Particle Physics and Cosmology - UCCS Home - · PDF file620 Particle Physics and Cosmology...

46

CHAPTER OUTLINE

46.1 The Fundamental Forces in Nature46.2 Positrons and Other Antiparticles46.3 Mesons and the Beginning of Particle Physics46.4 Classification of Particles46.5 Conservation Laws46.6 Strange Particles and Strangeness46.7 Making Elementary Particles and Measuring Their

46.8 Finding Patterns in the

Properties

Particles 46.9 Quarks46.10 Multicolored Quarks

46.12 The Cosmic Connection46.11 The Standard Model

Problems and Perspectives46.13

Particle Physics and Cosmology

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 The production of a single gamma ray could not satisfy the lawof conservation of momentum, which must hold true inthis—and every—interaction.

Q46.3 In the quark model, all hadrons are composed of smaller unitscalled quarks. Quarks have a fractional electric charge and a

baryon number of 13

. There are 6 types of quarks: up, down,

strange, charmed, top, and bottom. Further, all baryons contain3 quarks, and all mesons contain one quark and one anti-quark.Leptons are thought to be fundamental particles.

Q46.4 Hadrons are massive particles with structure and size. There are two classes of hadron: mesons andbaryons. 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 fundamentalparticles. Leptons interact via the weak force.

Q46.5 Baryons are heavy hadrons with spin 12

or 32

, are composed of three quarks, and have long

lifetimes. Mesons are light hadrons with spin 0 or 1, are composed of a quark and an antiquark, andhave short lifetimes.

Q46.6 Resonances are hadrons. They decay into strongly interacting particles such as protons, neutrons,and pions, all of which are hadrons.

Q46.7 The baryon number of a proton or neutron is one. Since baryon number is conserved, the baryonnumber 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 particlephysics.

619

620 Particle Physics and Cosmology

Q46.9 The decays of the muon, tau, charged pion, kaons, neutron, lambda, charged sigmas, xis, and omegaoccur by the weak interaction. All have lifetimes longer than 10 13− s. Several produce neutrinos;none produce photons. Several violate strangeness conservation.

Q46.10 The decays of the neutral pion, eta, and neutral sigma occur by the electromagnetic interaction.These are three of the shortest lifetimes in Table 46.2. All produce photons, which are the quanta ofthe electromagnetic force. All conserve strangeness.

Q46.11 Yes, protons interact via the weak interaction; but the strong interaction predominates.

Q46.12 You can think of a conservation law as a superficial regularity which we happen to notice, as aperson who does not know the rules of chess might observe that one player’s two bishops arealways 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 bothmatter and energy as fundamental constituents of the world. We buy and sell both of them. Inclassical physics you can also think of linear momentum, angular momentum, and electric charge asbasic stuffs of which the universe is made. In relativity we learn that matter and energy are notconserved separately, but are both aspects of the conserved quantity relativistic total energy.Discovered more recently, four conservation laws appear equally general and thus equallyfundamental: Conservation of baryon number, conservation of electron-lepton number,conservation of tau-lepton number, and conservation of muon-lepton number. Processes involvingthe 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 quantumnumbers of an isolated system.

Q46.13 No. Antibaryons have baryon number –1, mesons have baryon number 0, and baryons have baryonnumber +1. The reaction cannot occur because it would not conserve baryon number, unless somuch energy is available that a baryon-antibaryon pair is produced.

Q46.14 The Standard Model consists of quantum chromodynamics (to describe the strong interaction) andthe electroweak theory (to describe the electromagnetic and weak interactions). The Standard Modelis our most comprehensive description of nature. It fails to unify the two theories it includes, andfails 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.

Q46.15 All baryons and antibaryons consist of three quarks. All mesons and antimesons consist of two

quarks. Since quarks have spin quantum number 12

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 spin0 or 1.

Q46.16 Each flavor of quark can have colors, designated as red, green and blue. Antiquarks are coloredantired, antigreen, and antiblue. A baryon consists of three quarks, each having a different color. Byanalogy to additive color mixing we call it colorless. A meson consists of a quark of one color andantiquark with the corresponding anticolor, making it colorless as a whole.

Q46.17 In 1961 Gell-Mann predicted the omega-minus particle, with quark composition sss. Its discovery in1964 confirmed the quark theory.

Chapter 46 621

Q46.18 The Ξ− particle has, from Table 46.2, charge –e, spin 12

, B = 1, L L Le = = =µ τ 0, and strangeness –2.

