Chapter 30
Nuclear Energy and Elementary Particles
Problem Solutions
30.1
The num ber of fissions required is
N
E
Q
3.30 1010 J
1 MeV
208 MeV 1.60 10 13 J
The m ass of a single
9.92 1020
235
U atom is matom
235
total m ass of U required is
mtotal
30.2
Nmatom
9.92 1020 3.90 10
The energy released in the reaction
Q
m c2
m 235 U
92
2 mn
m 98 Zr
40
1
0
(235 u)(1.66 10
25
kg
235
92
n
3.87 10
U
98
40
4
27
kg
kg u) 3.90 10
25
kg , so the
0.387 g
1
Zr + 135
52 Te + 3 0 n is
m135 Te c 2
52
235.043 923 u 2 1.008 665 u
97.912 0 u 134.908 7 u 931.5 MeV u
192 MeV
30.3
The energy released in the reaction
Q
m c2
m235 U 11mn
92
m88 Sr
38
235.043 923 u 11 1.008 665 u
1
0
235
92
n
U
88
38
1
Sr + 136
54 Xe + 12 0 n is
m136 Xe c 2
54
87.905 614 u 135.907 220 u 931.5 MeV u
126 MeV
596
N uclear Energy and Elementary Particles
30.4
30.5
619
Three d ifferent fission reactions are possible:
144
54
1
0
n
235
92
U
90
38
Sr
144
54
Xe 210n
1
0
n
235
92
U
90
38
Sr
143
54
Xe 3 01 n
143
54
Xe
1
0
n
235
92
U
90
38
Sr
142
54
Xe 4 01 n
142
54
Xe
Xe
4 000 kg m3 . Thus, the m ass
(a) With a specific gravity of 4.00, the d ensity of soil is
of the top 1.00 m of soil is
m
kg
4 000 3
m
V
1.00 m 43 560 ft
2
1m
3.281 ft
2
1.62 107 kg
At a rate of 1 part per m illion, the m ass of uranium in this soil is then
m
106
mU
1.62 107 kg
106
16.2 kg
(b) Since 0.720% of naturally occurring uranium is
part (a) is
m235 U
7.20 10
92
30.6
3
mU
7.20 10
3
16.2 kg
235
92
235
92
U , the m ass of
0.117 kg
U in the soil of
117 g
At 40.0% efficiency, the useful energy obtained per fission event is
Eevent
0.400 200 MeV event 1.60 10
13
J MeV
1.28 10
11
J event
The num ber of fission events required each d ay is then
N
P t
Eevent
1.00 109 J s 8.64 104 s d
1.28 10
Each fission event consum es one
m
Nmatom
11
J event
235
92
6.75 1024 events d
U atom . The m ass of this num ber of
6.75 1024 events d
235 u 1.66 10
27
kg u
235
92
U atom s is
2.63 kg d
In contrast, a coal-burning steam plant prod ucing the sam e electrical pow er uses m ore
than 6 106 kg d of coal.
618
30.7
CH APTER 30
235
The m ass of
U in 1.0-kg of fuel is 0.017 kg, and the num ber of
m
matom
N
0.017 kg
235 u 1.66 10
27
235
U nuclei is
4.36 1022
kg u
At 208 MeV per fission event and 20% efficiency, the useful energy available from this
num ber of fission events is
4.36 1022 events
E
From Work
s
30.8
2.9 1011 J
1.0 105 N
(a) The m ass of
235
2.9 106 m
4.4 106 metric ton 103
3.1 1010 g
235 g mol
m
M
The num ber of
N
nN A
2.9 1011 J
0.20
2.9 103 km (or about 1800 miles)
(b) The num ber of m oles in the quantity of
n
J Mev
U in the reserve is
0.70
100
m 235 U
13
E , the d istance the ship can travel on this 1.0-kg of fuel is
Fdrag s
E
Fdrag
208 MeV event 1.60 10
235
kg
ton
235
103
g
kg
3.1 1010 g
U found above is
1.3 108 mol
U atom s in this reserve is
1.3 108 mol 6.02 1023 atoms mol
7.8 1031 atoms
(c) Assum ing all atom s und ergo fission and all released energy captured , the total
energy available is
E
N 208
MeV
event
7.8 1031 events
208
MeV
event
1.6 10 13 J
1 MeV
2.6 1021 J
(d ) At a consum ption rate of 1.5 1013 J s , the m axim um tim e this energy supply could
last is
t
E
P
2.6 1021 J
1 yr
13
1.5 10 J s 3.156 107 s
5.5 yr
N uclear Energy and Elementary Particles
619
(e) Fission is not sufficient to supply the w orld w ith energy at a price of $130 or less per
kilogram of uranium .
