ISM chapter45 tủ tài liệu training

45
Applications of Nuclear Physics
CHAPTER OUTLINE
45.1

Interactions Involving Neutrons

45.2

Nuclear Fission

45.3

Nuclear Reactors

45.4

Nuclear Fusion

45.5

Radiation Damage

45.6

Uses of Radiation

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ45.1

Answer (c). We compute the change in mass number A: 235 + 1 – 137
– 96 = 3. All the protons that start out in the uranium nucleus end up
in the fission product nuclei.

OQ45.2

Answer (d). The best particles to trigger a fission reaction of the
uranium nuclei are slow moving neutrons. Fast moving neutrons
may not stay in close proximity with a uranium nucleus long enough
to have a good probability of being captured by the nucleus so that a
reaction can occur. Positively charged particles, such as protons and
alpha particles, have difficulty approaching the target nuclei because
of Coulomb repulsion.

OQ45.3

Answer (c). The total energy released was

E = (17 × 103 ton) ( 4.2 × 109 J 1 ton ) = 7.1 × 1013 J
and according to the mass-energy equivalence, the mass converted
was

m=

E
7.1 × 1013 J
=
= 7.9 × 10−4 kg = 0.79 g  1 g
c 2 ( 3.00 × 108 m s )2

1146

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45

1147

OQ45.4

The ranking is (b) > (c) > (a) > (d). See Table 45.1 for the RBE factors.
Dose (a) is 1 rem. Dose (b) is (1 rad × 10) = 10 rem. Doses (c) and (d)
are (1 rad × 4 or 5) = 4 to 5 rem, but dose (d) is to the hands only (less
mass has absorbed the radiation). If we assume that (a) and (b) as
well as (c) were whole-body doses to many kilograms of tissue (more
mass has absorbed the radiation), we find the ranking stated.

OQ45.5

Answer (c). The function of the moderator is to slow down the
neutrons released by one fission so that they can efficiently cause
more fissions.

OQ45.6

The ranking is Q1 > Q2 > Q3 > 0. Because all of the reactions involve
108 nucleons, we can look just at the change in binding-energy-pernucleon as shown on the curve of binding energy. The jump from
lithium to carbon is the biggest jump (~ 5.4 → 7.7 MeV), and next the
jump from A = 27 to A = 54 (~ 8.3 → 8.8 MeV), which is near the peak

of the curve. The step up for fission from A = 108 to A = 54 (~ 8.7 →
8.8 MeV) is smallest. All the reactions result in an increase in
binding-energy-per-nucleon, so both of the fusion reactions
described and the fission reaction put out energy, so Q is positive for
all.
Imagine turning the curve of binding energy upside down so that it
bends down like a cross-section of a bathtub. On such a curve of total
energy per nucleon versus mass number it is easy to identify the
fusion of small nuclei, the fission of large nuclei, and even the alpha
decay of uranium, as exoenergetic processes. The most stable nucleus
is at the drain of the bathtub, with minimum energy.

OQ45.7

Answer (d). The particles lose energy by collisions with nuclei in the
bubble chamber to make their speed and their cyclotron radii
r = mv/qB decrease.

OQ45.8

Answer (b). The cyclotron radius is given by

r = mv qB = 2 ( 21 m2 v 2 ) qB = 2mK qB
K and B are the same for both particles, but the ratio m q is smaller
for the electron; therefore, the path of the electron has a smaller
radius, meaning the electron is deflected more.
OQ45.9

Answer (b). The nuclei must be energetic enough to overcome the
Coulomb repulsion between them so that they can get close enough

to fuse, and numerous enough for many collisions to occur in a short
period of time so that the reaction produces more energy than it
requires to operate.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1148

Applications of Nuclear Physics

ANSWERS TO CONCEPTUAL QUESTIONS
CQ45.1

The two factors presenting the most technical difficulties are the
requirements of a high plasma density and a high plasma
temperature. These two conditions must occur simultaneously.

CQ45.2

For the deuterium nuclei to fuse, they must be close enough to each
other for the nuclear forces to overcome the Coulomb repulsion of
the protons—this is why the ion density is a factor. The more time
that the nuclei in a sample spend in close proximity, the more nuclei
will fuse—hence the confinement time is a factor.

CQ45.3

The products of fusion reactors are generally not themselves
unstable, while fission reactions result in a chain of reactions which

almost all have some unstable products, because they have an excess
of neutrons.

