Transcript Exercise 1 - high school teachers at CERN

INTERPRETATION OF BUBBLE
CHAMBER IMAGES
Bubble Chamber
images look very
beautiful, but at
the same time
very complicated
to try to
interpret!
The picture on
the left shows the
tracks left by
charged particles
and antiparticles
Fermilab bubble chamber: 4.6 m in diameter in a 3 T magnetic field
HOW ARE TRACKS FORMED?
A moving charged particle interacts with the liquid along its path via the
Coulomb force it exerts on atomic electrons.
In this way it transfers enough energy to start the liquid boiling and thus
leaves behind it a trail of small bubbles.
However:
A photon has no charge and therefore does not interact via the Coulomb force.
Therefore it cannot leave a trail of bubbles
Q. What other particles would not leave a track in a bubble chamber ?
With so many complicated tracks to look at,
how can the paths of individual particles be
interpreted?
We can do one of the following -
1
Relate the movement of electrons produced by Compton Scattering with
the direction of the magnetic field
2
Relate the movement of the deflected particles with their charge and
the direction of the magnetic field
3
Relate the production of particle / antiparticle pairs (matter &
antimatter) to the energy of the photon which produced them
4
Apply conservation of momentum to a head-on collision between an
electron and a positron
1 Relate the movement of electrons produced by Compton
Scattering with the direction of the magnetic field applied
to the bubble chamber
Notice the trajectory of
the spiralling lone electron
indicated by the arrow this electron was knocked
out of the atom that
originally held it by a high
energy photon.
Questions
1.
Identify the other examples of this interaction in the picture above
2.
What is the direction of the magnetic field?
3.
Why doesn’t the photon leave a track ?
Answers
1. Other Compton electrons are indicated by the arrows (above)
2. From the rule for the force exerted by a magnetic field on a current - the field
is going into the page
3. The photon’s path is not visible because it does not carry any charge and so
does not cause the formation of bubbles in the chamber……

B
2 Relate the movement of deflected particles with their
charge and the direction of the magnetic field

B



F

B
y
Remember that the direction of the force on a moving
charged particle is found using the rule for the force
exerted by a magnetic field on a current .
So the electrons all turn clockwise in the above
diagram
z

qv

F
x
B
A
An electron ( e - ) and a positron (e + ) leave the tracks shown in the picture above
Question:
Which is which?

B

e-
B
A
e+
Answer:
The particle which deflects to the right (track B) is an electron, e-
The particle which deflects to the left (track A) is a positron, e+

B

3 Relate the production of particle / antiparticle pairs
(matter & antimatter) to the energy of the photon which
produced them
Two photons are produced at C
where a positron annihilates
with an electron. Just one
travels to D where it interacts
with a nucleus in the liquid and
materializes into an
electron/positron (e- e+) pair.
ee+
D
C
To a good approximation,
all of this photon’s energy
is shared by the e-/e+ pair.
Question
Calculate the kinetic energy of the e-/e+ pair, knowing that the momentum of the
photon is 265  31 MeV/c .
( me = 0.511 MeV/c2 )
Answer
E = pc
If the photon momentum
is 265 MeV/c
then E = 265 MeV
ee+
Converting to Joules,
D
C
1MeV = 1.6 x 10-13 J
E = 265 x 1.6 x 10-13
= 4.24 x 10-11 J = K(e-,e+)
4
Apply the Conservation of Momentum to a head-on
collision between an electron and a positron
E
e+
e-
At point E a rare event took place.
For a snooker
example of a
head on
collision, click
below
The positive track (e+) seems to change into a negative track (e- ).
Head-on Stop.avi
What happened was that the positron made a head-on collision
with an electron.
Question
What is the linear momentum of the electron if the incoming positron’s linear
momentum was 54  15 MeV/c? .
( me = 0,511 Mev/c2 )
(melecton= mpositron)
E
e+
e-
Answer
Visually, just before and just after E, the tracks seem to be more or less equally
curved – thus having the same momentum.
Since the track of the positron appears to have stopped at the point of collision, all
of its momentum must have been transferred to the electron.
Therefore the linear momentum of the electron must be about 54  15 MeV/c since
the mass of the electron is the same as that of the positron