CHAPTER 32 IONIZING RADIATION,
NUCLEAR ENERGY, AND
ELEMENTARY PARTICLES
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (c) The biologically equivalent dose (in rems) is the absorbed dose (in rads) times the relative
biological effectiveness. (See Equation 32.4.)
2. (d) Equation 32.3 defines the relative biological effectiveness as the dose of 200-keV X-rays
that produces a certain biological effect divided by the dose of radiation that produces the
same biological effect. Thus, since RBEA = 2 × RBEB, it takes a smaller dose of A to
produce the same biological effect as B, smaller by a factor of 2.
3. 9.0 × 10−3 rem
4. (b) Since electric charge must be conserved, we know that the number of protons must be the
same before and after the reaction takes place. Therefore, Z + 7 = 6 + 1, so Z = 0. We also
know that the total number of nucleons must be conserved, so the total number before and
after the reaction takes place must be the same. Therefore, A + 14 = 14 + 1, so A = 1. With a
single uncharged nucleon in the nucleus, the unknown species must be a neutron 01 n .
5. (a) The symbol for the α particle is 42 He , and the symbol for the proton p is 11 H . Therefore,
there are 2 + 13 = 15 protons present before the reaction takes place and 15 + 1 = 16 protons
present after the reaction takes place, which violates the conservation of electric charge.
There are 27 + 4 = 31 nucleons present before the reaction takes place and 1 + 31 = 32
nucleons present after the reaction takes place, which violates the conservation of nucleon
number.
6. (e) The compound nucleus is
236
92 U
for any X and Y. Since the total number of nucleons is
conserved, it follows that 1 + 235 = AX + AY + η, where η is the number of neutrons 01 n
produced by the reaction. Therefore, greater values of η lead to smaller values for AX and
AY. Since electric charge also is conserved, it follows that 0 + 92 = ZX + ZY + η(0).
Therefore, ZX + ZY = 92 for any X and Y.
Chapter 32 Answers to Focus on Concepts Questions
1625
7. (b) In a fission reactor each fission event, on average, must produce at least one neutron.
Otherwise there would be no neutrons to cause additional fission events, and it would not be
possible to establish a controlled chain reaction. If the fission products of the reaction each
have a binding energy per nucleon that is less than the binding energy per nucleon of the
starting nucleus, the reaction does not produce energy. It only produces energy when the
fission products have, on average, a binding energy per nucleon that is greater than the
binding energy per nucleon of the starting nucleus.
8. (a) In order for a fusion reaction to be potentially energy-producing, the binding energy per
nucleon of the starting nuclei must be smaller than the binding energy per nucleon of the
product nucleus. Since the maximum binding energy per nucleon occurs at a nucleon
number of about 60, the starting nuclei with nucleon numbers of 30 in reaction I have the
smaller binding energy per nucleon, as required.
9. (c) The energy produced by a fusion reaction is the mass defect ∆m (in u) for the reaction
times 931.5 MeV/u, since an energy of 931.5 MeV is equivalent to 1 u. The mass defect is
the total mass of the initial nuclei minus the total mass of the product nuclei. Thus, to obtain
the ranking, we need only calculate the mass defect for each reaction from the given masses
and rank the defects in descending order. The results are: reaction I (∆m = 0.0035 u),
reaction II (∆m = 0.0138 u), reaction III (∆m = 0.0053 u).
10. (d) The quark theory explains the electric charge that each hadron carries. An antiproton
carries a charge of −e, which is opposite the charge of +e that a proton carries. Only
possibility D shows a charge of −e + 13 e − 23 e − 23 e .
(
)
11. (e) Hubble’s law indicates that a galaxy located at a distance d from the earth is moving
away from the earth at a speed that is directly (not inversely) proportional to d.
1626
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
CHAPTER 32 IONIZING RADIATION,
NUCLEAR ENERGY, AND
ELEMENTARY PARTICLES
PROBLEMS
______________________________________________________________________________
1.
SSM WWW REASONING The biologically equivalent dose (in rems) is the product of
the absorbed dose (in rads) and the relative biological effectiveness (RBE), according to
Equation 32.4. We can apply this equation to each type of radiation. Since the biologically
equivalent doses of the neutrons and α particles are equal, we can solve for the unknown
RBE.
SOLUTION Applying Equation 32.4 to each type of particle and using the fact that the
biological equivalent doses are equal, we find that
( Absorbed dose )α RBEα = ( Absorbed dose )neutrons RBE neutrons
Biologically equivalent dose
of α particles
Biologically equivalent dose
of neutrons
Solving for RBEα and noting that (Absorbed dose)neutrons = 6(Absorbed dose)α, we have
RBEα =
6 ( Absorbed dose )α
( Absorbed dose )neutrons
RBE neutrons ) =
( 2.0 ) = 12
(
( Absorbed dose )α
( Absorbed dose )α
______________________________________________________________________________
2.
REASONING The absorbed dose (AD) in grays (Gy) is the energy absorbed divided by the
Energy absorbed
tumor mass: AD =
(Equation 32.2). Because both tumors receive the same
Mass
absorbed dose, the absorbed dose found in (a) will allow us to determine the energy
absorbed by the second tumor, again via Equation 32.2.
SOLUTION
a. Substituting the given values into Equation 32.2, we obtain
AD =
Energy absorbed
1.7 J
=
= 14 Gy
Mass
0.12 kg
b. For the second tumor, the absorbed dose is still AD = 14 Gy, but the mass is now 0.15 kg.
Solving Equation 32.2 for the energy absorbed, we obtain
Energy absorbed = ( AD ) Mass = (14 Gy )( 0.15 kg ) = 2.1 J
Chapter 32 Problems
3.
1627
REASONING AND SOLUTION
a. The biologically equivalent dose is equal to the product of the absorbed dose and the
RBE (see Equation 32.4):
Biologically equivalent dose = (Absorbed dose)(RBE) = ( 38 rad )(12 ) = 460 rem
b. According to the discussion at the end of Section 32.1, a person exposed to a whole-body,
single dose of 460 rem has a 50% chance of dying .
______________________________________________________________________________
4.
REASONING
a. The absorbed dose (AD) and the biologically equivalent dose (BED) are related by
Equation 32.4:
BED = AD × RBE
(32.4)
where AD is the absorbed dose measured in rads, and RBE is the relative biological
effectiveness of the cosmic rays. The cosmic rays are protons, for which the RBE is 10. We
note that the biologically equivalent dose is given in millirems, where 1 mrem = 10−3 rem.
b. The absorbed dose is related to the energy absorbed and the mass of the person by
Equation 32.2:
AD =
Energy absorbed
Mass
(32.2)
If the energy is measured in joules, and the mass in kilograms, the absorbed dose is given in
grays (Gy), where 1 Gy = 1 J/kg. To convert the absorbed dose found in (a) from rads to
grays, we will use the equivalence 1 rad = 0.01 Gy = 0.01 J/kg.
SOLUTION
a. Solving Equation (32.4) for the absorbed dose (AD), we find that
AD =
BED 24 × 10−3 rem
=
= 2.4 × 10−3 rad
RBE
10
b. Solving Equation (32.2) for the mass and using 1 rad = 0.01 Gy = 0.01 J/kg, we find that
Mass =
Energy absorbed
=
AD
(
1.9 ×10−3 J
= 79 kg
⎛ 0.01 J/kg ⎞
−3
)
2.4 ×10 rad ⎜
⎟
⎝ 1 rad ⎠
1628
5.
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
SSM REASONING AND SOLUTION According to Equation 32.2, the absorbed dose
(AD) is equal to the energy absorbed by the tumor divided by its mass:
AD =
Energy absorbed
=
Mass
⎛ 1.60 × 10 –19 J ⎞
⎟
1 eV
⎝
⎠
( 25 s ) (1.6 × 1010 s –1 )( 4.0 × 106 eV ) ⎜
0.015 kg
= 1.7 × 101 Gy = 1.7 × 103 rad
The biologically equivalent dose (BED) is equal to the product of the absorbed dose (AD)
and the RBE (see Equation 32.4):
BED = AD × RBE = (1.7 × 103 rad)(14) = 2.4 × 10 4 rem
(32.4)
______________________________________________________________________________
6.
