ch32 giancoli7e manual

8/15/2019 Ch32 Giancoli7e Manual

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32-1

Responses to Questions

1. A reaction between two nucleons that would produce a π − is p n p p .π −+ → + +

2. No, the decay is still impossible. In the rest frame of the proton, this decay is energetically impossible, because the proton’s mass is less than the mass of the products. Since it is impossible in the rest frame,it is impossible in every other frame as well. In a frame in which the proton is moving very fast, thedecay products must be moving very fast as well to conserve momentum. With this constraint, therewill still not be enough energy to make the decay energetically possible.

3. Antiatoms would be made up of ( ) − charged antiprotons and neutral antineutrons in the nucleus with

( )+ charged positrons surrounding the nucleus. If antimatter and matter came into contact, then the particle–antiparticle pairs would annihilate, converting their mass into energetic photons.

4. The photon signals the electromagnetic interaction.

5. ( a) Yes. If a neutrino is produced during a decay, then the weak interaction is responsible.

(b) No. For example, a weak interaction decay could produce a 0Z instead of a neutrino.

6. The neutron decay process also produces an electron and an antineutrino; the antineutrino will only be present in a weak interaction.

7. An electron takes part in the electromagnetic, weak, and gravitational interactions. A neutrino takes part in the weak and gravitational interactions. A proton takes part in the strong, electromagnetic,weak, and gravitational interactions.

8. The following chart shows charge and baryon conservation checks for many of the decays inTable 32–2 of the textbook.

Particle Decay Charge conservation Baryon conservationW

eW e v+ +→ +

(others are similar)

1 1 0+ = + + 0 0 0= +

0Z 0Z e e+ −→ +

(others are similar)

0 1 ( 1)= + + − 0 0 0= +

E LEMENTARY P ARTICLES 32


2/24


3/24

Elementary Particles 32-3


Particle Decay Charge conservation Baryon conservation

xi 0 0 0π Ξ → Λ + 0 0 0= + 1 1 0+ = + +

0 π − −Ξ → Λ + 1 0 ( 1)− = + − 1 1 0+ = + +

omega 0 π − −Ω → Ξ + 1 0 ( 1)− = + − 1 1 0+ = + + 0 K

− −Ω → Λ + 1 0 ( 1)− = + − 1 1 0+ = + +

0π − −Ω → Ξ + 1 ( 1) 0− = − + 1 1 0+ = + +

9. Since decays via the electromagnetic interaction are indicated by the production of photons, the decays

in Table 32–2 that occur via the electromagnetic interaction are those of the Higgs boson, the 0 ,π the0 ,Σ and the 0 .η

10. All of the decays listed in Table 32–2 with a neutrino or antineutrino as a decay product occur via the

weak interaction. These include the W, muon, tau, pion, kaon,0

LK , and neutron. In addition, theZ particle and the Higgs boson both decay via the weak interaction. In each case, include both the particle and the corresponding antiparticle.

11. Since the ∆ baryon has B 1,= it is made of three quarks. Since the spin of the ∆ baryon is 3/2, none ofthese quarks can be antiquarks. Thus, since the charges of quarks are either 2/3+ or 1/3,− the onlycharges that can be created with this combination are 1( 1/3 1/3 1/3),q = − = − − − 0 ( +2/3 1/3 1/3),= − −

1 ( 2/3 2/3 1/3),+ = + + − and 2 ( 2/3 2/3 2/3).+ = + + +

12. Based on the lifetimes shown in Table 32–4 of the textbook, the particle decays that occur via theelectromagnetic interaction are J/ (3097)ψ and (9460).

13. All of the particles in Table 32–4 except for J/ (3097),ψ (9460), and the sigma particles decay viathe weak interaction, based on their lifetimes.

14. Baryons are formed from three quarks or antiquarks, each of spin 12 or12 ,

− respectively. Any

combination of quarks and antiquarks will yield a spin magnitude of either 12 or32 . Mesons are

formed from two quarks or antiquarks. Any combination of two quarks or antiquarks will yield a spinmagnitude of either 0 or 1.

15. If a neutrinolet was massless, then it would not interact via the gravitation force; if it had no electricalcharge, then it would not interact via the electromagnetic force; if it had no color charge, then it wouldnot interact via the strong force; and if it does not interact via the weak force, then it would not interact

with matter at all. It would be very difficult to say that it even exists at all. However, a similarargument could be made for photons. Photons have no color, no mass, and no charge, but they do exist.

16. ( a) No. Leptons are fundamental particles with no known internal structure. Baryons are made up ofthree quarks.

(b) Yes. All baryons are hadrons.

(c) No. A meson is a quark–antiquark pair.

(d ) No. Hadrons are made up of quarks, and leptons are fundamental particles.


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32-4 Chapter 32


17. No. A particle made up of two quarks would have a particular color. Three quarks or a quark– antiquark pair is necessary for the particle to be white or colorless. A combination of two quarks andtwo antiquarks is possible, as the resulting particle could be white or colorless. The neutral pion can insome ways be considered a four-quark combination.

18. In the nucleus, the strong interaction with the other nucleons does not allow the neutron to decay.When a neutron is free, the weak interaction is the dominant force and can cause the neutron to decay.

19. No, the reaction ee p n v− + → + is not possible. The electron lepton number is not conserved: The

reactants have eL 1 0 1,= + = but the products have eL 0 1 1.= − = − Thus, this reaction is not possible.If the product were an electron neutrino (instead of an antineutrino), then the reaction would be possible.

20. The reaction 0 e p e v+ −Λ → + + proceeds via the weak force. We know that this is the case since an

antineutrino is emitted, which only happens in reactions governed by the weak interaction.

Responses to MisConceptual Questions

1. ( a) A common misconception is that all six quarks make up most of the known matter. However,charmed, strange, top, and bottom quarks are unstable. Protons and neutrons are made of up anddown quarks, so they make up most of the known matter.

2. ( b, e) Quarks combine together, forming mesons (such as the π meson) and baryons (such as the proton and the neutron). The electron and Higgs boson are fundamental particles and thereforecannot be made of quarks.

3. ( c) All of the fundamental forces act on a variety of objects, including our bodies. Although thestrong and weak forces are very short range, the electromagnetic force is a long-range force, justlike gravity. One reason we notice the “weak” gravity force more than the electromagnetic forceis that most objects are electrically neutral, so they do not have significant net electromagnetic

forces on them. It is true that the gravitational force between people and other objects of similarsize is too small for us to notice, but due to the huge mass of the Earth, we are always aware ofthe influence of the Earth’s gravitational force on us.

4. ( d ) A tau lepton has a tau lepton number of one. When it decays, the lepton number must be conserved,so it cannot decay into only hadrons; at least a tau neutrino would have to be one of the by-products.

5. ( d ) In a circular accelerator, the beam can repeatedly pass through the same physical location eachtime around. A single accelerating potential difference at one location can then be used to increasethe beam energy multiple times. In a linear accelerator, the beam passes each point only once, sothe beam path needs to be very long with many accelerating potential differences along its path.

