sm chapter46

46Particle Physics and Cosmology

CHAPTER OUTLINE

46.1 The Fundamental Forces in Nature

46.2 Positrons and Other Antiparticles46.3 Mesons and the Beginning of

Particle Physics46.4 Classifi cation of Particles46.5 Conservation Laws46.6 Strange Particles and

Strangeness46.7 Finding Patterns in the Particles46.8 Quarks46.9 Multicolored Quarks46.10 The Standard Model46.11 The Cosmic Connection46.12 Problems and Perspectives

ANSWERS TO QUESTIONS

Q46.1 Strong Force—Mediated by gluons. Electromagnetic Force—Mediated by photons. Weak Force—Mediated by W+, W− , and Z0 bosons. Gravitational Force—Mediated by gravitons.

*Q46.2 Answer (b). The electron and positron together have very little momentum. A 1.02-MeV photon has a defi nite chunk of momentum. Production of a single gamma ray could not satisfy the law of conservation of momentum, which must hold true in this—and every—interaction.

Q46.3 Hadrons are massive particles with structure and size. There are two classes of hadrons: mesons and baryons. Hadrons are composed of quarks. Hadrons interact via the strong force. Leptons are light particles with no structure or size. It is believed that leptons are funda-mental particles. Leptons interact via the weak force.

Q46.4 Baryons are heavy hadrons with spin 1

2 or

3

2 composed of three quarks. Mesons are light

hadrons with spin 0 or 1 composed of a quark and an antiquark.

*Q46.5 Answer (c). The muon has much more rest energy than the electron and the neutrinos together. The missing rest energy goes into kinetic energy.

*Q46.6 Answer (b). The z component of its angular momentum must be 3/2, 1/2, –1/2, or –3/2, in units of .

Q46.7 The baryon number of a proton or neutron is one. Since baryon number is conserved, the baryon number of the kaon must be zero.

Q46.8 Decays by the weak interaction typically take 10 10− s or longer to occur. This is slow in particle physics.

*Q46.9 Answer a, b, c, and d. Protons feel all these forces; but within a nucleus the strong interaction predominates, followed by the electromagnetic interaction.

591

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Q46.10 You can think of a conservation law as a superfi cial regularity which we happen to notice, as a person who does not know the rules of chess might observe that one player’s two bishopsare always on squares of opposite colors. Alternatively, you can think of a conservation law asidentifying some stuff of which the universe is made. In classical physics one can think of both matter and energy as fundamental constituents of the world. We buy and sell both of them. In classical physics you can also think of linear momentum, angular momentum, and electric charge as basic stuffs of which the universe is made. In relativity we learn that matter and energy are not conserved separately, but are both aspects of the conserved quantity relativistic total energy. Discovered more recently, four conservation laws appear equally general and thus equally fundamental: Conservation of baryon number, conservation of electron-lepton number, conservation of tau-lepton number, and conservation of muon-lepton number. Processes involving the strong force and the electromagnetic force follow conservation of strangeness, charm, bottomness, and topness, while the weak interaction can alter the total S, C, B, and T quantum numbers of an isolated system.

Q46.11 No. Antibaryons have baryon number –1, mesons have baryon number 0, and baryons have baryon number +1. The reaction cannot occur because it would not conserve baryon number, unless so much energy is available that a baryon-antibaryon pair is produced.

Q46.12 The Standard Model consists of quantum chromodynamics (to describe the strong interaction) and the electroweak theory (to describe the electromagnetic and weak interactions). The Standard Model is our most comprehensive description of nature. It fails to unify the two theories it includes, and fails to include the gravitational force. It pictures matter as made of six quarks and six leptons, interacting by exchanging gluons, photons, and W and Z bosons.

Q46.13 (a) and (b) All baryons and antibaryons consist of three quarks. (c) and (d) All mesons and

antimesons consist of two quarks. Since quarks have spin quantum number 1

2 and can be spin-up

or spin-down, it follows that the three-quark baryons must have a half-integer spin, while the two-quark mesons must have spin 0 or 1.

Q46.14 Each fl avor of quark can have colors, designated as red, green, and blue. Antiquarks are colored antired, antigreen, and antiblue. A baryon consists of three quarks, each having a different color. By analogy to additive color mixing we call it colorless. A meson consists of a quark of one color and antiquark with the corresponding anticolor, making it colorless as a whole.

Q46.15 The electroweak theory of Glashow, Salam, and Weinberg predicted the W+ , W− , and Z particles. Their discovery in 1983 confi rmed the electroweak theory.

*Q46.16 Answer (e). Both trials conserve momentum. In the fi rst trial all of the kinetic energy K1 + K

2 = 2K

1

is converted into internal energy. In the second trial we end up with a glob of twice the mass moving at half the speed, so it has half the kinetic energy of one original clay ball, K

1/2. Energy

K1/2 is converted into internal energy, one-quarter of that converted in trial one.

Q46.17 Hubble determined experimentally that all galaxies outside the Local Group are moving away from us, with speed directly proportional to the distance of the galaxy from us.

*Q46.18 Answer c, b, d, e, a, f, g. The temperature corresponding to b is on the order of 1013 K. That for hydrogen fusion d is on the order of 107 K. A fully ionized plasma can be at 104 K. Neutral atoms can exist at on the order of 3 000 K, molecules at 1 000 K, and solids at on the order of 500 K.

Q46.19 Before that time, the Universe was too hot for the electrons to remain in any sort of stable orbit around protons. The thermal motion of both protons and electrons was too rapid for them to be in close enough proximity for the Coulomb force to dominate.

592 Chapter 46


Particle Physics and Cosmology 593

*Q46.20 Answer (a). The vast gulfs not just between stars but between galaxies and especially between clusters, empty of ordinary matter, are important to bring down the average density of the Universe. We can estimate the average density defi ned for the Solar System as the mass of the sun divided by the volume of a lightyear-size sphere around it:

2 10

2 106 10 6 10

30

43

1620×

×= × = ×− −kg

m)kg/m3

3

π (223 g/cm3, ten million times larger than the critical

density 3H2/8pG = 6 × 10–30 g/cm3.

Q46.21 The Universe is vast and could on its own terms get along very well without us. But as the cos-mos is immense, life appears to be immensely scarce, and therefore precious. We must do our work, growing corn to feed the hungry while preserving our planet for future generations and preserving future possibilities for the universe. One person has singular abilities and opportunities for effort, faithfulness, generosity, honor, curiosity, understanding, and wonder. His or her place is to use those abilities and opportunities, unique in all the Universe.

SOLUTIONS TO PROBLEMS

Section 46.1 The Fundamental Forces in Nature

Section 46.2 Positrons and Other Antiparticles

P46.1 Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy E of the photon must be

E E= = ( ) = = × −2 2 938 3 1 876 6 3 00 10010. . . MeV MeV J

Thus, E hf= = × −3 00 10 10. J

f = ×× ⋅

= ×−

−

3 00 10

6 626 104 53 10

10

3423.

..

J

J s Hz

λ = = ××

= × −c

f

3 00 10

4 53 106 62 10

8

2316.

..

m s

Hz m

P46.2 The half-life of 14O is 70.6 s, so the decay constant is λ = = = −ln ln

.. .

