinite range,
b can be infinite and still some
(infinitely small) deflection
happens.
0
0

impact parameter b
For simplicity and beauty we may
expect a smooth function
connecting limiting regions ...
Which function it could be?
b  const . cotan
Dealing with angles makes us think about trigonometric functions for b (a). Which of them
touches infinity? Which of them grows to infinity at an angle of zero? At which angle
does it go through zero? How can we modify the argument to accommodate this?
a
2
29
Expert pages! It‘s a challenge (but you don´t need to understand them)!
To work with this formula quantitatively, let‘s consider a-gold scattering:
Instead of substituting here the mass
and velocity of the incoming a,
we may use its kinetic energy … mv2 =2 Ek
A typical kinetic energy of a
from an radioactive source is about
1 MeV = 1,60×10-13 J.
Universal constants
e = 1,60×10-19 C
0 = 8,85×10-12 Fm-1
Charges
of scattered objects
ZAu = 79
za = 2
2
ds  Zze 2 
1
3.23 10  27 m 2

 

2 
a
a
dW  8p0 mv  sin 4
sin 4
2
2
What is the probability that a particle will be deflected by more than 1 degree?
s
a 1
 ...
2
180

1
2p sin a da
sin 4
a
 2p ...
2

1
2
Substitution sin
180
a
2
x
2sin
a
2
cos
sin 4
a
a
2
da
 8p ...

1
2
2
 8p ...
2
180
sin 90

sin 0.5
cos
a a
2
sin 3
d
a
2 
2
sin 90
dx
2  1 



8
p
...
 5.33 10 22 m 2
3
2

x
 2 x  sin0.5
30
Expert pages! It‘s a challenge (but you don´t need to understand them)!
What is the distance b (experts speak about the impact parameter) corresponding to the
deflection angle 1°?
Zze 2
a
b
cotan
 1.14 10 13.cotan 0.5m  1.30 10-11 m
2
4p0 mv
2
These quantitative results have the meaning which we will discuss further.
Experts need some more training to get a working knowledge of the subject. Here are some
additional tasks or questions:
1. Could you prove the statements on energy and momentum balance in the case of the scattering
of a light particle on an infinitely heavy target?
2. Could you explain, why angular momentum is conserved in a central field?
3. Could you translate the calculated value of the differential cross section for scattering into
angles greater than 1° into an effective area, and compare this area to the geometrical cross
section of a sphere with a diameter of 10-10 m?
4. Could you repeat the calculation and find the cross section for particles being deflected to
angle less than 1°?
5. We have got the result for a point-like central charge (or one within a sphere less than b).
What would happen if b were smaller than the radius of the central charge (If the projectile
were aimed at the target better) ?
31
Rutherford´s experiments scattering a particles on a thin gold foil came at
a time when the atom was considered (due to J. J. Thomson) as something
like a plum pudding of positively charged matter with grains of negatively
charged light electrons.
- 6e
Thomson´s view of carbon:
6 negative light electrons and
a massive matter with a positive
charge of 6 units distributed
more or less uniformly
over the volume of the atom.
The whole atom is neutral and has
a diameter of about 10-10 m.
The calculation (if interested, see the expert pages)
shows that to get some non-negligible deflection you need
a big enough charge and a small enough distance of flight
to the scattering center … the a particle should be aimed
at a distance of 1/10 of the atomic radius and feel
the whole positive charge of a gold atom to be deflected
by 1°. But in the Thomson model the positive charge
is spread over the whole volume of the atom!
If the a particle passes through the distributed charge,
it is deflected much less! How it could be?
+2e
a (He++)
Rutherford realized that
a particles (helium atom
without electrons) were
slightly deflected when
passed through materials.
This was rather strange
from the point of view of
Thomson´s model
and inspired further
study … he and Geiger
did their famous
experiment ...
32
An experiment carried out by Geiger (1911) and Geiger & Marsden (1913)
3. … the gold scattering foil F
4. … and a zinc sulphide screen S
attached rigidly to a microscope M
6. The box was fixed
to a graduated circular
platform A,
which could be rotated
in the airtight joint C.
7. The microscope and zinc sulphide screen
rotated with the box, while the gold
foil and source remained fixed.
1. The apparatus consisted in essence
of a strong cylindrical metal box B
2. … which contained
a source of a particles R.
5. The source of a particles
was an a ray tube filled with
radon. A narrow beam of
a particles from the source R
was directed through the
diaphragm D to fall normally
on the scattering foil F.
8. The box was closed by a ground glass plate P
and could be evacuated through the tube T.
9. By rotating the platform A the a particles scattered in different directions could be
observed on the zinc sulphide screen. Observations were taken for scattering angles
between 5° and 150° and both silver and gold foils were used as scattering material.
Two sets of experiments were carried out, the first comparing angles from 15° to 150°
and the second angles from 5° to 30°.
33
Most a-particles scatter at small angles but there were also large
deflections visible in the experiment. Rutherford suggested a model with
small heavy nucleus and electrons around. Look at the comparison of the data
Number of counts
by Geiger & Marsden with the model:
1000
100
The gold data 10and
prediction
compared
1
0
Number of counts
1000000
10
20
30
Angle
100000
10000
1000
100
10
0
30
60
90
120
150
180
Angle
An excerpt from Radiations from Radioactive Substances by Sir Ernest Rutherford,
James Chadwick and C. D. Ellis, published by Cambridge University Pres,s, 1930
34
Hans Geiger
Ernest Rutherford
+
Light
negative
electrons
and his atom
A lot of
empty
space
Positively charged
nucleus with almost
all mass of the atom.
The data agreed well with the
Rutherford model. During many subsequent years physicists discovered how
electrons behave in atoms and what the nucleus is composed of and …
All experiments performed until now agree well with Rutherford´s model.
As usual in physics, we speak frequently about reality (atom is composed
from the nucleus and …) and not about the best available model. Please,
remember what we have done since we first discussed the idea of
a scattering experiment. Later we will meet more variants of this type of
experiment.
35
To be continued
36