6. ( b, e, f , g ) Atoms are not fundamental, because they are made up of protons, neutrons, and electrons.Protons and neutrons are not fundamental, as they are made up of quarks. The fundamental

particles are leptons (including electrons), quarks, and gauge bosons (including the photon andHiggs boson).

7. ( a) Unlike the electron, the positron has positive charge and a negative lepton number. Unlike thecharge and lepton number, the mass of the positron has the same sign (and magnitude) as themass of the electron.

8. ( e) A common misconception is that the strong force is a result of just the exchange of π mesons between the protons and neutrons. This is correct on the scale of the nucleons. However, when


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the quark composition of the protons, neutrons, and π mesons is considered at the elementary particle scale, it is seen that the transfer is due to the exchange of gluons. Therefore, bothanswers can be considered correct at different scales. Students who answer ( d ) should be givencredit for their answer as well.

9. ( c) Pions are not fundamental particles and are made up of quark and antiquark pairs. Leptons and bosons (including photons) are fundamental particles, but are not a constituent of protons andneutrons. Protons and neutrons are composed of up and down quarks.

10. ( d ) Quarks, gluons, neutrons, and the Higgs boson interact through the strong force. Electrons andmuons are charged particles and interact through the electromagnetic force. Neutrinos onlyinteract through the weak force.

Solutions to Problems

1. The total energy is given by Eq. 26–6a.

2 KE 0.938 GeV 4.65 GeV 5.59 GeV E mc= + = + =

2. Because the energy of the electrons is much greater than their mass, we have KE . E pc= = Combinethat with Eq. 32–1 for the de Broglie wavelength.

34 817

9 19(6.63 10 J s)(3.00 10 m/s)

; 4.4 10 m(28 10 eV)(1.60 10 J/eV)

h hc hc E pc p E

E λ

λ λ

−−

−

× ×= = → = → = = = ×

× ×

3. Use Eq. 32–2 to calculate the frequency.

197

272(1 60 10 C)(1.7 T)

1.3 10 Hz 13 MHz2 2 [4(1.67 10 kg)]

qB f

mπ π

−

−

. ×= = = × =

×

4. The time for one revolution is the period of revolution, which is the circumference of the orbit divided by the speed of the protons. Since the protons have very high energy, their speed is essentially thespeed of light.

35

82 2 (1.0 10 m)

2.1 10 s3.0 10 m/s

r T

π π υ

−×= = = ××

5. The frequency is related to the magnetic field by Eq. 32–2.

27 7

192 2 (1.67 10 kg)(3.1 10 Hz)

2.0 T2 1.60 10 C

qB mf f B

m qπ π

π

−

−

× ×= → = = =

×

6. ( a) The maximum kinetic energy is2 2 2

212KE .2

q B Rm

mυ = = Compared to Example 32–2, the charge

has been doubled and the mass has been multiplied by 4. These two effects cancel each other inthe equation, so the maximum kinetic energy is unchanged (8.653 MeV).

6 197

27KE

KE2 2(8.653 10 eV)(1.60 10 J/eV)

8.7 MeV 2.0 10 m/s4(1.66 10 kg)m

υ −

−

× ×= = = = ×

×


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32-6 Chapter 32


(b) The maximum kinetic energy is2 2 2

212KE .2

q B Rm

mυ = = Compared to Example 32–2, the charge

is unchanged and the mass has been multiplied by 2. Thus the kinetic energy will be half of whatit was in Example 32–2 (8.654 MeV).

6 197

27

KEKE

2 2(4.327 10 eV)(1.60 10 J/eV)4.3 MeV 2.0 10 m/s

2(1.66 10 kg)mυ

−

−

× ×= = = = ×

×

The alpha and the deuteron have the same charge to mass ratio, so they move at the same speed.

(c) The frequency is given by .2qB

f mπ

= Since the charge to mass ratio of both the alpha and the

deuteron is half that of the proton, the frequency for both the alpha and the deuteron will be halfthe frequency found in Example 32–2 for the proton (25.94 MHz).

13 MHz f =

7. From Eq. 30–1, the diameter of a nucleon is about 15nucleon 2.4 10 m.d −= × The 25-MeV alpha

particles and protons are not relativistic, so their momentum is given by KE2 . p m mυ = = The

wavelength is given by Eq. 32–1,KE

.2

h h p m

λ = =

3415

27 6 19

3415

p 27 6 19 p

KE

KE

6 63 10 J s2.88 10 m

2 2(4)(1.66 10 kg)(25 10 eV)(1.6 10 J/eV)

6.63 10 J s5.75 10 m

2 2(1.67 10 kg)(25 10 eV)(1.6 10 J/eV)

h

m

h

m

α α

λ

λ

−−

− −

−−

− −

. × = = = ×

× × ×

× = = = ×

× × ×

We see that nucleond α λ ≈

and p nucleon2 .d λ ≈

Thus, the alpha particle will be better for picking outdetails in the nucleus.

8. Because the energy of the protons is much greater than their mass, we have KE . E pc= = Combinethat with Eq. 32–1 for the de Broglie wavelength. That is the minimum size that protons of that energycould resolve.

34 819

12 19(6.63 10 J s)(3.0 10 m/s)

; 1.8 10 m(7.0 10 eV)(1.60 10 J/eV)

h hc hc E pc p E

E λ

λ λ

−−

−

× ×= = → = → = = = ×

× ×

9. If the speed of the protons is c, then the time for one revolution is found from uniform circular motion.The number of revolutions is the total time divided by the time for one revolution. The energy per

revolution is the total energy gained divided by the number of revolutions.

12 9 3

8

5

2 2 22

( )2 (1.0 10 eV 150 10 eV)2 (1.0 10 m)Engergy/revolution

(3.00 10 m/s)(20 s)

8.9 10 eV/rev 0.9 MeV/rev

r r r t ct T n

T c T r

E E r n ct

π π π υ

υ π π π

= → = = = =

∆ ∆ × − × ×= = =

×

= × ≈


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10. ( a) The magnetic field is found from the maximum kinetic energy as derived in Example 32–2.2 2 2 KE

KE2

2q B R m

Bm qR

= → = →

27 6 19

192(2 014)(1.66 10 kg)(12 10 eV)(1.60 10 J/eV) 0.7082 T 0.71 T(1.60 10 C)(1.0 m) B

− −

−

. × × ×= = ≈

×

(b) The cyclotron frequency is given by Eq. 32–2.19

627

(1.60 10 C)(0.7082 T)5.394 10 Hz 5.4 MHz

2 2 (2.014)(1.66 10 kg)

qB f

mπ π

−

−

×= = = × ≈

×

(c) The deuteron will be accelerated twice per revolution, so it will gain energy equal to twice itscharge times the voltage on each revolution.

619

19 312 10 eV

number of revolutions (1.60 10 J/eV)2(1.60 10 C)(22 10 V)

273 revolutions 270 revolutions

n −

−

×= = ×

× ×

= ≈

(d ) The time is the number of revolutions divided by the frequency (which is revolutions persecond).

56

273 revolutions5.1 10 s

5.394 10 rev/s

nt

f −∆ = = = ×

×

(e) If we use an average radius of half the radius of the cyclotron, then the distance traveled is theaverage circumference times the number of revolutions.

