serway1 (1)

Particle Physics and Cosmology 601

P46.25 (a) Ξ− → Λ0 + µ− + νµ
Baryon number: +1 → +1+ 0 + 0
Le : 0 → 0 + 0 + 0 Charge: −1 → 0 −1+ 0
Lτ : 0 → 0 + 0 + 0 Lµ : 0 → 0+1+1
Conserved quantities are: Strangeness: −2 → −1+ 0 + 0

(b) K 0 → 2π 0 B, charge, Le, and Lτ
S
0→0
Baryon number: 0→0 Charge: 0→0
Lµ : +1 → 0
Le : 0 → 0
Strangeness: B, charge, Le, Lµ , and Lτ
Lτ : 0 → 0
−1+1 → 0 + 0
Conserved quantities are: 0+0 → 0+0
−1+ 0 → −1+ 0
(c) K − + p → Σ0 + n Charge: S, charge, Le, Lµ , and Lτ
Baryon number: 0 +1 → 1+1 Lµ :
Le : 0 + 0 → 0 + 0 Strangeness: 0→0
Lτ : 0 + 0 → 0 + 0 0 → 0+0
Conserved quantities are: −1 → −1+ 0

(d) Σ0 + Λ0 + γ Charge: B, S, charge, Le, Lµ , and Lτ
Baryon number: +1 → 1+ 0 Lµ :
Le : 0 → 0 + 0 Strangeness: +1 −1 → +1 −1
Lτ : 0 → 0 + 0 0 + 0 → +1 −1
Conserved quantities are: 0+0 → 0+0

(e) e+ + e− → µ+ + µ− Charge: B, S, charge, Le, Lµ , and Lτ
Baryon number: 0 + 0 → 0 + 0
Lµ : −1+ 0 → 0 −1
Le : −1+ 1 → 0 + 0 Strangeness: 0+0 → 0+0
Lτ : 0 + 0 → 0 + 0 0 + 0 → +1 −1
Conserved quantities are:
B, S, charge, Le, Lµ , and Lτ
(f) p + n → Λ0 + Σ−

Baryon number: −1+1 → −1+1 Charge:
Le : 0 + 0 → 0 + 0
Lτ : 0 + 0 → 0 + 0 Lµ :
Conserved quantities are: Strangeness:

602 Chapter 46

P46.26 (a) K+ + p → ? + p

The strong interaction conserves everything.

Baryon number, 0+1→ B+1 so B = 0
so Q = +1
Charge, +1+1 → Q +1 so Le = Lµ = Lτ = 0
so S = 1
Lepton numbers, 0+0 → L+0

Strangeness, +1+ 0 → S + 0

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 → B + 0 so B = 1

Charge, −1 → Q −1 so Q = 0

Lepton numbers, 0→L+0 so Le = Lµ = Lτ = 0

Strangeness, −3 → S + 0 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 → B+0+0 so B = 0
Charge, +1 → Q +1+ 0 so Q = 0

Lepton numbers, Le, 0 → Le + 0 + 0 so Le = 0
Strangeness, Lµ , 0 → Lµ − 1+ 1 so Lµ = 0
Lτ , 0 → Lτ + 0 + 0 so Lτ = 0
1→S+0+0 so ∆S = ±1

(for weak interaction): S = 0

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

P46.27 Time-dilated lifetime:

T = γ T0 = 0.900 ×10−10 s = 0.900 ×10−10 s = 3.214 ×10−10 s
1− v2 c2 1− (0.960)2

( )( )distance = 0.960 3.00 ×108 m s 3.214 ×10−10 s = 9.26 cm

Particle Physics and Cosmology 603

P46.28 (a) ( )1.602 177 ×10−19 C (1.15 T) (1.99 m) 686 MeV
c
pΣ+ = eBrΣ+ = 5.344 288 × 10−22 (kg ⋅ m s) (MeV c) =

p = eBr = ( )1.602 177 ×10−19 C (1.15 T)(0.580 m) = 200 MeV
π π c
+ + 5.344 288 ×10−22 (kg ⋅ m s) (MeV 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:

pn cosφ + (199.961 581 MeV c) cos 64.5° = 686.075 081 MeV c (1)
(2)
∴ pn cosφ = 599.989 401 MeV c

pn sinφ = (199.961 581 MeV c)sin 64.5° = 180.482 380 MeV c

From (1) and (2): pn = (599.989 401 MeV c)2 + (180.482 380 MeV c)2

= 627 MeV c

( ) ( )(c) E = p c 2+ m c 2 2 = (199.961 581 MeV)2 + (139.6 MeV)2 = 244 MeV
π π π
+ + +

