ism_chapter_41

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1409 
Chapter 41 
Elementary Particles and the Beginning of the Universe 
 
Conceptual Problems 
 
1 • 
Similarities 
 
Differences 
Baryons and mesons are hadrons, i.e., they 
participate in the strong interaction. Both 
are composed of quarks. 
Baryons consist of three quarks and are 
fermions. Mesons consist of two quarks 
and are bosons. Baryons have baryon 
number +1 or −1. Mesons have baryon 
number 0. 
 
2 • 
Determine the Concept The muon is a lepton. It is a spin-½ particle and is a fermion. It 
does not participate in strong interactions. It appears to be an elementary particle like the 
electron. The pion is a meson. Its spin is 0 and it is a boson. It does participate in strong 
interactions and is composed of quarks. 
 
*3 • 
Determine the Concept A decay process involving the strong interaction has a very short 
lifetime (∼10−23 s), whereas decay processes that proceed via the weak interaction have 
lifetimes of order 10−10 s. 
 
4 • 
(a) True 
 
(b) False. There are two kinds of hadrons-baryons, which have spin 21 (or ,, 2523 and so 
on), and mesons, which have zero or integral spin. 
 
5 • 
False. Mesons have zero or integral spins. 
 
6 • 
Determine the Concept A meson has 2 quarks, a baryon has 3 quarks. 
 
7 • 
Determine the Concept No; from Table 41-2 it is evident that any quark-antiquark 
combination always results in an integral or zero charge. 
 
Chapter 41 
 
 
1410 
8 • 
(a) False. Leptons are not made up of quarks. 
 
(b) True 
 
(c) False. Electrons are leptons and leptons interact via the weak interaction. 
 
(d) True 
 
(e) True 
 
*9 • 
Determine the Concept No. Such a reaction is impossible. A proton requires three 
quarks. Three quarks are not available because a pion is made of a quark and an 
antiquark and the antiproton consists of three antiquarks. 
 
Estimation and Approximation 
10 •• 
Picture the Problem Assuming that the lifetime of a proton is 1032 y, one proton out of 
every 1032 protons should decay every year on average. Hence, we can estimate the 
expected time between proton-decays that occur in the water of a filled Olympic-size 
swimming pool by determining the number of protons N in the pool and dividing 1032 y by 
this number. 
 
The mean time between 
disintegrations is the ratio of the 
lifetime of the protons to the number 
of protons N in the pool: 
 
N
t y10
32
mean =∆ (1) 
The number of protons N in the pool 
is related to the mass of water in the 
pool Mwater, the molar mass of water 
mmolar, water, and the number of 
protons per molecule n: 
 
 watermolar,
A
water m
nN
M
N = 
 
Solve for N to obtain: 
 watermolar,
waterA
m
MnNN = 
 
Because the mass of the water is the 
product of its density and the 
volume of the pool: 
 
 watermolar,
poolwaterA
m
VnN
N
ρ= 
Elementary Particles and the Beginning of the Universe 
 
 
1411
Substituting for N in equation (1) 
yields: 
 
( )
poolwaterA
 watermolar,
32
 watermolar,
poolwaterA
32
mean
y10
y10
VnN
m
m
VnNt
ρ
ρ
=
=∆
 
 
Because each molecule of water has 
10 protons: 
 
molecule
protons10=n 
Substitute numerical values and evaluate ∆tmean: 
 
( )
( )( )( )
d8.21
y
d365.24y0598.0y0598.0
m2m25m100
m
kg10
mol
molecules1002.6
molecule
protons10
g10
kg1
mol
g18y10
3
323
3
32
mean
=×==
⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ ×⎟⎠
⎞⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×
=∆t
 
 
11 • 
Picture the Problem We can use 2nucleus
2
protonem rkqF = and 2nucleus2protongrav rGmF = to 
estimate the ratio of the electromagnetic and gravitational forces between two protons 
located in a nucleus. 
 
The electromagnetic force between 
two protons located in a nucleus is 
given by: 
 
2
nucleus
2
proton
em r
kq
F = 
 
The gravitational force between 
these same protons is given by: 
 
2
nucleus
2
proton
grav r
Gm
F = 
Divide Fem by Fgrav to obtain: 
 
2
proton
2
proton
2
nucleus
2
proton
2
nucleus
2
proton
grav
em
Gm
kq
r
Gm
r
kq
F
F == 
 
Substitute numerical values and evaluate Fem/Fgrav: 
 
Chapter 41 
 
 
1412 
( )
( )
36
227
2
2
11
219
2
2
9
grav
em 1024.1
kg1067.1
kg
mN1067.6
C1060.1
C
mN1099.8
×=
×⎟⎟⎠
⎞
⎜⎜⎝
⎛ ⋅×
×⎟⎟⎠
⎞
⎜⎜⎝
⎛ ⋅×
=
−−
−
F
F
 
 
Spin and Antiparticles 
 
*12 • 
Picture the Problem We can use both conservation of energy and momentum to explain 
why the energies of the two γ-rays must be equal. We can find the energy of each γ-ray in 
Table 41-1 and find their wavelengths using λ = hc/E. 
 