12distance 2 (1.0 m)(273) 860 mrnπ π = = =

11. Start with an expression from Section 32–1, relating the momentum and radius of curvature for a particle in a magnetic field, with q replaced by e.

eBr m eBr p eBr m

υ υ = → = → =

In the relativistic limit, / , p E c= so . E

eBr c

= To get the energy in electron volts, divide the energy by

the charge of the object.

E E eBr Brc

c e= → =

12. The energy released is the difference in the mass energy between the products and the reactant.

0 02 2 2 1115.7 MeV 939.6 MeV 135.0 MeV 41.1 M eVn E m c m c m cπ Λ∆ = − − = − − =

13. The energy released is the difference in the mass energy between the products and the reactant.2 2 2 139.6 MeV 105.7 MeV 0 33.9 MeVv E m c m c m c µ π µ + +

∆ = − − = − − =

14. Use Eq. 32–3 to estimate the range of the force based on the mass of the mediating particle.34 8

2 162 6 19

(6.63 10 J s)(3.00 10 m/s)3.98 10 m

2 2 2 (497.7 10 eV)(1.60 10 J/eV)

hc hcmc d

d mcπ π π

−−

−

× ×≈ → ≈ = ≈ ×

× ×


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32-8 Chapter 32


15. The energy required is the mass energy of the two particles.22 2(939.6 MeV) 1879.2 MeVn E m c= = =

16. The reaction is multistep and can be written as shown here. The energy

released is the initial rest energy minus the final rest energy of the proton and pion, using Table 32–2.

02

p( ) 1192.6 MeV 938.3 MeV 139.6 MeV 114.7 MeV E m m m cπ −Σ∆ = − − = − − =

17. Because the two protons are heading toward each other with the same speed, the total momentum ofthe system is 0. The minimum kinetic energy for the collision would result in all three particles at rest,

so the minimum kinetic energy of the collision must be equal to the mass energy of the 0 .π Each

proton will have half of that kinetic energy. From Table 32–2, the mass of the 0π is 2135.0 MeV/ .c

02KE KE proton proton2( ) 135.0 MeV 67.5 MeVm cπ = = → =

18. Because the two particles have the same mass and they are traveling toward each other with the samespeed, the total momentum of the system is 0. The minimum kinetic energy for the collision wouldresult in all four particles at rest, so the minimum kinetic energy of the collision must be equal to the

mass energy of the K K + −

pair. Each initial particle will have half of that kinetic energy. From

Table 32–2, the mass of each K +

and K −

is 2493.7 MeV/ .c

2 2 p or p K p or p K KE KE2( ) 2 493.7 MeVm c m c= → = =

19. The energy of the two photons (assumed to be equal so that momentum is conserved) must be thecombined mass energy of the proton and antiproton.

34 82 15

2 6 19(6.63 10 J s)(3.00 10 m/s)

2 2 2 1.32 10 m

(938.3 10 eV)(1.60 10 J/eV)

c hcmc hf h

mc

λ

λ

−−

−

× ×= = → = = = ×

× ×

20. ( a) 0 n π −Λ → + Charge conservation is violated, since 0 0 1.≠ − Strangeness is violated, since 1 0 0.− ≠ +

(b) 0 p K −Λ → + Energy conservation is violated, since

2 2 2 21115.7 MeV/ 938.3 MeV/ 493.7 MeV/ 1432.0 MeV/ .c c c c< + =

(c) 0 π π + −Λ → + Baryon number conservation is violated, since 1 0 0.≠ + Strangeness is violated, since 1 0 0.− ≠ +

Spin is violated, since 12 0 0.≠ +

21. The total momentum of the electron and positron is 0, so the total momentum of the two photons must be 0. Thus each photon has the same momentum, so each photon also has the same energy. The totalenergy of the photons must be the total energy of the electron–positron pair.

2 photonse /e pair

34 812 12

2 6 3 19

KE

KE

2( ) 2 2

(6 63 10 J s)(3.00 10 m/s)1.335 10 m 1.3 10 m

(0.511 10 eV 420 10 eV)(1.60 10 J/eV)

c E E mc hf h

hc

mc

λ

λ

+ −

− −−

= → + = = →

. × ×= = = × ≈ ×

+ × + × ×

2

0 0 +

p

γ

π −

Σ → Λ

↓ → +


9/24


10/24

32-10 Chapter 32


p p p p p p:+ → + + + This reaction is possible. All conservation laws are satisfied.

p p p e e p:+ ++ → + + + This reaction will not happen. Baryon number is not conserved (2 0), ≠

and lepton number is not conserved (0 2).≠ −

24. Since the pion decays from rest, the momentum before the decay is zero. Thus the momentum after thedecay is also zero, so the magnitudes of the momenta of the positron and the neutrino are equal. Wealso treat the neutrino as massless. Use energy and momentum conservation along with the relativisticrelationship between energy and momentum.

2 2 2 2 2 2 2 4 2e e e e e

2 2 4 2 2 2 4 2 2 2 2 4 2 4e e e e e e e

2 2 2 22 2 2e e1 1

e e e2 2

2 22 2 e1

e e2

KE

KE

; ( ) ( )

( ) 2 2

2 2

v v v vm c E E p p p c p c E m c E

E m c m c E m c E m c E E m c m c m c

m c m c E m c m c m c

m m

m cm c m c

π

π π π π π

π π π π

π

+ + + + + +

+ + + + + + + + + + + +

+ +

+ + + + +

+ +

+

+ + +

= + = → = → − =

− = − = − + → = +

= + → + = + →

= − +2

1

2 2

(0.511MeV/ )(0.511 MeV)(139.6 MeV) 0.511 MeV

2 2(139.6 MeV/ )

69.3 MeV

c

m cπ += − +

=

Here is an alternate solution, using the momentum.

+2

e

2 2 2 2 2e e e e e e e

2 2e

2 2 2

KE KE

KE KE

[ ( )] 139.6 MeV 0.511 MeV 139.1 MeV

( ) ( ) ; ;

139.1 MeV ( ) (0.511 MeV) (0.511 MeV)

139.6 ( ) (0.511 MeV) (139.6) 2(139.6)

v

v v v v

v

Q m m m c

E m c p c m c m c E p c p p p

Q pc pc

pc pc pc

π +

+ + + + + + +

+

= − + = − =

= − = + − = = = =

= + → = + − +

− = + → − + 2 2 2 2

2 2

eKE KE KE

( ) (0.511 MeV)

(139.6 MeV) (0.511MeV) 69.8 MeV2(139.6 MeV)

69.8 MeV; 139.1 MeV 69.8 MeV 69.3 MeVv v v

p c pc

pc

E Q +

= +

− = =

= = − = = − =

25. ( a) The Q-value is the mass energy of the reactants minus the mass energy of the products.

02 2 2

p( ) 1115.7 MeV (938.3 MeV 139.6 MeV) 37.8 MeVQ m c m c m cπ −Λ= − + = − + =

(b) Energy conservation for the decay gives the following.

0 02 2

p pm c E E E m c E π π − −Λ Λ= + → = −

Momentum conservation says that the magnitudes of the momenta of the two products are equal.

