case study: analyzing elementary particle trajectories...

Radioactivity and Elementary Particles 713NEL

13.613.6Case Study: Analyzing Elementary

Particle TrajectoriesThe Bubble ChamberThe bubble chamber was developed in 1952 by Donald Glaser (Figure 1), who won theNobel Prize in physics in 1960 for his invention. It was the most commonly used particledetector from about 1955 to the early 1980s. Although they are incompatible with thefunctions of modern particle accelerator systems, bubble chambers have provided physi-cists with a wealth of information regarding the nature of elementary particles and theirinteractions.

Many of the properties of fundamental particles may be determined by analyzing thetrails they create in a bubble chamber. Using measurements made directly on bubblechamber photographs (like the one on the front cover of this textbook), we can often iden-tify the particles by their tracks and calculate properties such as mass, momentum, andkinetic energy. In a typical experiment, a beam of a particular type of particle leaves anaccelerator and enters a bubble chamber, which is a large, liquid hydrogen-filled vessel,as described in Section 13.3. Figure 2 shows a bubble chamber apparatus.

The liquid hydrogen is in a “superheated” state and boils in response to the slightestdisturbance. As fast-moving subatomic particles stream through the liquid hydrogen,they leave characteristic “bubble trails” that can be photographed for analysis. Thechamber is placed in a strong magnetic field, which causes charged particles to travelin curved trajectories.

In the bubble chamber photographs we will examine, the magnetic field points out ofthe page so positively charged particles curve to the right (clockwise), and negativelycharged particles curve to the left (counterclockwise) (Figure 3).

Figure 1Donald Glaser

Figure 2Bubble chamber apparatus

Figure 3The particles enter the chamber fromthe bottom of the photograph. Thiswill be true of all bubble chamberphotos that appear in this casestudy. Note that to determine thesign of a particle’s charge, it is nec-essary to know in which directionthe particle travelled along the track.

714 Chapter 13 NEL

In this case study, we will be analyzing the following four subatomic particles: proton(p), kaon (K), pion (p), and sigma (�). Of these, the proton is positively charged and isa stable component of ordinary matter. The other three are short-lived and may be pos-itive, negative, or neutral (Table 1).

Table 1 Some Elementary Particles and Their Charges

Particle Name Charge

Positive Negative Neutral

proton p� no particle no particle

kaon K� K� K0

pion p� p� p0

sigma �� 0

Particle ReactionsWhen a beam of one type of particle collides with a target, various interactions areobserved. The photograph in Figure 4 shows a number of kaons entering the chamberfrom the bottom. Since the magnetic field B�� points out of the page, the counterclockwisecurvature of the tracks indicates that the particles are negatively charged. If we follow a K–

track, we will notice that many of these particles pass through the chamber withoutincident. However, some interact with a hydrogen nucleus (a proton) and start a numberof reactions. The simplest type of interaction between two particles is called elastic scat-tering, and can be considered analogous to a billiard ball collision. We represent theelastic scattering of a K– particle and a proton (p+) symbolically as follows:

K� � p� → K� � p�

Notice that in this case, no change occurs in the identities of the interacting particles—they collide and bounce off each other unaffected. However, in some cases, an incidentK– particle reacts in such a way that new particles are created. In many cases, the createdparticles live for only 10�10 s or less and then disintegrate (decay) into other particles whichmay themselves undergo further decay. We will focus on the �� production reaction,which may be symbolized as follows:

K� � p� → �� p�

Notice in the above equation that the K– particle and the proton disappear, and a sigmaminus particle (��) and a pi plus particle (p�) are created. The �� particle is the mostunstable product of the interaction and, in about 10�10 s, disintegrates into a p� par-ticle and a neutral particle we will call X0, which leaves no trail in the bubble chamber.(Note that unlike �� and p�, X0 is not the symbol of a known subatomic particle; it isa generic symbol we are using to denote an invisible, uncharged particle produced inthis event.) We symbolize the �� decay reaction as follows:

�� → p� � X0

The overall interaction can be shown as a two-step process:

(1)K� � p� → �� p�

(2)→ p� � X0

where (1) represents �� production and (2) symbolizes �� decay.