REASONING According to Equation 32.4, the biologically equivalent dose is equal to the
product of the absorbed dose and the RBE (relative biological effectiveness).
If the absorbed doses are the same, then the radiation with the larger RBE produces the
greater biological effect.
If the two types of radiation have the same RBE, then the radiation that produces the greater
absorbed dose produces the greater biological effect.
SOLUTION Using Equation 32.4, the biologically equivalent dose can be determined for
each type of radiation. The rankings are given in the last column of the table.
Radiation
Biologically Equivalent Dose =
(Absorbed dose)(RBE)
Ranking
γ rays
(20 × 10–3 rad)(1) = 20 × 10–3 rem
3
Electrons
(30 × 10–3 rad)(1) = 30 × 10–3 rem
2
Protons
(5 × 10–3 rad)(10) = 50 × 10–3 rem
1
4
(5 × 10–3 rad)(2) = 10 × 10–3 rem
______________________________________________________________________________
Slow neutrons
7.
REASONING When the cancerous growth absorbs energy from the radiation, the growth
heats up. According to the discussion in Section 12.7, the rise in temperature depends on the
heat absorbed, as well as on the mass and specific heat capacity of the growth. As we have
seen in Section 32.1, the energy absorbed from the radiation is equal to the product of the
absorbed dose and the mass of the growth.
Chapter 32 Problems
1629
SOLUTION When a substance, such as the cancerous growth, has a mass m and absorbs
an amount of energy Q, the change ΔΤ in its temperature is given by
ΔT =
Q
cm
(12.4)
where c is the specific heat capacity of the substance. The energy Q absorbed is equal to the
absorbed dose of the radiation times the mass m:
Q = ( Absorbed dose ) m
(32.2)
Substituting Equation 32.2 into Equation 12.4 gives
ΔT =
( Absorbed dose ) m Absorbed dose
Q
=
=
cm
cm
c
2.1 Gy
= 5.0 × 10−4 C°
4200 J/ ( kg ⋅ C° )
______________________________________________________________________________
=
8.
REASONING The relation between the rad and gray units is presented in Section 32.1 as
1 rad = 0.01 gray. If, for instance, we wanted to convert an absorbed dose of 2.5 grays into
rads, we would use the conversion procedure:
⎛ 1 rad ⎞
⎟ = 250 rad
⎝ 0.01 Gy ⎠
( 2.5 Gy ) ⎜
In general, the conversion relation is
⎛ 1 rad ⎞
Absorbed dose ( in rad ) = ⎡⎣ Absorbed dose ( in Gy ) ⎤⎦ ⎜
⎟
⎝ 0.01 Gy ⎠
(1)
SOLUTION
According to Equation 32.4, the relative biological effectiveness (RBE) is given by
RBE =
Biologically equivalent dose
Absorbed dose (in rad)
The absorbed dose (in rad) is related to the absorbed dose (in Gy) by Equation (1), so the
RBE can be expressed as
1630
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
RBE =
Biologically equivalent dose
⎛ 1 rad ⎞
⎡⎣ Absorbed dose ( in Gy ) ⎤⎦ ⎜
⎟
⎝ 0.01 Gy ⎠
The absorbed dose (in Gy) is equal to the energy E absorbed by the tissue divided by its
mass m (Equation 32.2), so the RBE can be written as
−2
Biologically equivalent dose
2.5 × 10 rem
RBE =
=
= 0.85
⎛ 6.2 × 10−3 J ⎞ ⎛ 1 rad ⎞
⎛ E ⎞ ⎛ 1 rad ⎞
⎟
⎜⎜
⎟⎟ ⎜
⎜ ⎟⎜
⎟
⎝ m ⎠ ⎝ 0.01 Gy ⎠
⎝ 21 kg ⎠ ⎝ 0.01 Gy ⎠
______________________________________________________________________________
9.
REASONING AND SOLUTION According to Equation 32.2, the energy absorbed is equal
to the product of the absorbed dose (AD) and the mass of the tumor:
Energy = AD × Mass = (12 Gy)(2.0 kg) = 24 J
This energy is carried by ΔN particles in time Δt, so that
⎛ 1.60 ×10−19 J ⎞
⎛ ΔN ⎞
6
(
)
Energy = ⎜
(850
s)
0.40
10
eV
×
⎜
⎟
⎟
1 eV
⎝ Δt ⎠
⎝
⎠
Therefore,
ΔN
= 4.4 × 1011 s –1
Δt
______________________________________________________________________________
10. REASONING According to Equation 32.2, the absorbed dose is the energy absorbed
divided by the mass of absorbing material:
Absorbed dose =
Energy absorbed
Mass of absorbing material
The energy absorbed in this case is the sum of three terms: (1) the heat needed to melt a
mass m of ice at 0.0 °C into liquid water at 0.0 °C, which is mLf , according to the definition
of the latent heat of fusion Lf (see Section 12.8); (2) the heat needed to raise the temperature
of liquid water by an amount ΔT, which is cmΔT, where c is the specific heat capacity and
ΔT is the change in temperature from 0.0 to 100.0 °C, according to Equation 12.4; (3) the
heat needed to vaporize liquid water at 100.0 °C into steam at 100.0 °C, which is mLv,
according to the definition of the latent heat of vaporization Lv (see Section 12.8). Once the
energy absorbed is determined, the absorbed dose can be determined using Equation 32.2.
Chapter 32 Problems
1631
SOLUTION Using the value of c = 4186 J/(kg⋅C°) for liquid water from Table 12.2 and the
values of Lf = 33.5 × 104 J/kg and Lv = 22.6 × 105 J/kg from Table 12.3, we find that
Absorbed dose
in grays
=
Energy mLf + cmΔT + mLv
=
= Lf + cΔT + Lv
mass
m
= 33.5 × 104 J/kg + ⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ (100.0 C° ) + 22.6 × 105 J/kg
= 3.01 × 106 J/kg
Using the fact that 0.01 Gy = 1 rad, we find that
(
)
⎛ 1 rad ⎞
Absorbed dose = 3.01 × 106 Gy ⎜
= 3.01 × 108 rad
⎟
⎝ 0.01 Gy ⎠
______________________________________________________________________________
11.
SSM REASONING The number of nuclei in the beam is equal to the energy absorbed
by the tumor divided by the energy per nucleus (130 MeV). According to Equation 32.2,
the energy (in joules) absorbed by the tumor is equal to the absorbed dose (expressed in
grays) times the mass of the tumor. The absorbed dose (expressed in rads) is equal to the
biologically equivalent dose divided by the RBE of the radiation (see Equation 32.4). We
can use these concepts to determine the number of nuclei in the beam.
SOLUTION The number N of nuclei in the beam is equal to the energy E absorbed by the
tumor divided by the energy per nucleus. Since the energy absorbed is equal to the
absorbed dose (in Gy) times the mass m (see Equation 32.2) we have
N=
[ Absorbed dose (in Gy)] m
E
=
Energy per nucleus
Energy per nucleus
We can express the absorbed dose in terms of rad units, rather than Gy units, by noting that
1 rad = 0.01 Gy. Therefore,
⎛ 0.01 Gy ⎞
Absorbed dose (in Gy) = Absorbed dose (in rad) ⎜
⎟
⎝ 1 rad ⎠
The number of nuclei can now be written as
N=
E
=
Energy per nucleus
⎛ 0.01 Gy ⎞
⎟m
⎝ 1 rad ⎠
Energy per nucleus
[ Absorbed dose (in rad)] ⎜
1632
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
We know from Equation 32.4 that the Absorbed dose (in rad) is equal to the Biologically
equivalent dose divided by the RBE, so that
⎡ Biologically equivalent dose ⎤ ⎛ 0.01 Gy ⎞
⎢⎣
⎥⎦ ⎜ 1 rad ⎟ m
RBE
E
⎝
⎠
=
N =
Energy per nucleus
Energy per nucleus
⎡180 rem ⎤ ⎛ 0.01 Gy ⎞
⎢⎣ 16 ⎥⎦ ⎜ 1 rad ⎟ ( 0.17 kg )
⎝
⎠
=
= 9.2 × 108
19
−
⎛ 1.60 × 10 J ⎞
130 × 106 eV ⎜
⎟⎟
⎜
1 eV
⎝
⎠
______________________________________________________________________________
(
)
12. REASONING The reaction specified is
2
22
1 H + 11 Na
→ ZA X + 42 He
This reaction must satisfy the conservation of nucleon number and the conservation of
electric charge. Using these laws, we will be able to identify the unknown species ZA X .