All of these are described by its quark composition dss (Table 46.5). The properties of the quarks

from Table 46.3 let us add up charge: − − − = −13

13

13

e e e e ; spin + − + =12

12

12

12

, supposing one of the

quarks is spin-down relative to the other two; baryon number 13

13

13

1+ + = ; lepton numbers, charm,

bottomness, and topness zero; and strangeness 0 1 1 2− − = − .

Q46.19 The electroweak theory of Glashow, Salam, and Weinberg predicted the W+ , W− , and Z particles.Their discovery in 1983 confirmed the electroweak theory.

Q46.20 Hubble determined experimentally that all galaxies outside the Local Group are moving away fromus, with speed directly proportional to the distance of the galaxy from us.

Q46.21 Before that time, the Universe was too hot for the electrons to remain in any sort of stable orbitaround protons. The thermal motion of both protons and electrons was too rapid for them to be inclose enough proximity for the Coulomb force to dominate.

Q46.22 The Universe is vast and could on its own terms get along very well without us. But as the cosmos isimmense, 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. One person hassingular 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 allthe 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 Eof the photon must be

E E= = = = × −2 2 938 3 1 876 6 3 00 10010. . . MeV MeV Ja f .

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

f3 00 104 53 10

6 62 108

2316.

..

m s Hz

m .


P46.2 The minimum energy is released, and hence the minimum frequency photons are produced, whenthe 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 theenergy.

Thus, EE

E hfmin .= = = =2

2938 30

0 MeV min .

Thus, fmin

. .

..=

×

× ⋅= ×

−

−

938 3 1 60 10

6 626 102 27 10

13

3423

MeV J MeV

J s Hz

a fe je j

λ = =×

×= × −c

fmin

.

..

3 00 102 27 10

1 32 108

2315 m s

Hz m .

P46.3 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 K 2 118= MeV .

Section 46.3 Mesons and the Beginning of Particle Physics

P46.4 The reaction is µ ν ν+ −+ → +e

muon-lepton number before reaction: − + = −1 0 1a f a felectron-lepton number before reaction: 0 1 1a f a f+ = .

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

ν µ and ν e .

Then µ ν νµ+ −+ → +e e .

P46.5 The creation of a virtual Z0 boson is an energy fluctuation ∆E = ×93 109 eV . It can last no longer

than ∆∆

tE

=2

and move no farther than

c thc

E∆

∆a f e je j

e j= =

× ⋅ ×

× ×

FHG

IKJ = × =

−

−− −

4

6 626 10 3 00 10

4 93 10

11 06 10 10

34 8

9 1918 18

π π

. .. ~

J s m s

eV

eV1.60 10 J

m m .

Chapter 46 623

P46.6 A proton has rest energy 938.3 MeV. The time interval during which a virtual proton could exist is at

most ∆t in ∆ ∆E t =2

. The distance it could move is at most

c tcE

∆∆

= =× ⋅ ×

×

−

−−

2

1 055 10 3 10

2 938 3 1 6 1010

34 8

1316

.

. .~

J s m s

J m

e je ja fe j

.

According to Yukawa’s line of reasoning, this distance is the range of a force that could be associatedwith the exchange of virtual protons between high-energy particles.

P46.7 By Table 46.2, 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 .

P46.8 The time interval for a particle traveling with the speed of light to travel a distance of 3 10 15× − m is

∆ = =××

=−

−tdv

3 103 10

1015

823 m

m s s~ .

P46.9 (a) ∆ = − −E m m m cn p ee j 2

From Table A-3, ∆ = − =E 1 008 665 1 007 825 931 5 0 782. . . .b ga f 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 2e j e j+ + + = .

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

Solving the algebra, pc = 1 19. MeV .

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

cx

xe = =

−=

1 190 511 1

2 332

..

. MeV MeV

where xvce= .

Solving, x x2 21 5 43= −e j . and xvce= = 0 919.

v ce = 0 919. .

Then m v m vp p e e e= γ : vm v cm cp

e e e

p= =

×

× ×

−

−

γ 1 19 1 60 10

1 67 10 3 00 10

13

27 8

. .

. .

MeV J MeV

m s

a fe je je j

vp = × =3 80 10 3805. m s km s .

(c) The electron is relativistic, the proton is not.


Section 46.4 Classification of Particles

P46.10 In ? + → ++ +p n µ , charge conservation requires the unknown particle to be neutral. Baryonnumber conservation requires baryon number = 0. The muon-lepton number of ? must be –1.

So the unknown particle must be ν µ .

P46.11 Ω Λ+ +→ +0 K KS0 → ++ −π π (or π π0 0+ )

Λ0 → + +p π n p e e→ + ++ ν

Section 46.5 Conservation Laws

P46.12 (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+ → +

P46.13 (a) Baryon number and charge are conserved, with values of 0 1 0 1+ = +

and 1 1 1 1+ = + in both reactions.