Then convert that relationship to energy using2 2 2 2 4

, E p c m c= +

with energy conservation.

0

2 2 p p

2 2 4 2 2 4 2 2 2 4 p p p

2 2 4 2 4 2 2 2 4 p p p p

2 4 2 4 2 4 2 2 2 p

p 2

( ) ( )

( )

( 2 )

(1115.7 MeV) (938.3 MeV) (139.6 MeV)943.7 MeV

2(1115.7 MeV)2

p p p c p c

E m c E m c m c E m c

E m c m c E m c E m c

m c m c m c E

m c

π π

π π π

π

π

− −

− − 0 −

0 0 −

0 −

Λ

Λ Λ

Λ

Λ

= → = →

− = − = − −

− = − + − →

+ − + −= = =


11/24



02

p

2 p p p

2

KE

KE

1115.7 MeV 943.7 MeV 172.0 MeV

943.7 MeV 938.3 M eV 5.4 MeV

172.0 MeV 139.6 MeV 32.4 MeV

E m c E

E m c

E m c

π

π π π

−

− − −

Λ= − = − =

= − = − =

= − = − =

26. The two neutrinos must move together, in the opposite direction of the electron, in order for theelectron to have the maximum kinetic energy, and thus the total momentum of the neutrinos will beequal in magnitude to the momentum of the electron. Since a neutrino is (essentially) massless, wehave .v v E p c= The muon is at rest when it decays. Use conservation of energy and momentum, alongwith their relativistic relationship.

e

e e e

e

2e e e e e

2 2 2 2 2 2 4e e e e e e

2 4 2 4e

4 2 2 2 2 4 2e e e e e e e2

2 4 2e

e

KE

KE

( )

( ) ( )

2 2

v v

v v v v v v

p p p

m c E E E E p c p c E p p c E p c

m c E p c m c E p c E m c

m c m cm c m c E E E m c E m cm c

m c m c

µ

µ µ µ µ

µ µ

µ µ µ

µ

µ

−

− − − − − −

− − − − − − − −

− −

− − − − − − − − −

−

− −

−

= +

= + + = + + = + + = + →

− = → − = = − →

+− + = − → = = + →

+=

4 2 22

e2(105.7 MeV) (0.511MeV)

(0.511MeV) 52.3 MeV2(105.7 MeV)2

m cm c µ

−

−

+− = − =

27. We use the uncertainty principle to estimate the uncertainty in rest energy.34

20 16(6.63 10 J s)

9420 eV 0.009 MeV2 2 (7 10 s)(1.60 10 J/eV)

h E

t π π

−

− −

× ∆ ≈ = = ≈

∆ × ×

28. We estimate the lifetime from the energy width and the uncertainty principle.

34 2119

3

6.63 10 J s2 10 s

2 2 1.60 10 J2 (300 10 eV)

1 eV

h h E t

t E π π π

− −−

× ∆ ≈ → ∆ ≈ = = × ∆ ∆ ×

×

29. Apply the uncertainty principle, which says that .2

h E

t π ∆ ≈

∆

3421

193

6 63 10 J s7.5 10 s

2 2 1 60 10 J2 (88 10 eV)

1 eV

h h E t

t E π π π

−−

−

. × ∆ ≈ → ∆ ≈ = = ×

∆ ∆ . ××

30. ( a) For B b u,− = we have

Charge: 1 23 31− = − − Spin: 1 12 20 = −

Baryon number: 1 13 30 = − Strangeness: 0 0 0= +

Charm: 0 0 0= + Bottomness: 1 1 0− = − +

Topness: 0 0 0= +


12/24

32-12 Chapter 32


(b) Because B+

is the antiparticle of B ,−

B b u .+ = The 0B still must have a bottom quark but

must be neutral. Therefore, 0B b d .= Because 0B is the antiparticle to 0B , we must have

0B b d .=

31. We find the energy width from the lifetime in Table 32–2 and the uncertainty principle.

(a)34

1919 19

6.63 10 J s5.1 10 s 1293 eV 1300 eV

2 2 (5.1 10 s)(1.60 10 J/eV)

ht E

t π π

−−

− −

× ∆ = × ∆ ≈ = = ≈

∆ × ×

(b)34

24 824 19

6 63 10 J s4.4 10 s 1.499 10 eV 150 MeV

2 2 (4 4 10 s)(1.60 10 J/eV)

ht E

t π π

−−

−

. × ∆ = × ∆ ≈ = = × ≈

∆ . × ×2

32. ( a) Charge: (0) ( 1) ( 1)= + + − Charge is conserved.

Baryon number: ( 1) ( 1) (0)+ = + + Baryon number is conserved.

Lepton number: (0) (0) (0)= +

Lepton number is conserved.Strangeness: ( 2) ( 1) (0)− ≠ − + Strangeness is NOT conserved.

Spin: ( ) ( )1 12 2 (0)= + Spin is conserved.

Energy: 2 2 2 21314.9 Mev/ 1189.4 Mev/ 139.6Mev/ 1329Mev/c c c c> + = Energy is NOT conserved.

The decay is not possible Neither strangeness nor energy is conserved ..

(b) Charge: ( 1) (0) ( 1) (0)− = + − + Charge is conserved.

Baryon number: ( 1) ( 1) (0) (0)+ = + + + Baryon number is conserved.

Lepton number: (0) (0) (0) (1)= + + Lepton number is NOT conserved.

Strangeness: ( 3) ( 1) (0) (0)− ≠ − + + Strangeness is NOT conserved.

Spin: ( ) ( ) ( )3 1 12 2 2(0)≠ + + Spin is NOT conserved.

Energy: 2 2 2 21672.5 Mev/ 1192.6Mev/ 139.6 Mev/ 0 1332.2 Mev/c c c c> + + = Energy is conserved.

The decay is not possible Lepton number, strangeness, and spin are not conserved ..

(c) Charge: (0) (0) (0) (0)= + + Charge is conserved.

Baryon number: (0) (0) (0) (0)= + + Baryon number is conserved.

Lepton number: (0) (0) (0) (0)= + + Lepton number is conserved.

Strangeness: ( 1) ( 1) (0) (0)− = − + + Strangeness is conserved.

Spin: ( ) ( )1 12 2 (1) ( 1)= + + − Spin is conserved, if the gammas haveopposite spins.

Energy: 2 2 21192.6 Mev/ 1115.7 Mev/ 0 0 1115.7 Mev/c c c> + + = Energy is conserved.

The decay is possible .