2 2

70 60 009 82

1 2T ss 1

The number of 14O nuclei remaining after fi ve minutes is

N N e et= = ( ) = ×− −( )( )−

010 0 009 82 300

10 5 26λ ..

s s1

1108

The number of these in one cubic centimeter of blood is

′ =⎛

⎝⎜

⎞

⎠⎟ =N N

1 005 2

..

cm

total volume of blood

3

66 101 00

20002 63 108 5×( )⎛

⎝⎜

⎞

⎠⎟ = ×.

. cm

cm

3

3

and their activity is R N= ′ = ( ) ×( ) = ×−λ 0 009 82 2 63 10 2 58 105 3. . . s Bq 1 ~~ Bq103

*P46.3 (a) The rest energy of a total of 6.20 g of material is converted into energy of electromagnetic radiation:

E mc= = × ×( ) = ×−2 3 8 2 146 20 10 3 10 5 58 10. . kg m s J

(b) 5 58 10 5 58 10014

1 00014 14. .× = ×

⎛⎝⎜

⎞⎠⎟ J J

$ .

kWh

k

⎛⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

⎛⎝⎜

⎞⎠⎟ = ×W

J s

1 h

3 600 s$ .2 17 107

All from two cents’ worth of stuff.


594 Chapter 46

P46.4 The minimum energy is released, and hence the minimum frequency photons are produced, when the proton and antiproton are at rest when they annihilate.

That is, E E= 0 and K = 0. To conserve momentum, each photon must carry away one-half the energy.

Thus EE

E hfmin .= = = =2

2938 30

0 MeV min

Thus, fmin

. .

.=

( ) ×( )×

−

−

938 3 1 60 10

6 626 10

13

3

MeV J MeV44

232 27 10 J s

Hz⋅( ) = ×.

λ = = ××

= × −c

fmin

.

..

3 00 10

2 27 101 32 10

8

2315 m s

Hz mm

P46.5 In γ → ++ −p p ,

we start with energy 2.09 GeV we end with energy 938.3 MeV + 938.3 MeV + 95.0 MeV + K2

where K2 is the kinetic energy of the second proton.

Conservation of energy for the creation process gives K2 118= MeV

Section 46.3 Mesons and the Beginning of Particle Physics

P46.6 The reaction is µ ν ν+ −+ → +e

muon-lepton number before reaction: −( ) + ( ) = −1 0 1

electron-lepton number before reaction: 0 1 1( )+ ( ) =

Therefore, after the reaction, the muon-lepton number must be –1. Thus, one of the neutrinos must be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino associated with electrons:

νµ and νe

Then µ ν νµ+ −+ → +e e

P46.7 The creation of a virtual Z 0 boson is an energy fl uctuation ∆E = ×91 109 eV. It can last no

longer than ∆∆

tE

= 2

and move no farther than

c thc

E∆

∆( ) = =

× ⋅( ) ×( )−

4

6 626 10 3 00 1034 8

π. . J s m s

44 91 10

1

1 06 1

9 19π ×( ) ×⎛⎝⎜

⎞⎠⎟

= ×

− eV

eV

1.60 10 J

. 00 1018 18− −= m m~.



*P46.8 (a) The particle’s rest energy is mc2. The time interval during which a virtual particle of this

mass could exist is at most ∆t in ∆ ∆E t = 2

= mc2∆t. The distance it could move is at most

c tc

mc∆ = =

× ⋅( ) ×( )−

2

6 626 10 2 998 102

34 8. . J s m/s

44 1 602 10

1 240 10

42 19

6

π πmc m.

.

×( ) = × ⋅−

−

J/eV

eV m

cc mc

mc

2 2

2

1240

4

98 7

= ⋅

= ⋅

eV nm

eV nm

π

.

According to Yukawa’s line of reasoning, this distance is the range of a force that could be associated with the exchange of virtual particles of this mass.

(b) The range is inversely proportional to the mass of the fi eld particle.

(c) Our rule describes the electromagnetic, weak, and gravitational interactions. For the electromagnetic and gravitational interactions, we take the limiting form of the rule with infi nite range and zero mass for the fi eld particle. For the weak interaction, 98.7 eV ⋅ nm/90 GeV ≈ 10–18 m = 10–3 fm, in agreement with the tabulated information. For the strong interaction, we do not have a separately measured mass for a gluon, so we cannot say that this rule defi nes the range.

(d) 98.7 eV⋅nm/938.3 MeV ≈ 10–1–9–6 m ~10–16 m

P46.9 From Table 46.2 in the chapter text M cπ 0 135 2= MeV .

Therefore, Eγ = 67 5. MeV for each photon

pE

cc= =γ 67 5. MeV

and fE

h= = ×γ 1 63 1022. Hz

P46.10 The time interval for a particle traveling with the speed of light to travel a distance of 3 10 15× − m is

∆ = = ××

=−

−td

v3 10

3 1010

15

823 m

m s s~


596 Chapter 46

P46.11 (a) ∆ = − −( )E m m m cn p e2

From the table of isotopic masses in Chapter 44,

∆ = −( ) ( ) =E 1 008 665 1 007 825 931 5 0 782. . . . MeV

(b) Assuming the neutron at rest, momentum conservation for the decay process impliesp pp e= . Relativistic energy for the system is conserved

m c p c m c p c m cp p e e n2 2 2 2 2 2 2 2 2( ) + + ( ) + =

Since p pp e= , 938 3 0 511 939 62 2 2 2. ( ) . ( ) .( ) + + ( ) + =pc pc MeV

Solving the algebra gives pc = 1 19. MeV

If p c m ce e e= =γ v 1 19. MeV, then γ ve

c

x

x= =

−=1 19

0 511 12 33

2

.

..

MeV

MeV where x

ce= v

Solving, x x2 21 5 43= −( ) . and xc

e= =v0 919.

ve c= 0 919.

Then m mp p e e ev v= γ : vv

pe e e

p

m c

m c= =

( ) ×( )−γ 1 19 1 60 10

1

13. .

.

MeV J MeV

667 10 3 00 1027 8×( ) ×( )− . m s

vp = × =3 80 10 3805. m s km s

(c) The electron is relativistic; the proton is not. Our criterion for answers accurate to three

signifi cant digits is that the electron is moving at more than one-tenth the speed of light and the proton at less than one-tenth the speed of light.

Section 46.4 Classifi cation of Particles

P46.12 In ? ,+ → ++ +p n µ charge conservation requires the unknown particle to be neutral. Baryon number conservation requires baryon number = 0. The muon-lepton number of ? must be –1.

So the unknown particle must be νµ .



Section 46.5 Conservation Laws

P46.13 (a) p p e+ → ++ −µ Le 0 0 0 1+ → +

and Lµ 0 0 1 0+ → − +

(b) π π− ++ → +p p charge − + → + +1 1 1 1

(c) p p p+ → + +π baryon number : 1 1 1 0+ → +

(d) p p p p n+ → + + baryon number : 1 1 1 1 1+ → + +

(e) γ π+ → +p n 0 charge 0 1 0 0+ → +

P46.14 (a) Baryon number and charge are conserved, with respective values of 0 1 0 1+ = + and 1 1 1 1+ = + in both reactions.

(b) Strangeness is conservednot in the second reaction.

P46.15 (a) π µ νµ− −→ + Lµ : 0 1 1→ −

(b) K+ +→ +µ νµ Lµ : 0 1 1→ − +

(c) νe p n e+ → ++ + Le : − + → −1 0 0 1

(d) νe n p e+ → ++ − Le : 1 0 0 1+ → +

(e) ν µµ + → ++ −n p Lµ : 1 0 0 1+ → +

(f ) µ ν νµ− −→ + +e e Lµ : 1 0 0 1→ + + and Le : 0 1 1 0→ − +

P46.16 Baryon number conservation allows the first and forbids the second .