30.9
The total energy required for one year is
E
2 000 kWh month 3.60 106 J kWh 12.0 months
8.64 1010 J
The num ber of fission events need ed w ill be
N
8.64 1010 J
E
Eevent
208 MeV 1.60 10
and the m ass of this num ber of
m
2.60 1021
Nmatom
1.01 10 3 kg
30.10
235
13
2.60 1021
J MeV
U atom s is
235 u 1.66 10
27
kg u
1.01 g
(a) At a concentration of c 3 mg m3 3 10 3 g m3 , the m ass of uranium d issolved in
the oceans covering tw o-third s of Earth’ s surface to an average d epth of hav 4 km
is mU
mU
cV
c( 23 A) hav
3 10
3
g
m3
c[ 23 (4 RE2 )] hav , or
2
4
3
6.38 106 m
2
4 103 m
4 1015 g
(b) Fissionable 235 U m akes up 0.7% of the m ass of uranium com puted above. If w e
assum e all of the 235 U is collected and caused to und ergo fission, w ith the release of
about 200 MeV per event, the potential energy supply is
E
number of
235
U atoms 200 MeV
mU
0.7
100 m235U
200 MeV
atom
and at a consum ption rate of P 1.5 1013 J s , the tim e this could supply the
w orld ’ s energy need s is t E P , or
618
CH APTER 30
t
200 MeV
mU
0.7
100 m 235U
P
atom
4 1015 g
0.7
100
235 u 1.66 10
27
200 MeV 1.6 10 13 J
1 MeV
1.5 1013 J s
1 kg
103 g
kg u
1 yr
3.156 107 s
5 103 yr
(c) The uranium com es fro d issolving rock and m inerals. Rivers carry such solutes into
the oceans, so the ocean’ s supply of uranium is stead ily replenished . Further, if
breed er reactors are used , the current ocean supply can last half a m illion years!
30.11
(a)
4
2
He
4
2
He
8
4
Be
(b)
8
4
Be
4
2
He
12
6
C
(c) The total energy released in this pair of fusion reactions is
Q
m c2
3 m4 He
3 4.002 602 u
30.12
m12 C c2
12.000 000 u 931.5 MeV u
The energy released in the reaction 11 H
Q
m c2
m1 H
1
m2 H
1
2
1
H
3
2
7.27 MeV
is
He
m3 He c 2
2
1.007 825 u 2.014 102 u 3.016 029 u 931.5 MeV u
30.13
The energy released in the reaction 12 H
Q
m c2
m2 H
1
m3 H
1
3
1
H
4
2
He
1
0
5.49 MeV
n is
mn c 2
m4 He
2
2.014 102 u 3.016 049 u 4.002 603 u 1.008 665 u 931.5 MeV u
17.6 MeV 1.60 10
13
J MeV
2.82 10
12
J
The total energy required for the year is
E
2 000 kWh month 12.0 months yr 3.60 106 J kWh
8.64 1010 J yr
N uclear Energy and Elementary Particles
619
so the num ber of fusion events need ed for the year is
8.64 1010 J yr
2.82 10 12 J event
3.06 1022 events yr
C
13
7
N
(b)
13
7
N
H
14
7
N
(d )
14
7
N
1
1
(f)
15
7
N
1
1
E
Q
N
30.14
30.15
(a)
1
1
12
6
(c)
13
6
C
(e)
15
8
O
H
1
1
15
7
N
0
1
e
13
6
H
H
0
1
C
e
15
8
O
12
6
C
4
2
He
With the d euteron and triton at rest, the total m om entum before reaction is zero. To
conserve m om entum , the neutron and the alpha particle m ust m ove in opposite
d irections w ith equal m agnitud e m om enta after reaction, or p
pn . N eglecting
relativistic effects, w e use the classical relationship betw een m om entum and kinetic
energy , KE p2 2m , and w rite 2m KE
2mn KEn , or KE (mn m )KEn .