Case Study 13.6

Recognizing Events On a Bubble ChamberPhotograph Bubble chamber photographs may appear confusing at first; however, they become easierto understand when you know what to look for. It is relatively easy to locate events in thephotograph. An event is a change in a particle’s characteristics or a particle interactionthat is visible in the photograph.

K� mesons travel with sufficiently low energy that many of them slow down andcome to rest. For a �� production reaction to take place, the K� meson and the protonmust be approximately 10�15 m apart. Although there is a force of attraction betweenthe positively charged proton and the negatively charged K� meson, the interaction ismore likely to occur when the K� particle is moving slowly or is at rest. We know thatthe initial momentum of the K�/p� system is zero. Thus, the law of conservation ofmomentum requires that the �� and p� particles have equal and opposite momenta.This makes the �� and p+ tracks appear to be a single track. The �� production reac-tion liberates energy because the combined rest mass of the K�/p� system is greaterthan the combined rest mass of the ��/p� system. Figure 4 is a bubble chamber pho-tograph that contains a �� production/decay event; the event is circled.

Figure 4A bubble chamber photograph showing a K�/p� interaction (circled event)

��

��K�

��

12

Figure 5Schematic diagram of the K�/p�

interaction in Figure 4

Figure 5 is a schematic drawing of this event; it shows the particle identities anddirections of motion. Notice how the �� and p� tracks appear to be a single track. Weinterpret this to mean that the �� production reaction occurred after the K� came torest (at point 1). You will also notice a sharp bend in the �� track less than 1.0 cmfrom where it was created (point 2). It is at this sharp bend that the �� particle decayedinto a p� meson and the invisible (neutral) X0 particle. The length of the �� track willbe short due to the � particle’s relatively short lifetime. Look at each of the photographsin Figures 6 and 7 and see if you can identify individual events; each photograph has oneor more events.

716 Chapter 13 NEL

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re 7

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obvi

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fthe

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aph.

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re 6

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obvi

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even

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rthe

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left-

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even

ts?


Quantitative Analysis of Bubble Chamber EventsCareful measurements made of the length, direction, and curvature of the tracks inbubble chamber photographs reveal much more information about the particles andtheir interactions than a simple visual inspection of the photographs provides. Quantitativeanalysis of the particles that are seen in the photographs can also lead to the discoveryof invisible particles such as X0.

This case study will be divided into parts A and B. In part A you will analyze the tra-jectory of a π� particle (pπ�) and determine its charge-to-mass ratio and rest mass. Youwill then evaluate the calculated rest mass by comparing its value to the accepted valueas given in a standard reference table (Table 2). In part B, you will determine the rest massof the invisible X0 particle that is formed in the �� event.

Part A: Determination of the Charge-to-Mass Ratio and theRest Mass of the p� Particle Formed in �� DecayRecall that a particle of charge q and mass m, moving perpendicular to a uniform magnetic

field of magnitude B, has a charge-to-mass ratio �mq� given by �

mq� � �

Bvr� .

Assuming that charged (long-lived) subatomic particles travel at the speed of light (c � 3.0 108 m/s) through a bubble chamber, we can determine the charge-to-mass ratio by measuring the radius of curvature r of the tracks. In the bubble chamberphotographs we will examine, the magnetic field points out of the page and has amagnitude B equal to 1.43 T. Therefore,

�m

q� � �

Bcr

�

� �3.00

(1

.4310

T

8

)rm/s

� (1 T � 1 kg/Cs)

�m

q� � �

1.98 10r

8 Cm/kg�

If the radius of curvature of the particle track is measured in metres, the charge-to-massratio will be calculated in coulombs per kilogram (C/kg). If the charge of the particle isknown, then its mass can also be calculated. Since all known (long-lived) subatomicparticles possess a charge q that is the same as that of an electron (e � 1.6 10�19 C),we can use this value to calculate the rest mass of the p– particle.