SOLUTION The conservation of nucleon number states that the total number of nucleons
present before and after the reaction takes place are the same. Therefore, we have
2 + 22 = A + 4
Before
or
A = 2 + 22 − 4 = 20
After
The conservation of electric charge states that the total number of protons present before and
after the reaction takes place are the same
1 + 11 = Z + 2
Before
or
Z = 1 + 11 − 2 = 10
After
Therefore, with A = 20 and Z = 10, we consult the periodic table on the inside of the back
cover and find that the unknown species ZA X is the nucleus of neon 20
10 Ne .
13. SSM WWW REASONING The reaction given in the problem statement is written in
the shorthand form:
17
8O
(γ , α n)
12
6C .
The first and last symbols represent the initial and
final nuclei, respectively. The symbols inside the parentheses denote the incident particles
or rays (left side of the comma) and the emitted particles or rays (right side of the comma).
Chapter 32 Problems
1633
SOLUTION In the shorthand form of the reaction, we note that the designation αn refers to
an α particle (which is a helium nucleus 42 He ) and a neutron ( 01 n ). Thus, the reaction is
17
8O
γ +
→
12
6C
+ 42 He + 01 n
______________________________________________________________________________
14. REASONING Protons and neutrons are nucleons, and the total number of nucleons before
the reaction is equal to the total number of nucleons after the reaction.
Only the proton and the electron have electrical charge. The neutron and the γ-ray photon
are electrically neutral. The total electric charge of the particles before the reaction is equal
to the total electric charge of particles after the reaction.
SOLUTION
a. The total number of nucleons before the reaction is A + 14. The total number of nucleons
after the reaction is 1 + 17. Setting these two numbers equal to each other yields A = 4. The
net electric charge before the reaction is Z + 7. The net electric charge after the reaction is
1 + 8. Setting these two numbers equal to each other yields Z = 2. The unknown particle is
A
ZX
= 42 He .
b. The total number of nucleons before the reaction is 15 + A. The total number of nucleons
after the reaction is 12 + 4. Setting these two numbers equal to each other yields A = 1. The
net electric charge before the reaction is 7 + Z. The net electric charge after the reaction is
6 + 2. Setting these two numbers equal to each other yields Z = 1. The unknown particle is
A
ZX
1
= 1H .
c. The total number of nucleons before the reaction is 1 + 27. The total number of nucleons
after the reaction is A + 1. Setting these two numbers equal to each other yields A = 27.
The net electric charge before the reaction is 1 + 13. The net electric charge after the
reaction is Z + 0. Setting these two numbers equal to each other yields Z = 14. The
unknown particle is
A
ZX
=
27
14 Si
.
d. The total number of nucleons before the reaction is 7 + 1. The total number of nucleons
after the reaction is 4 + A. Setting these two numbers equal to each other yields A = 4. The
net electric charge before the reaction is 3 + 1. The net electric charge after the reaction is
2 + Z. Setting these two numbers equal to each other yields Z = 2. The unknown particle is
A
ZX
4
= 2 He .
______________________________________________________________________________
1634
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
15. REASONING The conservation of nucleon number states that the total number of nucleons
(protons plus neutrons) before the reaction occurs must be equal to the total number of
nucleons after the reaction. This conservation law will allow us to find the atomic mass
number A of the unknown nucleus. The conservation of electric charge states that the net
electric charge of the particles before the reaction must be equal to the net charge after the
reaction. This conservation law will allow us to find the atomic number Z of the unknown
nucleus. With a knowledge of the atomic number, we can use the periodic table to identify
the element.
SOLUTION
a. The total number of nucleons before the reaction is 1 + 232 = 233. The total number of
nucleons after the reaction is A. Setting these two numbers equal to each other yields
A = 233 . The net electric charge before the reaction is 0 + 90 = 90. The net electric charge
after the reaction is Z. Setting these two numbers equal to each other yields Z = 90 . A
check of the periodic table shows that this element is Thorium (
b. The
233
90Th
−
nucleus subsequently undergoes β decay
first reaction is
233
90Th
A
→ ZX +
0
−1 e .
.
( e ) , as does its daughter.
0
−1
The
By employing an analysis similar to that used in part
(a), the unknown nucleus is found to be
decay according to the reaction
233
90Th)
233
91 Pa
see that the final unknown nucleus is
→
−
233
91 Pa . This daughter nucleus also undergoes β
A
0
Z X + −1 e . Using the analysis of part (a) again, we
233
92 U
.
______________________________________________________________________________
16. REASONING During each reaction, both the total electric charge of the nucleons and the
total number of nucleons are conserved. For each type of nucleus that participates in a
reaction, the atomic mass number A specifies the number of nucleons, and the atomic
number Z specifies the number of protons. We will use these conserved quantities to
calculate the atomic mass number and atomic number for each of the unknown entities,
thereby identifying them.
SOLUTION
a. In the reaction described by
argon
34
18 Ar
designate as
34
18 Ar
( n, α ) ? , a neutron
( 01 n ) induces a reaction in which the
nucleus is broken into an α particle ( 42 He ) and an unknown entity, which we
A
ZX.
The reaction process can be written as
34
1
18 Ar + 0 n
→ 42 He + ZA X
Chapter 32 Problems
1635
Conservation of the total number of nucleons gives 34 + 1 = 4 + A, or A = 34 + 1 − 4 = 31.
Conservation of the total electric charge yields 18 + 0 = 2 + Z, or Z = 18 − 2 = 16. The
atomic number of sulfur is Z = 16, so the unknown entity must be sulfur
b. The reaction
82
34 Se
31
16 S
.
( ?, n ) 82
35 Br can be written as
82
A
34 Se + Z X
→ 01 n + 82
35 Br
Since the total number of nucleons is conserved, we see that 82 + A = 1 + 82, or A = 1.
Conservation of the total electric charge indicates that 34 + Z = 0 + 35, or Z = 35 − 34 = 1.
The unknown entity has A = Z = 1, so it must be a proton
c. The reaction
(
)
58
40
57
28 Ni 18 Ar, ? 27 Co
1
1H
.
can be written as
58
40
28 Ni + 18 Ar
→ ZA X + 57
27 Co
Conservation of the total number of nucleons reveals that the atomic mass number of the
unknown entity is given by 58 + 40 = A + 57, or A = 98 − 57 = 41. Conservation of the total
electric charge reveals that the atomic number is given by 28 + 18 = Z + 27, or
Z = 46 − 27 = 19. Potassium (K) has an atomic number Z = 19, so the unknown entity is the
potassium nucleus
41
19 K
.
d. From the notation ? ( γ , α ) 168 O , we see that the reaction is induced by a γ-ray photon,
which has A = Z = 0. Therefore, the reaction process is
A
0
Z X + 0γ
→ 42 He + 168 O
The unknown entity, then, has A = 4 + 16 = 20, and Z = 2 + 8 = 10. The atomic number
Z = 10 corresponds to neon, so the unknown entity is neon
20
10 Ne
.