(b) Strangeness is conservednot in the second reaction.

P46.14 Baryon number conservation allows the first and forbids the second .

P46.15 (a) π µ ν µ− −→ + Lµ : 0 1 1→ −

(b) K + +→ +µ ν µ Lµ : 0 1 1→− +

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

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

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

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

Chapter 46 625

P46.16 Momentum conservation for the decay requires the pions to have equal speeds.

The total energy of each is497 7. MeV

2

so E p c mc2 = +2 2 2 2e j gives

248 8 139 62 2 2. . MeV MeVa f b g a f= +pc .

Solving, pc mvcmc

v c

vc

= = =−

FHGIKJ206

1

2

2 MeV γ

b gpc

mc v c

vc2 2

206 1

11 48= =

−

FHGIKJ =

MeV139.6 MeV b g

.

vc

vc

= − FHGIKJ1 48 1

2

.

andvc

vc

vc

FHGIKJ = − FHG

IKJ

LNMM

OQPP = − F

HGIKJ

2 2 2

2 18 1 2 18 2 18. . .

3 18 2 182

. .vcFHGIKJ =

sovc= =

2 183 18

0 828..

.

and v c= 0 828. .

P46.17 (a) p+ +→ +π π 0 Baryon number : 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 ν µ+ +→ + 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→− +


P46.18 (a) p e→ ++ γ

Baryon number: + → +1 0 0

∆B ≠ 0 , so baryon number conservation is violated.

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

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

02= +b g ,

becomes E p c E E Ee e e2 2

02 2

02= + = +γ γd i , , .

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

pγ =

−=

−=0

202

0

2 2

2938 3 0 511

2 938 3469, ,

,

. ..

MeV MeV MeV

MeVa f a f

a f .

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.

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

But, E EE

v ce e

e= =

−γ 0

0

21

,,

b g

so 10 511

1 09 102

0 3− FHGIKJ = = = × −v

c

E

Ee

e

, ..

MeV469 MeV

which yields: v c= 0 999 999 4. .

P46.19 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 ν e

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

continued on next page

Chapter 46 627

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

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

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 ν µ

and ντ .

Section 46.6 Strange Particles and Strangeness

P46.20 The ρ π π0 → ++ − decay must occur via the strong interaction.

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

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

(b) π − + → +p K 0Λ0 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 (a) µ γ− −→ +e Le : 0 1 0→ + ,

and Lµ : 1 0→

(b) n p e→ + +− ν e 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


P46.23 (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→


Charge, − → − +1 1 0

Strangeness is violated by one unit, but everything else is conserved. Thus, the reaction canoccur via the weak interaction , but not the strong or electromagnetic interaction.

(d) Ω Ξ− −→ +π 0

Baryon number, 1 1 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→


Charge, 0 0→

Strangeness, 0 0→

No conservation laws are violated, but photons are the mediators of the electromagneticinteraction. Also, the lifetime of the η is consistent with the electromagnetic interaction .

Chapter 46 629

P46.24 (a) Ξ Λ− −→ + +0 µ ν µ

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+ → + −



P46.25 (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 ν µ

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 .

Chapter 46 631

Section 46.7 Making Elementary Particles and Measuring Their Properties

*P46.26 (a) p eBrc cΣ Σ+ += =

×

× ⋅=

−

−

1 602 177 10 1 15 1 99

5 344 288 10686

19

22

. . .

.

C T m

kg m s MeV MeVe ja fa f

b g b g

p eBrc cπ π+ += =

×

× ⋅=

−

−

1 602 177 10 1 15 0

5 344 288 10200 MeV

19

22

. .

.

C T .580 m

kg m s MeV

e ja fa fb g b g

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

p c cn cos . cos . .ϕ + °=199 961 581 64 5 686 075 081 MeV MeVb g∴ =p cn cos .ϕ 599 989 401 MeV (1)

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

From (1) and (2): p c c cn = + =599 989 401 180 482 380 6272 2

. . MeV MeV MeVb g b g

(c) E p c m cπ π π+ + += + = + =e j e j b g a f2 2 2 2 2199 961 581 139 6 244. . MeV MeV MeV

E p c m cn n n= + = + =b g e j b g a f2 2 2 2 2626 547 022 939 6 1 130. . MeV MeV MeV

E E EnΣ+ += + = + =π 243 870 445 1 129 340 219 1 370. . MeV MeV MeV

(d) m c E p cΣ Σ Σ+ + += − = − =2 2 2 2 21 373 210 664 686 075 081 1 190d i b g b g. . MeV MeV MeV

∴ =+m cΣ 1 190 2 MeV

E m cΣ Σ+ += γ 2 , where γ = −FHGIKJ = =−

11 373 210 664

1 154 42

2

1 2vc

..

