44.10: a The energy is so high that the total energy of each particle is half of the available energy, 50 GeV... So twoa proton beams colliding would each need energy of 38.7 GeV to gi
Trang 144.1: a) 2
2 2
1
1
mc c
v mc
m9.10910 31 kg,so K 1.2710 14 J
b) The total energy of each electron or positron is EK mc2 1.1547mc2
J
10
46
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
10
46
9 14
c) Ehc λsoλhc Ehc (9.4610 14 J)2.10pm
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
E Kmc2 5.00MeV0.511MeV5.51MeV
We can calculate the speed of the positron from Eq 37.38
996 0 MeV
5.51
MeV 511 0 1 1
1
2 2
2 2
2
E
mc c
v mc
E
c v
44.3: Each photon gets half of the energy of the pion
ray
gamma m
10 8 1 Hz 10 7 1
s m 10 00 3 λ
Hz 10 7 1 s)
J 10 63 6 (
) eV J 10 6 1 ( eV 10 9 6 (
MeV 69 MeV) 511 0 ( 270 ( 2
1 ) 270 ( 2
1 2
1
14 22
8
22 34
19 7
2 e
2 γ
f c h
E f
c m c
m
44.4: a)
) s m 10 (3.00 kg) 10 (9.11 (207)
s) J 10 626 6 (
c m
h c m
hc E
hc
μ μ
1.1710 14 m0.0117pm
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
E 63(0.511MeV)32MeV
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
Trang 244.6: a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a
frequency of 2.271023 Hz and a wavelength of 1.3210 15 m b) The energy of each photon will be 938.3MeV830MeV1768MeV, with frequency 42.81022 Hz and wavelength 7.0210 16 m
44.8: He Be C 1n
0
12 6
9
4
4
We take the masses for these reactants from Table 43.2, and use Eq 43.23
reaction
exoergic an
is This MeV
701
5
) u MeV (931.5 u) 1.008665 u
12.000000 u
9.012182 u
002603
4
(
Q
2
7 3
10
5
1
MeV 79 2 ) u MeV (931.5 u) (0.002995 u;
0.002995
u 11.018607 u
4.002603 u
7.016004 He)
Li
(
u 11.021602 u
10.012937 u
1.008665 B)
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 Ea 226MeV
44.11: a)
q
mf q
m B m
B
T 18 1
C 10 60 1
Hz) 10 (9.00 ) u kg 10 (1.66 u) 01 2 ( 2
19
6 27
B
) u kg 10 66 1 ( u 01 2 ( 2
m) (0.32 T) (1.18 C) 10 60 1 ( 2
13 27
2 2
2 19 2
2 2
m
R B q K
s m 10 81 1 ) u kg 10 (1.66 u) (2.01
J) 10 47 5 ( 2 2
and
MeV 3.42 eV 10 3.42
7 27
13 6
m
K v
m
eBR R
m
eB
For three-figure precision, the relativistic form of the kinetic energy must be used,
, )mc
(γ
e
)mc (γ
V , )mc (γ
eV
Trang 344.13: a) 2 2 2( 2)
2
2 a
2mc mc
E
The mass of the alpha particle is that of a4
2He atomic mass, minus two electron masses But to 3 significant figures this is just
GeV
3.73 )
u GeV (0.9315 u)
(4.00 u
4.00
He)
(4
M
GeV
30.6 GeV 73 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) 1065.8,so 0.999999559
MeV 938.3
MeV 10
1000 3
c v
b) Nonrelativistic:
s rad 10 83
3 8
m
eB
Relativistic:
s rad 10 59 3
γ m
eB
2 a 2
2
,
mc
E E
mc
E m m Eq (44.11)
GeV) 938 0 ( 2
GeV)]
7 38 ( 2
m
E
b) For colliding beams the available energy E is that of both beams So twoa
proton beams colliding would each need energy of 38.7 GeV to give a total of 77.4 GeV
44.16: The available energy E must be a ( 2 ) 2,
m η ρ so Eq (44.10) becomes
MeV
1254 MeV)
3 938 ( 2 MeV)
2(938.3
MeV)) 2(938.3
MeV 3 547
(
2 2
) 2 (
or ), 2 ( 2 )
2
(
2
2 2
2 t
2 t
2 4
2
0
0
c m m
c m m
E
c m E c m c m
m
p p
p η
p p
p η
2 97 p) (
)
(Z
kg 10 63 1 J;
10 1.461 eV
10 2
91
0
25 2
8 9
m
m
c E m E
Trang 444.18: a) We shall assume that the kinetic energy of the is negligible In that case we 0
can set the value of the photon’s energy equal to Q.