P46.17 (a) p+ +→ +π π 0 Baryon number conservation is violated: 1 0 0→ +

(b) p p p p+ + + ++ → + + π 0 This reaction can occur .

(c) p p p+ + + ++ → + π Baryon number is violated: 1 1 1 0+ → +

(d) π µ νµ+ +→ + This reaction can occur .

(e) n p e0e→ + ++ − ν This reaction can occur .

(f ) π µ+ +→ + n Violates baryon number : 0 0 1→ +

Violates muon-lepton number : 0 1 0→ − +


598 Chapter 46

P46.18 Momentum conservation for the decay requires the pions to have equal speeds.

The total energy of each is 497 7. MeV

2

so E p c mc2 = + ( )2 2 2 2

gives

248 8 139 62 2 2

. . MeV MeV( ) = ( ) + ( )pc

Solving, pc m cmc

c c= = =

− ( )⎛⎝⎜

⎞⎠⎟206

1

2

2 MeV γ v

v

v

pc

mc c c2 2

206 1

11 4= =

− ( )⎛⎝⎜

⎞⎠⎟ = MeV

139.6 MeV v

v. 88

v vc c

= −⎛⎝⎜

⎞⎠⎟1 48 1

2

.

and v vc c

⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −2 2

2 18 1 2 18 2 18. . .vvc

⎛⎝⎜

⎞⎠⎟

2

3 18 2 182

. .vc

⎛⎝⎜

⎞⎠⎟ =

so vc

= =2 18

3 180 828

.

..

and v = 0 828. c

P46.19 (a) In the suggested reaction p e→ ++ γ

We would have for baryon numbers + → +1 0 0

∆B ≠ 0, so baryon number conservation would be violated.

(b) From conservation of momentum for the decay: p pe = γ

Then, for the positron, E p c Ee e e2 2

02= ( ) + ,

becomes E p c E E Ee e e2 2

02 2

02= ( ) + = +γ γ, ,

From conservation of energy for the system: E E Ep e0, = + γ

or E E Ee p= −0, γ

so E E E E Ee p p2

02

022= − +, , γ γ

Equating this to the result from above gives E E E E E Ee p pγ γ γ2

02

02

022+ = − +, , ,

or EE E

Ep e

p

γ =−

= ( ) − (02

02

0

2

2

938 3 0 511, ,

,

. . MeV MeV))( )

=2

2 938 3469

. MeV MeV

Thus E E Ee p= − = − =0 938 3 469 469, .γ MeV MeV MeV

Also, pE

c cγγ= = 469 MeV

and p pce = =γ

469 MeV

continued on next page



(c) The total energy of the positron is Ee = 469 MeV

But, E EE

ce e

e= =− ( )

γ 00

21

,,

v

so 10 511

1 09 102

0−⎛⎝⎜

⎞⎠⎟ = = = ×v

c

E

Ee

e

, ..

MeV

469 MeV−−3

which yields v = 0 999 999 4. c

P46.20 The relevant conservation laws are ∆Le = 0

∆Lµ = 0

and ∆Lτ = 0

(a) π π+ +→ + +0 e ? Le : 0 0 1→ − + Le implies Le = 1 and we have a νe

(b) ? p p+ → + +− +µ π Lµ: Lµ + → + + +0 1 0 0 implies Lµ = 1 and we have a νµ

(c) Λ0 → + +−p µ ? Lµ : 0 0 1→ + + Lµ implies Lµ = −1 and we have a νµ

(d) τ µ+ +→ + +? ? Lµ : 0 1→ − + Lµ implies Lµ = 1 and we have a νµ

Lτ : − → +1 0 Lτ implies L τ = −1 and we have a ντ

Conclusion for (d): Lµ = 1 for one particle, and Lτ = −1 for the other particle.

We have νµ

and ντ

Section 46.6 Strange Particles and Strangeness

P46.21 (a) Λ0 → + −p π Strangeness: − → +1 0 0 (strangeness is not conserved )

(b) π − + → +p K0Λ0 Strangeness: 0 0 1 1+ → − + ( 0 0= and strangeness is conserved )

(c) p p+ → +Λ Λ0 0 Strangeness: 0 0 1 1+ → + − ( 0 0= and strangeness is conserved )

(d) π π− − ++ → + Σp Strangeness: 0 0 0 1+ → − ( 0 1≠ − : strangeness is not conserved )

(e) Ξ Λ− −→ +0 π Strangeness: − → − +2 1 0 ( − ≠ −2 1 so strangeness is not conserved )

(f) Ξ0 → + −p π Strangeness: − → +2 0 0 ( − ≠2 0 so strangeness is not conserved )

P46.22 The ρ π π0 → ++ − decay must occur via the strong interaction.

The KS0 → ++ −π π decay must occur via the weak interaction.


600 Chapter 46

P46.23 (a) µ γ− −→ +e Le : 0 1 0→ + ,

and Lµ : 1 0→

(b) n p e→ + +− νe Le : 0 0 1 1→ + +

(c) Λ0 0→ +p π Strangeness: − → +1 0 0,

and charge: 0 1 0→ + +

(d) p e→ ++ π 0 Baryon number: + → +1 0 0

(e) Ξ0 0→ +n π Strangeness: − → +2 0 0

P46.24 (a) π η− + →p 2 violates conservation of baryon number as 0 1 0+ → , not allowed .

(b) K n− −+ → +Λ0 π

Baryon number, 0 1 1 0+ → + Charge, − + → −1 0 0 1

Strangeness, − + → − +1 0 1 0

Lepton number, 0 0→ The interaction may occur via the strong interaction since all are conserved.

(c) K− −→ +π π 0

Strangeness, − → +1 0 0

Baryon number, 0 0→ Lepton number, 0 0→ Charge, − → − +1 1 0

Strangeness conservation is violated by one unit, but everything else is conserved. Thus,

the reaction can occur via the weak interaction , but not the strong or electromagnetic interaction.

(d) Ω Ξ− −→ + π 0

Baryon number, 1 1 0→ + Lepton number, 0 0→ Charge, − → − +1 1 0

Strangeness, − → − +3 2 0

May occur by weak interaction , but not by strong or electromagnetic.

(e) η γ→ 2

Baryon number, 0 0→ Lepton number, 0 0→ Charge, 0 0→ Strangeness, 0 0→ No conservation laws are violated, but photons are the mediators of the electromagnetic

interaction. Also, the lifetime of the h is consistent with the electromagnetic interaction .