To conserve energy, it is necessary that the kinetic energies of the reaction prod ucts
satisfy the relation KEn KE Q 17.6 MeV . Then, using the result from above, w e
have KEn (mn m )KEn 17.6 MeV , or the kinetic energy of the em erging neutron m ust
be
KEn
30.16
17.6 MeV
1.008 665 u
1
4.002 603 u
14.1 MeV
(a) The energy released in the reaction 11 H
Q
m c2
m1 H
1
11
5
B
3( 44 He) is
m11 B 3m 4 He c 2
5
2
1.007 825 u 11.009 306 u 3 4.002 603 u
931.5 MeV u
8.68 MeV
(b) The proton and the boron nucleus both have positive charges. Thus, they m ust have
enough kinetic energy to overcome the repulsive Coulomb force one exerts on the other
and approach each other very closely.
618
30.17
CH APTER 30
N ote that pair prod uction cannot occur in a vacuum . It m ust occur in the presence of a
m assive particle w hich can absorb at least som e of the m om entum of the photon and
allow total linear m om entum to be conserved .
When a particle-antiparticle pair is prod uced by a photon having the m inim um possible
frequency, and hence m inim um possible energy, the nearby m assive particle absorbs all
the m om entum of the photon, allow ing both com ponents of the particle -antiparticle pair
to be left at rest. In such an event, the total kinetic energy afterw ard s is essentially zero,
and the photon need only supply th e total rest energy of the pair prod uced .
The m inim um photon energy required to prod uce a proton -antiproton pair is
Ephoton
Thus,
f
proton
2 938.3 MeV 1.60 10
3.00 10 10 J
6.63 10 34 J s
Ephoton
h
3.00 108 m s
4.52 1023 Hz
c
f
and
30.18
2 ER
13
J MeV
3.00 10
10
J
4.52 1023 Hz
6.64 10
16
m
0.664 fm
The total kinetic after the pair prod uction is
KEtotal
Ephoton
2 ER
proton
2.09 103 MeV 2 938.3 MeV
213 MeV
The kinetic energy of the antiproton is then
KE p
30.19
KEtotal
KE p
213 MeV 95.0 MeV 118 MeV
The total rest energy of the 0 is converted into kinetic energy of the photons. Since the
total m om entum w as zero before the d ecay, the tw o photons m ust go in opposite
d irections w ith equal m agnitud e m om enta (and hence equal energies). Thus, the rest
energy of the 0 is split equally betw een the tw o photons, giving for each photon
ER ,
Ephoton
0
2
Ephoton
pphoton
c
135 MeV
2
67.5 Mev
67.5 MeV c
and
f
Ephoton
h
67.5 MeV
1.60 10 13 J
1 MeV
6.63 10 34 J s
1.63 1022 Hz
N uclear Energy and Elementary Particles
30.20
619
Observe that the given reactions involve only m esons and baryons. With no leptons
before or after the reactions, w e d o not have to consid er the conservation law s
concerning the various lepton num bers. All interactions alw ays conserve both charge
and baryon num bers. The strong interaction also conserves strangeness. Conservation
of strangeness m ay be violated by the w eak interaction but never by m ore than one un it.
With these facts in m ind consid er the given interactions:
K0
Charge
Baryon num ber
Strangeness
K0
0
0
+1
total before
0
0
+1
+1
0
0
1
0
0
total after
0
0
0
This reaction conserves both charge and baryon num ber, but d oes violate strangeness by
one unit. Thus, it can occur via the weak interaction but not other interactions.
0
0
Charge
Baryon num ber
Strangeness
0
+1
1
total before
0
+1
1
+1
0
0
total after
0
0
0
1
0
0
This reaction fails to conserve baryon number and cannot occur via any interaction.
30.21
Reaction
(a)
p p
e
Le : 0 0
0 1 ; and L : 0 0
p
p
Charge, Q:
(c)
p p
p
Baryon Number, B:
1 1
1 0
Baryon Number, B:
1 1
1 1 1
p p
p p n
p
(e)
n
0
Charge, Q:
The relevant conservation law s are
(a)
(b) ? p
0
e
?
p
1 1
1 0
(b)
(d )
30.22
Conservation Law Violated
0 1
0,
Le
L
1 1
0 0
0 , and
Le
0
0
0 1 Le , so Le
L
0
L
0
0.
L
1 0 0 , so L
1
e
1
618
CH APTER 30
(c)
0
p
(d )
?
? ?