Calculating the Radius of Curvature by the Sagitta MethodFor a curved track, the radius of curvature r cannot be measured directly since the centreof the circular arc is not known. However, if we draw a chord on the track and measureboth the length of the chord l and the length of the sagitta s (the sagitta is the distancefrom the midpoint of an arc to the midpoint of its chord), the radius r can then be foundin terms of l and s (Figure 8). As you can see in Figure 8, the Pythagorean theorem givesthe following relation between the radius r, chord length l, and sagitta s:

r 2 � (r � s)2 � ��2l��

2

Solving for r :

r 2 � r 2 � 2r s � s2 � �l4

2�

2r s � s2 � �l4

2�

r � �8l2

s� � �

2s

�

Case Study 13.6

Table 2 Rest Masses for p� and ��

Particle Rest Mass m

(MeV/c2)

p� 139.6

�� 1197.4

The charge-to-mass ratio equation,

�mq

� = �Bvr� , was first introduced in

Chapter 8, Section 8.2 in the form

�me� = �

Bvr

� . In this case, e represents

the charge on an electron, 1.6 10�19 C (the elementarycharge). The form of the equationbeing used here uses the symbol qto represent the charge on theparticle being analyzed.

LEARNING TIP

718 Chapter 13 NEL

Summary: Part ATo determine the charge-to-mass ratio of the p� particle in �� decay, we use the followingsteps:

1. Identify a �� event on the bubble chamber photograph and isolate the p� track.

2. Determine the radius of curvature r by measuring the chord length l and sagitta s

and substituting into the equation r � �l8

2

s� � �

2s�.

3. Calculate the charge-to-mass ratio by substituting the value of r into the equation

�m

q� ��

1.98 10

r

8 Cm/kg�.

4. Since the charge q of the p� particle equals the elementary charge, 1.6 10�19 C,this value can be used to calculate the mass of the p� particle in kilograms fromthe charge-to-mass ratio.

r

s

C

l

Figure 8The sagitta method for calculatingthe radius of curvature r

It is best to draw the longest possible chord subtending the track arc in order toobtain the greatest possible accuracy for the value of r.

A sigma decay event is identified in the bubble chamber photograph in Figure 9.

(a) Calculate the rest mass of the p� particle.

(b) Evaluate your answer in (a) by comparing this calculated rest mass with the acceptedvalue (mp� � 2.48 10�28 kg). Calculate the percent difference.

Solution(a) Measurements taken from the schematic (Figure 10):

lp� � 6.8 cm

sp� � 0.20 cm

SAMPLE problem 1


Case Study 13.6

First, calculate the radius of curvature r for the p� particle:

r � �8l2

sp

p

�

��

sp

2��

� �8((60..820cm

cm)2

)� � �

0.202cm�

r � 29 cm, or 0.29 m

Now to calculate the charge-to-mass ratio of the p� particle, substitute the value of rinto the equation:

�mq

� � �1.98 10

r

8 Cm/kg�

� �1.98

0.

1

2

0

9

8 C

m

m/kg�

�mq

� � 6.6 108 C/kg

Notice that the charge-to-mass ratio of the p� particle (6.6 108 C/kg) is smallerthan that of the electron (1.76 1011 C/kg). Since the two particles carry the samecharge, we may predict that the p� particle has a larger mass.

We calculate the mass of the p� particle by substituting the value of the elementarycharge (e � 1.6 10�19 C) for q :

Since q � 1.6 10�19 C, then

�mq

� � 6.6 108 C/kg

m � �6.6 1

q08 C/kg�

� �61.6.6

11008

�

C

19

/kCg

�

m � 2.42 10�28 kg

The calculated mass of the p� particle is 2.42 10�28 kg. The mass is indeedlarger than the mass of the electron (9.1 10�31 kg).

