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Bubble Chamber Detective Part I. Altered from the original at http://teachers.web.cern.ch/teachers/archiv/HST2005/bubble_chambers/BCwebsite/index.htm. Physicists learnt about hundreds of new particles by examining the trails they made in bubble chambers. This analysis uses a few rules;

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Bubble chamber detective part i
Bubble Chamber DetectivePart I

Altered from the original at http://teachers.web.cern.ch/teachers/archiv/HST2005/bubble_chambers/BCwebsite/index.htm


  • Physicists learnt about hundreds of new particles by examining the trails they made in bubble chambers. This analysis uses a few rules;

  • The bubble chamber is filled with liquid hydrogen. The particles may hit the electrons or protons.

  • Trails are formed by hydrogen bubbles triggered by moving charged particles.

  • Charge is conserved. It is either +1 or –1.

  • A constant magnetic field causes causes circular motion: qvB = mv2/r.

  • Momentum is conserved: mv = qBr.


?

The interactions must be downstream.


V examining the trails they made in bubble chambers. This analysis uses a few rules;

F

  • 2) What is the direction of the magnetic field?

    • Into the page

    • Out of the page

    • Right

    • Left

    • Not enough information


A


A

Charge is conserved.


5) Why don’t we see the neutral particle heading toward point A?

a) Neutral particles don’t cause bubbles

b) The particle wasn’t moving

c) The particle was moving too fast

A


qvB = mv2/r

mv =qvr


Momentum is conserved. point A?

7) In what direction was the neutral particle traveling before it decayed?

a)

b)

c)


B point A?

  • 8) How many particles leave point B ?

    • Two

    • At least two

    • Three

    • At least three

We can only be sure of two. Measurements of momentum could confirm a third.


B point A?

  • 9) What are the charges of the two charged particles from B?

    • The right one is positive and the other is negative

    • The right one is negative and the other is positive

    • Both are negative

    • Both are positive

The right one curves left - it must be positive.

The other one’s track is too short to see a curve, but it should be negative to conserve the kaon’s charge.


B point A?

10) How can a negative kaon produce a positive and a negative particle?

The negative kaon interacted with a positive particle. That particle was not visible because it

a) was not moving.

b) was moving too fast.

c) was moving into the page

The chamber is filled with hydrogen, a source of many protons that are not moving.


C point A?

  • 11) What happened at C?

  • The negative particle

    • collided with a neutral particle.

    • collided with a positive particle

    • decayed into a negative and a neutral particle

The momentum of the charged particle has changed, so there must be another particle to conserve momentum. It must be neutral to conserve charge.

The liquid hydrogen contains lots of electrons and protons but no neutral particles to collide with.


D point A?

  • 12) What happened at D?

  • The neutral particle

    • collides with a neutral particle.

    • collides with a positive particle

    • decays into a positive and a negative particle

    • decays into two negative particles

The left particle is curving right, so it is negative. We can’t see which way the right particle is curving but it must be positive to conserve charge.

The net charge after is neutral so it must be neutral before.

The liquid hydrogen does not contain neutral particles.


C point A?

D

B

  • 13) How can momentum be conserved at B, C and D?

    • The kaon is moving so fast, that its momentum appears to be unaffected

    • There must be neutral particles produced opposite the apparent increase in momentum

Both are possible explanations.


E point A?

D

Which of the particles that travel between points D and E has the larger momentum?


E point A?

D

It can be shown that the total momentum of the two particles is directed along the line joining D and E.


The momentum is proportional to the radius of curvature: mv = qBr

The momentum is perpendicular to the radius vector.


E mv = qBr

Does the neutral particle that decays at point D come from point C or B ?

D

C

B

For justification of method, click here:

http://teachers.web.cern.ch/teachers/archiv/HST2005/bubble_chambers/BCwebsite/articles/V0_decays.pdf


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