MeV1 189.541 303 MeV

Solving for v, v c= 0 500. .

P46.27 Time-dilated lifetime:

T Tv c

= =×

−=

×

−= ×

− −−γ 0 2 2

10

2

100 900 10

1

0 900 10

1 0 9603 214 10

. .

( . ).

10 s s s

distance m s s cm= × × =−0 960 3 00 10 3 214 10 9 268 10. . . .e je j .


*P46.28 (a) Let Emin be the minimum total energy of the bombarding particle that is needed to inducethe reaction. At this energy the product particles all move with the same velocity. Theproduct particles are then equivalent to a single particle having mass equal to the total massof the product particles, moving with the same velocity as each product particle. Byconservation of energy:

E m c m c p cmin + = +22

32 2

32e j b g . (1)

By conservation of momentum: p p3 1=

∴ = = −p c p c E m c32

12 2

12 2b g b g e jmin . (2)

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

32 2 2

12 2e j e j .

Square both sides:

E E m c m c m c E m c

Em m m c

m

K E m cm m m m m c

m

m m m c

m

min min min

min

min min

22

22

2 23

2 2 21

2 2

32

12

22 2

2

12 3

212

22

1 22

2

32

1 22 2

2

2

2

2

2 2

+ + = + −

∴ =− −

∴ = − =− − −

=− +

e j e j e je j

e j b g

Refer to Table 46.2 for the particle masses.

(b) Kc c

cmin

. .

..=

−=

4 938 3 2 938 3

2 938 35 63

2 2 2 2

2

a f a fe j

MeV MeV

MeV GeV

2 2

(c) Kc c

cmin. . . .

.=

+ − +=

497 7 1 115 6 139 6 938 3

2 938 3768

2 2 2 2

2

b g a fa f

MeV MeV

MeV MeV

2 2

(d) Kc c

cmin. .

.=

+ −=

2 938 3 135 2 938 3

2 938 3280

2 2 2 2

2

a f a fa f

MeV MeV

MeV MeV

2 2

(e) Kc

cmin

. . .

..=

× − +LNM

OQP =

91 2 10 938 3 938 3

2 938 34 43

3 2 2 2

2

e j a fa f

MeV

MeV TeV

2

Chapter 46 633

Section 46.8 Finding Patterns in the Particles

Section 46.9 Quarks

Section 46.10 Multicolored Quarks

Section 46.11 The Standard Model

P46.29 (a) The number of protons

Np =×F

HGIKJFHG

IKJ = ×1 000

103 34 1026 g

6.02 10 molecules18.0 g

protonsmolecule

protons23

.

and there are Nn =×F

HGIKJFHG

IKJ = ×1 000 2 68 1026 g

6.02 10 molecules18.0 g

8 neutronsmolecule

neutrons23

. .

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. . .× + × = ×e j up quarks

and there are 2 2 68 10 3 34 10 8 70 1026 26 26. . .× + × = ×e j down quarks .

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

65 3 34 10 1026 28. ~×e j electrons

65 9 36 10 1026 29. ~×e j up quarks

65 8 70 10 1026 29. ~×e j down quarks .

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

P46.30 (a) proton u u d totalstrangeness 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 totalstrangeness 0 0 0 0 0baryon number 1 1/3 1/3 1/3 1charge 0 2e/3 –e/3 –e/3 0

P46.31 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 find m m m c c cu p n MeV meV MeV= − = − =13

213

2 938 939 6 3122 2 2e j e j .

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


P46.32 (a) K 0 d s totalstrangeness 1 0 1 1baryon number 0 1/3 –1/3 0charge 0 –e/3 e/3 0

(b) Λ0 u d s totalstrangeness –1 0 0 –1 –1baryon number 1 1/3 1/3 1/3 1charge 0 2e/3 –e/3 –e/3 0

P46.33 (a) π − + → +p K 0 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, andstrange 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

P46.34 In the first reaction, π − + → +p K 0 Λ0 , the quarks in the particles are: ud uud ds uds+ → + . There is anet 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 numberof each type of quark.

In the second reaction, π − + → +p K n0 , the quarks in the particles are: ud uud ds uds+ → + .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 ofeach type of quark.

Chapter 46 635

P46.35 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 1umissing.

The unknown particle is a neutron, udd.

Baryon and strangeness numbers are conserved.