Figure 9

��

��

K�

s ��

l ��

l��

Figure 10

720 Chapter 13 NEL

(b) Now calculate the percent difference between the calculated value and theaccepted value:

% difference � 100%

� 100%

% difference � 2.3%

(2.42 10�28 kg) � (2.48 10�28 kg)��

2.48 10�28 kg

calculated value � accepted value��

accepted value

PracticeUnderstanding Concepts

1. Given the bubble chamber photograph of a beam of kaons in Figure 11, calcu-late (a) the charge-to-mass ratio and (b) the rest mass of the p�particle.

Figure 11

Answers

1. (a) 7.9 108 C/kg

(b) 2.0 10�28 kg


Part B: Determination of the Identity of the X0 ParticleTo determine the identity of the invisible X0 particle, we must calculate the followingvalues:

A. the momentum of the p� particle (pp�)

B. the momentum of the �� particle (p��)

C. the angle v between the p� and �� momentum vectors at the point of decay

D. the rest mass of the X0 particle

We can use these values to compare the calculated rest mass to the rest masses of known(neutral) subatomic particles as they appear in a standard table, such as Table 3.

We will now describe each of the four calculations that need to be performed.

A. Calculating the Momentum of the p� ParticleA particle of charge q and momentum of magnitude p, moving perpendicular to a uniformmagnetic field of magnitude B, travels in a circular arc of radius r, given by the equation

r � �qpB�

We can rearrange the equation to solve for the magnitude of the momentum of a p�

particle:

p � qBr

The charge q on all known long-lived subatomic particles is constant (q � 1.6

10�19 C) and the value of B is also constant for a particular bubble chamber setting. Ifthe radius of curvature r is measured in centimetres, the momentum equation becomes:

(6.86 MeV/c cm)r

This equation may be used to calculate the momentum of the p– particle using thebubble chamber photographs in this case study.

B. Calculating the Momentum of the �� Particle The momentum of the �� particle cannot be determined from its radius of curvaturebecause the track is much too short. Instead, we make use of the known way that a par-ticle loses momentum as a function of the distance it travels. In each event in which theK� particle comes to rest before interacting, the law of conservation of energy requiresthat the �� particle have a specific momentum of 174 MeV/c. The relatively massive ��

particle loses energy rapidly, so its momentum at the point of its decay is significantlyless than 174 MeV/c, even though it travels only a short distance.

It is known that a charged particle’s range d, which is the distance it travelled beforecoming to rest, is approximately proportional to the fourth power of its initial momentum,(i.e., d ∝ p4). For a �� particle travelling in liquid hydrogen, the constant of propor-tionality is such that a particle of initial momentum 174 MeV/c has a maximum rangeof 0.597 cm, before it decays and becomes invisible. However, in most cases, the �� par-ticle decays before it comes to rest, so the distance it travels, l��, is less than its max-imum range, d0 = 0.597 cm. Therefore, the equation

d � 0.597 cm ��174

p

M��

eV/c��

4

will give the correct range for the �� particle.

Case Study 13.6

Table 3 Rest Masses of Several Known (Neutral) Subatomic Particles

Particle Rest Mass (MeV/c2)

p0 135.0

K0 497.7

n0 939.6

�0 1115.7

�0 1192.6

Y0 1314.9

722 Chapter 13 NEL

The difference between the maximum range d0 and the actual �� particle tracklength l�� is called the residual range. Thus, the relationship between the residual range(d0 � l��) and the momentum of the �� particle is

d0 � 0.597 ��174

p

M��

eV/c��

4

p�� 174 MeV/c �4 �0.59

d70�cm� ��0.5

l

9�

7��

cm��Since d0 � 0.597 cm

p�� 174 MeV/c �4 1 � �0.�5

l9�7�

cm��Note that when l �� 0.597 cm, p

�� 0 as would be expected.

C. Measuring the Angle v between the p� and �� Momentum Vectors at the Point of Decay

The angle v between the p� and �� momentum vectors at the point of decay can be meas-ured directly by drawing tangents to the p� and �� tracks at the point where the ��decayoccurs (Figure 12). We then measure the angle between the tangents using a protractor.