P46.25 (a) Ξ Λ− −→ + +0 µ νµ

Baryon number: + → + + +1 1 0 0 Charge: − → − +1 0 1 0

Le : 0 0 0 0→ + + Lµ : 0 0 1 1→ + +

Lτ : 0 0 0 0→ + + Strangeness: − → − + +2 1 0 0

Conserved quantities are: B L Le, , charge, and τ

(b) KS0 → 2 0π

Baryon number: 0 0→ Charge: 0 0→ Le : 0 0→ Lµ : 0 0→

Lτ : 0 0→ Strangeness: + →1 0

Conserved quantities are: B L L Le, , , charge, and µ τ

(c) K − + → +p nΣ0

Baryon number: 0 1 1 1+ → + Charge: − + → +1 1 0 0

Le : 0 0 0 0+ → + Lµ : 0 0 0 0+ → +

Lτ : 0 0 0 0+ → + Strangeness: − + → − +1 0 1 0

Conserved quantities are: S L L Le, , , charge, and µ τ

(d) Σ Λ0 0+ + γ

Baryon number: + → +1 1 0 Charge: 0 0→ Le : 0 0 0→ + Lµ : 0 0 0→ +

Lτ : 0 0 0→ + Strangeness: − → − +1 1 0

Conserved quantities are: B S L L Le, , , , charge, and µ τ

(e) e e+ − + −+ → +µ µ

Baryon number: 0 0 0 0+ → + Charge: + − → + −1 1 1 1

Le : − + → +1 1 0 0 Lµ : 0 0 1 1+ → + − Lτ : 0 0 0 0+ → + Strangeness: 0 0 0 0+ → +


(f) p n+ → + −Λ Σ0

Baryon number: − + → − +1 1 1 1 Charge: − + → −1 0 0 1

Le : 0 0 0 0+ → + Lµ : 0 0 0 0+ → + Lτ : 0 0 0 0+ → + Strangeness: 0 0 1 1+ → + −



602 Chapter 46

P46.26 (a) K p ? p+ + → +

The strong interaction conserves everything.

Baryon number, 0 1 1+ → +B so B = 0

Charge, + + → +1 1 1Q so Q = +1

Lepton numbers, 0 0 0+ → +L so L L Le = = =µ τ 0

Strangeness, + + → +1 0 0S so S = 1

The conclusion is that the particle must be positively charged, a non-baryon, with

strangeness of +1. Of particles in Table 46.2, it can only be the K+ . Thus, this is an

elastic scattering process.

The weak interaction conserves all but strangeness, and ∆S = ±1.

(b) Ω− −→ + ? π

Baryon number, + → +1 0B so B = 1

Charge, − → −1 1Q so Q = 0

Lepton numbers, 0 0→ +L so L L Le = = =µ τ 0

Strangeness, − → +3 0S so ∆S =1: S = −2

The particle must be a neutral baryon with strangeness of –2. Thus, it is the Ξ0 .

(c) K ? + +→ + +µ νµ

Baryon number, 0 0 0→ + +B so B = 0

Charge, + → + +1 1 0Q so Q = 0

Lepton numbers, L Le e, 0 0 0→ + + so Le = 0

L Lµ µ, 0 → − +1 1 so Lµ = 0

L Lτ τ, 0 → + +0 0 so Lτ = 0

Strangeness, 1 0 0→ + +S so ∆S = ±1

(for weak interaction): S = 0

The particle must be a neutral meson with strangeness = ⇒0 0π .

P46.27 Time-dilated lifetime:

T Tc

= = ×−

= ×−

− −

γ 0 2 2

100 900 10

1

0 900 10

1 0

. .

(

10 s s

v .. ).

9603 214 10

2

10= × − s

distance m s s= ×( ) ×( ) =−0 960 3 00 10 3 214 108 10. . . 99 26. cm



P46.28 (a) p eBrΣ Σ+ += =

×( ) ( ) (−1 602 177 10 1 15 1 9919. . . C T m))× ⋅( ) ( )

=−5 344 288 10

68622. kg m s MeV

MeV

c c

p eBrπ π+ += =

×( )( )−1 602 177 10 1 15 019. . C T .580 m(( )× ⋅( ) ( )

=−5 344 288 10

222. kg m s MeV

00 MeV

c c

(b) Let φ be the angle made by the neutron’s path with the path of the Σ+ at the moment of decay. By conservation of momentum:

p cn cos . cos . .φ + ( ) ° =199 961 581 64 5 686 075 081 MeV MeV c

∴ =p cn cos .φ 599 989 401 MeV (1)

p cn sin . sin . .φ = ( ) ° =199 961 581 64 5 180 482 380 MeV MeV c (2)

From (1) and (2): p c cn = ( ) + ( )=

599 989 401 180 482 380

62

2 2. . MeV MeV

77 MeV c

(c) E p c m cπ π π+ + += ( ) + ( ) = ( ) +

2 2 2 2199 961 581 139. . MeV 66 244

2 MeV MeV( ) =

E p c m cn n n= ( ) + ( ) = ( ) +2 2 2 2626 547 022 939 6. . MeV MeeV MeV( ) =2

1130

E E EnΣ+ += + = + =π

243 870 445 1129 340 219 1. . MeV MeV 3370 MeV

(d) m c E p cΣ Σ Σ+ + += − ( ) = ( ) −2 2 2 2

1 373 210 664 686 0. . MeV 775 081 11902

MeV MeV( ) =

∴ =+m cΣ

1190 2 MeV

E m cΣ Σ+ += γ 2 , where γ = −

⎛

⎝⎜

⎞

⎠⎟ =

−

11 373 210 6642

2

1 2vc

. MeV

1189.541 3033 MeV=1 154 4.

Solving for v, we fi nd v = 0 500. .c


604 Chapter 46

P46.29 (a) Let Emin be the minimum total energy of the bombarding particle that is needed to induce the reaction. At this energy the product particles all move with the same velocity. The prod-uct particles are then equivalent to a single particle having mass equal to the total mass of the product particles, moving with the same velocity as each product particle. By conserva-tion of energy:

E m c m c p cmin + = ( ) + ( )22

32 2

3

2 (1)

By conservation of momentum: p p3 1=

∴( ) = ( ) = − ( )p c p c E m c3

2

1

2 21

2 2

min (2)

Substitute (2) in (1): E m c m c E m cmin min+ = ( ) + − ( )22

32 2 2

12 2

Square both sides:

E E m c m c m c E m cmin min min2

22

22 2

32 2 2

122+ + ( ) = ( ) + − ( ))

∴ =− −( )

∴ = − =

2

32

12

22 2

2

12

2E

m m m c

m

K E m c

min

min min

mm m m m m c

m

m m m c32

12

22

1 22

2

32

1 2

22

2

− − −( ) =− +( )⎡

⎣⎤⎦

22

22m

Refer to Table 46.2 for the particle masses.

(b) Kc

min

. .=

( )⎡⎣ ⎤⎦ − ( )⎡⎣ ⎤⎦4 938 3 2 938 32 2 2 MeV MeV2 22

MeV GeV

c

c

2

22 938 35 63

..( ) =

(c) Kc

min

. . . .=

+( ) − +( )497 7 1115 6 139 6 938 32 2 2

MeV 2 MMeV

MeV MeV

2 c

c

2

22 938 3768

.( )=

(d) Kc

min

. .=

( ) +⎡⎣ ⎤⎦ − ( )⎡⎣ ⎤⎦2 938 3 135 2 938 32 2 2 MeV2 MeV

MeV MeV

2 c

c

2

22 938 3280

.( )=

(e) Kc

min

. . .=

×( ) − +( )⎡⎣

⎤⎦

⎡ 91 2 10 938 3 938 33 2 2 2 MeV2

⎣⎣⎢⎤⎦⎥

( )=

2 938 34 43

2..

MeV TeV

c



Section 46.7 Finding Patterns in the Particles

Section 46.8 Quarks

Section 46.9 Multicolored Quarks

Section 46.10 The Standard Model

P46.30 (a) The number of protons

Np = ×⎛

⎝⎜

⎞

⎠⎟1 000

10 g

6.02 10 molecules

18.0 g

23 pprotons

molecule protons

⎛⎝⎜

⎞⎠⎟ = ×3 34 1026.

and there are Nn = ×⎛

⎝⎜

⎞

⎠⎟1 000 g

6.02 10 molecules

18.0 g

8 n23 eeutrons

molecule neutrons

⎛⎝⎜

⎞⎠⎟ = ×2 68 1026.