(a) 
energy. same thehave they Hence,
magnitude. momentum same thehavemust photons twoenergy the and
momentumboth conserve To E/c. isphoton a of momentum The zero.
bemust momentum final the, thereforezero; is momentum initial The
 
 
(b) From Table 41-1: 
 
MeV6.139=γE 
(c) The wavelength of each γ ray is 
given by: EE
hc fmMeV1240 ⋅==λ 
 
Substitute numerical values and 
evaluate λ: fm88.8MeV6.139
fmMeV1240 =⋅=λ 
 
13 • 
Picture the Problem In each case, the required energy is given by 22mcE = where m is 
mass of each particle produced in the pair-production reaction. These masses can be 
found in Tables 41-1 and 41-3. 
 
(a) For γ → π + + π –: ( )
MeV2.279
MeV/6.13922 222
=
== cccmE π
 
 
(b) For γ → p + p –: ( )
MeV1877
MeV/3.93822 222p
=
== cccmE
 
 
(c) For γ → µ – + µ +: ( )
MeV3.211
MeV/659.10522 222
=
== cccmE µ
 
 
Elementary Particles and the Beginning of the Universe 
 
 
1413
The Conservation Laws 
 
14 • 
Picture the Problem We need to check for conservation of energy, charge, baryon 
number, and lepton number. 
 
(a) Energy conservation: 
. violatedison conservati
energy , Because np mm < 
 
Charge conservation: 
+e → 0 + e + 0 = +e 
 conserved.
 is charge decay, after the and
before is chargenet theBecause e+
 
 
Baryon number: 
+1 → +1 + 0 + 0 = +1 conserved. isnumber baryon decay,
 after the and before 1 is Because +B
 
 
Lepton number; electrons: 
0 → 0 + 0 + 0 = 0 
conserved. is electrons
for number lepton thedecay,
after the and before 0 Because e =L
 
 
energy. ofon conservati sit violate because allowednot is process The 
 
(b) Energy conservation: 
 violated.ison conservati
energy ,m Because pn −+< πmm 
 
Charge conservation: 
0 → +e + (−e) = 0 
 conserved.
is charge decay, after the and
before 0 is chargenet theBecause
 
 
Baryon number: 
1 → 1 + 0 = 1 
 conserved.
isnumber baryon decay,
after the and before 0 Because =B
 
 
Lepton number; electrons: 
0 → 0 + 0 = 0 
 
conserved. isnumber lepton decay,
after the and before 0 L Because =
 
 
allowed.not isdecay thisconserved,not isenergy Because 
Chapter 41 
 
 
1414 
(c) 
momentum. conserve toemitted
 bemust rays more)(or two violated;ison conservati Momentum γ
 
 
(d) Energy conservation: conserved. isEnergy 
 
Charge conservation: 
+1 + (−1) → 0 + 0 = 0 
 conserved. is charge
 decay, after the and before
 ero is chargenet theBecause z
 
 
Baryon number: 
+1 + (−1) → 0 + 0 = 0 
 
conserved. isnumber baryon decay,
 after the and before 0 Because =B
 
 
Lepton number; electrons: 
0 → 0 + 0 + 0 = 0 
conserved. is electrons
for number lepton thedecay,
after the and before 0 Because e =L
 
 
 process. allowedan is this violated,are lawson conservati theof none Because 
 
(e) Energy conservation: 
conserved. is
energy , Because
enp ++> mmmCharge conservation: 
0 + 1 → 0 + 1 = 1 
 conserved. is charge
 decay, after the and before
 one is chargenet theBecause
 
 
Baryon number: 
0 + 1 → 1 + 0 = 0 
 
conserved. isnumber baryon decay,
 after the and before 1 Because =B
 
 
Lepton number; electrons: 
−1 + 0 → 0 + (−1) = −1 
conserved. is electrons
for number lepton thedecay,
after the and before 1 Because e −=L
 
 
 process. allowedan is this violated,are lawson conservati theof none Because 
 
Elementary Particles and the Beginning of the Universe 
 
 
1415
15 • 
Picture the Problem The decay will occur via the strong interaction if strangeness is 
conserved. If ∆S = ±1, it will occur via the weak interaction. If S changes by more than 1, 
the decay will not occur. 
 