( )En = ( pnc)2 + mnc2 2 = (626.547 022 MeV)2 + (939.6 MeV)2 = 1130 MeV

EΣ+ = E + + En = 243.870 445 MeV +1129.340 219 MeV = 1 370 MeV
π

pΣ+ c 2 = (1 373.210 664 MeV)2 − (686.075 081 MeV)2 = 1190 MeV
( )(d) E2
mΣ+ c2 = Σ+ −

∴ mΣ+ = 1190 MeV c2

EΣ+ = γ mΣ+ c2 , where γ = ⎛ − v2 ⎞−1 2 = 1 373.210 664 MeV = 1.154 4
⎜1 c2 ⎟ 1189.541 303 MeV
⎝ ⎠

Solving for v, we find v = 0.500c .

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:

( ) ( )Emin + m2c2 = m3c2 2 + p3c 2 (1)
(2)
By conservation of momentum: p3 = p1

m1c2 2
( )( ) ( )∴ 2
p3c 2 = p1c = E2 −
min

( ) ( )Emin + m2c2 = 2 2
Substitute (2) in (1): m3c2 + E2 − m1c2
min

Square both sides:

( ) ( ) ( )E2 2= 2 2
min
+ 2Emin m2c2 + m2 c 2 m3c2 + E2 − m1c2
min

( )∴ Emin =
m32 − m12 − m22 c2
2m2
c2 = ⎡⎣m32 − ⎤
( ) ( )∴ Kmin = Emin − m1c2 = 2 ⎦ c2
m32 − m12 − m22 − 2m1m2 m1 + m2
2m2
2m2

Refer to Table 46.2 for the particle masses.

Kmin = ⎡⎣4 (938.3)⎤⎦2 MeV2 c2 − ⎡⎣2 (938.3)⎦⎤2 MeV2 c2 5.63 GeV
=
2 938.3 MeV c2
( )(b)

(c) Kmin = (497.7 +1115.6)2 MeV2 c2 − (139.6 + 938.3)2 MeV2 c2 768 MeV
2 (938.3) MeV c2 =

(d) K min = ⎡⎣2 (938.3) +135⎤⎦2 MeV2 c2 − ⎡⎣2 (938.3)⎦⎤2 MeV2 c2 280 MeV
2 (938.3) MeV c2 =

⎡ 2 − ⎡⎣(938.3 + ⎤ ⎤
Kmin = ⎢⎣ ⎦ ⎦⎥
( )(e) 938.3)2 MeV2 c2
91.2 ×103

2 (938.3) MeV c2 = 4.43 TeV

Particle Physics and Cosmology 605

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 g ⎛ 6.02 × 1023 molecules ⎞ ⎛ 10 protons ⎞ = 3.34 × 1026 protons
⎜ 18.0 g ⎟ ⎝⎜ molecule ⎠⎟
⎝ ⎠

and there are Nn = 1 000 g ⎛ 6.02 × 1023 molecules ⎞ ⎛ 8 neutrons ⎞ = 2.68 ×1026 neutrons
⎜ 18.0 g ⎟ ⎝⎜ molecule ⎟⎠
⎝ ⎠

So there are for electric neutrality 3.34 ×1026 electrons

The up quarks have number ( )2 3.34 ×1026 + 2.68 ×1026 = 9.36 ×1026 up quarks

and there are ( )2 2.68 ×1026 + 3.34 ×1026 = 8.70 ×1026 down quarks

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

( )65 3.34 ×1026 ~1028 electrons
( )65 9.36 ×1026 ~1029 up quarks
( )65 8.70 ×1026 ~1029 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 00
baryon number
charge 1 1/3 1/3 1/3 1

e 2e/3 2e/3 –e/3 e

(b) neutron u d d total
0 0 0 00
strangeness 1 1/3 1/3 1/3 1
baryon number 0 2e/3 –e/3 –e/3 0
charge