One particle m ust be
30.23
? p
L
0
0
0 1 L , so L
L
0
0
1 L , so L
1
L
0
0 L , so L
1
1
w ith L
1
1, w hile the other is
w ith L
n
Conservation of charge
Q e
0 e or
Q 0
Conservation of Baryon num ber
B 1
1 0 or
B 0
Conservation of electron-lepton num ber
Le 0
0 0 or
Le
Conservation of m uon-lepton num ber
L
0
0 1 or
L
Conservation of tau -lepton num ber
L
0
0 0 or
L
The particle having these properties is an antim uon -neutrino. It is the
30.24
1
(a)
p
K
Baryon num ber, B :
0 1
0 1
B 0
Charge, Q :
1 1
1 1
Q 0
Baryon num ber, B :
0 1
0 1
B 0
Charge, Q :
1 1
1 1
Q 0
p
0
1
0
.
N uclear Energy and Elementary Particles
619
(b) Strangeness is conserved in the first reaction:
Strangeness, S :
0 0
1 1
S
0
The second reaction d oes not conserve strangeness:
Strangeness, S :
0 0
0 1
S
1
The second reaction cannot occur via the strong or electromagnetic interactions .
(c) If one of the neutral Kaons w ere also prod uced in the second reaction, giving
K0 , then strangeness w ould no longer be violated :
p
Strangeness, S :
0 0
0 1 +1
S
0
Because the total m ass of the prod uct particles in this reaction w ould be greater
than that in the first reaction [see part (a)], the total incid ent energy of the reacting
particles would have to be greater for this reaction that for the first reaction .
30.25
0
(a)
Charge:
1
Baryon num ber:
1
Lepton num bers,
Strangeness:
0 1 0
1 0 0
Q 0
B 0
Le :
0
0 0 0
Le
0
L :
0
0 1 1
L
0
L:
0
0 0 0
L
0
2
1 0 0
S
0
618
CH APTER 30
(b) K0
2
0
Charge:
0
0 0
Q 0
Baryon num ber:
0
0 0
B 0
Le :
0
0 0
Le
0
L :
0
0 0
L
0
L:
0
0 0
L
0
Lepton num bers,
Strangeness:
(c)
K
p
0
1
0
n
Charge:
1 1
Baryon num ber:
Lepton num bers,
Strangeness:
S
0 0
0 0
Q 0
B 0
0 1
1 1
Le :
0 0
0 0
Le
0
L :
0 0
0 0
L
0
L:
0 0
0 0
L
0
1 0
1 0
S
0
N uclear Energy and Elementary Particles
0
(d )
0
Charge:
0
0 0
Q 0
Baryon num ber:
1
1 0
B 0
Le :
0
0 0
Le
0
L :
0
0 0
L
0
L:
0
0 0
L
0
Lepton num bers,
Strangeness:
(e)
1
1 0
S
Charge:
1 1
1 1
Q 0
Baryon num ber:
0 0
e
0
e
Lepton num bers,
Strangeness:
Le :
1 1
0 0
0 0
B 0
Le
0
L :
0 0
1 1
L
0
L:
0 0
0 0
L
0
0 0
0 0
S
0
619
618
CH APTER 30
(f)
p n
0
Charge:
1 0
Baryon num ber:
1 1
Lepton num bers,
Strangeness:
0 1
1 1
Q 0
B 0
Le :
0 0
0 0
Le
0
L :
0 0
0 0
L
0
L:
0 0
0 0
L
0
0 0
1 1
S
0
30.26
proton u
u
strangeness
0
0
0
1
1
baryon
1
num ber
3
3
2
e
3
2
e
3
charge
e
d
total
0
0
1
1
3
e3
e
neutron
strangeness
0
baryon
1
num ber
charge
0
u
0
1
3
2e 3
d
0
1
3
e3
d total
0
0
1
1
3
e3 0
N uclear Energy and Elementary Particles
30.27
The num ber of w ater m olecules in one liter ( mass
1 000 g
18.0 g mol
N
6.02 1023 molecules mol
619
1 000 g ) of w ater is
3.34 1025 molecules
Each m olecule contains 10 protons, 10 electrons, and 8 neutrons. Thus, there are
Ne 10 N
and
Nn
8N
3.34 1026 electrons , N p 10 N
3.34 1026 protons ,
2.68 1026 neutrons
Each proton contains 2 up quarks and 1 d ow n qu ark, w hile each neutron ha s 1 up quark
and 2 d ow n quarks. Therefore, there are
Nu
2 Np
Nn
Nd
Np
2 Nn
9.36 1026 up quarks , and
8.70 1026 down quarks
30.28
K 0 Particle
strangeness
baryon num ber
1
0
charge
0
30.29
0
d
0
13
e3
total
u
1
13
1
0
strangeness
baryon num ber
1
1
0
13
e3
0
charge
0
2e 3
d
0
13
e3
Com pare the given quark states to the entries in Table 30.4:
(a) suu
30.30
s
Particle
K0
(b)
ud
(d )
ssd
(c)
sd
(a)
uud : charge
2
e
3
2
e
3
1
e
3
e . This is the antiproton .