(a) List the strangeness of Ω– , Ξ0, 
and π –: 
 
Ω–: S = −3 
Ξ0: S = −2 
π –: S = 0 
 
Determine ∆S: 
 
( ) 132 +=−−−=∆S 
 n.interactio weak the viaproceedcan reaction the1, Because +=∆S 
 
(b) List the strangeness of Ξ0, p, π –, 
and π 0: 
 
Ξ0: S = −2 
p: S = 0 
π –: S = 0 
π 0: S = 0 
 
Determine ∆S: 
 
( ) 220 +=−−=∆S 
 allowed.not isreaction the2, Because +=∆S 
 
(c) List the strangeness of Λ0, 
p + , and π –: 
 
Λ0: S = −1 
p +: S = 0 
π –: S = 0 
 
Determine ∆S: 
 
( ) 110 +=−−=∆S 
 n.interactio weak the viaproceedcan reaction the1, Because +=∆S 
 
16 • 
Picture the Problem The decay will occur via the strong interaction if strangeness is 
conserved. If ∆S = ±1, it will occur via the weak interaction. If S changes by more than 1, 
the decay will not occur. 
 
(a) List the strangeness of Ω– , Λ0, 
and K–: 
 
Ω–: S = −3 
Λ0: S = −1 
K–: S = −1 
 
Determine ∆S: 
 
( ) 1311 +=−−−−=∆S 
Chapter 41 
 
 
1416 
 n.interactio weak the viaproceedcan reaction the1, Because +=∆S 
 
(b) List the strangeness of Ξ0, p, and 
π –: 
 
Ξ0: S = −2 
p: S = 0 
π –: S = 0 
 
Determine ∆S: 
 
( ) 220 +=−−=∆S 
 allowed.not isreaction the2, Because +=∆S 
 
17 • 
Picture the Problem The decay will occur via the strong interaction if strangeness is 
conserved. If ∆S = ±1, it will occur via the weak interaction. If S changes by more than 1, 
the decay will not occur. 
 
(a) List the strangeness of Ω– , Λ0, 
eν , and e –: 
 
Ω–: S = −3 
Λ0: S = −1 
eν : S = 0 
e –: S = 0 
 
Determine ∆S: 
 
( ) 231 +=−−−=∆S 
 allowed.not isreaction the2, Because +=∆S 
 
(b) List the strangeness of Σ+, p, and 
π0: 
 
Σ+: S = −1 
p: S = 0 
π0: S = 0 
 
Determine ∆S: 
 
( ) 110 +=−−=∆S 
 n.interactio weak the viaproceedcan reaction the1, Because +=∆S 
 
18 • 
Picture the Problem We can decide whether the given decays of the τ particle are 
possible by determining whether energy conservation is satisfied and whether 
conservation of both the τ and µ lepton numbers is satisfied. 
 
(a) 
numbers.lepton and both the of
on conservati andon conservatienergy satisfiesIt allowed. isdecay first The
µτ 
 
Elementary Particles and the Beginning of the Universe 
 
 
1417
(b) 
numbers.lepton and
 conservenot doesit because allowednot is schemedecay second The
µ
τ
 
 
(c) The total kinetic for the decay 
in (a) is: 
 
22
tot cmcmK µτ −= 
From Table 41-3 we have: 
 
2MeV/1784 cm =τ 
and 
2MeV/659.105 cm =µ 
 
Substitute numerical values and 
evaluate Ktot: 
( ) ( )
MeV1678
MeV/106MeV/1784 2222tot
=
−= ccccK
 
 
Remarks: Note that the kinetic energy of the individual decay products cannot be 
determined from the decay scheme alone. 
 
19 •• 
Picture the Problem Examination of the decay products will reveal whether all the final 
products are stable. A decay process is allowed if energy, charge, baryon number, and 
lepton number are conserved. 
 
(a) No; the neutron is not stable: 
 
eepn ν++→ −+ 
(b) Add the reactions to obtain: 
 
µµ νννν 22 3e3ep ee ++++++→Ω −++− 
 
(c) Charge conservation: 
100003111 −=++++−+→− conserved. is charge decay, the
after and before 1 Because −=Q
 
 
Baryon number: 
1 → 1 + 0 + 0 + 0 + 0 + 0 + 0 = 1 conserved. is numberbaryon decay, the
after and before 1 Because =B
 
 
Lepton number: 
0 → 0 −1 + 3 + 1 − 3 − 2 + 2 = 0 
 conserved. is electrons
for number lepton thedecay, the
after and before 0 Because e =L
 
 
Chapter 41 
 
 
1418 
Strangeness: 
−3 → 0 
 
n.interactio
 weak the viaallowed isdecay each 
 and 1,decay baryon each in 
 However, conserved.not is sStrangenes
+=∆S
 
*20 •• 
Picture the Problem A decay process is allowed if energy, charge, baryon number, and 
lepton number are conserved. 
 