606 Chapter 46

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

mp = 2 mu + md (1)

and mn = mu + 2 md (2)
Solving simultaneously,

1 = 13⎣⎡2
( )( )mu=3 c2 − 939.6 meV c2 ⎦⎤ = 312 MeV c2
we find 2 mp − mn 938 MeV

and from either (1) or (2), md = 314 MeV c2 .

P46.33 (a) K0 d s total

strangeness 1 0 11
baryon number 0 1/3 –1/3 0
charge 0 –e/3 e/3 0

(b) Λ0 u d s total

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

0 2e/3 –e/3 –e/3 0

P46.34 In the first 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 → K0 + n, 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 + Λ0 ud + uud → ds + uds
In terms of constituent quarks:

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

(c) K− + p → K+ + K0 + Ω− us + uud → us + ds + sss or 1 → 1
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,

continued on next page

Particle Physics and Cosmology 607

(d) p + p → K0 + p + 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 = Λ0 or Σ0

P46.36 Σ0 + p → Σ+ + 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− charge = ⎛⎜⎝− 2 ⎞ + ⎛⎝⎜− 2 ⎞ + ⎛ 1 ⎞ = −e . This is the antiproton .
P46.38 (a) uud: 3 e⎠⎟ 3 e⎠⎟ ⎝⎜ 3 e⎟⎠

(b) udd: charge = ⎝⎛⎜− 2 ⎞ + ⎛ 1 ⎞ + ⎛ 1 ⎞ = 0 . This is the antineutron .
3 e⎠⎟ ⎝⎜ 3 e⎟⎠ ⎝⎜ 3 e⎟⎠

Section 46.11 The Cosmic Connection

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

(a) v = Hr = 17 × 10−3 m 1.85 m ⎛ 1 ly ⎞ ⎛ c m s ⎞ ⎛ 1 yr 7 ⎞
s ⋅ ly ⎜ c ⋅1 yr ⎟ ⎜ 3 × 10 8 ⎟ ⎝⎜ 3.156 × 10 s ⎠⎟
⎝ ⎠ ⎝ ⎠

= 17 ×10−3 m 1.85 m = 1.80 ×10−18 11.85 m = 3.32 ×10−18 ms
s ⋅ 9.47 ×1015 m s

This is unobservably small.

(b) v = Hr = 1.80 ×10−18 1 3.84 ×108 m = 6.90 ×10−10 m s again too small to measure
s

608 Chapter 46

P46.40 Section 39.4 says f = fobserver source 1+ va c
1− va c

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

Then, c = c 1− v c and λ′ = λ 1+ v c
λ′ λ 1+ v c 1−v c

P46.41 (a) λ′ = λ 1+ v c 510 nm = 434 nm 1+ v c
(b) 1−v c 1−v c

1.182 = 1+ v c = 1.381
1−v c

1+ v = 1.381−1.381 v 2.38 v = 0.381
cc c

v = 0.160 or v = 0.160c = 4.80 ×107 m s
c

v = HR : R= v = 4.80 ×107 m s = 2.82 ×109 ly
H 17 ×10−3 m s⋅ ly

P46.42 (a) λn′ = λn 1+ v c = (Z +1) λn 1+ v c = (Z +1)2
1− v c
1−v c

v (Z + 1)2 ⎛ v ⎞⎟⎠(Z + 1)2 ( )⎛ v ⎞ = Z2 +2Z
c ⎜⎝ c c ⎠⎟
⎜⎝
1+ = − Z2 +2Z+2

v= c ⎛ Z Z2 + 2Z 2 ⎞
⎝⎜ 2+ 2Z + ⎠⎟

(b) R= v = c ⎛ Z2 + 2Z ⎞
H H ⎝⎜ Z 2 + 2Z + 2⎠⎟

P46.43 v = HR ( )1.7 ×10−2 m s

H=
ly

( )(a) v 2.00 ×106 ly = 3.4 ×104 m s λ′ = λ 1+ v c = 590 (1.000 113 3) = 590.07 nm