So there are for electric neutrality 3 34 1026. × electrons

The up quarks have number 2 3 34 10 2 68 10 9 36 1026 26 26. . .×( )+ × = × up quarks

and there are 2 2 68 10 3 34 10 8 70 1026 26 26. . .×( )+ × = × down quarkss

(b) Model yourself as 65 kg of water. Then you contain:

65 3 34 10 1026 28. ~×( ) electrons

65 9 36 10 1026 29. ~×( ) up quarks

65 8 70 10 1026 29. ~×( ) down quarks

Only these fundamental particles form your body. You have no strangeness, charm, topness, or bottomness.

P46.31 (a) proton u u d total

strangeness 0 0 0 0 0baryon number 1 1/3 1/3 1/3 1charge e 2e/3 2e/3 –e/3 e

(b) neutron u d d total

strangeness 0 0 0 0 0baryon number 1 1/3 1/3 1/3 1charge 0 2e/3 –e/3 –e/3 0


606 Chapter 46

P46.32 Quark composition of proton = uud and of neutron = udd. Thus, if we neglect binding energies, we may write

m m mp u d= +2 (1)

and m m mn u d= + 2 (2)

Solving simultaneously,

we fi nd m m m c cu p n MeV meV= −( ) = ( ) −⎡⎣ ⎤1

32

1

32 938 939 62 2. ⎦⎦ = 312 2 MeV c

and from either (1) or (2), m cd MeV= 314 2 .

P46.33 (a)K0 d s total

strangeness 1 0 1 1baryon number 0 1/3 –1/3 0charge 0 –e/3 e/3 0

(b) Λ0 u d s total

strangeness –1 0 0 –1 –1baryon number 1 1/3 1/3 1/3 1charge 0 2e/3 –e/3 –e/3 0

P46.34 In the fi rst reaction, π − + → +p K0 Λ0 , the quarks in the particles are ud uud ds uds.+ → + There is a net of 1 up quark both before and after the reaction, a net of 2 down quarks both before and after, and a net of zero strange quarks both before and after. Thus, the reaction conserves the net number of each type of quark.

In the second reaction, π − + → +p K n,0 the quarks in the particles are ud uud ds udd.+ → +In this case, there is a net of 1 up and 2 down quarks before the reaction but a net of 1 up, 3 down, and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net number of each type of quark.

P46.35 (a) π − + → +p K0 0Λ In terms of constituent quarks: ud uud ds uds+ → +

up quarks: − + → +1 2 0 1, or 1 1→ down quarks: 1 1 1 1+ → + , or 2 2→ strange quarks: 0 0 1 1+ → − + , or 0 0→

(b) π+ + ++ → +p K Σ du uud us uus+ → +

up quarks: 1 2 1 2+ → + , or 3 3→ down quarks: − + → +1 1 0 0, or 0 0→ strange quarks: 0 0 1 1+ → − + , or 0 0→

(c) K p K K− + −+ → + +0 Ω us uud us ds sss+ → + + up quarks: − + → + +1 2 1 0 0, or 1 1→ down quarks: 0 1 0 1 0+ → + + , or 1 1→ strange quarks: 1 0 1 1 3+ → − − + , or 1 1→




(d) p p K p0+ → + + ++π ? uud uud ds uud ud+ → + + + ?

The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks.

up quarks: 2 2 0 2 1+ = + + +? (has 1 u quark)

down quarks: 1 1 1 1 1+ = + − +? (has 1 d quark)

strange quarks: 0 0 1 0 0+ = − + + +? (has 1 s quark)

quark composition = =uds or 0Λ Σ0

P46.36 Σ Σ0 + → + ++p γ X

dds uud uds 0 ?+ → + +

The left side has a net 3d, 2u, and 1s. The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing.

The unknown particle is a neutron, udd.

Baryon and strangeness numbers are conserved.

P46.37 Compare the given quark states to the entries in Tables 46.4 and 46.5:

(a) suu = +Σ

(b) ud = −π

(c) sd K0=

(d) ssd = −Ξ

P46.38 (a) uud: charge = −⎛⎝⎜

⎞⎠⎟ + −

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ = −2

3

2

3

1

3e e e e . This is the antiproton .

(b) udd: charge = −⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =2

3

1

3

1

30e e e . This is the antineutron .

Section 46.11 The Cosmic Connection

*P46.39 We let r in Hubble’s law represent any distance.

(a)

v = = ×

⋅ ⋅⎛

⎝⎜

⎞

⎠⎟

×−Hr

c

c17 10 1 85

1

1 3 13 m

s ly m

ly

yr.

00

1

3 156 10

17 10

8 7

3

m s

yr

s

m

s

⎛

⎝⎜

⎞

⎠⎟

×⎛⎝⎜

⎞⎠⎟

= × −

.

⋅⋅ ×= × =−

9 47 101 85 1 80 10

11 85 3 32

1518

.. . . .

m m

s m ×× −10 18 m s

This is unobservably small.

(b) v = = × × = ×− −Hr 1 80 101

3 84 10 6 90 1018 8 10. . .s

m m s again too small to measure


608 Chapter 46

P46.40 Section 39.4 says f fc

ca

a

observer source= +−

1

1

vv

The velocity of approach, va, is the negative of the velocity of mutual recession: v v.a = −

Then, c c c

c′= −

+λ λ1

1

vv

and ′ = +−

λ λ 1

1

vv

c

c

P46.41 (a) ′ = +−

λ λ 1

1

vv

c

c 510 434

1

1 nm nm= +

−vv

c

c

1 181

11 3812. .= +

−=v

vc

c

1 1 381 1 381+ = −v vc c

. . 2 38 0 381. .vc

=

vc

= 0 160. or v = = ×0 160 4 80 107. .c m s

(b) v = HR : RH

= = ×× ⋅

= ×−

v 4 80 10

17 102 82 10

7

39.

. m s

m s ly ly

P46.42 (a) ′ = +−

= +( )λ λ λn n n

c

cZ

1

11

vv

1

11

2+−

= +( )vv

c

cZ

1 1 12 2+ = +( ) −

⎛⎝⎜

⎞⎠⎟ +( )v v

cZ

cZ

vc

Z Z Z Z⎛⎝⎜

⎞⎠⎟ + +( ) = +2 22 2 2

v = ++ +

⎛⎝⎜

⎞⎠⎟

cZ Z

Z Z

2

2

2

2 2

(b) RH

c

H

Z Z

Z Z= = +

+ +⎛⎝⎜

⎞⎠⎟

v 2

2

2

2 2

P46.43 v = HR H =×( )−1 7 10 2. m s

ly

(a) v 2 00 10 3 4 106 4. .×( ) = × ly m s ′ = +−

= ( ) =λ λ 1

1590 1 000 113 3 590 07

vv

c

c. . nm

(b) v 2 00 10 3 4 108 6. .×( ) = × ly m s ′ = +−

=λ 5901 0 011 33

1 0 011 33597

.

. nm

(c) v 2 00 10 3 4 109 7. .×( ) = × ly m s ′ = +−

=λ 5901 0 113 3

1 0 113 3661

.