(a) Energy conservation: 
ed.not violat ison conservati
energy ,22 Because n µπ mmm +> 
 
Charge conservation: 
0 → +1 + (−1) + 0 + 0 = 0 
conserved.
 is charge decay, after the and
before 0 is charge total theBecause
 
 
Baryon number: 
1 → 0 + 0 + 0 + 0 = 0 
 violated.isnumber 
baryon ofon conservati 0, to1 from
 changesnumber baryon Because
+ 
 
Lepton number: 
0 → 0 + 0 + 1 + (−1) = 0 
conserved. is muons
fornumber lepton thedecay, the
after and before 0 Because =µL
 
 
number.baryon ofon conservati sit violate because allowednot is process The 
 
(b) Energy conservation: 
ed.not violat ison conservati
energy ,2 Because emm >π 
 
Charge conservation: 
0 → +1 + (−1) + 0 + 0 = 0 
conserved.
 is charge decay, after the and
before 0 is charge total theBecause
 
 
Baryon number: 
0 → 0 + 0 + 0 + 0 = 0 
conserved.
 isnumber baryon thedecay,
 after the and before 0 Because =B
 
 
Elementary Particles and the Beginning of the Universe 
 
 
1419
Lepton number: 
0 → 0 + 0 + 0 + 0 = 0 conserved. isnumber lepton thedecay,
after the and before 0 Because e =L
 
allowed. is and lawson conservati all satisfiesdecay The 
 
Quarks 
 
21 • 
Picture the Problem For each quark combination we can determine the baryon number 
B, the charge Q, and the strangeness S and then use Table 41-1 to find a match and 
complete the following table. 
 
 Combination B Q S hadron 
(a) uud 1 +1 0 p+ 
(b) udd 1 0 0 n 
(c) uus 1 +1 −1 Σ+ 
(d) dds 1 −1 −1 Σ− 
(e) uss 1 0 −2 Ξ 0 
(f) dss 1 −1 −2 Ξ − 
 
22 • 
Picture the Problem For each quark combination we can determine the baryon number 
B, the charge Q, and the strangeness S and then use Table 41-1 to find a match and 
complete the following table. 
 
 Combination B Q S hadron 
(a) du 0 +1 0 π+ 
(b) du 0 −1 0 π− 
(c) su 0 +1 +1 K+ 
(d) su 0 −1 −1 K− 
 
23 • 
Determine the Concept From Table 41-2 we see that to satisfy the conditions of charge = 
+2 and zero strangeness, charm, topness, and bottomness, the quark combination must be 
uuu. 
 
24 • 
Picture the Problem Because K + and K0 are mesons, they consist of a quark and an 
antiquark. We can use Tables 41-1 and 41-2 to find combinations of quarks with the 
correct values for electric charge, baryon number, and strangeness for these particles. 
 
Chapter 41 
 
 
1420 
(a) For K + we need: Q = +1 
B = 0 
S = +1 
 
. is properties e with thesquarks ofn combinatioA su 
 
(b) For K0 we need: Q = 0 
B = 0 
S = +1 
 
. is properties e with thesquarks ofn combinatioA sd 
 
25 • 
Determine the Concept Because D + and D− are mesons, they consist of a quark and an 
antiquark. 
 
(a) B = 0, so we must look for a combination of quark and antiquark. Because it has 
charm of +1, one of the quarks must be c. Because the charge is +e, the antiquark must be 
.d The possible combination for D+ is .dc . 
 
(b) Because D− is the antiparticleof D+, the quark combination is .dc 
 
26 • 
Picture the Problem We can use Tables 41-1 and 41-2 to find combinations of quarks 
with the correct values for electric charge, baryon number, and strangeness for these 
particles. Because K – and 0K are mesons, they consist of a quark and an antiquark. 
 
(a) For K – we need: Q = −1 
B = 0 
S = −1 
 
. is properties e with thesquarks ofn combinatioA su 
 
(b) For 0K we need: Q = 0 
B = 0 
S = −1 
 
. is properties e with thesquarks ofn combinatioA sd 
 
Elementary Particles and the Beginning of the Universe 
 
 
1421
Remarks: An alternative solution could take advantage of our results in Problem 20 
for the antiparticles K + and K 0. 
 
*27 •• 
Picture the Problem Because Λ0, p – , and Σ– are baryons, they are made up of three 
quarks. We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct 
values for electric charge, baryon number, and strangeness for these particles. 
 
(a) For Λ0 we need: Q = 0 
B = +1 
S = −1 
 
. is conditions thesesatisfiesn that combinatioquark The uds 
 
(b) For p – we need: Q = −1 
B = −1 
S = +1 
 
. is conditions thesesatisfiesn that combinatioquark The duu 
 
(c) For Σ– we need: Q = −1 
B = +1 
S = −1 
 
. is conditions thesesatisfiesn that combinatioquark The dds 
 
28 •• 
Picture the Problem Because n , Ξ0 , and Σ+ are baryons, they are made up of three 
quarks. We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct 
values for electric charge, baryon number, and strangeness for these particles. 
 
