Tài liệu Physics exercises_solution: Chapter 44 pdf

44.1: a)
2
22
2
1547.01
1
1
mc
cv
mcK













J1027.1sokg,10109.9
1431 
 Km
b) The total energy of each electron or positron is

22
1547.1 mcmcKE
J.1046.9

14

The total energy of the electron and positron is converted into the total
energy of the two photons. The initial momentum of the system in the lab frame is zero
(since the equal-mass particles have equal speeds in opposite directions), so the final
momentum must also be zero. The photons must have equal wavelengths and must be
traveling in opposite directions. Equal
λ
means equal energy, so each photon has energy
J.1046.9
14

c)
pm10.2)J1046.9(λsoλ
14


hcEhchcE
The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also
have kinetic energy, the energy of each photon is greater, so its wavelength is less.
44.2:
The total energy of the positron is

MeV.5.51MeV0.511MeV00.5
2
 mcKE
We can calculate the speed of the positron from Eq. 37.38
.996.0
MeV5.51
MeV511.0

11
1
2
2
22
2
2





















E
mc

c
vmc
E
c
v
44.3:
Each photon gets half of the energy of the pion
ray.gammam108.1
Hz107.1
sm1000.3
λ
Hz107.1
s)J1063.6(
)eVJ106.1()eV109.6(
MeV69MeV)511.0()270(
2
1
)270(
2
1
2
1
14
22
8
22
34
197
2
e

2
γ












f
c
h
E
f
cmcmE

44.4: a)
)sm10(3.00kg)10(9.11(207)
s)J10626.6(
λ
831
34
2






cm
h
cm
hc
E
hc
μμ

pm.0.0117m1017.1
14


In this case, the muons are created at rest (no kinetic energy). b) Shorter wavelengths
would mean higher photon energy, and the muons would be created with non-
zero kinetic
energy.
44.5: a)
eee
63207270 mmmmmm









MeV.32MeV)511.0(63



E
b) A positive muon has less mass than a positive pion, so if the decay from muon to
pion was to happen, you could always find a frame where energy was not conserved. This
cannot occur.
44.6:
a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a
frequency of
Hz1027.2
23

and a wavelength of
m.1032.1
15

b) The energy of
each photon will be
MeV,1768MeV830MeV3.938


with frequency
Hz108.42
22

and wavelength
m.1002.7

16

44.7:
J.1020.7)sm1000.3()kg400kg400()(
19282
 cmE
44.8:
nCBeHe
1
0
12
6
9
4
4
2

We take the masses for these reactants from Table 43.2, and use Eq. 43.23
reaction.exoergicanisThisMeV.701.5
)uMeV(931.5u)1.008665u12.000000u9.012182u002603.4(





Q
44.9:
HeLiBn
4
2

7
3
10
5
1
0

MeV79.2)uMeV(931.5u)(0.002995u;0.002995
u11.018607u4.002603u7.016004He)Li(
u11.021602u10.012937u1.008665B)n(
4
2
7
3
10
5
1
0



m
m
m
The mass decreases so energy is released and the reaction is exoergic.
44.10:
a) The energy is so high that the total energy of each particle is half of the
available energy, 50 GeV. b) Equation (44.11) is applicable, and
MeV.226
a

E
44.11: a)
q
mf
q
m
B
m
Bq


2


T18.1
C1060.1
Hz)10(9.00)ukg10(1.66u)01.2(2
19
627






B
B

b)
J1047.5

)ukg1066.1()u01.2(2
m)(0.32T)(1.18C)1060.1(
2
13
27
22219
222







m
RBq
K
.sm1081.1
)ukg10(1.66u)(2.01
J)1047.5(22
and
MeV3.42eV103.42
7
27
13
6








m
K
v
44.12: a)
)cs.m1012.3)bs.1097.32
77

m
eBR
R
m
eB
f



For three-
figure precision, the relativistic form of the kinetic energy must be used,
,)mc(γeV
2
1 V.10115
1
so1so
6
2
2




.
e
)mc(γ
V,)mc(γeV
44.13: a)
)(2
222
a
mcEmcE
m


2
2
2
a
2
mc
mc
E
E
m

The mass of the alpha particle is that of a
4
2
He atomic mass, minus two electron masses.
But to 3 significant figures this is just