P46.36 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

P46.37 (a) uud : charge = −FHGIKJ + −FHG

IKJ +FHGIKJ = −

23

23

13

e e e e . This is the antiproton .

(b) udd : charge = −FHGIKJ +FHGIKJ +FHGIKJ =

23

13

13

0e e e . This is the antineutron .

Section 46.12 The Cosmic Connection

P46.38 Section 39.4 says f fv cv c

a

aobserver source=

+−

11

.

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

Then,c c v c

v c′=

−+λ λ

11

and ′ =+−

λ λ11

v cv c

.

P46.39 (a) ′ =+−

λ λ11

v cv c

510 43411

nm nm=+−

v cv c

1 1811

1 3812. .=+−

=v cv c

1 1 381 1 381+ = −vc

vc

. . 2 38 0 381. .vc=

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

(b) v HR= : RvH

= =×

× ⋅= ×−

4 80 1017 10

2 82 107

39.

. m s

m s ly ly


P46.40 (a) ′ =+−

= +λ λ λn n nv cv c

Z11

1a f 11

1 2+−

= +v cv c

Za f

1 1 12 2+ = + − FHGIKJ +

vc

Zvc

Za f a f vc

Z Z Z ZFHGIKJ + + = +2 22 2 2e j

v cZ Z

Z Z=

++ +

FHG

IKJ

2

22

2 2

(b) RvH

cH

Z ZZ Z

= =+

+ +

FHG

IKJ

2

22

2 2

P46.41 v HR= H =× −1 7 10 2. m s

lye j

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

= =λ λ11

590 1 000 113 3 590 07v cv c

. .b g nm

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

=λ 5901 0 011 331 0 011 33

597..

nm

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

=λ 5901 0 113 31 0 113 3

661..

nm

P46.42 (a) Wien’s law: λmax .T = × ⋅−2 898 10 3 m K .

Thus, λmax. .

. .=× ⋅

=× ⋅

= × =− −

−2 898 10 2 898 101 06 10 1 06

3 33 m K m K

2.73 K m mm

T .

(b) This is a microwave .

*P46.43 We suppose that the fireball of the Big Bang is a black body.

I e T= = × ⋅ = ×− −σ 4 8 4 61 5 67 10 2 73 3 15 10( ) . . . W m K K W m2 4 2e ja fAs a bonus, we can find the current power of direct radiation from the Big Bang in the section of theuniverse observable to us. If it is fifteen billion years old, the fireball is a perfect sphere of radiusfifteen billion light years, centered at the point halfway between your eyes:

P = = = × ××FHG

IKJ ×−IA I r( ) . .4 3 15 10 4 15 10

3 103 156 102 6 9 2 8 2

7 2π π W m ly

m s1 ly yr

s yr2e ja fe j e j

P = ×7 98 1047. W.

Chapter 46 637

P46.44 The density of the Universe is

ρ ρπ

= =FHGIKJ1 20 1 20

38

2

. .cH

G.

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

M rH

Gr

H rG

= FHGIKJ =FHGIKJFHGIKJ =ρ π

ππ

43

1 2038

43

0 60032

32 3

..

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

12

02mvGmM

rGmM

R− = − so

12

0 6000

0 6002 22 3 2 3

mH rGm

rH r

GGmR

H rG

−FHG

IKJ = −

FHG

IKJ

. .

− = −0 100 0 600. .rR

so R r= 6 00. .

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

P46.45 (a) k T m cpB ≈ 2 2

so Tm c

kp

≈ =×

×FHG

IKJ−

−2 2 938 3

1 38 10

1 60 1010

2

23

1313

B

MeV

J K

J1 MeV

K.

.

.~

a fe j

(b) k T m ceB ≈ 2 2

so Tm ck

e≈ =×

×FHG

IKJ−

−2 2 0 511

1 38 10

1 60 1010

2

23

1310

B

MeV

J K

J1 MeV

K.

.

.~

a fe j

*P46.46 (a) The Hubble constant is defined in v HR= . The distance R between any two far-separatedobjects opens at constant speed according to R v t= . Then the time t since the Big Bang isfound from

v H vt= 1 = H t tH

=1

.

(b)1 1

17 103 10

1 76 10 17 63

810

H=

× ⋅

×FHG

IKJ = × =− m s ly

m s1 ly yr

yr billion years. .


*P46.47 (a) Consider a sphere around us of radius R large compared to the size of galaxy clusters. If thematter M inside the sphere has the critical density, then a galaxy of mass m at the surface ofthe sphere is moving just at escape speed v according to

K Ug+ = 012

02mvGMm

R− = .