(a) For n we need: Q = 0 
B = −1 
S = 0 
 
. is conditions thesesatisfiesn that combinatioquark The ddu 
 
(b) For Ξ0 we need: Q = 0 
B = +1 
S = −2 
Chapter 41 
 
 
1422 
. is conditions thesesatisfiesn that combinatioquark The uss 
 
(c) For Σ+ we need: Q = +1 
B = +1 
S = −1 
 
. is conditions thesesatisfiesn that combinatioquark The uus 
 
29 •• 
Picture the Problem Because Ω– and Ξ– are baryons, they are made up of three quarks. 
We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values 
for electric charge, baryon number, and strangeness for these particles. 
 
(a) For Ω– we need: Q = −1 
B = +1 
S = −3 
 
. is conditions thesesatisfiesn that combinatioquark The sss 
 
(b) For Ξ– we need: Q = −1 
B = +1 
S = −2 
 
. is conditions thesesatisfiesn that combinatioquark The ssd 
 
30 •• 
Picture the Problem We can use Table 41-2 to identify the properties of the particles 
made up of the given quarks. 
 
(a) For ddd: 1−=Q 
1+=B 
0=S 
 
(b) For cu : 0=Q 
0=B 
0=S 
1charm −= 
Elementary Particles and the Beginning of the Universe 
 
 
1423
(c) For bu : 1+=Q 
0=B 
0=S 
1bottomness −= 
 
(d) For sss : 1+=Q 
1−=B 
3+=S 
 
The Evolution of the Universe 
 
*31 • 
Picture the Problem We can use Hubble’s law to find the distance from the earth to this 
galaxy. 
 
The recessional velocity of galaxy is 
related to its distance by Hubble’s 
law: 
 
Hrv = 
Solve for r: 
H
vr = 
 
Substitute numerical values and 
evaluate r: 
( ) ( )( )
y1026.3
y10
km/s 23
km/s1030.025
y10
km/s 23
0.025
8
6
5
6
⋅×=
⋅
×=
⋅
=
c
cc
cr
 
 
32 • 
Picture the Problem We can use Hubble’s law to find the speed of the galaxy. 
 
The recessional velocity of galaxy is 
related to its distance by Hubble’s 
law: 
 
Hrv = 
Substitute numerical values and evaluate v: 
 
( ) ccc
c
v 920.0
km/s 103.00
y1012
y10
km/s 23
5
9
6 =⎟⎠
⎞⎜⎝
⎛
×⋅×⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅= 
 
Chapter 41 
 
 
1424 
33 •• 
Picture the Problem We can substitute for f ′ and f0, using v = fλ, in Equation 39-16b to 
show that the relativistic wavelength shift is .
/1
/1
0 cv
cv' −
+= λλ 
 
From Equation 39-16b: 
cv
cvff'
/1
/1
0 +
−= 
 
Express f ′ and f0 in terms of λ′ and 
λ0: 'f' λ
c= and 
0
0
c
λ=f 
 
Substitute for f ′ and f0 to obtain: 
cv
cvcc
/1
/1
' 0 +
−= λλ 
 
Solve for λ′: 
cv
cv' −
+=
1
1
0λλ 
 
*34 •• 
Picture the Problem Using Hubble’s law, we can rewrite the equation from Problem 31 
as 
cHr
cHr'
/1
/1
0 −
+= λλ . 
 
From Problem 33 we have: 
cv
cv'
/1
/1
0 −
+= λλ 
 
Use Hubble’s law to relate v to r: Hrv = 
 
Substitute for v to obtain: 
cHr
cHr'
/1
/1
0 −
+= λλ 
 
(a) For r = 5×106 c⋅y: 
 ( )( )( )( )
nm6.656
km/s103
yc105
yc10
km/s 231
km/s103
yc105
yc10
km/s 231
nm3.656
5
6
6
5
6
6
=
×
⋅×⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅−
×
⋅×⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅+='λ 
 
(b) For r = 50 ×106 c⋅y: 
 
Elementary Particles and the Beginning of the Universe 
 
 
1425
( )( )( )( )
nm8.658
km/s103
yc1050
yc10
km/s 231
km/s103
yc1050
yc10
km/s 231
nm3.656
5
6
6
5
6
6
=
×
⋅×⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅−
×
⋅×⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅+='λ 
 
(c) For r = 500 ×106 c⋅y: 
 ( )( )( )( )
nm0.682
km/s103
yc10500
yc10
km/s 231
km/s103
yc10500
yc10
km/s 231
nm3.656
5
6
6
5
6
6
=
×
⋅×⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−
×
⋅×⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅+='λ 
 
(d) For r = 5×109 c⋅y: 
 ( )( )( )( )
nm0.983
km/s103
yc105
yc10
km/s 231
km/s103
yc105
yc10
km/s 231
nm3.656
5
9
6
5
9
6
=
×
⋅×⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−
×
⋅×⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅+='λ 
 
General Problems 
 
35 • 
 Determine the Concept 
 
(a) It must be a meson, and it must consist of a quark and its antiquark. 
 