GeV.3.73)uGeV(0.9315u)(4.00u4.00He)(
4
2
M
GeV.30.6GeV73.3
GeV)2(3.73
GeV)0.16(
So
2

m
E
b) For colliding beams of equal mass, each has half the available energy, so each
has 8.0 GeV.
44.14: a)
.999999559.0so,8.1065
MeV938.3
MeV101000
3
cvγ 


b) Nonrelativistic:
.srad1083.3
8

m
eB

Relativistic:

.srad1059.3
1
5

γm
eB

44.15: a) With
2
2
a
2
2
,
mc
E
EmcE
mm

Eq. (44.11).
So
GeV.3190
GeV)938.0(2
GeV)]7.38(2[
2

m
E
b) For colliding beams the available energy
a

E
is that of both beams. So two
proton beams colliding would each need energy of 38.7 GeV to give a total of 77.4 GeV.
44.16: The available energy
a
E
must be
,)2(
2
0
cmm
ρ
η

so Eq. (44.10) becomes

MeV.1254MeV)3.938(2
MeV)2(938.3
MeV))2(938.3MeV3.547(
2
2
)2(
or),2(2)2(
2
2
22
t
2
t
242

0
0







cm
m
cmm
E
cmEcmcmm
p
p
p
η
ppp
η
44.17: Section 44.3 says
.GeV2.91)Z(
20
cm 

2.97p)()(Z
kg1063.1J;101.461eV102.91
0
25289




mm
cEmE
44.18: a) We shall assume that the kinetic energy of the
0

is negligible. In that case we
can set the value of the photon’s energy equal to Q.

.MeV77MeV)11161193(
photon
EQ 
b) The momentum of this photon is
smkg104.1
)sm10(3.00
)eVJ10(1.60eV)10(77
20
8
186
photon






c
E
p

To justify our original assumption, we can calculate the kinetic energy of a
0

that has
this value of momentum

MeV.77MeV2.7
MeV)2(1116
MeV)(77
22
2
2
22
0


Q
mc
E
m
p
K
Thus, we can ignore the momentum of the
0

without introducing a large error.
44.19:
.)(
0


mmMm
p


Using Table (44.3):

MeV.116
MeV135.0MeV938.3MeV1189)(
2

 cmE
44.20: From Table (44.2),
MeV.2.105)2(
2
e
 cmmm
v

44.21:
Conservation of lepton number.
a)
110:,11: 

eue
LLvve


so lepton numbers are not conserved.
b)
110: 


eτe
Lvveτ

11: 
τ
L
so lepton numbers are conserved.
c)
.γ

e

Lepton numbers are not conserved since just one lepton is
produced from zero original leptons.
d)
,110:γn 

ee
Lep
so the lepton numbers are conserved.
44.22:
a) Conserved: Both the neutron and proton have baryon number 1, and the
electron and neutrino have baryon number 0. b) Not conserved: The initial baryon
number is 1 +1 = 2 and the final baryon number is 1. c) Not conserved: The proton has
baryon number 1, and the pions have baryon number 0. d) Conserved: The initial and
final baryon numbers are 1+1 = 1+1+0.
44.23:
Conservation of strangeness:
a)

.


vK 

Strangeness is not conserved since there is just one strange
particle, in the initial states.
b)
.n
0



pK
Again there is just one strange particle so strangeness
cannot be conserved.
c)
,011:
00


SKK

so strangeness is conserved.
d)
,0110:
00


SKp


so strangeness is conserved.
44.24:
a) Using the values of the constants from Appendix F,
,
050044.137
1
1029660475.7
4
3
0
2


c
e


or
1371
to three figures.
b) From Section 38.5

hε
e
v
0
2
1
2


but notice this is just
.
2
asrewritingclaimedas
4
0
2














π
h
c
c
ε
e




44.25:
1
)s(ms)(J
m)(J
1
2











c
f

and thus
c
f

2
is dimensionless. (Recall
2
f
has units of energy times distance.)