1−v c

( )(b) v 2.00 ×108 ly = 3.4 ×106 m s λ′ = 590 1+ 0.011 33 = 597 nm
1− 0.011 33

( )(c) v 2.00 ×109 ly = 3.4 ×107 m s λ′ = 590 1+ 0.113 3 = 661 nm
1− 0.113 3

Particle Physics and Cosmology 609

*P46.44 (a) What we can see is limited by the finite age of the Universe and by the finite 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)p r3 where
r = 13.7 billion light-years. The rate of volume increase is

dV = d 4 π r 3 dr ⎛ 3 ×108 m 3.156 × 107 s ⎞2
dt 3 dt 4π ⎜13.7 ×109 s ⎟
= 4 π 3r2 = 4πr2c = ly ⎠ 3 ×108 m
dt 3 ⎝ 1 ly s

= 6.34 × 1061 m3/s

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

( )V = 4 πr3 = 4 π 1.496 ×1011 m 3 = 1.40 ×1034 m3
33
m = ρV = 6 ×10−28 kg m3 1.40 ×1034 m3 = 8.41×106 kg

(b) By Gauss’s law, the dark matter would create a gravitational field 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

8.41×106 kg = 4.23×10−24 of the mass of the Sun. It is immeasurably small .
1.99 ×1030 kg

P46.46 (a) Wien’s law: λmaxT = 2.898 ×10−3 m ⋅ K

Thus, λmax = 2.898 ×10−3 m⋅K = 2.898 ×10−3 m ⋅ K = 1.06 ×10−3 m = 1.06 mm
T 2.73 K
.

(b) This is a microwave .

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

( )I = eσT 4 = (1) 5.67 ×10−8 W m2 ⋅ K4 (2.73 K)4 = 3.15 ×10−6 W m2

As a bonus, we can find 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 fireball is a perfect sphere of
radius fourteen billion light years, centered at the point halfway between your eyes:

⎛ 3 × 10 8 m s ⎞ 2
⎜ ⎟
2 2

s yr
( ) ( ) ( )P = IA = I (4πr2 ) =
3.15 ×10−6 W m2 (4π ) 14 ×109 ly ⎝ 1 ly yr ⎠ 3.156 ×107

P = 7 ×1047 W

610 Chapter 46

P46.48 The density of the Universe is

ρ = 1.20ρc = 1.20 ⎛ 3H 2 ⎞
⎜ 8π G ⎟
⎝ ⎠

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

M = ρ ⎛ 4 π r 3 ⎞ = 1.20 ⎛ 3H 2 ⎞ ⎛ 4 π r 3 ⎞ = 0.600 H 2r3
⎝ 3 ⎠ ⎜⎝ 8π G ⎟⎠ ⎝ 3 ⎠ 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 mv2 − GmM = 0 − GmM so 1 mH 2r2 − Gm ⎛ 0.600 H 2r3 ⎞ = 0 − Gm ⎛ 0.600 H 2r3 ⎞
2 r ⎜⎝ G ⎟⎠ R ⎝⎜ G ⎠⎟
2r R

−0.100 = −0.600 r so R = 6.00r
R

The Universe will expand by a factor of 6.00 from its current dimensions.

P46.49 (a) kBT ≈ 2mpc2

T ≈ 2mpc2 = 2 (938.3 MeV) ⎛1.60 ×10−13 J ⎞ ~ 1013 K
kB ⎜⎟
1.38 ×10−23 J K ⎝ 1 MeV ⎠
( )so

(b) kBT ≈ 2mec2

T ≈ 2mec2 = 2 (0.511 MeV) ⎛1.60 ×10−13 J ⎞ ~ 1010 K
kB ⎜⎟
1.38 ×10−23 J K ⎝ 1 MeV ⎠
( )so