(b) The π0 is its own antiparticle. 
 
(c) The Ξ 0 is a baryon; it cannot be its own antiparticle; the antiparticle is the 
Ξ0 = .ssu 
 
36 •• 
Picture the Problem Examination of the decay products will reveal whether all the final 
products are stable. A decay process is allowed if energy, charge, baryon number, and 
lepton number are conserved. 
 
(a) stable. are products final theYes, 
 
(b) Add the reactions to obtain: 
 
Chapter 41 
 
 
1426 
γννν µµ 2 ep e0 +++++→Ξ −+ 
 
(c) Charge conservation: 
00 =+→ −+ ee 
conserved. is Charge 
 
Baryon number: 
1 → 1 + 0 = 1 
 
conserved. isnumber Baryon 
 
Lepton number: 
0 → 0 + 1 − 1 + 1 − 1 = 0 
 
conserved. isnumber Lepton 
 
Strangeness: 
−2 → 0 
 
1. decays
first two in the becausen interactio
 weak the viaallowed isreaction the
 However, conserved.not is sStrangenes
+=∆S
 
 
(d) . ,Ξ onconservatienergy violating theof mass
rest an thegreater th be wouldproductsdecay theof massesrest theNo;
0 
 
*37 •• 
Picture the Problem The π o particle is composed of two quarks,uu . Hence, the 
reaction π o → γ + γ is equivalent to uu → γ + γ . 
 
(a) photons. in the resulting annihilate u andu The 
 
(b) momentum.linear conserve torequired are photons moreor Two 
 
38 •• 
Picture the Problem A decay process is allowed if energy, charge, baryon number, and 
lepton number are conserved. 
 
(a) Energy conservation: 
ed.not violat ison conservati
energy , Because p πmmm +>Λ 
 
Charge conservation: 
0 → 1 −1 = 0 
 
 
ed.not violat ison conservati
 charge decay, after the and
 before 0 is charge total theBecause
 
 
Elementary Particles and the Beginning of the Universe 
 
 
1427
Baryon number: 
1 → 1 + 0 = 1 
 conserved.
 isnumber baryon number,
baryon in change no is thereBecause
 
 
Lepton number: 
0 → 0 + 0 = 0 conserved. isnumber lepton sides,both on 0 isnumber lepton Because
 
 
allowed. is and lawson conservati all satisfieddecay The 
 
(b) Energy conservation: 
conserved.not is
energy ,m Because pn +<Σ mm 
 
Charge conservation: 
−1 → 0 − 1 = −1 
 
conserved. is charge change,
not does charge total theBecause
 
 
Baryon number: 
+1 → 1 − 1 = 0 
 
conserved.not isnumber baryon 
 0, to1 from changes Because +B
 
 
Lepton number: 
0 → 0 + 0 = 0 conserved. isnumber lepton 
 sides,both on 0 is Because L
 
 
allowed.not isit 
number,baryon andon conservatienergy both atesdecay viol theBecause
 
 
(c) Energy conservation: conserved. isEnergy 
 
Charge conservation: 
−1 → − 1 + 0 + 0 = −1 
 
conserved. is charge change,
not does charge total theBecause
 
 
Baryon number: 
0 → 0 + 0 + 0 = 0 
 
conserved. isnumber baryon 
 change,not does Because B
 
 
Lepton number: 
1 → 1 − 1 + 1 = 1 conserved. isnumber lepton 
 change,not does Because L
 
 
allowed. is and lawson conservati all satisfieddecay The 
Chapter 41 
 
 
1428 
Remarks: The decay in Part (c) is the decay process for the muon µ− (see Example 
41-2). 
 
*39 •• 
Picture the Problem We can systematically determine Q, B, S, and s for each reaction 
and then use these values to identify the unknown particles. 
 
(a) For the strong reaction: p + π − → Σo + ? 
 
Charge number: +1 − 1 = 0 + Q ⇒ Q = 0 
 
Baryon number: +1 + 0 = +1 + B ⇒ B = 0 
 
Strangeness: 0 + 0 = −1 + S ⇒ S = +1 
 
Spin: 
 
+ 21 + 0 = + 21 + s ⇒ s = 0 
.K kaon, theis particle that theindicate properties These 0 
 
(b) For the strong reaction: p + p → π + + n + K+ + ? 
 