17. REASONING AND SOLUTION
a. We note that 01 n is a neutron (n) and 11 H is a proton (p), so the reaction can be written as
14
7N
( n, p ) 146 C
.
b. This reaction can be written as
238
92 U
( n, γ ) 239
92 U
.
c. We note that 21 H is a deuteron (d), so the reaction can be written as
24
12 Mg
( n, d ) 23
11 Na
.
______________________________________________________________________________
1636
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
18. REASONING AND SOLUTION The reaction can be written as
A
Z
X+
63
29 Cu
→
62
29 Cu
+ 11 H +
1
0n
Therefore, A + 63 = 62 + 1 + 1, so that we find A = 1. In addition, Z + 29 = 29 + 1, so that
Z = 1. Thus, the unknown particle AZ X is a proton 11 H . Therefore, the nucleus formed
temporarily has
Z = 29 + 1 = 30
This nucleus is zinc,
64
30 Zn
and
A = 63 + 1 = 64
.
______________________________________________________________________________
19. SSM REASONING Energy is released from this reaction. Consequently, the combined
mass of the daughter nucleus
the parent nucleus
14
7N
and
12
6C
2
1
and the α particle 42 He is less than the combined mass of
H . The mass defect is equivalent to the energy released. We
proceed by determining the difference in mass in atomic mass units and then use the fact
that 1 u is equivalent to 931.5 MeV (see Section 31.3).
SOLUTION The reaction and the atomic masses are as follows:
2
1H
2.014 102 u
+
14
7N
14.003 074 u
12
6C
→
12.000 000 u
+
4
2 He
4.002 603 u
The mass defect Δm for this reaction is
Δ m = 2.014 102 u + 14.003 074 u – 12.000 000 u – 4.002 603 u = 0.014 573 u
Since 1 u = 931.5 MeV, the energy released is
⎛ 931.5 MeV ⎞
(0.014 573 u) ⎜
⎟ = 13.6 MeV
1u
⎝
⎠
______________________________________________________________________________
20. REASONING AND SOLUTION During a fission reaction, neutrons are produced in
addition to the other fission products. We can see from the reaction that the missing nucleon
number is
A = 235 + 1 – 93 – 141 = 2
Thus, the nucleons produced are 2 neutrons .
______________________________________________________________________________
Chapter 32 Problems
1637
21. SSM REASONING The rest energy of the uranium nucleus can be found by taking the
atomic mass of the 235
92 U atom, subtracting the mass of the 92 electrons, and then using the
fact that 1 u is equivalent to 931.5 MeV (see Section 31.3). According to Table 31.1, the
mass of an electron is 5.485 799 × 10–4 u. Once the rest energy of the uranium nucleus is
found, the desired ratio can be calculated.
SOLUTION The mass of
92 electrons, we have
Mass of
235
92 U
235
92
U is 235.043 924 u. Therefore, subtracting the mass of the
−4
nucleus = 235.043 924 u − 92(5.485 799 ×10 u) = 234.993 455 u
The energy equivalent of this mass is
⎛ 931.5 MeV ⎞
⎟ = 2.189 × 10 5 MeV
(234.993 455 u) ⎜
⎝
⎠
1u
Therefore, the ratio is
200 MeV
= 9.0 × 10 –4
5
2.189 × 10 MeV
______________________________________________________________________________
22. REASONING The fission reaction is
1
235
132
101
0 n + 92 U → 50 Sn + 42 Mo +
η
( 01n )
where η is the number of neutrons produced. This reaction must satisfy the conservation of
nucleon number. Using this conservation law, we will be able to determine η.
SOLUTION The conservation of nucleon number states that the total number of nucleons
present before and after the reaction takes place are the same. Therefore, we have
1 + 235 = 132 + 101 + η (1)
Before
or
η = 1 + 235 − 132 − 101 = 3
After
23. REASONING The energy released can be found from the mass defect of the reaction.
According to the discussion in Sections 31.3 and 31.4, the mass defect is equal to the sum of
the individual masses before the reaction minus the sum of the masses after the reaction.
Since the mass of each nucleus is given in atomic mass units (u), we can find the energy
released (in MeV) from the mass defect by using the relation 1 u = 931.5 MeV.
1638
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
SOLUTION The reaction is
1
235
0n +
92 U
1.009 u 235.044 u
→
140
54 Xe
+
139.922 u
94
1
38 Sr + 2 0 n
93.915 u 2(1.009 u )
The sum of the masses before the reaction is 1.009 u + 235.044 u = 236.053 u. The sum of
the masses after the reaction is 139.922 u + 93.915 u + 2(1.009 u) = 235.855 u. The mass
defect is
Δm = 236.053 u − 235.855 u = 0.198 u
The energy released (in MeV) is
⎛ 931.5 MeV ⎞
Energy = ( 0.198 u ) ⎜
⎟ = 184 MeV
1u
⎝
⎠
______________________________________________________________________________
24. REASONING According to Equation 6.10b, the power P is the energy E being produced
divided by the time t:
E
P=
t
The energy is the number N of fissions times the energy per fission EPF, or E = NEPF.
Substituting this result into the expression for power gives
P=
E NEPF
=
t
t
(1)
SOLUTION Using Equation (1), we obtain
P=
NEPF
t
2.0 ×1019 )( 2.0 × 102 MeV )
(
=
1.0 s
Since we are asked for the power in watts (1 W = 1 J/s), it is necessary to convert the energy
units of MeV into joules. For this purpose we note that 1 MeV = 1 × 106 eV and
1 eV = 1.60 × 10−19 J. We find, then, that
2.0 × 1019 ) ( 2.0 ×102
(
P=
1.0 s
)
MeV ⎛ 1×106 eV
⎜⎜
⎝ 1 MeV
⎞ ⎛ 1.60 ×10−19
⎟⎟ ⎜⎜
1 eV
⎠⎝
J⎞
8
⎟⎟ = 6.4 ×10 W
⎠
Chapter 32 Problems
1639
25. REASONING AND SOLUTION In order to find the mass of the two fragments we need to
know the equivalent mass of 225.0 MeV of energy. We know that 1 u = 931.5 MeV.
Therefore, the mass equivalent of 225.0 MeV is
1u
⎛
⎞
mequiv = (225.0 MeV) ⎜
⎟ = 0.2415 u
⎝ 931.5 MeV ⎠
By balancing each side of the fission reaction we can find the mass of the fragments, i.e.,
mfragments = mU-235 + mneutron − m3 neutrons − mequiv
= 235.043 924 u + 1.008 665 u − 3(1.008 665 u) − 0.2415 u = 232.7851 u
______________________________________________________________________________
235
92 U
26. REASONING The mass of a single uranium
atom is m = 235 u (see Appendix F),
where 1 u = 1.66×10−27 kg. The total mass M of uranium
235
92 U
consumed is the mass m
multiplied by the number N of uranium atoms:
M = Nm
(1)
To determine the number N of uranium 235
92 U atoms needed to generate the total energy
output Etot of the United States in one year, we will divide Etot by the energy
E = 2.0×102 MeV released by a single fission:
N =
E tot
(2)
E
Because 1 eV = 1.6×10−19 J, we have that
(
1 MeV = 1.0 ×106 eV
)
1.6 ×10−19 J
= 1.6 ×10−13 J
1 eV
SOLUTION Substituting Equation (2) into Equation (1) yields
M = Nm =
E tot m
E
(3)
Substituting the given values and the conversion factors into Equation (3), we obtain the
mass of uranium 235
92 U that would be needed to supply the energy output of the United
States for one year:
M =
E tot m
E
=
( 9.3 × 1019 J ) ( 235 u ) ⎛⎜ 1.66 × 10 −27
2.0 × 10 2 MeV
⎜
⎝
1u
kg ⎞ ⎛ 1 MeV
⎞
6
⎟⎟ ⎜
⎟ = 1.1 × 10 kg
−13
J ⎠
⎠ ⎝ 1.6 × 10
1640
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
27. SSM REASONING AND SOLUTION
a. The number of nuclei in one gram of U-235 can be obtained as follows:
1 gram of U-235 =
( 2351 mol) ( 6.02 ×1023 nuclei/mol ) = 2.56 ×1021 nuclei
Each nucleus yields 2.0 × 102 MeV of energy, so we have
⎛ 1.60 ×10−19 J ⎞
21
10
E = ( 2.0 ×108 eV ) ⎜
⎟ ( 2.56 × 10 ) = 8.2 × 10 J
1 eV
⎝
⎠
b. If 30.0 kWh of energy are used per day, the total energy use per year is
⎛ 3.60 × 106
Etotal = ( 30.0 kWh/d ) ⎜
⎝ 1 kWh
J ⎞ ⎛ 365 d ⎞
= 3.94 × 1010 J/yr
⎟⎜
⎟
⎠ ⎝ 1 yr ⎠
The amount of U-235 needed in a year, then, is
3.94 × 1010 J
= 0.48 g
8.2 × 1010 J/g
______________________________________________________________________________
m=
28. REASONING A mass m of water that passes through the core absorbs an amount of
heat Q. According to Equation 12.4, Q = cmΔT , where c = 4420 J/(kg·C°) is the specific
heat capacity of the water and ΔT = T − T0 is the difference between the final temperature
Q
T = 287 °C and initial temperature T0 = 216 °C of the water. We will use P =
t
(Equation 6.10b) to determine the amount Q of heat that the water must absorb from the
core in a time t in order to keep the core from heating up, where P = 5.6×109 W is the
thermal output of the core. These relations will allow us to determine the mass of water that
passes through the core each second.
