Bubble Chamber Lab Answers (Stanford)

Qualitative Part

1. Are the tracks straight or curved? Use a ruler!

The tracks, while they appear to be straight, all are slightly curved. This can be found be taking a straight edge towards the first part of a path. The path will always curve away from the straight edge.

2. Are there different amounts of curvature to the tracks? What might this represent?

Yes, there are different amounts of curvature, which might represent changes in the quantitative momentum of the different particles, since a larger momentum would result in a larger radius. For instance, the two oppositely charged particles created from the decay of a neutral one will curve in opposite directions, indicating that different charge creates different curvature.

3. Why might some particles not leave a track?

If the particles do not leave a track, they might not have a charge, which would leave no trail of hydrogen bubbles. Evidence for this includes how two particles can seemingly appear out of nowhere. Those two particles are oppositely charged, indicating that there was a neutrally charged particle that decayed into oppositely charged ones, demonstrating that the neutral particle left no visible path.

4. Why might some tracks just fade out?

Some of the tracks simply fade out because the particles lose energy forming bubbles, so might lose enough that there is no remaining kinetic energy, or there also might be an angle to the track, and the particle might be moving in a direction away from what is seen. For instance, the particles tend to be easier to detect at the beginning, creating a brighter, more continuous path, and fade away, indicating that the particle is either gradually losing energy or gradually moving away.

5. How can two particle tracks spontaneously appear from what looks like nothing?

Two particle tracks can spontaneously appear from what looks like nothing because they are the decay products of something neutral. It must be noted that, when that happens, the two particle tracks curve in opposite directions, and so imply that the charges are opposite.

6. What do the tiny corkscrews represent?

The tiny corkscrews show the path that electrons take as they are created from the pion proton collision. The electrons spiral because as they lose energy, the radius of the circular path that they take decreases. Electrons take a circular path because they have small amounts of momentum, so have a smaller radius.

7. Do you see any evidence of collisions? If so, what are they?

Yes, when a path forks into two, it indicates a collision. For instance, when a spiral comes off of a path, it indicates that a collision has occurred and an energetic electron has been produced. Also, when a slightly curved path forks into two that curve away from each other, two oppositely charged particles have been created from the collision of a charged particle and another particle.

Part Two

1. In Picture 1, locate the point where a collision may be occurring.
A collision may be occurring at the point at which the alpha particle becomes a beta particle and a pion particle.

2. Using Picture 2, measure the radius r for the negatively charged pion, alpha and beta particles. At this stage, the identities of alpha and beta are not known. Before you do this, explain in detail how you will measure the radius. Your precision is very important to the accuracy of this activity.

In order to measure the radius of the particles, the formula R=(h2+c2/4)/(2h) where h=height of the arc and c= the horizontal length of the curve.

• C_pion=61.8cm
• H_pion=5.8cm
• R_pion= 329.97cm
• C_alpha= 64.0cm
• H_alpha= 4.2cm
• R_alpha= 488.14cm
• C_beta=35.0cm
• H_beta=3.9cm
• R_beta= 157.54

3. Construct a coordinate system (x-y axes) at the vertex of the collision. Using the equation p = 0.3 B r, determine the momentum (magnitude and direction) of each particle at the vertex, in units of MeV/c.

• B= 30 kGauss
• Magnitude P_pion= .3*30kG*329.97cm=2969.73 MeV/c
• Direction P_pion= 127.4 degrees
• Magnitude P_alpha=.3*30kG*488.14cm=4393.26MeV/c
• Direction P_alpha=158 degrees
• Magnitude P_beta=.3*30kG*157.54cm=1417.86MeV/c
• Direction P_beta=28.5 degrees

4. What is the total momentum before the collision and the total momentum after the collision? Is momentum conserved?

• X momentum before= 2969.73 MeV/c *Cos(127.6degrees)=-1811.52MeV/c
• Y momentum before= 2969.73 MeV/c *Sin(127.6degrees)= 2352.89 MeV/c
• X momentum after= 4393.26MeV/c *Cos(158degrees)+1417.86MeV/c*Cos(28.5o)=-2827.32MeV/c
• Y momentum after= 4393.26MeV/c *Sin(158degrees) + 1417.86MeV/c*Sin(28.5degrees)=-2322.29 MeV/c

Momentum is not conserved.

5. What does the negative charge and curvature of the pion imply about the particle labeled beta?
The negative charge and the curvature of the pion imply that the particle labeled betashould be negative, since the pion is decaying into i However, if both are negative, than the charge of the alpha should be neutral. However, in that circumstance the alpha would not have shown up in the bubble chamber. Therefore, the collision could take place between the pion and a particle moving through the third dimension, and therefore not capable of being seen on the bubble chamber.

6. Write a brief conclusion about your results of this activity.
The results of this lab show that momentum is conserved with the particles given, and the charge does not automatically seem to be conserved. There are simple explanations for both of those phenomenon, though. First, with regards to momentum, there must be at least one more particle, theoretically neutral since a track is not shown, that was part of the decay scheme and would be the remaining part of the momentum or there was at least one neutral particle that was part of the collision. The lost momentum could also be found by movement in the third dimension, which isn’t documented by the bubble chamber. The same hypothesis applies to be the conservation of charge, since a positive and a negative particle would have produced a positive particle and a negative particle as a result.

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