Charge number: +1 + 1 = +1 + 0 + 1 + Q ⇒ Q = 0 
 
Baryon number: +1 + 1 = 0 + 1 + 0 + B ⇒ B = +1 
 
Strangeness: 0 + 0 = 0 + 0 + 1 + S ⇒ S = −1 
 
Spin: 
 
+ 21 + 21 = 0 + 21 + 0 + s ⇒ s = + 21 
.baryon or the either the is particle that theindicate properties These 00 ΛΣ 
 
(c) For the strong reaction: p + K − → Ξ− + ? 
 
Charge number: +1 − 1 = −1 + Q ⇒ Q = +1 
 
Baryon number: +1 + 0 = +1 + B ⇒ B = 0 
 
Strangeness: 0 − 1 = −2 + S ⇒ S = −1 
 
Spin: 
 
+ 21 + 0 = + 21 + s ⇒ s = 0 
.K kaon, theis particle that theindicate properties These + 
 
40 •• 
Picture the Problem We can systematically determine Q, B, S, and s for the reaction and 
then use these values to identify the unknown particle. The Q value for the reaction is 
given by ( ) 2cmQ ∆−= and the expression for the threshold energy for the reaction is 
given in the problem statement. 
Elementary Particles and the Beginning of the Universe 
 
 
1429
 
(a) For the strong reaction: p + p → Λo + Ko + p + ? 
 
Charge number: +1 + 1 = 0 + 0 + 1 + Q ⇒ Q = +1 
 
baryon number: +1 + 1 =+1 + 0 + 1 + B ⇒ B = 0 
 
strangeness: 0 + 0 = −1 + 1 + 0 + S ⇒ S = 0 
 
spin: 
 
+ 21 + 21 = + 21 + 0 + 21 + s ⇒ s = 0 
. pion, a is particleunknown that theindicate properties These +π 
 
(b) The reaction is: p + p → Λo + Ko + p + π + 
 
The Q-value for the reaction is: 
 ( ) ( )[ ] 2pKpp oo cMMMMmmQ ++++−+= Λ π 
 
Use Table 41-1 to find the mass-energy values: 
 
( ) ( )[ ] MeV815MeV139.6938.3497.71116938.3938.3 −=+++−+=Q 
Because Q < 0, the reaction is endothermic. 
 
(c) The threshold energy for this reaction is: 
 
( )+oo pKpp
p
th 2 π
MMMMmm
m
QK +++++−= Λ 
 
Using Table 41-1 to find the mass-energy values, substitute numerical values and 
evaluate Kth: 
 
( ) ( )
GeV984.1MeV1984
MeV139.6938.3497.71116938.3938.3
MeV 938.32
MeV 815
th
==
+++++−−=K
 
 
41 •• 
Picture the Problem We can solve the equation derived in Problem 31 for the 
recessional velocity of the galaxy and then use Hubble’s equation to estimate the distance 
to the galaxy. 
 
(a) From Problem 31 we have: 
cv
cv'
/1
/1
0 −
+= λλ 
 
Chapter 41 
 
 
1430 
Solve for the radicand: 
 
2
0/1
/1
⎟⎟⎠
⎞
⎜⎜⎝
⎛=−
+
λ
λ'
cv
cv
 
 
Substitute numerical values for λ′ 
and λ0: 935.4nm 656.3
nm 1458
/1
/1 2 =⎟⎠
⎞⎜⎝
⎛=−
+
cv
cv
 
 
Simplify to obtain: 
 c
v
c
v +=⎟⎠
⎞⎜⎝
⎛ − 114.953 
and 
3.9535.953 =
c
v
 
 
Solve for v: ( )
km/s1099.1m/s1099.1
m/s103664.0664.0
58
8
×=×=
×== cv
 
 
(b) From the Hubble equation we 
have: H
vr = 
 
Substitute numerical values and 
evaluate r: y1065.8
y10
km/s 23
km/s101.99 9
6
5
⋅×=
⋅
×= c
c
r 
 
 42 ••• 
Picture the Problem We can use the masses of the parent and daughters to find the total 
kinetic energy of the decay products under the assumption that the Λ0 is initially at rest. 
Application of conservation of energy and the definition of kinetic energy will yield the 
ratio of the kinetic energy of the pion to the kinetic energy of the proton. Finally, we can 
use our results in (a) and (b) to find the kinetic energies of the proton and the pion for this 
decay. 
 