P46.50 (a) The Hubble constant is defined in v = HR. The distance R between any two far-separated

objects opens at constant speed according to R = v t. Then the time t since the Big Bang is

found from

v = H vt 1 = Ht t= 1
H

(b) 1 = 1 ⎛ 3 × 108 m s ⎞ = 1.76 ×1010 yr = 17.6 billion years
H 10−3 ⎜ yr ⎟
17 × m s ⋅ ly ⎝ 1 ly ⎠

Particle Physics and Cosmology 611

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 mv2 − GMm = 0
2R

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 . Then
dt
⎛ ⎞2 R T
⎜ dR ⎟ = 2 GM dR = R−1/2 2 GM R dR =
∫ 2GM ∫ d t
⎝dt ⎠ R dt 0 0

R3/2 R = T 2 R3/2 = 2 GM T
3
2 GM t T = 2 R3 2 = 2
32 0
0 3 2 GM 3

R
2GM R

From above, 2GM = v
R
so
Now Hubble’s law says T=2R
So 3v

v = HR

T=2 R = 2
3 HR 3H

T = 2 ⎛ 3×108 m s ⎞ = 1.18 ×1010 yr = 11.8 billion years
17 ×10−3 ⎜ ⎟
( )(b)
3 m s ⋅ ly ⎝ 1 ly yr ⎠

P46.52 In our frame of reference, Hubble’s law is exemplified by v1 = H R1 and v2 = H R2 . From

( )these we may form the equations −v1 = −H R1 and v2 − v1 = H R2 − R1 . These equations

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

( ) ( )find −v1 = H −R1 and as she looks at cluster two to find v2 − v1 = H R2 − R1 .

Section 46.12 Problems and Perspectives

( ( )( ) )L =
*P46.53 (a) G= 1.055 ×10−34 J ⋅ s 6.67 ×10−11 N ⋅ m2 kg2 = 1.61×10−35 m
(b) c3 3.00 ×108 m s 3

The Planck time is given as T = L = 1.61×10−35 m = 5.38 ×10−44 s ,
c 3.00 ×108 m s

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 fireball 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 fire hide all information from still earlier times.

612 Chapter 46

Additional Problems

P46.54 We find the number N of neutrinos:

( )1046 J = N (6 MeV) = N 6 ×1.60 ×10−13 J

N = 1.0 ×1058 neutrinos

The intensity at our location is

1.0 ×1058 ⎛ 1 ly ⎞2
1.7 ×105 ly 2 ⎜⎝⎜ ms ⎟⎟⎠ = 3.1×1014
( ) ( )( )N = N= m−2
4π
A 4πr2
3.00 ×108 3.16 ×107 s

The number passing through a body presenting 5 000 cm2 = 0.50 m2

( )⎛ 1⎞
m2 ⎠⎟
⎜⎝
is then 3.1 × 1014 0.50 m2 = 1.5 ×1014

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

{ ) ) }( (v = 170 000 yr = 1 = 1
1.86 ×
c 170 000 yr + 10 s 1 + 10 s ⎡⎣ 1.7 × 105 yr 3.156 × 107 s yr ⎤⎦ 1 + 10 −12

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

1+1.86 ×10−12 2 −1
1+1.86 ×10−12 2
( )mc2 = E = E 1− v2
( ) ( )γ c2
= 10 MeV 1− 1 2 = 10 MeV
1+1.86 ×10−12

( ) ( )mc2 ≈ 10 MeV 2 1.86 ×10−12 = 10 MeV 1.93×10−6 = 19 eV
1

Then the upper limit on the mass is

m = 19 eV or m = 19 eV ⎛ u ⎞ = 2.1× 10−8 u
c2 c2 ⎜ 10 ⎟
⎝ 931.5 × 6 eV c2 ⎠

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

(b) µ− → π − + νe 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)⎣⎡4π 2 ⎤ 1023
1.5 × 1011 ⎦ W = 1.1 × W

Over 109 years, the Sun emits 3.6 × 1039 J in neutrinos. This represents an annihilated mass

m c2 = 3.6 × 1039 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.