Calculating the Mass of the Invisible X0 Particle Although we cannot see bubble chamber tracks left by the X0 particle, its mass can be cal-culated from measurements made directly on bubble chamber photographs. To do this,we use the values of pp�, p

��, and angle v, determined in the previous three parts, to cal-culate the energy and momentum of the particles that participate in the � decay process(p� and ��). We then use the law of conservation of energy and the law of conserva-tion of momentum to calculate the energy and momentum of the invisible X0 particle.These values are used to calculate the mass of the invisible particle. Finally, the identityof the particle can be determined by comparing the calculated mass to a table that liststhe masses of known particles.

Since the particles that participate in the � decay event travel at speeds close to the speedof light c , all calculations must be done using relativistic equations. The two equationswe will use are

E � mc2 and E2 � p 2c2 + m2c4

where E is the total energy of the particle in MeV, m is the rest mass of the particle inMeV/c2, and p is the momentum of the particle in MeV/c .

When performing calculations, use the values for the rest masses of the �� and p� par-ticles shown in Table 2.

Remember that the invisible particle X0, whose mass we want to find, is produced inthe �� decay process according to the equation

�� → p� � Xo

The law of conservation of momentum requires the X0 particle to have a momentump�

X0 given by the equation

p�X0 � p�

�� p�p�

The relativistic equation E 2 �p2c2 � m2c4 was introduced in

Chapter 11 as, p � .

With a little effort, you should beable to convert one into theother. Assume that m0 � m inthis case.

mv��

�1 � �cv2

2��

LEARNING TIP

��

��

K�

�

Figure 12Measuring the angle v between thep� and �� momentum vectors


Case Study 13.6

The law of conservation of energy requires the Xo particle to have energy EX0 given by

EX0 � E�� Ep�

Since E 2 � p2c2 + m2c4, then

mc2 � �E 2 � p�2c2�

Therefore, we calculate the rest mass m of the X0 particle as follows:

mX0 � �(E�� Ep�)2�� ( p��c2 � pp��c2)2�

Expanding the (p�

�c2 � pp�c2)2 term we obtain

mX0c2 � �(E�� Ep�)2�� ( p�

��2c4 � 2�p�

�pp��c4 � p�p�2c4)�

This equation can be used to calculate the rest mass of the invisible X0 particle frommeasurements made on a bubble chamber photograph. The X0 particle can be identi-fied by comparing its calculated rest mass with the masses of known particles in a ref-erence table such as Table 3.

Summary: Part BTo measure the rest mass of the invisible X0 particle formed in a � decay event, you willneed to use all of the concepts and equations discussed in part B of this case study. Hereare the steps you should follow:

1. Analyze the bubble chamber tracks on a bubble chamber photograph and iden-tify a �� decay event.

2. (a) Identify the p� track.(b) Draw the longest chord possible subtending the p� track arc.(c) Measure the length of the arc lp� and the saggita sp� in centimetres. Use

these values to calculate the radius of curvature rp� of the p� track, andthen substitute this value into the equation pp� � (6.86 MeV/ccm)r to calculate the momentum of the p� particle in MeV/c .

(d) Substitute the value of pp� and the value of mp� (140.0 MeV/c2) into theequation Ep�

2 � pp�2c2 � mp�

2c4 to calculate the total energy Ep� of thep� particle in MeV.

3. (a) Identify the �� track.(b) Measure the �� particle track length, l

��, in centimetres, and substitute this

value into the equation d � 0.597 cm ��174

p

M��

eV/c��

4

to calculate the

�� particle’s momentum in MeV/c.

(c) Substitute the value of p�

� and the value of m�

� (1197.0 MeV/c2) into theequation E

��

2 � p�

�2c2 � m

��

2c4 to calculate the total energy E�

� of the�

� particle in MeV.

4. Draw tangents to the p� and �� trajectories and use a protractor to measurethe angle v.

724 Chapter 13 NEL

The bubble chamber photograph in Figure 13 shows a circled �� decay event.