(b) udd : charge
2
e
3
1
e
3
1
e
3
0 . This is the antineutron .
s
1
13
e3
total
1
1
0
618
30.31
CH APTER 30
The reaction is
0
X , or on the quark level, uds uud
p
uus 0 ?
The left sid e has a net 3u, 2d, and 1s . The right sid e has 2u, 0d, and 1s , leaving
1u and 2 d m issing.
The unknown particle is a neutron, udd . Baryon and strangeness num bers are conserved .
30.32
(a)
0
p
(b)
e
is forbid d en by conservation of charge
is forbid d en by conservation of electron-lepton number
conservation of energy , and conservation of muon-lepton number
(c)
30.33
is forbid d en by conservation of baryon number
p
To the reaction for nuclei, 11 H
sid es to obtain 11 Hatom
3
2
3
2
4
2
Heatom
4
2
He
Heatom
0
1
He
0
1
e
e
0
1
e
e
, w e ad d three electrons to both
e
. Then w e use the m asses of the
neutral atom s to com pute
m c2
Q
m1 H
1
m3 He
2
m4 He
2
2 me c 2
1.007 825 u 3.016 029 u 4.002 603 u 2 0.000 549 u
931.5 MeV u
18.8 MeV
30.34
For the particle reaction,
and Le
30.35
(a)
2 , the lepton num bers before the event are L
1 . These values m ust be conserved by the reaction so one of the em erging
neutrinos m ust have L
and
e
1 w hile the other has Le
1 . The em erging particle are
e
Violates conservation of muon-lepton number
e
and also conservation of electron-lepton number
(b) n
p e
e
Violates conservation of electron-lepton number
1
N uclear Energy and Elementary Particles
(c)
0
p
(d ) p
e
0
Violates conservation of charge
0
Violates conservation of electron-lepton number
619
and also conservation of baryon number
(e)
0
n
0
Violates conservation of strangeness number by 2 units
Even w eak interactions only violate this rule by 1 unit.
30.36
Assum ing a head -on collision, the total m om entum is zero both before and after the
reaction p p p
X . Therefore, since the proton and the pion are at rest after
reaction, particle X m ust also be left at rest.
Particle X m ust be a neutral baryon in ord er to conserve charge and baryon num ber in
the reaction. The rest energy this particle is
or
E0 X
2 E0 p
70.4 MeV
E0 p
E0
E0 p
E0
140.8 MeV
E0 X
938.3 MeV 139.6 MeV 140.8 MeV 939.5 MeV
Particle X is a neutron .
30.37
If a neutron starts w ith kinetic energy KEi 2.0 MeV and loses one half of its kinetic
energy in each collision w ith a m od erator atom , its kinetic energy after n collisions w ill
be KE f KEi 2n .
The kinetic energy associated w ith particle in a gas at tem perature T
(see Chapter 10 of the textbook) is
KE f
3
k BT
2
3
1.38 10
2
23
J K 293 K
1 eV
1.60 10 19 J
20.0 C 293 K
0.0379 eV
Thus, the num ber of collisions the neutron m ust m ake before it reaches the energy
associated w ith a room tem perature gas is n ln 2 ln( KEi KE f ) , or
n
1
2.0 106 eV
ln
ln 2
0.0379 eV
26
618
30.38
CH APTER 30
(a) The num ber of d euterons in one kilogram of d euterium is
N
m
matom
1 kg
2.01 u 1.66 10
27
kg u
3.00 1026
Each occurrence of the reaction 21 D 21 D 23 He 01 n consum es tw o d euterons and
releases Q 3.27 MeV of energy. The total energy available from the one kilogram
of d euterium is then
E
N
Q
2
3.00 1026
2
3.27 Mev
1.60 10 13 J
1 MeV
7.85 1013 J
(b) At a rate of eight cents per kilow att-hour, the value of this energy is
value
E rate
7.85 1013 J
$0.08
kWh
1 kWh
3.60 106 J
$1.74 106
$1 740 000
(c) Deuterium m akes up four-tw entieths or one-fifth of the m ass of a heavy w ater
m olecule. Thus, five kilogram s of heavy w ater is necessary to obtain one kilogram
of d euterium . The cost for this w ater (5 kg)($300 kg) $1500 .