13/24



33. The expected lifetime of the virtual W particle is found from the Heisenberg uncertainty principle.34

27 279 19

6 63 10 J s 1 eV8.20 10 s 8 10 s

2 (80.385 10 eV) 1.60 10 J E t t

E π

−− −

−

. × ∆ ∆ ≈ → ∆ ≈ = × = × ≈ ×

∆ × ×

U U

34. ( a) The 0Ξ has a strangeness of −2, so it must contain two strange quarks. In order to make a neutral

particle, the third quark must be an up quark. So 0 u s s .Ξ =

(b) The−Ξ has a strangeness of −2, so it must contain two strange quarks. In order to make a

particle with a total charge of −1, the third quark must be a down quark. So, d s s .−Ξ =

35. ( a) The neutron has a baryon number of 1, so there must be three quarks. The charge must be 0, asmust be the strangeness, the charm, the bottomness, and the topness. Thus, u d d .n =

(b) The antineutron is the antiparticle of the neutron, so u dd .n =

(c) The 0Λ has a strangeness of −1, so it must contain an s quark. It is a baryon, so it must containthree quarks. And it must have charge, charm, bottomness, and topness equal to 0. Thus

0 u d s .Λ =

(d ) The 0Σ has a strangeness of +1, so it must contain an s quark. It is a baryon, so it must containthree quarks. And it must have charge, charm, bottomness, and topness equal to 0. Thus

0u d s .Σ =

36. ( a) The combination uu d has charge = +1, baryon number = +1, and strangeness, charm,

bottomness, and topness all equal to 0. Thus, u u d p . =

(b) The combination u u s has charge = −1, baryon number = −1, strangeness = +1, and charm,

bottomness, and topness all equal to 0. Thus, u u s .−= Σ

(c) The combination u s has charge = −1, baryon number = 0, strangeness = −1, and charm,

bottomness, and topness all equal to 0. Thus, u s K .−=

(d) The combination du has charge = −1, baryon number = 0, and strangeness, charm, bottomness,

and topness all equal to 0. Thus, d u .π −=

(e) The combination c s has charge = −1, baryon number = 0, strangeness = −1, charm = −1, and

bottomness and topness of 0. Thus, c s D . S −=

37. To form the 0 D meson, we must have a total charge of 0, a baryon number of 0, a strangeness of 0,and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must havea c quark with a charge of 23 .e

+ To have a neutral meson, there must be another quark with a charge

of 23 .e− To have a baryon number of 0, that second quark must be an antiquark. The only candidate

with those properties is an anti-up quark. Thus, 0 c u . D =


14/24

32-14 Chapter 32


38. To form the S D+

meson, we must have a total charge of +1, a baryon number of 0, a strangeness of +1,and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must havea c quark with a charge of 23 .e

+ To have a total charge of +1, there must be another quark with a

charge of13 .e

+ To have a baryon number of 0, that second quark must be an antiquark. To have a

strangeness of +1, the other quark must be an antistrange. Thus, c s .S D+ =

39. Here is a Feynman diagram for the reaction 0 p n.π π − + → + There are

other possibilities, since the 0π also can be represented as a d d

combination or as a mixture of d d and u u combinations.

40. Since leptons are involved, the reaction n pv µ µ −+ → + is a weak

interaction. Since there is a charge change in the lepton, a W bosonmust be involved in the interaction. If we consider the neutron as having

emitted the boson, then it is a W ,−

which interacts with the neutrino. If

we consider the neutrino as having emitted the boson, then it is a W ,+

which interacts with the neutron.

41. The total energy is the sum of the kinetic energy and the mass energy.The wavelength is found from the relativistic momentum.

2 9 6 10

2 2 2 2 2 2

34 817

1910 2 6 2

KE 15 10 eV 938.3 10 eV 1.5938 10 eV 16GeV

( ) ( )

(6.63 10 J s)(3.00 10 m/s) 17.8 10 m

(1.60 10 J/eV)(1.5938 10 eV) (938.3 10 eV)

E mc

h h hc p E mc E mc

c

λ

−−

−

= + = × + × = × ≈

= = =− −

× ×= = ×

×× − ×

42. To find the length in the lab, we need to know the speed of the particle that is moving relativistically.Start with Eq. 26–5a.

22 22 2

2

KE

KE

1 1 11 1 1 0 7585

1 / 950 MeV1 11777 MeV

mc c c c

cmc

υ

υ

= − → = − = − = .

− + +

13130

lab 2 2 2

8 13 4lab lab

2.91 10 s4.465 10 s

1 / 1 (0 7585)

(0 7585)(3.00 10 m/s)(4.465 10 s) 1.02 10 m

t t

c

x t

υ

υ

−−

− −

∆ ×∆ = = = ×

− − .

∆ = ∆ = . × × = ×


15/24



43. By assuming that the initial kinetic energy is approximately 0, the total energy released is the massenergy of the annihilating pair of particles.

(a) 2total 2 2(0.511 MeV) 1.022 MeV E mc= = =

(b)2

total 2 2(938.3 MeV) 1876.6 MeV E mc= = =

44. ( a) At an energy of 4.0 TeV, the protons are moving at practically the speed of light. From uniformcircular motion, we find the time for the protons to complete one revolution around the ring.Then the total charge that passes any point in the ring during that time is the charge of the entiregroup of stored protons. The current is then the total charge divided by the period.

14 19 8

3

2 2 2

(2 10 protons)(1 60 10 C/proton)(3 0 10 m/s)0.355A 0.4 A

2 2 (4.3 10 m)

R R RT

T c

Ne Nec I

T R

π π π υ υ

π π

−

= → = =

× . × . ×= = = = ≈

×

(b) The 4.0 TeV is equal to the KE of the proton beam. We assume that the car would not be moving

relativistically.21

beam car beam 2

12 14 19 beam

KE KE KE

KE2 2(4.0 10 eV/proton)(2 10 protons)(1.60 10 J/eV)413 m/s

1500 kg

400 m/s

m

m

υ

υ −

= → = →

× × ×= = =

≈

Our assumption that the car is not relativistic is confirmed. This is about 900 mi/h.

45. These protons will be moving at essentially the speed of light for the entire time of acceleration. Thenumber of revolutions is the total gain in energy divided by the energy gain per revolution. Then thedistance is the number of revolutions times the circumference of the ring, and the time is the distance

of travel divided by the speed of the protons.12 9

56

5 3 10 10

101

8

(4.0 10 eV 450 10 eV) 4.44 10 rev

/rev 8.0 10 eV/rev

(2 ) (4 44 10 )2 (4 3 10 m) 1.199 10 m 1.2 10 m

1.199 10 m4.0 10 s

3.00 10 m/s

E N

E

d N R

d t

c

π π

∆ × − ×= = = ×

∆ ×

= = . × . × = × ≈ ×

×= = = ×

×

46. ( a) For the reaction 0 0 p K p ,π π − + → + + the conservation laws are as follows:

Charge: 1 1 0 1 0− + ≠ + + Charge is NOT conserved.

The reaction is not possible, because charge is not conserved .

Also, we note that the reactants would have to have significant kinetic energy to be able to

“create” the 0K .

(b) For the reaction 0 0K p ,π − + → Λ + the conservation laws are as follows:

Charge: 1 1 0 0− + = + Charge is conserved.

Spin: 1 12 20 0+ = + Spin is conserved.


16/24

32-16 Chapter 32


Baryon number: 0 1 1 0+ = + Baryon number is conserved.

Lepton number: 0 0 0 0+ = + Lepton number is conserved.

Strangeness: 1 0 1 0− + = − + Strangeness is conserved.

The reaction is possible, via the strong interaction .