SOLUTION Solving Q = cmΔT (Equation 12.4) for m yields
m=
Q
c ΔT
(1)
Q
(Equation 6.10b) for Q yields Q = Pt . Substituting this result into
t
Equation (1), we find that
Solving P =
Chapter 32 Problems
m=
1641
Q
Pt
=
cΔT cΔT
(2)
With t = 1.0 s, we obtain the mass of water that passes through the core each second:
(
)
5.6 × 109 W (1.0 s )
Pt
m=
=
= 1.8 × 10 4 kg
cΔT ⎡ 4420 J/ ( kg ⋅ C ) ⎤ ( 287 C − 216 C )
⎣
⎦
29. SSM WWW REASONING We first determine the total power generated (used and
wasted) by the plant. Energy is power times the time, according to Equation 6.10, and given
the energy, we can determine how many kilograms of 235
92 U are fissioned to produce this
energy.
SOLUTION Since the power plant produces energy at a rate of 8.0 ×108 W when
operating at 25 % efficiency, the total power produced by the power plant is
(8.0 ×108 W ) 4 = 3.2 ×109 W
The energy equivalent of one atomic mass unit is given in the text (see Section 31.3) as
1 u = 1.4924 × 10 –10 J = 931.5 MeV
Since each fission produces 2.0 × 10 2 MeV of energy, the total mass of
235
92 U
required to
generate 3.2 × 10 9 W for a year (3.156 × 107 s) is
Power times time gives
energy in joules
(3.2 ×109 J/s )(3.156 ×107 s )
⎛ 931.5 MeV
×⎜
−10
⎝ 1.4924 ×10
⎞
⎟
J⎠
Converts joules to
MeV. See Sec. 31.3.
⎛ 1.0 235
⎞
92 U nucleus
⎜
⎟
⎜ 2.0 ×102 MeV ⎟
⎝
⎠
Converts MeV to
number of nuclei
⎛
⎞
0.235 kg
⎜
⎟ = 1200 kg
⎜ 6.022 ×1023 235 U nuclei ⎟
92
⎝
⎠
Converts number of nuclei to
kilograms
1642
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
30. REASONING Each generation of fissions lasts for an average time of tavg = 1.2×10−8 s, and
each generation has a higher fission rate and power output than the previous generation. The
total time T it takes for the power output to reach the stated level is equal to the number N of
generations times the average time of a generation:
T = N tavg
(1)
The power output P from a single generation of fissions is directly proportional to the
number of fissions in that generation. Thus, the power output in the critical state is
25×103 W = P0 = K(1.00), where K is the proportionality constant. The power output of the
first generation of fissions in the supercritical state is P1 = K(1.00)(1.01) = 1.01 P0. The
power output of the second generation of fissions in the supercritical state is
P2 = K(1.00)(1.01)(1.01) = (1.01)2 P0. We can see, then, that the power output of the Nth
generation of fissions in the supercritical state is
PN = (1.01) N P0
(2)
where PN = 3300 MW = 3300×106 W. Equations (1) and (2) will allow us to determine the
total elapsed time T.
SOLUTION Solving for the term in Equation (2) containing N, we obtain
(1.01) N =
PN
(3)
P0
Taking the natural logarithm of both sides of Equation (3) and solving for N yields
⎛P ⎞
N ln (1.01) = ln ⎜ N ⎟
⎝ P0 ⎠
or
N=
ln ( PN P0 )
ln (1.01)
Substituting Equation (4) into Equation (1), we find that
T = N tavg =
ln ( PN P0 ) tavg
ln (1.01)
⎛ 3300 ×106 W ⎞
−8
ln ⎜
⎟ (1.2 ×10 s )
3
= ⎝ 25 ×10 W ⎠
= 1.4 ×10−5 s
ln (1.01)
(4)
Chapter 32 Problems
1643
31. REASONING The reaction is
2
2
1H + 1H
2.0141 u 2.0141 u
→
3
1
2 He + 0 n
3.0160 u 1.0087 u
The energy produced by a fusion reaction is the mass defect Δm (in u) for the reaction times
931.5 MeV/u, since 931.5 MeV is the energy equivalent of 1 u. The mass defect is the total
mass of the initial nuclei minus the total mass of the product nuclei.
SOLUTION Using the given masses, we obtain
⎛ 931.5 MeV ⎞
Energy = ( Δm ) ⎜
⎟
1u
⎝
⎠
⎛ 931.5 MeV ⎞
= ( 2.0141 u + 2.0141 u − 3.0160 u − 1.0087 u ) ⎜
⎟ = 3.3 MeV
1u
⎝
⎠
32. REASONING If energy is released during a reaction, the total rest energy of the particles
after the reaction must be less than the total rest energy before the reaction. Since energy
and mass are equivalent, the total mass of the particles after the reaction is less than the total
mass of the particles before the reaction.
According to Einstein’s relation between mass and energy, Equation 28.5, the difference Δm
in total masses is related to the energy ΔE0 released by the reaction by Δm = ΔE0/c2, where c
is the speed of light in a vacuum.
SOLUTION The difference between the total mass before the reaction and the total mass
after the reaction is
Δm = 1.0078 u + 1.0087 u − 2.0141 u = 0.0024 u
Total mass before reaction
Total mass
after reaction
Note that the γ-ray photon has no rest mass, so that we can ignore it in our calculations.
Since 1 u = 931.5 MeV, the energy released is
⎛ 931.5 MeV ⎞
E = (0.0024 u) ⎜
⎟ = 2.2 MeV
1u
⎝
⎠
______________________________________________________________________________
1644
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
33. REASONING The energy released can be found from the mass defect of the reaction.
According to the discussion in Sections 31.3 and 31.4, the mass defect is equal to the sum of
the individual masses before the reaction minus the sum of the masses after the reaction.
Since the mass of each atom is given in atomic mass units (u), we can find the energy
released (in MeV) from the mass defect by using the relation 1 u = 931.5 MeV.