Q(11931116)MeV77MeVEphoton
b) The momentum of this photon is
s m kg 10 4.1 )
s m 10 (3.00
) eV J 10 (1.60 eV) 10
8
18 6
c
E
p
To justify our original assumption, we can calculate the kinetic energy of a that has 0 this value of momentum
MeV) 2(1116
MeV) (77 2
2
2 2
2 2
mc
E m
p K
Thus, we can ignore the momentum of the without introducing a large error.0
44.19: mM()m p m 0 Using Table (44.3):
MeV
116
MeV 135.0 MeV
938.3 MeV
1189 )
m m c
44.21: Conservation of lepton number.
a) :11, :011
e u
v
so lepton numbers are not conserved
b) :011
e τ
e v L v
e
τ
L τ:11
so lepton numbers are conserved
c) eγ Lepton numbers are not conserved since just one lepton is
produced from zero original leptons
d) n γ :011,
e
e L e
p 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
Trang 544.23: Conservation of strangeness:
a) K v Strangeness is not conserved since there is just one strange particle, in the initial states
b) nK p0 Again there is just one strange particle so strangeness cannot be conserved
c) K K 0 0 S:110,so strangeness is conserved
d) pK 0 0 S:0110, so strangeness is conserved
44.24: a) Using the values of the constants from Appendix F,
, 050044
137
1 10
29660475
7 4
3 0
2
c
e
or 1137to three figures
b) From Section 38.5
h
ε
e
v
0
2
1 2
2 as rewriting claimed
as
4 0
2
π
h c
c ε
) s (m s) (J
m) (J
1
2
c
f
and thus
c
f
2
is dimensionless (Recall f 2has units of energy times distance.)
Trang 644.26: a)
The particle has Q1 (as its label suggests) and S 3 Its appears as a
“hole”in an otherwise regular lattice in the SQ plane The mass difference between each Srow 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
3
1 3
1 3
e Q
0 0 0 0
1 ) 1 ( 0 0
; 1 3
1 3
1 3 1
C S B
3
2 3
2
e Q
; 0 3
1 3
B
1 0 1
; 0 0 0
C S
0 ) 0 ( 3
; 0 ) 0 ( 3
; 1 3
1 3
; 1 3
1 3
C S
B e
Q
1 ) 1 ( 0
; 0 0 0
; 0 3
1 3
1
; 1 3
2 3
1
C S
B e
Q
Trang 744.28: a) S 1 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 u s 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 d d s c) S 2 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 u d d
b) So nuddis not its own antiparticle
c) ψ c soψ c cψ so the ψ is its own antiparticle.
44.4 for masses)
0
0 1
1 1
p , 1n ,
0
1
1 uud udd 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
) λ λ ( 1
0
0
s
s
Now we use Eq 44.13 to obtain
1
1 1
1 1
c v c v
v c
v c z
b) Solving the above equation for we obtain
3846 0 1 5 1
1 5 1 1 ) 1 (
1 ) 1 (
2
2 2
2
z
z
Thus, v0.3846 c1.15108 m s
c) We can use Eq 44.15 to find the distance to the given galaxy,
Mpc 10 6 1 Mpc) ) s m ( 10 1 7 (
) s m 10 15 1
4 8 0
H
v
r
H r
v
s km 10 04 1 s km 10 0 3
s km 10 04 1 s km 10 0 3 λ
λ
5 5
5 5
v c
v c
s
Mly ) s km ( 20
s m 10 00
0
H
c
represents looking back in time so far that the light has not been able to reach us
Trang 844.35: a)
658.5590.0 1 (2.998 0 m s)
1 0 590 5 658 1
) λ λ (
1 ) λ λ
2
2 2
0
2
v
s s
3.280107m s
Mly m 10 2.0
s m 10 3.280
4 7 0
H
v
r
44.36: Squaring both sides of Eq (44.13) and multiplying by
givesλ2( )
0 c v
v
c λ2s(cv), and solving this for v gives Eq (44.14).