(c) For the reaction 0K n ,π γ + ++ → Σ + + the conservation laws are as follows:

Charge: 1 0 1 0 0+ = + + Charge is conserved.

Spin: 1 12 20 0 1+ = − + + Spin is conserved.

Baryon number: 0 1 1 0 0+ = + + Baryon number is conserved.

Lepton number: 0 0 0 0 0+ = + + Lepton number is conserved.

Strangeness: 1 0 1 0 0+ ≠ − + + Strangeness is NOT conserved.

The reaction is not possible via the strong interaction because strangeness is not conserved.It is possible via the weak interaction.

(d ) For the reaction0 0

K ,π π π + +

→ + + the conservation laws are as follows:Charge: 1 0 0 1= + + Charge is conserved.

Spin: 0 0 0 0= + + Spin is conserved.

Baryon number: 0 0 0 0= + + Baryon number is conserved.

Lepton number: 0 0 0 0= + + Lepton number is conserved.

Strangeness: 1 0 0 0≠ + + Strangeness is NOT conserved.

The reaction is not possible via the strong interaction because strangeness is not conserved.It is possible via the weak interaction.

(e) For the reaction ee ,vπ + +→ + the conservation laws are as follows:

Charge: 1 1 0= + Charge is conserved.

Spin: 1 12 20 = − + Spin is conserved.

Baryon number: 0 0 0= + Baryon number is conserved.

Lepton number: 0 1 1= − + Lepton number is conserved.

Strangeness: 0 0 0 0 0+ = + + Strangeness is conserved.

The reaction is possible, via the weak interaction.

47. ( a) For the reaction p K ,π − + −+ → + Σ the conservation laws are as follows:

Charge: 1 1 1 1− + = − Charge is conserved.

Spin: 1 12 2

0 0+ = + Spin is conserved.

Baryon number: 0 1 0 1+ = + Baryon number is conserved.

Lepton number: 0 0 0 0+ = + Lepton number is conserved.

Strangeness: 0 0 1 1+ = − Strangeness is conserved.

The reaction is possible, via the strong interaction .


17/24



(b) For the reaction p K ,π + + ++ → + Σ the conservation laws are as follows:

Charge: 1 1 1 1+ = + Charge is conserved.

Spin: 1 12 20 0+ = + Spin is conserved.

Baryon number: 0 1 0 1+ = + Baryon number is conserved.Lepton number: 0 0 0 0+ = + Lepton number is conserved.

Strangeness: 0 0 1 1+ = − Strangeness is conserved.

The reaction is possible, via the strong interaction.

(c) For the reaction 0 0 0 p K ,π π − + → Λ + + the conservation laws are as follows:

Charge: 1 1 0 0 0− + = + + Charge is conserved.

Spin: 1 12 20 0 0+ = + + Spin is conserved.


Lepton number: 0 0 0 0 0+ = + + Lepton number is conserved.

Strangeness: 0 0 1 1 0+ = + +2 Strangeness is conserved.

The reaction is possible, via the strong interaction.

(d ) For the reaction 0 0 p ,π π + + → Σ + the conservation laws are as follows:

Charge: 1 1 0 0+ ≠ + Charge is NOT conserved.

The reaction is not possible, because charge is not conserved.

(e) For the reaction e p p e ,vπ − −+ → + + the conservation laws are as follows:

Charge: 1 1 1 1 0− + = − + Charge is conserved.

Spin: 1 1 1 12 2 2 20+ = + − Spin is conserved.


Lepton number: 0 0 0 1 1+ = + − Lepton number is conserved.

Strangeness: 0 0 0 0 0+ = + + Strangeness is conserved.

The reaction is possible, via the weak interaction .

Note that we did not check mass conservation, because in a collision, there is always somekinetic energy brought into the reaction. Thus the products can be heavier than the reactants.

48. The π − is the antiparticle of the ,π + so the reaction is .v µ π µ − −→ + The conservation rules are

as follows:

Charge: 1 1 0− = − + Charge is conserved.Baryon number: 0 0 0= + Baryon number is conserved.

Lepton number: 0 1 1= − Lepton number is conserved.

Strangeness: 0 0 0= + Strangeness is conserved.

Spin: 1 12 20 = − Spin is conserved.


18/24

32-18 Chapter 32


49. Use Eq. 32–3 to estimate the mass of the particle based on the given distance.34 8

2 1118 19

(6 63 10 J s)(3.0 10 m/s) 11.98 10 eV 200 GeV

2 2 (10 m) 1 60 10 J/eV

hcmc

d π π

−

− −

. × ×≈ = = × ≈ . ×

This value is of the same order of magnitude as the mass of the W . ±

50. The Q-value is the mass energy of the reactants minus the mass energy of the products.

For the first reaction, 0 p p p p :π + → + +

0 02 2 2 2

p p2 (2 ) 135.0 MeVQ m c m c m c m cπ π = − + = − = −

For the second reaction, p p p n :π ++ → + +

2 2 2 2 2 2 2 p p n p n2 ( )

938.3 M eV 939.6 MeV 139.6 MeV 140.9 MeV

Q m c m c m c m c m c m c m cπ π + += − + + = − −

= − − = −

51. The fundamental fermions are the quarks and electrons. In a water molecule there are 2 hydrogenatoms consisting of 1 electron and 1 proton each, and 1 oxygen atom, consisting of 8 electrons,8 protons, and 8 neutrons. Thus there are 18 nucleons, consisting of 3 quarks each, and 10 electrons.The total number of fermions is thus 18 3 10× + = 64 fundamental fermions.

52. We assume that the interaction happens essentially at rest, so that there is no initial kinetic energy or

momentum. Thus the momentum of the neutron and the momentum of the 0π will have the same

magnitude. From energy conservation we find the total energy of the 0 .π

0

0

2 2 2 p n n

2 2 2 p n n

KE

KE( ) 139.6 MeV 938.3 MeV (939.6 MeV 0.60 MeV)137.7 MeV

m c m c E m c

E m c m c m c

π π

π π

−

−

+ = + + →

= + − + = + − +=

From momentum conservation, we can find the mass energy of the 0 .π We utilize Eq. 26–9 to relatemomentum and energy.

0 0 0 0 0 0

0 0 0

0

2 2 2 2 4 2 2 4 2 4 2 2 2 4n n n n n n

2 2 2 2 4 2 2 2 1/2n

2

( ) ( )

[(137.7 MeV) (939.6 MeV MeV) (939.6 MeV) ]

133.5 MeV 133.5 MeV/

p p p c p c E m c E m c m c E E m c

m c E E m c

m c

π π π π π π

π π π

π

= → = → − = − → = − + →

= − + = − + 0.60 +

= → =

The reference value is 135 0 MeV..

53. ( a) First we use the uncertainty principle, Eq. 28–1. The energy is so high that we assume , E pc=

so . E

pc

∆∆ =

34 8 915

31 19

2 2

(6 63 10 J s)(3.00 10 m/s) (1 GeV/10 eV)2 10 GeV

2 2 (10 m) (1.60 10 J/eV)

h E h x p x

c

hc E

x

π π

π π

−

− −

∆∆ ∆ ≈ → ∆ ≈ →

. × ×∆ ≈ = = ×

∆ ×


19/24



Next, we use de Broglie’s wavelength formula. We take the de Broglie wavelength as the unificationdistance.