CQ45.4

The advantage of a fission reaction is that it can generate much more
electrical energy per gram of fuel compared to fossil fuels. Also,
fission reactors do not emit greenhouse gases as combustion
byproducts like fossil fuels—the only necessary environmental
discharge is heat. The cost involved in producing fissile material is
comparable to the cost of pumping, transporting, and refining fossil
fuel.
The disadvantage is that some of the products of a fission reaction
are radioactive—and some of those have long half-lives. The other
problem is that there will be a point at which enough fuel is spent
that the fuel rods do not supply power economically and need to be
replaced. The fuel rods are still radioactive after removal. Both the
waste and the “spent” fuel rods present serious health and
environmental hazards that can last for tens of thousands of years.
Accidents and sabotage involving nuclear reactors can be very
serious, as can accidents and sabotage involving fossil fuels.

CQ45.5

Fusion of light nuclei to a heavier nucleus releases energy. Fission of
a heavy nucleus to lighter nuclei releases energy. Both processes are
steps towards greater stability on the curve of binding energy, Figure
44.5. The energy release per nucleon is typically greater for fusion,
and this process is harder to control.

CQ45.6

The excitation energy comes from the binding energy of the extra
nucleon.

CQ45.7

Advantages of fusion: high energy yield, no emission of greenhouse
gases, fuel very easy to obtain, reactor cannot go supercritical like a
fission reactor and low amounts of radioactive waste.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45

1149

Disadvantages: requires high energy input to sustain reaction,
lithium and helium are scarce, and neutrons released by the reaction
cause structural damage to reactor housing.
CQ45.8

For each additional dynode, a larger applied voltage is needed, and
hence a larger output from a power supply—“infinite” amplification
would not be practical. Nor would it be desirable: the goal is to
connect the tube output to a simple counter, so a massive pulse
amplitude is not needed. If you made the detector sensitive to
weaker and weaker signals, you would make it more and more
sensitive to background noise.

CQ45.9

The hydrogen nuclei in water molecules have mass similar to that of
a neutron, so that they can efficiently rob a fast-moving neutron of
kinetic energy as they scatter it. A neutron bouncing off a more
massive nucleus would lose less energy, so it would continue to
travel through the shield. Once the neutron is slowed down, a
hydrogen nucleus can absorb it in the reaction n + 11 H → 21 H.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 45.1

Interactions Involving Neutrons

Section 45.2

Nuclear Fission

*P45.1

The energy consumed by a 100-W lightbulb in a 1.0-h time period is

⎛ 3600 s ⎞
E = PΔt = ( 100 J/s ) ( 1.0 h ) ⎜
= 3.6 × 105 J
⎝ 1 h ⎟⎠
The number of fission events, yielding an average of 208 MeV each,
required to produce this quantity of energy is

n=
P45.2

E
3.6 × 105 J ⎛ 1 MeV ⎞
=
= 1.1 × 1016
−13 ⎟
⎜
208 MeV 208 MeV ⎝ 1.60 × 10 J ⎠

The mass of U-235 producing the same amount of energy as 1 000 kg of
coal is

⎛ 1 MeV ⎞
m = ( 3.30 × 1010 J ) ⎜
⎝ 1.60 × 10−13 J ⎟⎠
235 g
⎛ 1 U-235 nucleus ⎞ ⎛
⎞
×⎜
⎜
⎟
23
⎝
⎠ ⎝ 6.02 × 10 nucleus ⎟⎠
200 MeV
= 0.403 g

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1150
P45.3

P45.4

Applications of Nuclear Physics
Three different fission reactions are possible:
1
0

n+

235
92

U→

90
38

Sr +

144
54

Xe + 2 01 n

144

54

Xe

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

If the electrical power output of 1.00 GW is 40.0% of the power derived
from fission reactions, the power output of the fission process is
1.00 GW
= ( 2.50 × 109 J s ) ( 8.64 × 10 4 s d ) = 2.16 × 1014 J d
0.400

The number of fissions per day is

( 2.16 × 10

14

⎞
1 eV
⎛ 1 fission ⎞ ⎛
J d)⎜
⎟
6
−19 ⎟
⎜
⎝ 200 × 10 eV ⎠ ⎝ 1.60 × 10 J ⎠
= 6.75 × 1024 d −1

This also is the number of 235U nuclei used, so the mass of 235U used per
day is

(6.75 × 10

24

⎛
⎞
235 g mol
nuclei d ) ⎜
23
⎝ 6.02 × 10 nuclei mol ⎟⎠

= 2.63 × 103 g d = 2.63 kg d

In contrast, a coal-burning steam plant producing the same electrical
power uses more than 6 × 106 kg/d of coal.
P45.5

First, the thorium is bombarded:
1
0

233
n  + 232
   90 Th   →       90 Th

Then, the thorium decays by beta emission:
233
   90

Th   →    233
Pa +  −10 e + ν
   91

Protactinium-233 has more neutrons than the more stable
protactinium-231, so it too decays by beta emission:
233
   91

P45.6

(a)