. nm



*P46.44 (a) What we can see is limited by the fi nite age of the Universe and by the fi nite speed of light. We can see out only to a look-back time equal to a bit less than the age of the Universe. Every year on your birthday the Universe also gets a year older, and light now in transit from still more distant objects arrives at Earth. So the radius of the visible Universe expands at the speed of light, which is dr/dt = c = 1 ly/yr.

(b) The volume of the visible section of the Universe is (4/3)pr3 where r = 13.7 billion light-years. The rate of volume increase is

dV

dt

d r

dtr

dr

dtr c= = = = ×

43

343

2 2 93 4 4 13 7 10π π π π . lly

3 s

1 ly3

ms m

s

× ×⎛

⎝⎜

⎞

⎠⎟ ×

10 3 156 1010

8 7 2

8.

= 6.34 × 1061 m3/s

*P46.45 (a) The volume of the sphere bounded by the Earth’s orbit is

V r

m V

= = ×( ) = ×

=

4

3

4

31 496 10 1 40 103 11 3 34π π

ρ

. . m m3

== × × = ×−6 10 1 40 10 8 41 1028 34 6 kg m m kg3 3. .

(b) By Gauss’s law, the dark matter would create a gravitational fi eld acting on the Earth to accelerate it toward the Sun. It would shorten the duration of the year in the same way that 8 41 106. × kg of extra material in the Sun would. This has the fractional effect of

8 41 10

1 99 104 23 10

6

3024.

..

××

= × − kg

kg of the mass of the Sun. It is immeasurably small .

P46.46 (a) Wien’s law: λmax .T = × ⋅−2 898 10 3 m K

Thus, λmax

. .= × ⋅ = × ⋅ =− −2 898 10 2 898 10

13 3 m K m K

2.73 KT.. .06 10 1 063× =− m mm

. (b) This is a microwave .

P46.47 We assume that the fi reball of the Big Bang is a black body.

I e T= = × ⋅( )( ) = ×−σ 4 8 41 5 67 10 2 73 3 15( ) . . . W m K K2 4 110 6− W m2

As a bonus, we can fi nd the current power of direct radiation from the Big Bang in the portion of the Universe observable to us. If it is fourteen billion years old, the fi reball is a perfect sphere of radius fourteen billion light years, centered at the point halfway between your eyes:

P = = = ×( )( ) ×( )−IA I r( ) .4 3 15 10 4 14 102 6 9π π W m ly2 228 2

7 23 103 156 10

×⎛

⎝⎜

⎞

⎠⎟ ×( ) m s

1 ly yr s yr.

P = ×7 1047 W


610 Chapter 46

P46.48 The density of the Universe is

ρ ρπ

= =⎛

⎝⎜

⎞

⎠⎟1 20 1 20

3

8

2

. .c

H

G

Consider a remote galaxy at distance r. The mass interior to the sphere below it is

M rH

Gr= ⎛

⎝⎞⎠ =

⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠ =ρ π

ππ4

31 20

3

8

4

3

0 6032

3.. 00 2 3H r

G

both now and in the future when it has slowed to rest from its current speed v = Hr. The energy of this galaxy-sphere system is constant as the galaxy moves to apogee distance R:

1

202m

GmM

r

GmM

Rv − = − so 1

2

0 6000

0 6002 22 3 2 3

mH rGm

r

H r

G

Gm

R

H r

G−

⎛⎝⎜

⎞⎠⎟

= −. .⎛⎛⎝⎜

⎞⎠⎟

− = −0 100 0 600. .r

R so R r= 6 00.

The Universe will expand by a factor of 6 00. from its current dimensions.

P46.49 (a) k T m cpB ≈ 2 2

so Tm c

kp≈ = ( )

×( )×

−

2 2 938 3

1 38 10

1 60 12

23B

MeV

J K

.

.

. 0010

1313

−⎛

⎝⎜

⎞

⎠⎟

J

1 MeV K~

(b) k T m ceB ≈ 2 2

so Tm c

ke≈ = ( )

×( )×

−

2 2 0 511

1 38 10

1 60 12

23B

MeV

J K

.

.

. 0010

1310

−⎛

⎝⎜

⎞

⎠⎟

J

1 MeV K~

P46.50 (a) The Hubble constant is defi ned in v = HR. The distance R between any two far-separated objects opens at constant speed according to R t= v . Then the time t since the Big Bang is found from

v v= H t 1 = Ht tH

= 1

(b) 1 1

17 10

3 101 76

3

8

H=

× ⋅×⎛

⎝⎜

⎞

⎠⎟ = ×− m s ly

m s

1 ly yr. 110 17 610 yr billion years= .



P46.51 (a) Consider a sphere around us of radius R large compared to the size of galaxy clusters. If the matter M inside the sphere has the critical density, then a galaxy of mass m at the surface of the sphere is moving just at escape speed v according to

K Ug+ = 0 1

202m

GMm

Rv − =

The energy of the galaxy-sphere system is conserved, so this equation is true throughout the

history of the Universe after the Big Bang, where v = dR

dt. Then

dR

d t

GM

R

⎛

⎝⎜

⎞

⎠⎟ =

22

dR

d tR GM= −1 2 2/ R dR GM d t

R T

0 02∫ ∫=

R

GM tR

T3 2

003 2

2/

= 2

323 2R GM T/ =

TR

GM

R

GM R= =2

3 2

2

3 2

3 2

From above, 2GM

R= v

so TR= 2

3 v

Now Hubble’s law says v = HR

So TR

HR H= =2

3

2

3

(b) T =× ⋅( )

×⎛

⎝⎜

⎞

⎠⎟ =−

2

3 17 10

3 10

11 1

3

8

m s ly

m s

ly yr. 88 10 11 810× = yr billion years.

P46.52 In our frame of reference, Hubble’s law is exemplifi ed by v R1 1= H and

v R2 2= H . From

these we may form the equations − = − v R1 1H and

v v R R2 1 2 1− = −( )H . These equations

express Hubble’s law as seen by the observer in the fi rst galaxy cluster, as she looks at us to

fi nd − = −( ) v R1 1H and as she looks at cluster two to fi nd

v v R R2 1 2 1− = −( )H .

Section 46.12 Problems and Perspectives

*P46.53 (a) LG

c= =

× ⋅( ) × ⋅(− −

3

34 111 055 10 6 67 10. . J s N m kg2 2 ))×( )

= × −

3.00 10 m s m

8 3351 61 10.

(b) The Planck time is given as TL

c= = ×

×= ×

−−1 61 10

5 38 1035

44.. ,

m

3.00 10 m s s

8

which is approximately equal to the duration of the ultra-hot epoch.

(c) Yes. The uncertainty principle foils any attempt at making observations of things when the age of the Universe was less than the Planck time. The opaque fi reball of the Big Bang, measured as the cosmic microwave background radiation, prevents us from receiving visible light from things before the Universe was a few hundred thousand years old. Walls of more profound fi re hide all information from still earlier times.


612 Chapter 46

Additional Problems

P46.54 We fi nd the number N of neutrinos:

10 6 6 1 60 1046 13 J MeV J= ( ) = × ×( )−N N .

N = ×1 0 1058. neutrinos

The intensity at our location is

N

A

N

r= = ×

×( ) ×4

1 0 10

4 1 7 10

1

3 00 102

58

5 2π.

. .π ly

ly88 7

2

14 2

3 16 103 1 10

m s s m

( ) ×( )⎛

⎝⎜⎜

⎞

⎠⎟⎟ = × −

..