34 8 9 1631 19

(6 63 10 J s)(3.00 10 m/s) (1 GeV/10 eV) 1 10 GeV(10 m) (1.60 10 J/eV)

h h p Ec

hc E

λ

λ −

− −

= = →

. × ×= = = ××

Both energies are reasonably close to 1610 GeV. This energy is the amount that could be violated inconservation of energy if the universe were the size of the unification distance.

(b) From Eq. 13–8, we have 32 . E kT =

25 1928 293

2 232 2(10 eV)(1.6 10 J/eV)

7.7 10 K 10 K 3 3(1.38 10 J/K)

E E kT T

k

−

−

×= → = = = × ≈

×

54. The Q-value is the mass energy of the reactants minus the mass energy of the products.

0 02 2 2 2 p K ( ) 139 6 MeV 938.3MeV (1115 7 MeV 497 6 MeV)

535.4MeV

Q m c m c m c m cπ − Λ= + − + = . + − . + .

= −

We consider the products to be one mass 0 0 2K 1613.3 MeV/m m cΛ= + = since they have the same

velocity. Energy conservation gives the following: 2 p . E m c E π − + = Momentum conservation says

that the incoming momentum is equal to the outgoing momentum. Then convert that relationship to

energy using the relativistic relationship that 2 2 2 2 4 . E p c m c= +

2 2 2 2 4 2 2 4M M

2 2 4 2 2 2 4 2 2 2 4 2 4 p p p

2 4 2 4 2 4 p 2

2 p

KE

( ) ( )

( ) 2

2

M p p p c p c E m c E M c

E m c E m c M c E E m c m c M c

M c m c m c E m c

m c

π π π π

π π π π π

π π π π

− − − −

− − − − −

−

− − −

= → = → − = − →

− = + − = + + − →

− −= = + →

2 4 2 4 2 4 p 2

2 p

2 2 2

KE2

(1613.3 MeV) (139.6 MeV) (938.3 MeV)(139.6 MeV) 767.8MeV

2(938.3 MeV)

M c m c m cm c

m cπ

π π

−

− −

− −= −

− −= − =

55. Since there is no initial momentum, the final momentum must add to zero. Thus each of the pions musthave the same magnitude of momentum and therefore the same kinetic energy. Use energyconservation to find the kinetic energy of each pion.

2 2 2 2 p pKE KE2 2 2 938 3 MeV 139 6 MeV 798.7 MeVm c m c m c m cπ π π π = + → = − = . − . =

56. The Q-value is the energy of the reactants minus the energy of the products. We assume that one of theinitial protons is at rest and that all four final particles have the same speed and therefore the samekinetic energy, since they all have the same mass. We consider the products to be one mass p4m=

since they all have the same speed.2 2 2 2 2

p p p p2 4 2 2Q m c m c m c Mc m c= − = − = −


20/24

32-20 Chapter 32


Energy conservation gives the following, where thKE is the threshold energy.

2 2 2th p pKE KE( ) M M m c m c E Mc+ + = = +

Momentum conservation says that the incoming momentum is equal to the outgoing momentum. Then

convert that relationship to energy using the relativistic relationship 2 2 2 2 4 . E p c m c= +

2 2 2 2 2 4 2 2 2 4 p M p M th p p

2 2 2 4 2 4 2 2 2 4 2 4th th p p p th th p p p

KE KE

KE KE KE KE

( ) ( ) ( ) ( )

2 4 4 (4 )

M p p p c p c m c m c Mc M c

m c m c m c m c m c m c

= → = → + − = + − →

+ + − = + + − →

2 2 2 4 2 4 2 2 4th p th p p p th p p

2th p

KE KE KE

KE

2 4 4 16 2 12

6 3

m c m c m c m c m c m c

m c Q

= + − → = →

= =

57. We use 0λ to represent the actual wavelength and λ to define the approximate wavelength. The

approximation is to ignore the mass in the expression for the total energy, 2KE . E mc= + We also use

Eqs. 26–5a, 26–6b, 26–9, and 32–1.2

2 2 2 2 2 2 2 2 2 2

0 0 01/2 1/22 2

KE KEKE

KE KEKE KE

KE KE

( ) ( ) ( ) 1 2

; ; 1.01 1.012 2

1 1

mc p c E mc mc mc

h hc hc hc hc p mc mc

λ λ λ λ λ

= − = + − = +

= = = > = → = →

+ +

2 810 10

2

34 817 17

10 19

2

2 2

2

2

KE

KE

KE

KE

2 2(9.38 10 eV)9.333 10 eV 9.3 10 eV

0.0201(1.01)

(6.63 10 J s)(3.00 10 m/s)1.332 10 m 1.3 10 m

(9.33 10 eV)(1.60 10 J/eV)

11

1 /

9.31 1 1

mc

hc

mcc

c cmc

λ

υ

υ

−− −

−

−

×= = = × ≈ ×

−1

× ×= = = × ≈ ×

× ×

= − → −

= − + = −

210

833 10 eV

1 0.999959.38 10 eV

c

− ×

+ = ×

58. As mentioned in Example 32–8, the 0π can be considered as either u u or d d. There are variousmodels to describe this reaction. Four are shown here.

Another model that shows the 0π as a combination of both u u and d d is also shown.


21/24



59. ( a) To conserve charge, the missing particle must be neutral. To conserve baryon number, themissing particle must be a meson. To conserve strangeness, charm, topness, and bottomness, themissing particle must be made of up and down quarks and antiquarks only. With all this

information, the missing particle is 0 .π

(b) This is a weak interaction since one product is a lepton. To conserve charge, the missing particlemust be neutral. To conserve the muon lepton number, the missing particle must be anantiparticle in the muon family. With this information, the missing particle is . v µ

60. A relationship between total energy and speed is given by Eq. 26–6b.22 2 2 2

2 222 2

2 22 8

12

1 / 11 /

9.38 10 eV1 1 1 (to 8 digits)

7.0 10 eV

mc mc mc E c

E E cc

mcc

c E

υ υ

υ

υ υ

= → − = → − = → −

×= − = − = → = ×

Use the binomial expansion to express the answer differently.

2 28 89 91

212 129 38 10 eV 9 38 10 eV

1 1 1 9.0 10 (1 9 0 10 )7 0 10 eV 7 0 10 eV ccυ

υ . × . ×

= − ≈ − = − × → = − . × . × . ×

This is about 3 m/s slower than the speed of light.

61. According to Section 32–1, the energy involved in the LHC’s collisions will reach 14 TeV. Use thatwith the analysis used in Example 32–1. From Eq. 30–1, nuclear sizes are on the order of

151.2 10 m.−×

34 820

12 19

205

15nucleus

(6 63 10 J s)(3.00 10 m/s)8.9 10 m

(14 10 eV)(1 60 10 J/eV)

8.9 10 m 17.4 10

13,5001 2 10 m

hc E

r

λ

λ

−−

−

−−

−

. × ×= = = ×

× . ×

×= = × ≈

. ×

62. ( a) The nucleus undergoes two beta decays, so in the nucleus 2 neutrons change into protons. Thedaughter nucleus must have 2 more protons than the parent but the same number of nucleons.