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 vdRdt

= . Then

dRd t

GMR

FHGIKJ =

22 dR

d tR GM= −1 2 2/ R dR GM d t

R T

0 02z z=

RGM t

RT3 2

003 2

2/

=23

23 2R GM T/ =

TR

GMR

GM R= =

23 2

23 2

3 2

.

From above,2GM

Rv=

so TRv

=23

.

Now Hubble’s law says v HR= .

So TR

HR H= =

23

23

.

(b) T =× ⋅

×FHG

IKJ = × =

−

2

3 17 10

3 101

1 18 10 11 83

810

m s ly

m s ly yr

yr billion yearse j

. .

*P46.48 In our frame of reference, Hubble’s law is exemplified by v R1 1= H and v R2 2= H . From these wemay form the equations − = −v R1 1H and v v R R2 1 2 1− = −Hb g. These equations express Hubble’slaw as seen by the observer in the first galaxy cluster, as she looks at us to find − = −v R1 1Hb g and asshe looks at cluster two to find v v R R2 1 2 1− = −Hb g.

Section 46.13 Problems and Perspectives

P46.49 (a) LG

c= =

× ⋅ × ⋅

×= ×

− −−

3

34 11

335

1 055 10 6 67 101 61 10

. ..

J s N m kg

3.00 10 m s m

2 2

8

e je je j

(b) This time is given as TLc

= =××

= ×−

−1 61 105 38 10

3544.

. m

3.00 10 m s s8 ,

which is approximately equal to the ultra-hot epoch.

(c) Yes.

Chapter 46 639

Additional Problems

P46.50 We find the number N of neutrinos:

10 6 6 1 60 1046 13 J MeV J= = × × −N Na f e j.

N = ×1 0 1058. neutrinos

The intensity at our location is

NA

Nr

= =×

× × ×

FHGG

IKJJ = × −

41 0 10

4 1 7 10

1

3 00 10 3 16 103 1 102

58

5 2 8 7

2

14 2

π.

. . ..

π ly

ly

m s s m

e j e je j.

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. . .×FHGIKJ = ×

1m

m22e j

or ~1014 .

*P46.51 A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years. Thehypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds:

c v170 000 10 yr 170 000 yr sb g b g= +

vc=

+=

+ × ×=

+ × −

170 000170 000 10

1

1 10

11 1 86 10 12

yr yr s s 1.7 10 yr 3.156 10 s yr5 7e je j .

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

mcE

Evc

22

2 12 2

12 2

12 21 10 11

1 1 86 1010

1 1 86 10 1

1 1 86 10= = − = −

+ ×=

+ × −

+ ×−

−

−γ MeV MeV

.

.

.e je je j

mc212

6101 86 10

110 1 93 10 19≈

×= × =

−− MeV

2 MeV eV

..

e j e j .

Then the upper limit on the mass is

mc

=19

2 eV

mc c

=×

FHG

IKJ = × −19

931 5 102 1 102 6 2

8 eV u eV

u.

. .

P46.52 (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 .


P46.53 The total energy in neutrinos emitted per second by the Sun is:

0 4 4 1 5 10 1 1 1011 2 23. . .a f e jπ ×LNM

OQP = ×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.

P46.54 p p p+ → + ++π X

We suppose the protons each have 70.4 MeV of kinetic energy. From conservation of momentum forthe collision, particle X has zero momentum and thus zero kinetic energy. Conservation of systemenergy then requires

M c M c M c M c K M c KXp p p p p2 2 2 2 2+ + = + + +π e j e j

M c M c K M cX2 2 22 938 3 2 70 4 139 6 939 5= + − = + − =p p MeV MeV MeV MeVπ . . . .a f

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

*P46.55 (a) If 2N particles are annihilated, the energy released is 2 2Nmc . The resulting photon

momentum is pEc

Nmcc

Nmc= = =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 particlewhich also has initial rest energy mc2 (but no momentum initially).

E p c mc2 2 2 2 2= + e j

Thus 2 2 2 2 2 2 2mc p c mce j e j= + .

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

first particle. Thus 4 2 2 2 2 2 2mc p c mce j e j= + , p mc2 2 2

3= e j . So p mc= 3 .

This process is repeated N times (annihilate N2

protons and N2

antiprotons). Thus the total

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

p Nmc= 3

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

Chapter 46 641

P46.56 (a) ∆ ∆ ≈E t , and ∆ = =××

= ×−

−trc

1 4 103 10

4 7 1015

824.

. m

m s s

∆ ≈∆

=× ⋅×

= ××

FHG

IKJ = ×

−

−−

−Et

1 055 104 7 10

2 3 101 60 10

1 4 1034

2411

132.