The number passing through a body presenting 5 000 0 50 cm m2 2= .

is then 3 1 10 0 50 1 5 1014 14. . .×⎛⎝⎜

⎞⎠⎟ ( ) = ×

1

m m

22

or ~ 1014

P46.55 A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years. The hypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds:

c 170 000 10 yr 170 000 yr s( ) = +( )v

vc

=+

=+ ×

170 000

170 000 10

1

1 10

yr

yr s s 1.7 10 y5 rr 3.156 10 s yr7( ) ×( )⎡⎣ ⎤⎦ =

+ × −

1

1 1 86 10 12.

For the neutrino we want to evaluate mc2 in E mc= γ 2:

mcE

Ec

22

2 12 21 10 1

1

1 1 86 1010= = − = −

+ ×( )=

−γv

MeV .

MMeV1 1 86 10 1

1 1 86 10

12 2

12 2

+ ×( ) −

+ ×( )−

−

.

.

mc2

12

6101 86 10

110 1 93 10≈

×( ) = ×( ) =−

− MeV2

MeV.

. 119 eV

Then the upper limit on the mass is

mc

= 192

eV or m

c c=

×⎛

⎝⎜

⎞

⎠⎟ = × −19

931 5 102 1 10

2 6 28 eV u

eV u

..

P46.56 (a) π π− + → Σ +p + 0 is forbidden by charge conservation

(b) µ π ν− −→ + e is forbidden by energy conservation

(c) p → + ++ + −π π π is forbidden by baryon number conservation

P46.57 The total energy in neutrinos emitted per second by the Sun is:

0 4 4 1 5 10 1 1 1011 2 23. . .( ) ×( )⎡⎣

⎤⎦ = ×π W W

Over 109 years, the Sun emits 3 6 1039. × J in neutrinos. This represents an annihilated mass

m c2 393 6 10= ×. J

m = ×4 0 1022. kg

About 1 part in 50 000 000 of the Sun’s mass, over 109 years, has been lost to neutrinos.



P46.58 p p p+ → + ++π X

The protons each have 70.4 MeV of kinetic energy. In accord with conservation of momentum for the collision, particle X has zero momentum and thus zero kinetic energy. Conservation of system energy then requires

M c M c M c M c K M c KXp p p p p2 2 2 2 2+ + = +( ) + +( )π

M c M c K M cX2 2 22 938 3 2 70 4 13= + − = + ( ) −p p MeV MeVπ . . 99 6 939 5. .MeV MeV=

X must be a neutral baryon of rest energy 939.5 MeV. Thus X is a neutron .

P46.59 (a) If 2N particles are annihilated, the energy released is 2 2Nmc . The resulting photon

momentum is pE

c

Nmc

cNmc= = =2

22

. Since the momentum of the system is conserved,

the rocket will have momentum 2Nmc directed opposite the photon momentum.

p Nmc= 2

(b) Consider a particle that is annihilated and gives up its rest energy mc2 to another particle which also has initial rest energy mc2 (but no momentum initially).

E p c mc2 2 2 2 2= + ( )

Thus 2 2 2 2 2 2 2mc p c mc( ) = + ( )

where p is the momentum the second particle acquires as a result of the annihilation of the

fi rst particle. Thus 4 2 2 2 2 2 2mc p c mc( ) = + ( ) , p mc2 2 2

3= ( ) . So p mc= 3 .

This process is repeated N times (annihilate N

2 protons and

N

2 antiprotons). Thus the total

momentum acquired by the ejected particles is 3Nmc, and this momentum is imparted to the rocket.

p Nmc= 3

(c) Method (a) produces greater speed since 2 3Nmc Nmc> .

P46.60 By relativistic energy conservation in the reaction, E m cm c

ce

eγ + =

−2

2

2 2

3

1 v (1)

By relativistic momentum conservation for the system, E

c

m

ceγ =

−3

1 2 2

v

v (2)

Dividing (2) by (1), XE

E m c ce

=+

=γ

γ2

v

Subtracting (2) from (1), m cm c

X

m c X

Xe

e e22

2

2

2

3

1

3

1=

−−

−

Solving, 13 3

1 2= −

−X

X and X = 4

5 so E m ceγ = =4 2 042 . MeV


614 Chapter 46

P46.61 m cΛ2 1115 6= . MeV Λ0 → + −p π

m cp2 938 3= . MeV m cπ

2 139 6= . MeV

The difference between starting rest energy and fi nal rest energy is the kinetic energy of the products.

K Kp + =π 37 7. MeV and p p pp = =π

Applying conservation of relativistic energy to the decay process, we have

938 3 938 3 139 6 1392 2 2 2 2 2. . .( ) + −⎡

⎣⎢⎤⎦⎥+ ( ) + −p c p c .. .6 37 7

⎡⎣⎢

⎤⎦⎥ = MeV

Solving the algebra yields

p c p cpπ = =100 4. MeV

Then, K m c m cp p p= ( ) + ( ) − =2 2 2 2100 4 5 35. . MeV

Kπ = ( ) + ( ) − =139 6 100 4 139 6 32 32 2. . . . MeV

P46.62 p p p n+ → + + +π

The total momentum is zero before the reaction. Thus, all three particles present after the reaction may be at rest and still conserve system momentum. This will be the case when the incident protons have minimum kinetic energy. Under these conditions, conservation of energy for the reaction gives

2 2 2 2 2m c K m c m c m cp p p n+( ) = + + π

so the kinetic energy of each of the incident protons is

Km c m c m c

pn p=

+ −=

+ −( )2 2 2

2

939 6 139 6 938 3π . . . MeV

2== 70 4. MeV

P46.63 Σ → +0 0Λ γ

From Table 46.2, m cΣ = 1192 5 2. MeV and m cΛ =1115 6 2. MeV

Conservation of energy in the decay requires

E E K Eo0, ,Σ = +( ) +Λ Λ γ or 1192 5 1115 62

2

. . MeV MeV= +⎛

⎝⎜

⎞

⎠⎟+p

mEΛ

Λγ

System momentum conservation gives p pΛ = γ , so the last result may be written as

1192 5 1115 62

2

. . MeV MeV= +⎛

⎝⎜⎜

⎞

⎠⎟⎟+

p

mEγ

γΛ

or 1192 5 1115 62

2 2

2. . MeV MeV= +

⎛

⎝⎜⎜

⎞

⎠⎟⎟+

p c

m cEγ

γΛ

Recognizing that m cΛ2 1115 6= . MeV and p c Eγ γ=

we now have 1192 5 1115 62 1115 6

2

. ..

MeV MeV MeV

= +( )

+E

Eγγ

Solving this quadratic equation gives Eγ = 74 4. MeV



P46.64 The momentum of the proton is qBr = ×( ) ⋅( )( )−1 60 10 0 250 1 3319. . . C kg C s m

pp = × ⋅−5 32 10 20. kg m s cpp = × ⋅ = × =− −1 60 10 1 60 10 99 811 11. . . kg m s J Me2 2 VV

Therefore p cp = 99 8. MeV

The total energy of the proton is E E cpp = + ( ) = ( ) + ( ) =02 2 2 2

938 3 99 8 944. . MeV

For the pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2-particle decay).

p cπ = 99 8. MeV E0 139 6π = . MeV

E E cpπ = + ( ) = ( ) + ( ) =02 2 2 2139 6 99 8 172. . MeV

Thus E Etotal after total before Rest energy= =

Rest energy of unknown particle = + =944 172 1116MeV MeV MeV (This is a Λ0 particle!)