44.37: a) ( H) ( H) (3He)
2
2 1
1
M
electron masses
MeV 5.494 )
u V Me (931.5 u) 10 8 89 5 ( ) (
u 10 5.898 u
3.16029 u
2.014102 u
7825 00 1
3 2
3
c m E
m
2
3 2
m
u 091 022 0
u 4.002603 u
3.016029 u
6649 008 1
E (m)c2 (0.022091u)(931.5MeV u)20.58MeV
44.38: 3m(4He) m(12C)7.80103u, or 7.27 MeV
e
v n p
endoergic is
and MeV 0.783 u)
V Me (931.5 u) 10 40 8 ( ) (
u 10 8.40 u
1.008665 u
1.007276 u
86 00054 0
4 2
4
c m E
m
O He
8 4
12
6
m u, or 7.16 MeV, an exoergic reaction
1 2
λ so , m 10 90 2
m m
m
m 10 9.66 K
3000
K 2.728 m)
10
062
1
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 G / c3 are
L
T T
L L
T M
E T)
L (
) M L (E T)
3
2
s) m 10 (3.00 2
) kg m N 10 (6.673 s) J 10
3 8
2 2 11 34
3
2
π c
G
b) Fixed target; equal mass particles,
TeV
10 1.04 MeV 10
1.04
938.3MeV MeV)
2(938.3
MeV) 10
(1.4 2
5 11
2 7
2 2 2
mc
E
m
Trang 944.44: 652MeV.
λ
, λ
2 p
2
m c hc K hc m c
K
44.45: The available energy must be the sum of the final rest masses: (at least)
MeV
136.0
MeV 135.0 MeV)
0.511 ( 2
e
m c m c
For alike target and beam particles: 0.511
MeV) 2(0.511
MeV) (136.0 2
2 2
e 2 e
2
c m
E E
e m
MeV 10 81 1 )
MeV 10 81 1 ( So MeV
10 81
1
e 4
44.46: In Eq.(44.9),
, )
( and
, with
and , )
p
2
MeV
904
139.6MeV MeV)
2(938.3
MeV) (938.3 MeV)
(139.6 MeV)
497.7 MeV
(1193
) ( 2
) ( ) (
2 2
2
2 2
p
2 2 p 2 2 2
a
c m
c m c
m E
K
44.47: The available energy must be at least the sum of the final rest masses
) ) ((
)
)
((
) ( 2 MeV 2103 )
MeV 7 493 ( 2 MeV 1116 )
( ) ( )
(
2 2 2
2
p
p
2 a 2
2 2
c m c
m
c m E
c m c m c
m
E
K
K K
) 3 938 ( 2
) 7 493 ( ) 3 938 ( ) 2103 ( )
( 2
) ) ((
) )
2 p
2 2 2
2 p
2
c m
c m c
m E
K
E K 1759MeV(m K)c2 K
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
0 0 0 or 0
b) 3 142.3MeV 2,
0
m η so the kinetic energy of the 0 mesons is 142.3 MeV For the other reaction, ( ) 2 133.1MeV
0
m m m m c
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
Trang 1044.50: 1.5 10 s.
J/eV) 10
(1.6 eV) 10 (4.4
s) J 10
19 6
34
E
p
)
MeV
32.0
MeV) 2(493.7 MeV
1019.4
Each kaon gets half the energy so the kinetic energy of the K is 16.0 MeV.
b) Since the mass is greater than the energy left over in part (a), it could not have 0 been produced in addition to the kaons
c) Conservation of strangeness will not allow K or K
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
s 10 6 7
s J 10 054 1 s
10 6
21
34
t E
10 8 2 MeV 3097
MeV 087
2
c
m
E
ψ
44.54: a) The number of protons in a kilogram is
10 6.7 molecule) protons
2 ( mol
kg 10 0 18
mol molecules 10
6.023 )
kg
00
1
3
23
Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay The energy per decay is 2 938.3MeV 1.503 10 10J,
energy deposited in a year, per kilogram, is
(1y)(1.50 10 J)
y 10 1.0
(2) ln ) 10 7 6
18 25
rad
70 0 Gy
10
0
7 3
b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem
Trang 1144.55: a) 2 2 2
Ξ
) ( m c m c m 0 c m c
1321MeV1116MeV139.6MeVE65MeV
b) Using (nonrelativistic) conservation of momentum and energy: P0 0P f
0 0 0
m
m v
v m
v
m
π π
π π
Also K 0 K π E from part (a)
π π
π
m K
v m m
m K
v m
0 0 0 0 0
2
1 2
MeV
8 57 MeV 2 7 65
MeV 2 7 MeV 6 139
V Me 1116 1
MeV 65
0
π
π
K
m m
E K
So the fractions of energy carried off by the particles are 0.11
65
2
7 for the and 0 0.89 for the
R
HR R
dt dR HR dt
dR presumed to be the same for
all points on the surface b) For constant , HR Hr
dt
dR dt
dr c) See part (a),
0
R
dt
dR
dt
dR
0
is a differential equation, the solution to which, for constant is ) 0 ,
0 0
t H
e R R(t
H where R is the value of R at 0 t 0 This equation may be solved by separation of variables, as ln(R) H0
dt
d R
dt
dR and integrating both
sides with respect to time e) A constant H0 would mean a constant critical density, which is inconsistent with uniform expansion