SOLUTION The reaction is
2
+ 21 H
1H
2.014 102 u 2.014 102 u
→
3
+ 11 H
1H
3.016 050 u 1.007 825 u
The sum of the masses before the reaction is 2.014 102 u + 2.014 102 u = 4.028 204 u. The
sum of the masses after the reaction is 3.016 050 u + 1.007 825 u = 4.023 875 u. The mass
defect is
Δm = 4.028 204 u − 4.023 875 u = 0.004 329 u
The energy released (in MeV) is
⎛ 931.5 MeV ⎞
Energy = ( 0.004 329 u ) ⎜
⎟ = 4.03 MeV
1u
⎝
⎠
______________________________________________________________________________
34. REASONING The conservation of nucleon number states that the total number of nucleons
(protons plus neutrons) before the reaction occurs must be equal to the total number of
nucleons after the reaction. This conservation law will allow us to find the atomic mass
number A of the unknown particle Y. The conservation of electric charge states that the net
electric charge of the particles before the reaction must be equal to the net charge after the
reaction. This conservation law will allow us to find the atomic number Z of the unknown
particle X.
SOLUTION
a. The total number of nucleons before the reaction is 1 + A. The total number of nucleons
after the reaction is 3. Setting these two numbers equal to each other yields A = 2 . The
net electric charge before the reaction is Z + 1. The net electric charge after the reaction is
1. Setting these two numbers equal to each other yields Z = 0 . The nucleon
1
ZX
1
= 0 n is a neutron . The nucleon
A
1Y
2
= 1 H is a deuterium nucleus .
b. The sum of the atomic masses before the reaction is 1.0087 u + 2.0141 u = 3.0228 u. The
sum of the atomic mass after the reaction is 3.0161 u. The difference between the sums is
3.0228 u – 3.0161 u = 0.0067 u. This mass difference is equivalent to an energy of
⎛ 931.5 MeV ⎞
⎟ = 6.2 MeV
1u
⎝
⎠
______________________________________________________________________________
( 0.0067 u ) ⎜
Chapter 32 Problems
1645
35. SSM REASONING To find the energy released per reaction, we follow the usual
procedure of determining how much the mass has decreased because of the fusion process.
Once the energy released per reaction is determined, we can determine the amount of
gasoline that must be burned to produce the same amount of energy.
SOLUTION The reaction and the masses are shown below:
3 21 H
4
2 He
→
3( 2.0141 u)
4.0026 u
+
1
1H
1.0078 u
+
1
0n
1.0087 u
The mass defect is, therefore,
3(2.0141 u) – 4.0026 u – 1.0078 u – 1.0087 u = 0.0232 u
Since 1 u is equivalent to 931.5 MeV, the released energy is 21.6 MeV, or since it is shown
in Section 31.3 that 931.5 MeV = 1.4924 × 10 –10 J , the energy released per reaction is
⎛ 1.4924 × 10 –10 J ⎞
–12
J
⎟ = 3.46 × 10
⎝ 931.5 MeV ⎠
(21.6 MeV ) ⎜
To find the total energy released by all the deuterium fuel, we need to know the number of
deuterium nuclei present. Noting that 6.1 × 10−6 kg = 6.1 × 10−3 g, we find that the number
of deuterium nuclei is
(6.1× 10
–3
⎛ 6.022 × 10 23 nuclei/mol ⎞
21
g) ⎜
⎟ = 1.8 × 10 nuclei
2.0141 g/mol
⎝
⎠
Since each reaction consumes three deuterium nuclei, the total energy released by the
deuterium fuel is
1
3
(3.46 × 10 –12 J/nuclei)(1.8 × 10 21 nuclei) = 2.1 × 10 9 J
If one gallon of gasoline produces 2.1 × 10 9 J of energy, then the number of gallons of
gasoline that would have to be burned to equal the energy released by all the deuterium fuel
is
⎛ 1.0 gal ⎞
2.1×109 J ⎜
⎟ = 1.0 gal
⎝ 2.1×109 J ⎠
______________________________________________________________________________
(
)
1646
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
36. REASONING The conservation of linear momentum states that the total momentum of an
isolated system remains constant. Here, we have the following fusion reaction:
2
3
1H + 1H
2.0141 u 3.0161 u
→
4
1
2 He + 0 n
4.0026 u 1.0087 u
in which the two initial nuclei are assumed to be at rest. Therefore, the initial total
momentum before the fusion occurs is zero, since the momentum vector is the mass times the
velocity vector. As a result of momentum conservation, we can conclude that the final total
momentum of the neutron 01 n and the α particle 42 He is also zero. This means that the
momentum vector of the neutron points opposite to the momentum vector of the α particle
and each has the same magnitude. Using p to denote the magnitude of the momentum, we
have that
pn = pα
(1)
The magnitude p of the momentum is the mass m times the speed v, since we are ignoring
relativistic effects. Thus, according to Equation (1), we have
mn vn = mα vα
or
vn = vα
mα
mn
From the masses given along with the reaction, we see that mα > mn. Therefore, the speed of
the neutron is greater than the speed of the α particle.
Using the facts that p = mv and KE = 12 mv 2 , we have
KE = 12 mv 2 =
m 2v 2 p 2
=
2m
2m
(2)
The total energy released in the reaction is E, and it is all in the form of kinetic energy of the
two product nuclei. Therefore, it follows that
E = KEα + KE n
or
KE n = E − KEα
(3)
SOLUTION Using Equation (2) to substitute into Equation (3), we have
KE n = E − KEα = E −
pα2
2mα
But pn = pα, according to Equation (1), so that Equation (4) becomes
(4)
Chapter 32 Problems
KE n = E −
pn2
2mα
1647
(5)
According to Equation (2), pn2 = 2mn ( KE n ) , so that Equation (5) can be rewritten as
2m ( KE n )
m ( KE n )
pn2
=E− n
=E− n
KE n = E −
mα
2mα
2mα
Solving for KEn reveals that
KE n =
E
17.6 MeV
=
= 14.1 MeV
1 + mn / mα 1 + (1.0087 u ) / ( 4.0026 u )
Using Equation (3), we find that
KEα = E − KE n = 17.6 MeV − 14.1 MeV = 3.5 MeV
The particle with the greater kinetic energy has the greater speed. Thus, the neutron has the
greater speed, as expected.
37. REASONING AND SOLUTION
a. The number n of atoms of hydrogen and its isotopes in 1 kg of water (molecular mass =
18 u) is
2 ( 6.02 × 1023 /mol ) (1.00 ×103 g )
n=
= 6.68 × 1025
18.015 g/mol
If deuterium makes up 0.015% of hydrogen in this number of atoms, the number N of
deuterium atoms in 1 kg of water is
N = (1.5 × 10–4)(6.68 × 1025) = 1.0 × 1022
b. Each deuterium nucleus provides 7.2 MeV of energy, so the energy from 1 kg of water is
⎛ 1.60 ×10−19
E = (1.0 ×1022 )( 7.2 ×106 eV ) ⎜
1 eV
⎝
J⎞
10
⎟ = 1.15 × 10 J
⎠
To supply 1.1 × 1020 J of energy, we would need
1.1× 1020 J
= 9.6 × 109 kg
10
1.15 × 10 J/kg
______________________________________________________________________________
m=
1648
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
38. REASONING AND SOLUTION
(1.007 825 u), deuterium
2
H
1
We will use the atomic mass of hydrogen
(2.014 102 u), helium
0
e
1
3
He
2
(3.016 030 u), helium
4
2
1
H
1
He
(0.000 549 u), and an electron (0.000 549 u). The mass defect
(4.002 603 u), a positron
Δm for two reactions of type (1) is
Δm = 2 (1.007 825 u + 1.007 825 u – 2.014 102 u – 0.000 549 u – 0.000 549 u) = 0.000 900 u
In this result, one of the two values of 0.000 549 u accounts for the positron. The other is
present because the two hydrogen atoms contain a total of two electrons, whereas a single
deuterium atom contains only one electron. The mass defect for two reactions of type (2) is
Δm = 2 (1.007 825 u + 2.014 102 u – 3.016 030 u) = 0.011 794 u
The mass defect for one reaction of type (3) is
Δm = 3.016 030 u + 3.016 030 u – 4.002 603 u – 1.007 825 u – 1.007 825 u = 0.013 807 u
The total mass defect for the proton-proton cycle is the sum of three values just calculated:
Δmtotal = 0.000 900 u + 0.011 794 u + 0.013 807 u = 0.026 501 u
Since 1 u = 931.5 MeV, the energy released is
( 0.026 501 u ) ⎛⎜ 931.5 MeV ⎞⎟ = 24.7 MeV
1u
⎝
⎠
______________________________________________________________________________
39. REASONING The energy released in the decay is the difference between the rest energy of
−
−
the π − and the sum of the rest energies of the μ and νμ . The rest energies of π − and μ
can be found in Table 32.3, and the rest energy of νμ is approximately zero.