44.26:
a)
The


particle has
1


Q
(as its label suggests) and
.3


S
Its appears as a
“hole”in an otherwise regular lattice in the
QS

plane. The mass difference between
each
S
row is around 145
MeV
(or so). This puts the


mass at about the right spot. As
it turns out, all the other particles on this lattice had been discovered already and it was
this “hole” and mass regularity that led to an accurate prediction of the properties of the


!
b) See diagram. Use quark charges
3
1
and,
3
1
,
3
2



 sdu
as a guide.
44.27: a)
:uds
;0
3
1
3
1
3
2
















eQ

.0000
1)1(00
;1
3
1
3
1
3
1



C
S
B
b)
:
u

c
;0
3
2
3
2



e
Q
;0
3
1
3
1









B

.101
;000





C
S
c) ddd:
.0)0(3;0)0(3
;1
3
1
3;1
3
1
3


















CS
B
e
Q
d)
:cd
.1)1(0;000
;0
3
1
3
1
;1
3
2
3
1





















CS
B
e
Q
44.28: a)
1

S
indicates the presence of one
s
antiquark and no s quark. To have
baryon number 0 there can be only one other quark, and to have net charge + e that quark
must be a u, and the quark content is
.
s
u
b) The particle has an
s
antiquark, and for a
baryon number of –1 the particle must consist of three antiquarks. For a net charge of –e,
the quark content must be
.sdd

c)
2


S
means that there are two s quarks, and for
baryon number 1 there must be one more quark. For a charge of 0 the third quark must be
a u quark and the quark content is uss.
44.29:
a) The antiparticle must consist of the antiquarks so:

.n ddu
b) So
udd

n
is not its own antiparticle
c)
ψ
c
c
ψ
c
c
ψ



so
so the

ψ
is its own antiparticle.
44.30:
MeV5906MeV))2(1777MeV9460()2(
2
γ
 cmm
τ
(see Sections 44.3 and
44.4 for masses).
44.31: In


decay,
e
v

nep
1
0
0
1
1
1

,n,p
1
0
1
1

udduud 
so in


decay a u quark changes to a d quark.
44.32:
a) Using the definition of
z
from Example 44.9 we have that
.
λ
λ
λ
)λλ(
11
0
0
0
s
s
z 


Now we use Eq. 44.13 to obtain
.
1
1
1
1
1












c
v
c
v
vc
vc
z
b) Solving the above equation for

we obtain
.3846.0
15.1
15.1
1)1(
1)1(
2
2
2
2








z
z

Thus,
.sm1015.13846.0
8
 cv
c) We can use Eq. 44.15 to find the distance to the given galaxy,
Mpc106.1
Mpc))sm(101.7(
)sm1015.1(
3
4
8
0




H
v
r
44.33: a)

.skm1004.1)Mly5210)(Mly)skm(20(
5
0
 rHv
b)
.44.1
skm1004.1skm100.3
skm1004.1skm100.3
λ
λ
55
55
0







vc
vc
s
44.34: From Eq. (44.15),
.Mly105.1
Mly)skm(20
sm1000.3
4
8
0




H
c
r
b) This distance
represents looking back in time so far that the light has not been able to reach us.
44.35: a)


 
s)m01998.2(
10.5905.658
10.5905.658
1)
λλ(
1)
λλ(
8
2
2
2
0
2
0




















cv
s
s

s.m10280.3
7

b)
Mly.1640
Mlym102.0
sm103.280
4
7
0





H
v
r
44.36:
Squaring both sides of Eq. (44.13) and multiplying by
 )(λgives
2
0
vcvc ),(λ
2
vc
s

and solving this for v gives Eq. (44.14).
44.37: a)
)He()H()H(
3
2
2
1
1
1
MMMm 
where atomic masses are used to balance
electron masses.

.MeV5.494)uVMe(931.5u)10889.5()(

u105.898u3.16029u2.014102u782500.1
32
3




cmE
m
b)
He)()He(
4
2
3
2n
MMmm 

u091022.0
u4.002603u3.016029u6649008.1





MeV.20.58u)VMe(931.5u)102209.0()(
2
 cmE
44.38:
3124
1080.7)C()He(3


 mm
u, or 7.27 MeV.
44.39:
,0assumingso
ee
vvnpe






mmmmmm
endoergic.isandMeV0.783u)VMe(931.5u)10.408()(
u108.40u1.008665u1.007276u8600054.0
42
4




cmE
m
44.40:
3
OHeC
1069.7
16
8

4
2
12
6

 mmm
u, or 7.16 MeV, an exoergic reaction.
44.41: For blackbody radiation


121
λλλso,m1090.2λ
21
3
mmmm
TTKT
.m109.66
K3000
K2.728
m)10062.1(
73 

44.42:
a) The dimensions of

are energy times time, the dimensions of G
are energy times
time per mass squared, and so the dimensions of
3
/ cG