Thus the daughter nucleus is 9642 Mo . The reaction would be96 96 040 42 1Zr Mo 2 e .

−−→ +

(b) Since the reaction is neutrinoless, lepton conservation would be violated in this decay. Checkingisotope mass values at www.nist.gov/pml/data/comp.cfm shows that the reaction is energetically

possible.


22/24

32-22 Chapter 32


(c) If the 9640 Zr would undergo two beta decays simultaneously and thereby emit 2 electron

antineutrinos, then it could decay to 9642 Mo without violating any conservation laws.

63. The width of the “bump” in Fig. 32–19 is about2

5 GeV/ .c Use that value with the uncertainty principle to estimate the lifetime of the Higgs boson.34

259 19

(6 63 10 J s) 11 10 s

2 2 2 (5 10 eV) (1.60 10 J/eV)

h h E t t

E π π π

−−

−

. × ∆ ∆ ≈ → ∆ ≈ = ≈ ×

∆ × ×

Solutions to Search and Learn Problems

1. ( a) The two major classes of fundamental particles are quarks and leptons.

(b) Quarks: up, down, strange, charm, bottom, and top.

Leptons: electron, muon, tau, electron neutrino, muon neutrino, and tau neutrino.

(c) Gravity, electromagnetic, weak nuclear, and strong nuclear.

(d ) Gravity is carried by the graviton; the electromagnetic force is carried by the photon; the weak

nuclear force is carried by the 0W , W , and Z+ −

bosons; the strong nuclear force is carried bythe gluon. The gravitational force is much weaker than the other three forces.

2. ( a) Hadrons interact via the strong nuclear force (as well as the other three fundamental forces) andare made up of quarks.

(b) Baryons are hadrons, are made of three quarks, and have a baryon number of either +1 or −1.(c) Mesons are hadrons, are made of one quark and one antiquark, and have a baryon number of 0.

3. The conservation laws that must hold are the conservation of momentum, angular momentum (spin),mass–energy, charge, baryon number, and lepton number. Momentum can be conserved by the two

decay products having equal but opposite momentum in the rest frame of the parent particle. There areno measurements given to enable us to check conservation of momentum, so we assume that it holds.The other conservation laws are evaluated for each decay. Spin can of course be positive or negative,so our “check” means seeing if there is a way for spins to add up, using either positive or negativevalues for the spin. For the mass–energy to be conserved, the mass of the initial particle must begreater than the mass of the resulting particles, with the remaining mass becoming kinetic energy of

the particles. Those values are in 2GeV/ ,c but units are omitted here.

For the reaction t W b:+→ +

Spin: ( )1 12 21= + − Spin is conserved.Mass–energy: 173 80.4 4.18> + Mass–energy is conserved.

Charge: ( )2 13 31= + − Charge is conserved.Baryon number: 1 13 30

= + Baryon number is conserved.

Lepton number: 0 0 0= + Lepton number is conserved.


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For the reaction t W b:−→ +

Spin: ( )1 12 21= + − Spin is conserved.Mass–energy: 173 80 4 4 18> . + . Mass–energy is conserved.

Charge: ( ) ( )2 13 31− = − + Charge is conserved.

Baryon number: ( )1 13 30− = + − Baryon number is conserved.Lepton number: 0 0 0= + Lepton number is conserved.

For the reaction W u d :+ → +

Spin: 1 12 21= + Spin is conserved.

Mass–energy: 80 4 0 0023 0 0048. > . + . Mass–energy is conserved.

Charge: 2 13 31= + Charge is conserved.

Baryon number: ( )1 13 30

= + − Baryon number is conserved.

Lepton number: 0 0 0= + Lepton number is conserved.

For the reaction W :v µ µ − −→ +

Spin: 1 12 21= + Spin is conserved.

Mass–energy: 80 4 0 1057 0. > . + Mass–energy is conserved.

Charge: 1 ( 1) 0− = − + Charge is conserved.

Baryon number: 0 0 0= + Baryon number is conserved.

Lepton number: 0 1 ( 1)= + − Lepton number is conserved.

4. ( a) Each tau will carry off half of the released kinetic energy. The kinetic energy is the difference inmass–energy of the two tau particles and the original Higgs boson.

02 3 3

H

12KE

( 2 ) 125 10 MeV 2(1777 MeV) 121.4 10 MeV 121.4 GeV

60.7 GeV

Q m m c

Q

τ

τ

= − = × − = × =

= =

(b) The total charge must be zero, so one will have a positive charge and the other will have anegative charge.

(c) No. The mass of the two Z bosons is greater than the mass of the initial Higgs boson, so thedecay would violate the conservation of energy.

5. ( a) We work in the rest frame of the isolated electron, so that it is initially at rest. Energyconservation gives the following.

2 2e e e e eKE KE KE 0m c m c E E E γ γ γ = + + → = − → = =

Since the photon has no energy, it does not exist, so it has not been emitted.

Here is an alternate explanation: Consider the electron in its rest frame. If it were to emit a single photon of energy , E γ then the photon would have a momentum / . p E cγ γ = For momentum to

be conserved, the electron would move away in the opposite direction as the photon with thesame magnitude of momentum. Since the electron is moving (relative to the initial frame ofreference), the electron will now have kinetic energy in that frame of reference, with a total


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32-24 Chapter 32

energy given by Eq. 26–9, 2 2 2 2 2 2( ) ( ) ( ) . E mc p c mc E γ γ = + = + With the added kinetic

energy, the electron now has more energy than it had at rest, and the total energy of the photonand electron is greater than the initial rest energy of the electron. This is a violation of theconservation of energy, so an electron cannot emit a single photon.

(b) For the photon exchange in Fig. 32–8, the photon exists for such a short time that the Heisenberguncertainty principle allows energy conservation to be violated during the exchange. So if t ∆ isthe duration of the interaction and E ∆ is the amount by which energy is not conserved during theinteraction, then as long as , E t ∆ ∆ > the process can happen.

6. Because the energy of the protons is much greater than their mass, we have KE . E pc= = Combinethis with the expression from Problem 11 that relates the momentum and radius of curvature for acharged particle in a synchrotron.

12

3 8(in eV) (7.0 10 eV)

(in T) 5.5 T(4.25 10 m)(3.00 10 m/s)

E B

rc

×= = =

× ×

Alternate derivation: In Chapter 26, Problem 46, it is stated that2

/m r q Bγ υ υ = for a relativisticcharge executing circular motion in a magnetic field. It is also stated in Eq. 26–4 that . p mγ υ = Combine those relationships with the fact that E pc= for highly relativistic particles.

2

12 19

19 3 8(7 0 10 eV)(1.60 10 J/eV)

5.5 T(1.60 10 C)(4.25 10 m)(3.00 10 m/s)

m p E q B qB qB

r r cr

E B

qrc

γ υ υ

−

−

= → = → = →

. × ×= = =

× × ×

ch32 giancoli7e manual

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