SOLUTION The rest energies of the particles are π − (139.6 MeV ) , μ − (105.7 MeV ) and
νμ ( ≈ 0 MeV ) . The energy released is 139.6 MeV – 105.7 MeV = 33.9 MeV .
______________________________________________________________________________
40. REASONING AND SOLUTION The lambda particle contains three different quarks, one
of which is the up quark u, and contains no antiquarks. Therefore, the remaining two quarks
must be selected from the down quark d, the strange quark s, the charmed quark c, the top
quark t, and the bottom quark b. Since the lambda particle has an electric charge of zero
and since u has a charge of +2e/3, the charges of the remaining two quarks must add up to a
total charge of –2e/3. This eliminates the quarks c and t as choices, because they each have
Chapter 32 Problems
1649
a charge of +2e/3. We are left, then, with d, s, and b as choices for the remaining two
quarks in the lambda particle. The three possibilities are as follows:
(1) u,d, s
( 2) u, d,b
(3) u, s,b
______________________________________________________________________________
41. REASONING AND SOLUTION The additional matter comes from the kinetic energy of
the incoming protons. We can find the amount by considering the difference in rest energies
as follows:`
⎛ 931.5 MeV ⎞
ΔE = (1.007 276 u) ⎜
⎟ + 139.6 MeV + 1116 MeV + 497.7 MeV
1u
⎝
⎠
⎛ 931.5 MeV ⎞
− 2(1.007 276 u) ⎜
⎟ = 815 MeV
1u
⎝
⎠
______________________________________________________________________________
42. REASONING The energies (E1, E2) of the γ-ray photons come from the total relativistic
energy E of the neutral pion, which loses both its rest energy and its kinetic energy when it
disintegrates. Therefore, we have that
E1 + E2 = E
(1)
v2
c2
(Equation 28.4), where c = 3.00×108 m/s is the speed of light in a vacuum, v = 0.780c is the
speed of the neutral pion, and mc 2 = E0 = 135.0 MeV (Equation 28.5) is the rest energy of
The total relativistic energy of the neutral pion is found from E = mc 2
1−
the neutral pion.
SOLUTION Solving Equation (1) for E2, we obtain
E2 = E − E1
Substituting E = mc 2
E2 = E − E1 =
1−
(2)
v2
(Equation 28.4) into Equation (2) yields
c2
mc 2
v2
1− 2
c
− E1 =
135.0 MeV
1−
( 0.780c )
c2
2
− 192 MeV = 24 MeV
1650
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
43. SSM WWW REASONING The momentum of a photon is given in the text as
p = E / c (see the discussion leading to Equation 29.6). This expression applies to any
massless particle that travels at the speed of light. In particular, assuming that the neutrino
has no mass and travels at the speed of light, it applies to the neutrino. Once the momentum
of the neutrino is determined, the de Broglie wavelength can be calculated from
Equation 29.6 (p = h/λ).
SOLUTION
a. The momentum of the neutrino is, therefore,
p=
⎞ ⎛ 1.4924 ×10−10
E ⎛
35 MeV
=⎜
⎜
⎟
c ⎝ 3.00 ×108 m/s ⎠ ⎜⎝ 931.5 MeV
J⎞
−20
kg ⋅ m/s
⎟⎟ = 1.9 ×10
⎠
where we have used the fact that 1.4924 × 10–10 J = 931.5 MeV (see Section 31.3).
b. According to Equation 29.6, the de Broglie wavelength of the neutrino is
h
6.63 ×10−34 J ⋅ s
=
= 3.5 ×10 −14 m
20
−
p 1.9 ×10
kg ⋅ m/s
______________________________________________________________________________
λ=
44. REASONING The kinetic energies of the electron and its antiparticle are negligible, so we
may assume that they are at rest before their annihilation. Therefore, the initial total linear
momentum of the system of the two particles is zero. Assuming that the system is isolated,
the principle of conservation of linear momentum holds and indicates that the final total
momentum of the system (that of the two γ-ray photons) is also zero. Since photons cannot
be at rest, the only way for the total momentum of the two photons to be zero is for the
photons to have momenta of equal magnitudes and opposite directions. We conclude, then,
E
(Equation 29.6), where E is the
that each photon has a momentum of magnitude p =
c
energy of either photon and c is the speed of light in a vacuum.
Before annihilation, each of the two particles has a negligible amount of kinetic energy, so
the only energy each possesses is its rest energy E0 = mc 2 (Equation 28.5), where m is the
mass of an electron (and also the mass of its antiparticle). From the energy conservation
principle, we know that the total rest energy 2E0 of the electron and its antiparticle is equal
to the total energy 2E of the two γ-ray photons. Therefore, we have that 2E0 = 2E, or
E0 = E = mc 2
(1)
Chapter 32 Problems
SOLUTION Substituting Equation (1) into p =
1651
E
(Equation 29.6), we obtain
c
E mc 2
= mc = ( 9.11×10−31 kg ) ( 3.00 ×108 m/s ) = 2.73 ×10−22 kg ⋅ m/s
=
c
c
______________________________________________________________________________
p=
45. REASONING AND SOLUTION In order for the proton to come within a distance r of the
second proton, it must overcome the Coulomb repulsive force. Therefore, the kinetic energy
of the incoming proton must equal the Coulombic potential energy of the system.
ke2 ( 8.99 × 109 N ⋅ m 2 /C2 )(1.6 × 10−19 C )
KE = EPE =
=
= 2.9 × 10−14 J
−
15
r
8.0 ×10
m
2
1 eV
⎛
⎞
KE = ( 2.9 × 10−14 J ) ⎜
= 1.8 × 105 eV = 0.18 MeV
−19 ⎟
J⎠
⎝ 1.60 ×10
______________________________________________________________________________
46. REASONING AND SOLUTION The absorbed dose (AD) is given by Equation 32.2,
AD =
Energy absorbed
Mass of absorbing material
Therefore, it follows that
Energy absorbed = AD × Mass = (2.5 × 10–5 Gy)(65 kg) = 1.6 × 10 –3 J
______________________________________________________________________________
47. SSM REASONING The K− particle has a charge of −e and contains one quark and one
antiquark. Therefore, the charge on the quark and the charge on the antiquark must add to
give a total of −e. Any quarks or antiquarks that cannot possibly lead to this charge are the
ones we seek.
SOLUTION
a. Any quark that has a charge of + 23 e (u, c, or t) cannot be present in the K− particle,
because if it were, then the antiquark that is present would need to have a charge of − 53 e to
give a total charge of −e. Since there are no antiquarks with a charge of − 53 e , we conclude
that
the K− particle does not contain u, c, or t quarks
1652
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
b. Any antiquark that has a charge of + 13 e ( d , s , or b ) cannot be present in the K−
particle, because if it were, then the quark that is present would need to have a charge of
− 43 e to give a total charge of −e. Since there are no quarks with a charge of − 43 e , we
conclude that
the K− particle does not contain d , s , or b antiquarks
Note: the K− particle contains the s quark and the u antiquark.
______________________________________________________________________________
48. REASONING The reaction
proton
10
5B
( 11H ) , can be written as
(α , p ) ZA X , where “α” is an α particle ( 42 He )
10
4
1
5 B + 2 He → 1 H
and “p” is a
+ ZA X
The total number of nucleons (protons plus neutrons) is conserved during any nuclear
reaction. This conservation law will allow us to determine the atomic mass number A.