(a) Calculate the rest mass of the invisible X0 particle formed in the interaction.

(b) Identify the X0 particle by comparing its calculated rest mass with the masses of theknown particles in Table 3.

Solution(a) To calculate the rest mass of the X0 particle, we will have to measure lp�, sp�, l

��,

and v directly on the bubble chamber photograph. See the schematic diagram of the�� event in Figure 14.

SAMPLE problem 2

5. Substitute the values for pp�, Ep�, p�

�, E�

�, and v into the equation

mX0c4 � �(E�

� �� Ep�)2�� (p�

��2c 4 � 2�p�

�pp��c4 � p�p�2c4)� to calculate the

mass of X0 in MeV/c2.

6. Determine the identity of particle X0 by comparing its calculated rest mass withthe masses of known particles in a reference table such as Table 3.

Figure 13Bubble chamber photograph

l�

�


Case Study 13.6

Measurements taken from the schematic (Figure 14):

lp� � 8.0 cm mp� � 140.0 MeV/c2

sp� � 0.30 cm m�

� � 1197.0 MeV/c2

l�

� � 0.6 cm v � 137°

1. Calculate the radius of curvature of the p� trajectory.

rp� � �8

lpsp

�

�

2� � �

sp2

�

�

� �8((80..030

cmcm

)2

)� � �

0.302cm�

rp� � 26.8 cm

2. Calculate the momentum of the p� particle.

pp� � 6.86 rp�

� (6.86 MeV/ccm�)(26.8 cm�)

pp� � 184 MeV/c

3. Calculate the energy of the p� particle.

Ep�2 � pp�

2c 2 � mp�2c 4

� (184 MeV/c)2c 2 + (140.0 MeV/c 2)c 4

� (3.4 104 MeV2/c 2)c 2 + (1.96 104 MeV2/c 4)c 4

Ep�2 � 5.4 104 MeV

Ep� � 2.3 102 MeV

�

��

K�

s ��

l ��

Figure 14Schematic of the eventin Figure 13

726 Chapter 13 NEL

• Some properties of fundamental particles can be identified by analyzing bubblechamber photographs.

• In the analysis of the p and � particles, we determine the momenta of the twoparticles and the angle v between the momentum vectors at the point of decay.This leads to calculation of the mass of the invisible particle, which can then becompared with a table of known masses to identify the particle.

Case Study: Analyzing ElementaryParticle TrajectoriesSUMMARY

4. Calculate the momentum of the �� particle.

p�

� � 174 MeV/c �41 � �

0.5�l

9�

7

�

cm��

� 174 MeV/c �41 � �0

0.�5.697

cmcm��

� 174 MeV/c �4 1 � 1�p

�� 0 MeV/c

5. Calculate the energy of the �� particle.

E��

2 � p�

�2c 2 � m�

�2c 4

E��

2 � 0 MeV2 + (1197.0 MeV2/c 4)c 4

E�� 1.197 103 MeV

6. Measure the angle between the p� and �� trajectories (p� and �� momentum vectors) v = 137o

7. Calculate the rest mass of the invisible X0 particle.

mX0c 2 � �(E�� Ep�)2 �� ( p

��2c�4 � 2p

��pp�c�4 � pp��

2c 4)�

� �(1.1970� 103�MeV �� 2.3 � 102 M�eV)2 ��0 � 0� � (3.4� 104� MeV2�/c 4)c 4��

� �(9.7 � 102 M�eV)2 �� (3.4 � 104 M�eV2)�

� �(9.4 � 105 M�eV2) �� (3.4 � 104 M�eV2)�

mX0c 2 � 9.5 102 MeV

mX0 � 9.5 102 MeV/c2

(b) The X0 particle’s rest mass (9.5 102 MeV/c2) corresponds best with the rest massof the n0 (nu) particle (940 MeV/c2). Therefore, we assume that the invisible neutralparticle produced in the sigma event is the n0 particle

case study: analyzing elementary particle trajectories...

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