Particle Physics and Cosmology 613

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 pc2 + Mπ c2 + M X c2 = M pc2 + Kp + M pc2 + Kp

M X c2 = M pc2 + 2Kp − Mπ c2 = 938.3 MeV + 2(70.4 MeV) − 139.6 MeV = 939.5 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 2Nmc2 . The resulting photon
momentum is p = E = 2Nmc2 = 2Nmc. Since the momentum of the system is conserved,
cc
the rocket will have momentum 2Nmc directed opposite the photon momentum.

p = 2Nmc

(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).

( )E2 = p2c2 + mc2 2
( ) ( )Thus 2mc2 2 = p2c2 + mc2 2

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

( ) ( ) ( )first particle. Thus 4 mc2 2 = p2c2 + mc2 2 , p2 = 3 mc2 2 . So p = 3mc.

NN
This process is repeated N times (annihilate protons and antiprotons). Thus the total

22
momentum acquired by the ejected particles is 3Nmc, and this momentum is imparted to
the rocket.

p = 3Nmc

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

P46.60 By relativistic energy conservation in the reaction, Eγ + mec2 = 3mec2 (1)
By relativistic momentum conservation for the system, 1− v2 c2 (2)

Eγ = 3mev
c 1− v2 c2

Dividing (2) by (1), X = Eγ Eγ = v
+ mec2 c

Subtracting (2) from (1), mec2 = 3mec2 − 3mec2 X
1− X2 1− X2

Solving, 1 = 3− 3X and X= 4 so Eγ = 4mec2 = 2.04 MeV
1− X2 5

614 Chapter 46

P46.61 mΛc2 = 1115.6 MeV Λ0 → p +π−
mpc2 = 938.3 MeV mπ c2 = 139.6 MeV

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

K p + Kπ = 37.7 MeV and pp = pπ = p

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

⎡ (938.3)2 + p2c2 − 938.3⎥⎦⎤ + ⎡ (139.6)2 + p2c2 −139.6⎦⎥⎤ = 37.7 MeV
⎣⎢ ⎢⎣

Solving the algebra yields
pπ c = ppc = 100.4 MeV

Then, ( )K p = mpc2 2 + (100.4)2 − mpc2 = 5.35 MeV

Kπ = (139.6)2 + (100.4)2 − 139.6 = 32.3 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 mpc2 + K p = mpc2 + mnc2 + mπ c2

so the kinetic energy of each of the incident protons is

Kp = mn c 2 + mπ c2 − mpc2 = (939.6 +139.6 − 938.3) MeV = 70.4 MeV
2
2

P46.63 Σ0 → Λ0 + γ

From Table 46.2, mΣ = 1192.5 MeV c2 and mΛ = 1115.6 MeV c2

Conservation of energy in the decay requires

( )E0,Σ = Eo,Λ + KΛ + Eγ or 1192.5 MeV = ⎛ MeV + pΛ 2 ⎞ Eγ
⎜1115.6 2mΛ ⎟+
⎝ ⎠

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

1192.5 MeV = ⎛ MeV + pγ 2 ⎞ Eγ
⎝⎜⎜1115.6 2mΛ ⎠⎟⎟ +

or 1192.5 MeV = ⎛ MeV + pγ 2c2 ⎞ Eγ
⎜⎜⎝1115.6 2 mΛ c2 ⎟⎠⎟ +

Recognizing that mΛc2 = 1115.6 MeV and pγ c = Eγ
we now have
1192.5 MeV = 1115.6 MeV + Eγ 2 MeV) + Eγ

2 (1115.6

Solving this quadratic equation gives Eγ = 74.4 MeV

Particle Physics and Cosmology 615

P46.64 The momentum of the proton is ( )qBr = 1.60 ×10−19 C (0.250 kg C ⋅ s)(1.33 m)
pp = 5.32 × 10−20 kg ⋅ m s
cpp = 1.60 ×10−11 kg ⋅ m2 s2 = 1.60 ×10−11 J = 99.8 MeV

Therefore pp = 99.8 MeV c Ep = E02 + (cp)2 = (938.3)2 + (99.8)2 = 944 MeV
The total energy of the proton is

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

pπ = 99.8 MeV c E0π = 139.6 MeV

Eπ = E02 + (cp)2 = (139.6)2 + (99.8)2 = 172 MeV

Thus Etotal after = Etotal before = Rest energy

Rest energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV (This is a Λ0 particle!)
Mass = 1116 MeV c2 .