are

.L
L
T
T
L
L
T
M
E
T)L(
)ML(ET)(E
2
2
2
1/2
3
2



































b)
.m101.616
s)m10(3.002
)kgmN10(6.673s)J10(6.626
35
21
38

221134
3
2
1




















πc
G

44.43: a)
TeV14TeV)7(2E
a



b) Fixed target; equal mass particles,
TeV.101.04MeV101.04
938.3MeV
MeV)2(938.3
MeV)10(1.4
2
511
27
2
2
2




mc
mc
E
E
a
m
44.44:
MeV.526
λ
,
λ
2
p

2
p
 cm
hc
K
hc
cmK
44.45:
The available energy must be the sum of the final rest masses: (at least)
MeV.136.0
MeV135.0MeV)0.511(2
2
22
ea
0


 cmcmE
π
For alike target and beam particles:
511.0
MeV)2(0.511
MeV)(136.0
2
2
2
e
2
e
2

a
 cm
cm
E
E
e
m
.MeV1081.1)MeV1081.1(SoMeV.1081.1MeV
42
e
44
 cmK
44.46:
In Eq.(44.9),
,)(and,withand,)(
2
p
2
a
00
KcmEmmmMcmmE
π
m
πK



MeV.904
139.6MeV
MeV)2(938.3

MeV)(938.3MeV)(139.6MeV)497.7MeV(1193
)(
2
)()(
222
2
2
p
22
p
222
a










cm
cm
cmcmE
K
44.47:
The available energy must be at least the sum of the final rest masses.
.))(())((
)(2.MeV2103)MeV7.493(2MeV1116)()()(

2222
p
p
2
a
222
a
0
cmcm
c
mEcmcmcmE
K
KK





So
MeV
)3.938(2
)7.493()3.938()2103(
)(2
))(())((
222
2
p
2222
p
2

a






cm
cmcmE
E
K
K

.)(MeV1759
2
KcmE
KK


So the threshold energy K = 1759 MeV– 493.7 MeV=1265 MeV.
44.48:
a) The decay products must be neutral, so the only possible combinations are


0000
or
b)
,VMe3.1423
2
0

0
cmm
η


so the kinetic energy of the
0

mesons is 142.3
MeV. For the other reaction,
.MeV1.133)(
2
0
0


cmmmmK


44.49: a) If the


decays, it must end in an electron and neutrinos. The rest energy of

(139.6 MeV) is shared between the electron rest energy (0.511 MeV) and kinetic energy
(assuming the neutrino masses are negligible). So the energy released is 139.6 MeV –
0.511 MeV = 139.1 MeV.
b) Conservation of momentum leads to the neutrinos carrying away most of the
energy.
44.50:

s.101.5
J/eV)10(1.6eV)10(4.4
s)J10(1.054
22
196
34








E

44.51: a)
222
p
2
)()()()( cmcmcmcmE
K
K




MeV.32.0
MeV)2(493.7MeV1019.4




Each kaon gets half the energy so the kinetic energy of the

K
is 16.0 MeV.
b) Since the
0

mass is greater than the energy left over in part (a), it could not have
been produced in addition to the kaons.
c) Conservation of strangeness will not allow
.



KorK
44.52: a) The baryon number is 0, the charge is
e

, the strangeness is 1, all lepton
numbers are zero, and the particle is
.

K
b) The baryon number is 0, the charge is
e

,
the strangeness is 0, all lepton numbers are zero, and the particle is

.


c) The baryon
numbers is –1, the charge is 0, the strangeness is zero, all lepton numbers are 0, and the
particle is an antineutron. d) The baryon number is 0, the charge is
e

, the strangeness
is 0, the muonic lepton number is –1, all other lepton numbers are 0, and the particle is
.


44.53:
keV87J1039.1
s106.7
sJ10054.1
s106.7
14
21
34
21











t
Et

.108.2
MeV3097
MeV087.0
5
2



cm
E
ψ
44.54:
a) The number of protons in a kilogram is
.106.7molecule)protons2(
molkg100.18
molmolecules106.023
)kg00.1(
25
3
23













Note that only the protons in the hydrogen atoms are considered as possible sources of
proton decay. The energy per decay is
,J10503.1MeV3.938
102
p

cm
and so the
energy deposited in a year, per kilogram, is












J)1050.1()y1(
y101.0

(2)ln
)107.6(
10
18
25
rad.70.0Gy100.7
3


b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem.
44.55: a)
222
Ξ
2
)()()()(
0
cmcmcmcmE
π