Mass MeV= 1116 2c .

P46.65 π µ νµ− −→ + : From the conservation laws for the decay,

m c E Eπ µ ν2 139 6= = +. MeV [1]

and p pµ ν= , E p cν ν= : E p c p cµ µ ν2 2 2 2 2105 7 105 7= ( ) + ( ) = ( ) + ( ). .MeV MeV

or E Eµ ν2 2 2

105 7− = ( ). MeV [2]

Since E Eµ ν+ = 139 6. MeV [1]

and E E E Eµ ν µ ν+( ) −( ) = ( )105 7 2. MeV [2]

then E Eµ ν− = ( ) =105 7

139 680 0

2.

..

MeV

MeV [3]

Subtracting [3] from [1], 2 59 6Eν = . MeV and Eν = 29 8. MeV

P46.66 The expression e dEE k T− B gives the fraction of the photons that have energy between E andE dE+ . The fraction that have energy between E and infi nity is

e dE

e dE

e dE k T

e

E k T

E

E k T

E k T

E

E

−∞

−∞

−∞

−

∫

∫

∫=

−( )B

B

BB

0

kk T

E k T

E

E k T

E k T

dE k T

e

ee

B

B

B

B

B−( )= =∞

− ∞

− ∞−

∫0

0

We require T when this fraction has a value of 0.0100 (i.e., 1.00%)

and E = = × −1 00 1 60 10 19. . eV J

Thus, 0 010 01 60 10 1 38 1019 23

.. .= − ×( ) ×( )− −

eTJ J K

or ln ..

0 010 01 60 10 119

23( ) = − ××( ) = −

−

−

J

1.38 10 J K T

..16 104× K

T giving T = ×2 52 103. K


616 Chapter 46

P46.67 (a) This diagram represents the annihilation of an electron and an antielectron. From charge and lepton-number conservation at either vertex, the

exchanged particle must be an electron, e− .

(b) This is the tough one. A neutrino collides with a neutron, changing it into a proton with release of a muon. This is a weak interaction. The exchanged particle has charge +e and is a W+ .

P46.68 (a) The mediator of this weak interaction is a Z boson0 .

(b) The Feynman diagram shows a down quark and its antiparticle annihilating each other. They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge. In this case the product particle

is a photon .

For conservation of both energy and momentum in the collision we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a suffi ciently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.68.

Depending on the color charges of the d and d quarks, the ephemeral particle could also be

a gluon , as suggested in the discussion of Figure 46.13(b).

P46.69 (a) At threshold, we consider a photon and a proton colliding head-on to produce a proton and

a pion at rest, according to p p+ → +γ π 0 . Energy conservation gives

m c

u cE m c m cp

p

2

2 2

2 2

1−+ = +γ π

Momentum conservation gives m u

u c

E

cp

10

2 2−− =γ .

Combining the equations, we have

m c

u c

m c

u c

u

cm c m cp p

p

MeV

2

2 2

2

2 2

2 2

1 1

938 3 1

−+

−= + π

. ++( )−( ) +( )

= +u c

u c u c1 1938 3 135 0. . MeV MeV

so u

c= 0 134.

and Eγ = 127 MeV

(b) λmax .T = ⋅2 898 mm K

λmax

..= ⋅ =2 898

1 06mm K

2.73 Kmm


FIG. P46.67

FIG. P46.68



(c) E hfhc

γ λ= = = ⋅

×= ×

−

−

1 2401 17 10

9 eV 10 m

1.06 10 m3. −−3 eV

(d) In the primed reference frame, the proton is moving to the right at ′ =u

c0 134. and

the photon is moving to the left with hf ′ = ×1 27 108. eV. In the unprimed frame,hf = × −1 17 10 3. eV. Using the Doppler effect equation from Section 39.4, we have for the speed of the primed frame

1 27 10

1

11 17 10

1 1 71 10

8 3

22

. .

.

× = +−

×

= − ×

−

−

vv

v

c

c

c Then the speed of the proton is given by

u

c

u c c

u c= ′ +

+ ′= + − ×

+

−vv1

0 134 1 1 71 10

1 0 134 12

22. .

. −− ×( ) = − ×−−

1 71 101 1 30 10

22

22

..

And the energy of the proton is

m c

u c

p2

2 2 22 21

938 3

1 1 1 30 106 19

−=

− − ×( )= ×

−

.

..

MeV110 938 3 10 5 81 1010 6 19× × = ×. . eV eV

ANSWERS TO EVEN PROBLEMS

P46.2 ~103 Bq

P46.4 2 27 1023. × Hz; 1 32. fm

P46.6 νµ and νe

P46.8 (b) The range is inversely proportional to the mass of the fi eld particle. (c) Our rule describes the electromagnetic, weak, and gravitational interactions. For the electromagnetic and gravita-tional interactions, we take the limiting form of the rule with infi nite range and zero mass for the fi eld particle. For the weak interaction, 98.7 eV ⋅ nm/90 GeV ≈ 10–18 m = 10–3 fm, in agreement with the tabulated information. (d) ~10–16 m

P46.10 ~10 23− s

P46.12 νµ

P46.14 (b) The second violates strangeness conservation.

P46.16 The second violates conservation of baryon number.

P46.18 0 828. c

P46.20 (a) νe (b) νµ (c) νµ (d) ν νµ τ+

P46.22 See the solution.

P46.24 (a) not allowed; violates conservation of baryon number (b) strong interaction (c) weak interaction (d) weak interaction (e) electromagnetic interaction

P46.26 (a) K+ (b) Ξ0 (c) p 0


618 Chapter 46

P46.28 (a) 686 MeV

c and 200 MeV

c (b) 627 MeV c (c) 244 MeV, 1130 MeV, 1 370 MeV

(d) 1190 2 MeV c , 0 500. c

P46.30 (a) 3.34 × 1026 e−, 9.36 × 1026 u, 8.70 × 1026 d (b) ~1028 e−, ~1029 u, ~1029 d. I have zero strangeness, charm, topness, and bottomness.

P46.32 mu = 312 MeV/c2 m

d = 314 MeV/c2


P46.36 a neutron, udd

P46.38 (a) –e, antiproton b) 0, antineutron


P46.42 (a) v = ++ +

⎛⎝⎜

⎞⎠⎟

cZ Z

Z Z

2

2

2

2 2 (b)

c

H

Z Z

Z Z

2

2

2

2 2

++ +

⎛

⎝⎜

⎞

⎠⎟

P46.44 (a) What we can see is limited by the fi nite age of the Universe and by the fi nite speed of light. We can see out only to a look-back time equal to a bit less than the age of the Universe. Every year on your birthday the Universe also gets a year older, and light now in transit from still more distant objects arrives at Earth. So the radius of the visible Universe expands at the speed of light, which is 1 ly/yr. (b) 6.34 × 1061 m3/s

P46.46 (a) 1 06. mm (b) microwave

P46.48 6 00.

P46.50 (a) See the solution. (b) 17 6. Gyr


P46.54 ~1014

P46.56 (a) charge (b) energy (c) baryon number

P46.58 neutron

P46.60 2 04. MeV

P46.62 70.4 MeV

P46.64 1 116 MeV/c2

P46.66 2.52 × 103 K

P46.68 (a) Z0 boson (b) gluon or photon


sm chapter46

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