P46.65 π − → µ− + νµ : From the conservation laws for the decay, [1]
mπ c2 = 139.6 MeV = Eµ + Eν
and pµ = pν , Eν = pν c : [2]
or ( )Eµ2 = pµc 2 + (105.7 MeV)2 = ( pν c)2 + (105.7 MeV)2 [1]
Since [2]
and Eµ2 − Eν2 = (105.7 MeV)2 [3]
then
Subtracting [3] from [1], Eµ + Eν = 139.6 MeV

( )( )Eµ + Eν Eµ − Eν = (105.7 MeV)2

Eµ − Eν = (105.7 MeV)2 = 80.0

139.6 MeV

2Eν = 59.6 MeV and Eν = 29.8 MeV

P46.66 The expression e−E kBT dE 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−E kBT dE
e−E kBT −dE kBT e−E kBT ∞

∫ ∫ ( )=E E E −E kBT

∞∞ = e = e−E kBT ∞
e−E kBT dE e−E kBT −dE kBT 0

00

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

and E = 1.00 eV = 1.60 ×10−19 J

Thus, ( ) ( )0.010 0 = e− 1.60×10−19 J 1.38×10−23 J K T

ln(0.010 0) = − 1.60 × 10−19 J = − 1.16 × 104 K giving T = 2.52 × 103 K
1.38 × 10−23 J K T
( )or T

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+ .

FIG. P46.67

P46.68 (a) The mediator of this weak interaction is a Z0 boson .

(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 . FIG. P46.68

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 sufficiently 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 + γ → p + π 0 . Energy conservation gives

mp c 2 c2 + Eγ = mpc2 + mπ c2
1− u2

Momentum conservation gives mpu − Eγ = 0.
1− u2 c2 c

Combining the equations, we have

mpc2 + mp c 2 c2 u = mpc2 + mπ c2
1− u2 c2 1− u2 c

938.3 MeV(1+ u c) = 938.3 MeV +135.0 MeV
(1− u c)(1+ u c)

so u = 0.134
c

and Eγ = 127 MeV

(b) λmaxT = 2.898 mm ⋅ K

λmax = 2.898 mm ⋅K = 1.06 mm
2.73 K

continued on next page

Particle Physics and Cosmology 617

(c) Eγ = hf = hc = 1 240 eV ⋅10−9 m = 1.17 ×10−3 eV
λ 1.06 ×10−3 m
(d) In the primed reference frame, the proton is moving to the right at u′ = 0.134 and
c
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 ×108 = 1+ v c 1.17 ×10−3
1−v c

v = 1 −1.71×10−22
c

Then the speed of the proton is given by

u′ c+ v c 0.134 +1−1.71×10−22
1+ u′v c2 1+ 0.134 1−1.71×10−22
( )u= = = 1−1.30 ×10−22

c

And the energy of the proton is

mpc2 = 938.3 MeV = 6.19 ×1010 × 938.3×106 eV = 5.81×1019 eV

( )1− u2 c2 1− 1−1.30 ×10−22 2

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 field 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 infinite range and zero mass for the
field 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 0.828c

P46.20 (a) νe (b) νµ (c) νµ (d) νµ +ντ
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 and 200 MeV (b) 627 MeV c (c) 244 MeV, 1130 MeV, 1 370 MeV
cc

(d) 1190 MeV c2, 0.500c

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 md = 314 MeV/c2
P46.34 See the solution.

P46.36 a neutron, udd

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

P46.40 See the solution.
P46.42
P46.44 ⎛ Z2 + 2Z ⎞ c ⎛ Z2 + 2Z ⎞
(a) v = c ⎝⎜ Z2 + 2Z + 2 ⎟⎠ (b) ⎜ ⎟
H ⎝ Z 2 + 2Z + 2 ⎠

(a) What we can see is limited by the finite age of the Universe and by the finite 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

P46.52 See the solution.

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