MeV.65MeV139.6MeV1116MeV1321





E

b) Using (nonrelativistic) conservation of momentum and energy:


f
PP 0
0
.
0
0
00














v
m
m
vvmvm
π
πππ

Also
EKK
π



0
from part (a).
So


























π
π
ππ
m
m
Kvm
m
m
KvmK
0
000
0
00
1
2
1
2
1
22
MeV.8.57MeV2.765
MeV2.7
MeV6.139
VMe1116
1
MeV65
1

0
0










π
π
K
m
m
E
K
So the fractions of energy carried off by the particles are
11.0
65
2.7

for the
0

and
0.89 for the
.



44.56: a) For this model,
,so, H
R
HR
R
dtdR
HR
dt
dR

presumed to be the same for
all points on the surface. b) For constant
,

.HrHR
dt
dR
dt
dr


c) See part (a),
.
0
R
dtdR
H


d) The equation
RH
dt
dR
0

is a differential equation, the solution to
which, for constant
,)is
0
00
tH
eRR(tH 
where
0
R
is the value of R at
0

t
. This equation
may be solved by separation of variables, as
0
)(ln HR
dt
d
R
dtdR

and integrating b

oth
sides with respect to time. e) A constant
0
H
would mean a constant critical density,
which is inconsistent with uniform expansion.
44.57: From Pr.(44.56):
.


r
RRr

So
.0since
11
2

dt
dθ
dt
dr
θ
dt
dθ
θ
r
dt
dr
θ

dt
dR
So
.
1111
0
rHr
dt
dR
Rdt
dr
v
dt
dr
rdt
dr
R
θdt
dR
R








Now















dt
dR
θ
dθ
d
dt
dR
R
r
d
θ
d
d
θ
dv
0
K

dt
dR


where
K
is a constant.
.
11
.0since
0
t
K
Kt
dt
dR
R
H
dt
d
t
K
R
K
dt
dR














So the current value of the Hubble constant is
T
1
where T is the present age of the
universe.
44.58: a) For mass m, in Eq.
.
1
soand,,)23.37(
2
cm0
cm0
0cm
cvv
vv
vvvvu
m






For mass
.so,0,,
cmcm
vvvvuM
M



b) The condition for no net momentum in the
center of mass frame is
MmMMmm
γandγvMγvmγ where,0
correspond to the velocities
found in part (a). The algebra reduces to
,)(
00 Mmm
γγγ



where
,,
cm0
0
c
v
c
v





and the condition for no net momentum becomes



M
m γγ)(
00


.
)/(1
and,
1
1
,
2
0
0
cm
2
0
0
0
0
cvMm
mv
v

Mm
m
m
γ
M
or
γM
M











c) Substitution of the above expression into the expressions for the velocities found
in part (a) gives the relatively simple forms
.
Mm
γ
m
γv,v
Mm
M
γvv
M

γ
m




0
0000
0
After some more algebra,
0
0
0
0
22
2222
γ
γ
M
γ
γ
m
mMMm
mM
γ,
mMMm
Mm
γ







, from which
.γ2γ
0
22
mMMmMm
Mm


This last expression, multiplied by
,
2
c
is the available energy
a
E
in the center of mass
frame, so that
,2)()(
2
2(
22222
2
0
22222
4
0

222
a
m
EMcMcmc
)c)(m
γMc()(Mc)(mc
)cmM
γMmE



which is Eq. (44.9).
44.59:
00
n


a)
22
n
22
)()()()(
00
cmcmcmcmE



MeV.4.41
MeV0.135MeV6.939MeV1116





b)
Using conservation of momentum and kinetic energy; we know that the momentum of
the neutron and pion must have the same magnitude,

pp
n

.2)(
2)(
)()(
)()(
22222
22222
2222
22222
EcmKcmKcmKKK
cmKcmKcm
cmcpcm
cmcpcmcmEK
nnn
nn
nnn
nnnnnn







(
.222)(2)
2222222222

KcmEKcEmKcmEKcmKcm
nnn

Collecting terms we find :
2222
2)222( cEmEcmEcmK
nn



MeV.62.35
.
)MeV6.939(2)MeV4.41(2)MeV0.135(2
)MeV6.939)(MeV4.41(2)MeV4.41(
2






K
K
So the fractional energy carried by the pion is

,86.0
4
.
41
62.35

and that of the neutron is 0.14.