..

..

J s s

J1 MeV

J MeVe j

mE

cc c=

∆≈ ×2

2 2 2 21 4 10 10. ~ MeV MeV

(b) From Table 46.2, m cπ2 139 6= . MeV a pi - meson .

P46.57 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 final rest energy is the kinetic energy of theproducts.

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 6 139 6 37 72 2 2 2 2 2. . . . .a f a f+ −LNM

OQP + + −LNM

OQP =p c p c 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 35e j a f. . MeV

Kπ = + − =139 6 100 4 139 6 32 32 2. . . .a f a f MeV .

P46.58 By relativistic energy conservation in the reaction, E m cm c

v ce

eγ + =

−

22

2 2

3

1. (1)

By relativistic momentum conservation for the system,E

cm v

v ceγ =

−

3

1 2 2. (2)

Dividing (2) by (1), XE

E m cvce

=+

=γ

γ2 .

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 =

45

so E m ceγ = =4 2 042 . MeV .


P46.59 Momentum of proton is qBr = × ⋅−1 60 10 0 250 1 3319. . . C kg C s me jb ga fp m sp = × ⋅−5 32 10 20. kg c pp = × ⋅ = × =− −1 60 10 1 60 10 99 811 11. . . kg m s J MeV2 2 .

Therefore, p cp = 99 8. MeV .

The total energy of the proton is E E cpp = + = + =02 2 2 2983 3 99 8 944b g a f a f. . MeV .

For 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 172b g a f a f. . MeV

Thus, E Etotal after total before Rest energy= = .

Rest Energy of unknown particle = + =944 172 1 116 MeV MeV MeV (This is a Λ0 particle!)

Mass MeV= 1 116 2c .

P46.60 Σ → +0 0Λ γ

From Table 46.2, m cΣ = 1 192 5 2. MeV and m cΛ = 1115 6 2. MeV .

Conservation of energy in the decay requires

E E K Eo0, ,Σ = + +Λ Λe j γ or 1 192 5 1 115 62

2

. . MeV MeV= +FHG

IKJ +

pm

EΛ

Λγ .

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

1 192 5 1 115 62

2

. . MeV MeV= +FHG

IKJ+

p

mEγγ

Λ

or 1 192 5 1 115 62

2 2

2. . MeV MeV= +FHG

IKJ+

p c

m cEγγ

Λ

.

Recognizing that m cΛ2 1 115 6= . MeV and p c Eγ γ=

we now have 1 192 5 1 115 62 1 115 6

2

. ..

MeV MeV MeV

= + +E

Eγγb g .

Solving this quadratic equation, Eγ = 74 4. MeV .

P46.61 p p p n+ → + + +π

The total momentum is zero before the reaction. Thus, all three particles present after the reactionmay be at rest and still conserve system momentum. This will be the case when the incident protonshave 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+ = + +e j π

so the kinetic energy of each of the incident protons is

Km c m c m c

pn p

=+ −

=+ −

=2 2 2

2939 6 139 6 938 3

70 4π . . ..

a f MeV2

MeV.

Chapter 46 643

P46.62 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= + = +d i a f b g a f. . MeV MeV

or E Eµ ν2 2 2105 7− = . MeVa f . [2]

Since E Eµ ν+ = 139 6. MeV [1]

and E E E Eµ ν µ ν+ − =d id i a f105 7 2. MeV [2]

then E Eµ ν− = =105 7139 6

80 02.

..

MeV MeV

a f. [3]

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

P46.63 The expression e dEE k T− B gives the fraction of the photons that have energy between E and E dE+ .The fraction that have energy between E and infinity is

e dE

e dE

e dE k T

e dE k T

e

ee

E k T

E

E k T

E k T

E

E k T

E k TE

E k T

E k T

−∞

−∞

−∞

−∞

− ∞

− ∞−

zz

zz

=

−

−

= =

B

B

B

B

B

B

B

B

B0 0

0

b g

b g.

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

. =− × ×− −

eT J J Ke j e j

or ln .. .

0 010 01 60 10 1 16 1019

23

4

b ge j

= −×

×= −

×−

−

J

1.38 10 J K

K

T T giving T = ×2 52 103. K .

P46.64 (a) This diagram represents the annihilation of anelectron and an antielectron. From charge andlepton-number conservation at either vertex, the

exchanged particle must be an electron, e− .

(b) This is the tough one. A neutrino collides with aneutron, changing it into a proton with release of amuon. This is a weak interaction. The exchanged

particle has charge +e and is a W+ .