The total electric charge (number of protons) is also conserved during any nuclear reaction.
This conservation law will allow us to determine the atomic number Z. From a knowledge of
Z, we can use the periodic table to identify the element X.
SOLUTION The conservation of the total number of nucleons states that the total number
of nucleons before the reaction (10 + 4) is equal to the total number after the reaction
(1 + A):
10 + 4 = 1 + A
or A = 13
The conservation of total electric charge states that the number of protons before the
reaction (5 + 2) is equal to the total number after the reaction (1 + Z):
5 + 2 = 1+ Z
or
Z= 6
By consulting a periodic table (see the inside of the back cover), we find that the element
whose atomic number is 6 is carbon:
X = C (or carbon)
____________________________________________________________________________________________
Chapter 32 Problems
1653
49. REASONING AND SOLUTION The energy of the neutron after the first collision is
(1.5 × 106 eV)(0.65). After the second collision it is (1.5 × 106 eV)(0.65)(0.65). Thus,
after the nth collision it is
(1.5 × 106 eV)(0.65)n = 0.040 eV
Solving for n gives
⎛ 0.040 eV ⎞
n log (0.65 ) = log ⎜
⎟ = –7.57
⎝ 1.5 × 10 6 eV ⎠
n=
or
–7.57
= 40.4
log ( 0.65 )
Therefore, 40 collisions will reduce the energy to something slightly greater than 0.040 eV,
and to reduce the energy to at least 0.040 eV, 41 collisions are needed.
______________________________________________________________________________
50. REASONING According to Equation 32.2 the absorbed dose is given by
Absorbed dose =
Energy absorbed
Mass of absorbing material
According to Equation 12.5, the amount of energy Q needed to boil a mass m of liquid water
is
Q = mLv
where Lv = 22.6 × 105 J/kg is the latent heat of vaporization (see Table 12.3). Substituting
this expression for Q into the expression for the absorbed dose, we find that
Absorbed dose =
mLv
Energy absorbed
=
= Lv
m
Mass of absorbing material
(1)
Note that a value for the mass m of the water is not needed, since it is eliminated
algebraically from Equation (1). Furthermore, we see that the absorbed dose has the same
units as does the latent heat, namely, 1 J/kg = 1 Gy. To obtain the absorbed dose in rads, we
will use the fact that 1 rad = 0.01 Gy.
SOLUTION Using Equation (1) and converting units, we obtain
⎛ 1 rad ⎞
⎛ 1 rad ⎞
5
8
Absorbed dose (in rads) = Lv ⎜
⎟ = 22.6 ×10 J/kg ⎜
⎟ = 2.26 ×10 rad
⎝ 0.01 Gy ⎠
⎝ 0.01 Gy ⎠
(
)
1654
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
51. SSM REASONING To find the energy released per reaction, we follow the usual
procedure of determining how much the mass has decreased because of the fusion process.
Once the energy released per reaction is determined, we can determine the mass of lithium
10
6
J.
3 Li needed to produce 3.8 × 10
SOLUTION The reaction and the masses are shown below:
2
1H
+
2.014 u
6
3 Li
6.015 u
→ 2
4
2 He
2(4.003 u)
The mass defect is, therefore, 2.014 u + 6.015 u – 2(4.003 u) = 0.023 u . Since 1 u is
equivalent to 931.5 MeV, the released energy is 21 MeV, or since the energy equivalent of
one atomic mass unit is given in Section 31.3 as 1 u = 1.4924 ×10−10 J = 931.5 MeV ,
⎛ 1.4924 × 10 –10 J ⎞
–12
(21 MeV ) ⎜
J
⎟ = 3.4 × 10
⎝ 931.5 MeV ⎠
In 1.0 kg of lithium
6
3 Li ,
there are
⎛ 1.0 × 103 g ⎞ ⎛ 6.022 × 1023 nuclei/mol ⎞
26
(1.0 kg of 63 Li) ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = 1.0 × 10 nuclei
1.0
kg
6.015
g/mol
⎝
⎠ ⎝
⎠
Therefore, 1.0 kg of lithium 63 Li would produce an energy of
(3.4 × 10 –12 J/nuclei)(1.0 × 10 26 nuclei)= 3.4 × 10 14 J
If the energy needs of one household for a year is estimated to be 3.8 ×1010 J , then the
amount of lithium required is
3.8 × 10 10 J
14
= 1.1 × 10 –4 kg
3.4 × 10 J/kg
______________________________________________________________________________
52. REASONING We first determine the energy released by 1.0 kg of
235
92 U
. Using the data
given in the problem statement, we can then determine the number of kilograms of coal that
must be burned to produce the same energy.
SOLUTION The energy equivalent of one atomic mass unit is given in the text (see
Section 31.3) as
1 u = 1.4924 × 10 –10 J = 931.5 MeV
Chapter 32 Problems
Therefore, the energy released in the fission of 1.0 kg of
(1.0 kg of
235
92 U)
235
92 U
1655
is
⎛ 1.0 × 10 3 g/kg ⎞ ⎛ 6.022 × 10 23 nuclei ⎞
⎟
⎜
⎟ ⎜
1.0 mol
⎠
⎝ 235 g/mol ⎠ ⎝
⎛ 2.0 × 10 2 MeV ⎞ ⎛ 1.4924 × 10 –10 J ⎞
13
× ⎜
⎟ ⎜
⎟ = 8.2 × 10 J
nuclei
⎠ ⎝ 931.5 MeV ⎠
⎝
When 1.0 kg of coal is burned, about 3.0 × 107 J is released; therefore the number of
kilograms of coal that must be burned to produce an energy of 8.2 × 1013 J is
⎛ 1.0 kg ⎞
mcoal = 8.2 ×1013 J ⎜
= 2.7 ×106 kg
7 ⎟
⎝ 3.0 ×10 J ⎠
______________________________________________________________________________
(
)
53. REASONING AND SOLUTION The binding energy per nucleon for a nucleus with
A = 239 is about 7.6 MeV per nucleon, according to Figure 32.9 The nucleus fragments into
two pieces of mass ratio 0.32 : 0.68. These fragments thus have nucleon numbers
A1 = (0.32)(A) = 76
and
A2 = (0.68)(A) = 163
Using Figure 32.9 we can estimate the binding energy per nucleon for A1 and A2. We find
that the binding energy per nucleon for A1 is about 8.8 MeV, representing an increase of
8.8 MeV – 7.6 MeV = 1.2 MeV per nucleon. Since there are 76 nucleons present, the
energy released for A1 is
76 × 1.2 MeV = 91 MeV
Similarly, for A2, we see that the binding energy increases to 8.0 MeV per nucleon, the
difference being 8.0 MeV – 7.6 MeV = 0.4 MeV per nucleon. Since there are 163 nucleons
present, the energy released for A2 is
163 × 0.4 MeV = 70 MeV
The energy released per fission is E = 91 MeV + 70 MeV = 160 MeV .
______________________________________________________________________________
1656
IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
54. REASONING AND SOLUTION Using Equation 32.1,
Exposure =
q
( 2.58 × 10 ) m
–4
we can get the amount of charge produced in 1 kilogram of dry air when exposed to 1.0 R of
X-rays as follows:
q = (2.58 × 10–4)(1.0 R)(1 kg) = 2.58 × 10–4 C
Alternatively, we know that 1.0 R deposits 8.3 × 10–3 J of energy in 1 kg of dry air. Thus,
8.3 × 10–3 J of energy produces 2.58 × 10–4 C. To find how much energy E is required to
produce 1.60 × 10–19 C, we set up a ratio as follows:
E
1.60 × 10 –19 C
=
8.3 × 10 –3 J 2.58 × 10 –4 C
or
E = 5.1 × 10 –18 J
Converting to eV, we find
E = (5.1 × 10–18 J)(1 eV)/(1.6 × 10–19 J) = 32 eV
______________________________________________________________________________