(d ) Whether it w ould be cost-effective d epend s on how m uch it cost to fuse the
d euterium and how m uch net energy w as prod uced . If the cost is nine-tenths of the
value of the energy prod uced , each kilogram of deuterium w ould still yield a profit
of $174 000 .
N uclear Energy and Elementary Particles
30.39
(a) The num ber of m oles in 1.0 gal of w ater is
m
M
n
V
M
1.0
g
cm3
1.0 gal
103 cm3
1L
3.786 L
1 gal
2.1 102 mol
18 g mol
so the num ber of hyd rogen atom s (2 per w ater m olecule) w ill be
NH
2 nN A
2 2.1 102 mol 6.02 1023 mol
1
2.5 1026
and one of every 6 500 of these is a d euteron. Thus, the num ber of num ber of
d euterons contained in 1.0 gal of w ater is
ND
NH 6 500 2.5 1026 6 500 3.9 1022 deuterons
and the available energy is
E
3.9 1022 deuterons 1.64 MeV deuteron 1.6 10
13
J MeV
1.0 1010 J
(b) At a consum ption rate of P 1.0 104 J s , the tim e that this could supply a
person’ s energy need s is
t
30.40
E
P
1.0 1010 J
1d
4
1.0 10 J s 86 400 s
Einput
The total energy input required is
or
Einput
100 000 103 J s
12 d
P
Erequired
efficiency
100 d 86 400 s d
0.30
efficiency
2.9 1015 J
At 208 MeV per fission event, the num ber of fissions of
N
Einput
208 Mev
2.9 1015 J
1 MeV
208 Mev 1.60 10-13 J
The m ass of this num ber of
m
N
M
NA
235
t
235
U nu clei need ed is
8.7 1025
U atom s w ill be
8.7 1025 atoms
6.02 1023 atoms mol
235
g
mol
1 kg
103 g
34 kg
619
618
30.41
CH APTER 30
Conserving energy in the d ecay
E
, w ith the
initially at rest gives
ER, , or
E
E
E
[1]
139.6 MeV
Since the total m om entum w as zero before the d ecay, conservation of m om entum
requires the m uon and antineutrino go off in opposite d irections w ith equ al m agnitud e
m om enta, or p
p . The relativistic relation betw een total energy and m om entum of a
particle (Equation 26.10 in the textbook) then gives for the antineutrino: E
p E c . Applying the sam e equation to the m uon, w e obtain
E2
p c
2
ER2,
pc
2
ER2,
E2
p c , or
ER2 ,
or
E2
E2
E
E
E
E
ER2,
105.7 MeV
2
and
E
E
105.7 MeV
E
2
[2]
E
Substituting Equation [1] into [2] gives E
E
E
E
(105.7 MeV)2 139.6 MeV , or
[3]
80.0 MeV
Subtracting Equation [3] from Equation [1] yield s 2E
E
30.42
59.6 MeV , and
29.8 MeV
The reaction
p
K0
0
is ud uud
ds uds at the qu ark level. There is
a net 1u and 2d quarks both before and after the reaction . This reaction conserves the net
num ber of each type quark.
For the reaction
p
K0
n , or ud uud
ds udd , there is a net
1u and 2d before the reaction and 1u, 3d, and 1 antistrange quark afterwards . This reaction
d oes not conserve the net num ber of each type quark.
30.43
To com pute the energy released in each occurrence of the reaction
4p
2
N uclear Energy and Elementary Particles
619
w e ad d four electrons to each sid e to prod uce neutral atom s and obtain
. Then, recalling that the neutrino and the p hoton both
4( 11 Hatom ) 42 Heatom 2e 2
have zero rest m ass, and using the atom ic m asses from Append ix B in the t extbook gives
Q
m c2
4 m1 H
1
4 1.007 825 u
25.7 Mev
atom
m 4 He
2
2me c 2
atom
4.002 603 u 2 0.000 549 u
1.60 10 13 J
1 MeV
4.11 10
12
931.5 MeV u
J
With each occurrence of this reaction consum ing four protons, the rate of proton
consum ption required to prod uce the pow er outp ut of the Sun is
rate
P
energy per proton
3.85 1026 J s
Q 4
3.85 1026 J s
1.03 10 12 J proton
3.74 1038 proton s