γ γ

(a) (b)

FIG. P46.64


P46.65 (a) The mediator of this weak interaction is a

Z boson0 .

(b) The Feynman diagram shows a down quark andits antiparticle annihilating each other. They canproduce a particle carrying energy, momentum,and angular momentum, but zero charge, zerobaryon number, and, it may be, no color charge. Inthis case the product particle is a photon .

FIG. P46.65

For conservation of both energy and momentum in the collision we would expect twophotons; but momentum need not be strictly conserved, according to the uncertaintyprinciple, if the photon travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.65. Depending on the color charges ofthe d and d quarks, the ephemeral particle could also be a gluon , as suggested in the

discussion of Figure 46.14(b).

*P46.66 (a) At threshold, we consider a photon and a proton colliding head-on to produce a proton anda 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

uc

m c m c

u c

u c u c

p pp

MeV MeV MeV

2

2 2

2

2 2

2 2

1 1

938 3 1

1 1938 3 135 0

−+

−= +

+

− += +

π

.. .

b gb gb g

souc= 0 134.

and Eγ = 127 MeV .

(b) λmax .T = ⋅2 898 mm K

λmax.

.=⋅

=2 898

1 06 mm K

2.73 K mm

(c) E hfhc

γ λ= = =

⋅

×= ×

−−1 240

1 17 109

3 eV 10 m1.06 10 m

eV-3 .

continued on next page

Chapter 46 645

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

uc

0 134. 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 thespeed of the primed frame

1 27 1011

1 17 10

1 1 71 10

8 3

22

. .

.

× =+−

×

= − ×

−

−

νν

cc

vc

Then the speed of the proton is given by

uc

u c cu c

=′ +

+ ′=

+ − ×

+ − ×= − ×

−

−−ν

ν10 134 1 1 71 10

1 0 134 1 1 71 101 1 30 102

22

2222. .

. ..

e j.

And the energy of the proton is

m c

u c

p MeV eV eV

2

2 2 22 2

10 6 19

1

938 3

1 1 1 30 106 19 10 938 3 10 5 81 10

−=

− − ×= × × × = ×

−

.

.. . .

e j.

ANSWERS TO EVEN PROBLEMS

P46.2 2 27 1023. × Hz; 1 32. fm P46.22 (a) electron lepton number and muonlepton number;(b) electron lepton number;P46.4 ν µ and ν e(c) strangeness and charge;(d) baryon number; (e) strangeness

P46.6 ~10 16− m

P46.24 see the solutionP46.8 ~10 23− s

P46.26 (a) 686 MeV

c and

200 MeVc

;P46.10 ν µ

(b) 627 MeV c ;P46.12 (a) electron lepton number and muon

lepton number; (b) charge;(c) 244 MeV , 1 130 MeV , 1 370 MeV ;(d) 1 190 2 MeV c , 0 500. c

(c) baryon number; (d) baryon number;(e) charge

P46.28 (a) see the solution; (b) 5 63. GeV ;(c) 768 MeV ; (d) 280 MeV ; (e) 4 43. TeV

P46.14 the second violates conservation of baryonnumber

P46.30 see the solution

P46.16 0 828. cP46.32 see the solution

P46.18 (a) see the solution;P46.34 see the solution

(b) 469 MeV ; 469 MeV

c for both;

P46.36 (a) Σ+ ; (b) π − ; (c) K 0 ; (d) Ξ−(c) 0 999 999 4. c

P46.38 see the solutionP46.20 see the solution


P46.40 (a) v cZ Z

Z Z=

++ +

FHG

IKJ

2

22

2 2; (b)

cH

Z ZZ Z

2

22

2 2+

+ +

FHG

IKJ

P46.56 (a) ~102 2 MeV c ; (b) a pi - meson

P46.58 2 04. MeV

P46.42 (a) 1 06. mm; (b) microwaveP46.60 74 4. MeV

P46.44 6 00.P46.62 29 8. MeV

P46.46 (a) see the solution; (b) 17 6. GyrP46.64 (a) electron-position annihilation; e− ;

(b) a neutrino collides with a neutron,producing a proton and a muon; W+

P46.48 see the solution

P46.50 ~1014

P46.66 (a) 127 MeV ; (b) 1 06. mm; (c) 1 17. meV ;(d) 5 81 1019. × eVP46.52 (a) charge; (b) energy; (c) baryon number

P46.54 neutron

Particle Physics and Cosmology - UCCS Home - · PDF file620 Particle Physics and Cosmology...

Documents

Transcript of Particle Physics and Cosmology - UCCS Home - · PDF file620 Particle Physics and Cosmology...