(a) The total kinetic energy of the 
decay products is given by: 
( ) 2ptot cmmmK π−−= Λ 
 
 
Substitute numerical values (see Table 41-1) and evaluate Ktot: 
 
MeV1.38MeV6.139MeV3.938MeV1116 2222tot =⎟⎠
⎞⎜⎝
⎛ −−= c
ccc
K 
 
(b) The ratio of the kinetic energies 
is given by: 
 
2
pp
2
2
pp2
1
2
2
1
p vm
vm
vm
vm
K
K πππππ == 
Elementary Particles and the Beginning of the Universe 
 
 
1431
Use conservation of momentum 
(nonrelativistic) to obtain: 
 
ppvmvm =ππ ⇒ 
π
π
m
m
v
v p
p
= 
Substitute for the ratio of the speeds 
to obtain: 
ππ
ππ
m
m
m
m
m
m
K
K p
2
p
pp
=⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
 
Substitute numerical values and 
evaluate the ratio of the kinetic 
energies: 
72.6MeV6.139
MeV3.938
2
2
p
==
c
c
K
Kπ 
 
(c) Express the total kinetic energy 
in terms of Kπ and Kp: 
 
totp KKK =+ π (1) 
Use our results in (a) and (b) to 
obtain: 
 
MeV1.3872.6 pp =+ KK 
Solve for Kp: MeV94.4p =K 
 
Substitute in equation (1) to obtain: MeV2.33ptot =−= KKKπ 
 
*43 ••• 
Picture the Problem The total kinetic energy of the decay products is the rest energy of 
the Σ0 particle. We can find the momentum of the photon from its energy and use the 
conservation of momentum to calculate the kinetic energy of the Λ0. 
 
(a) The total kinetic energy of the 
decay products is given by: 
( ) 2tot cmK Σ= 
 
 
Substitute numerical values (see 
Table 41-1) and evaluate Ktot: 
 
MeV1193MeV1193 22tot =⎟⎠
⎞⎜⎝
⎛= c
c
K 
(b) The momentum of the photon is 
given by: 
 
c
cmE
c
E
p
2
Λ−== γγ 
 
Substitute numerical values and evaluate 
pγ: 
 
c
c
c
cp
MeV0.77
MeV1116MeV1193 22
=
⎟⎠
⎞⎜⎝
⎛−
=γ 
Chapter 41 
 
 
1432 
(c) The kinetic energy of the Λ0 is 
given by: 
 
Λ
Λ
Λ = m
pK
2
2
 
or, because γpp =Λ , 
Λ
Λ = m
p
K
2
2
γ 
 
Substitute numerical values and 
evaluate KΛ: 
 MeV66.2MeV11162
MeV0.77
2
2
=
⎟⎠
⎞⎜⎝
⎛
⎟⎠
⎞⎜⎝
⎛
=Λ
c
cK 
 
(d) A better estimate of the energy 
of the photon is: 
 
ΛΛ −−= KcmEE 2γ 
Substitute numerical values and evaluate Eγ: 
 
MeV3.74MeV66.2MeV1116MeV1193 22 =−⎟⎠
⎞⎜⎝
⎛−= c
c
Eγ 
 
The improved estimate of the momentum 
of the photon is: ccc
Ep MeV3.74MeV3.74 === λγ 
 
44 ••• 
Picture the Problem The solution strategy is outlined in the problem statement. 
 
(a) Express ∆t = t2 − t1 in terms of u2 
and u1: 
( )
21
21
12
12 uu
uux
u
x
u
xttt −=−=−=∆ 
 
Noting that u1u2 ≈ c2, we have: 
 2c
uxt ∆≈∆(1) 
where ∆u = u1 − u2 
 
(b) From Equation 39-25 we have: 212222
11 ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−=
E
mc
E
mc
c
u
 
 
Expand binomially to obtain: 22
2
11 ⎟⎟⎠
⎞
⎜⎜⎝
⎛−=
E
mc
c
u
 
 
Elementary Particles and the Beginning of the Universe 
 
 
1433
(c) Express u1 − u2 in terms of E1, 
E2, and mc2: 
 
( )
( ) ( )
2
2
2
1
2
2
2
1
22
2
1
2
2
22
2
1
21
2
11
EE
EEmcc
EE
mcuu
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=−
 
 
Substitute numerical values and evaluate ∆u: 
 
( ) ( )[ ]
( ) ( ) c
c
c
c
u 1222
22
2
2
2
1050.7
MeV5MeV202
MeV5MeV20eV20
−×=
−⎟⎠
⎞⎜⎝
⎛
=∆ 
 
Use equation (1) to evaluate ∆t: ( )( )
s2.40
y
Ms31.56y10275.1
105.7y107.1
6
2
125
=
××=
×⋅×≈∆
−
−
c
cct
 
 
(d) Using mc2 = 40 eV for the rest energy of a neutrino: 
 
( ) ( )[ ]
( ) ( ) c
c
c
c
u 1122
22
2
2
2
1000.3
MeV5MeV202
MeV5MeV20eV40
−×=
−⎟⎠
⎞⎜⎝
⎛
=∆ 
 
Use equation (1) to evaluate ∆t: ( )( )
s161
y
Ms31.56y101.5
103y107.1
6
2
115
=
××=
×⋅×≈∆
−
−
c
cct
 
 
Remarks: Note that the spread in the arrival time for neutrinos from a supernova 
can be used to estimate the mass of a neutrino.