Pa   →    233
U+
   92

0
−1

e+ν

The energy released is equal to the Q value, given by
Q = ( Δm) c 2 = [ mn + MU-235 − MBa-141 − MKr-92 − 3mn ] c 2

with
Δm = [1.008 665 u + 235.043 923 u − 140.914 4 u

−91.926 2 u − 3 ( 1.008 665 u )] = 0.185 993 u

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45

1151

Then,

Q = ( 0.185 993 u )( 931.5 MeV u ) = 173 MeV
(b)

The fraction of rest energy transformed is

f =
P45.7

Δm 0.185 993 u
=
= 7.88 × 10−4 = 0.078 8%
mi
236.05 u

The energy released in the reaction
is

1
0

n+

235
92

U →

88
38

Sr +

136

54

Xe + 12 10 n

Q = ( Δm) c 2 = ⎡ m 235 U − 11mn − m 88 Sr − m136 Xe ⎤ c 2
38
54
⎣ 92
⎦
= ⎡⎣ 235.043 923 u − 11( 1.008 665 u )

−87.905 614 u − 135.907 220 u ]( 931.5 MeV u )

= 126 MeV

P45.8

In N collisions, the energy is reduced from 2.00 MeV to 0.039 eV:
N

( 2.00 × 10 eV ) ⎛⎜⎝ 21 ⎞⎟⎠ ≤ 0.039 eV
6

N

0.039
⎛ 1⎞
⎜⎝ ⎟⎠ ≤
2
2.00 × 106

⎛ 1⎞
⎛ 0.039 ⎞
N ln ⎜ ⎟ ≤ ln ⎜
⎝ 2⎠
⎝ 2.00 × 106 ⎟⎠
⎛ 2.00 × 106 ⎞
N ln ( 2 ) ≥ ln ⎜
⎝ 0.039 ⎟⎠
which gives
N ≥ 25.6 → N = 26

P45.9

The mass defect is
Δm = ( mn + MU ) − ( MZr + MTe + 3mn )
Δm = [ 1.008 665 u + 235.043 923 u

− 97.912 7 u − 134.916 5 u − 3 ( 1.008 665 u ) ⎤⎦

= 0.197 393 u

The energy equivalent is

⎛ 931.5 MeV c 2 ⎞
Δmc 2 = ( 0.197 393 u ) c 2 ⎜
⎟⎠ = 184 MeV
⎝
u

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1152
P45.10

Applications of Nuclear Physics
(a)

At a concentration of c = 3 mg/m3 = 3 × 10–3 g/m3, the mass of
uranium dissolved in the oceans covering two-thirds of Earth’s
surface to an average depth of havg = 4 km is

mU = cV = c ( 23 A ) ⋅ hav = c ⎡⎣ 23 ( 4π RE2 ) ⎤⎦ ⋅ hav
or
2
g ⎞ ⎛ 2⎞
⎛
mU = ⎜ 3 × 10−3
4π ( 6.38 × 106 m ) ( 4 × 103 m )
⎟
⎜
⎟
3
⎝
m ⎠ ⎝ 3⎠

= 4 × 1015 g
(b)

Fissionable 235U makes up 0.700% of the mass of uranium

computed above. If we assume all of the 235U is collected and
caused to undergo fission, with the release of about 200 MeV per
event, the potential energy supply is

E = ( number of
=

235

U atoms )( 200 MeV )

0.700 ⎛ mU
⎜
100 ⎝ m 235U

⎞
⎟ ( 200 MeV )
atom ⎠

and at a consumption rate of P = 1.5 × 1013 J/s, the time interval
this could supply the world’s energy needs is Δt = E/P, or

Δt =
=

0.700 ⎛ mU
⎜
100 ⎝ m 235U

⎞ ( 200 MeV )

⎟
P
atom ⎠

⎛ 1 kg ⎞ ⎤
0.700 ⎡
4 × 1015 g
⎢
⎥
100 ⎢⎣ ( 235 u )( 1.66 × 10−27 kg u ) ⎜⎝ 103 g ⎟⎠ ⎥⎦
⎡⎛ 200 MeV ⎞ ⎛ 1.60 × 10−13 J ⎞ ⎛
1 yr
× ⎢⎜
⎜
13
7
⎜
⎟
⎟
⎝
⎣⎝ 1.50 × 10 J s ⎠ ⎝ 1 MeV ⎠ 3.16 × 10

⎞⎤
⎟
s ⎠ ⎥⎦

= 5 × 103 yr
(Compare this value to that in part (b) of Problem 17, which is a
more realistic estimate of the time interval for the uranium that
can be extracted reasonably from the Earth.)

(c)

The uranium comes from rocks and minerals dissolved in water
and carried into the ocean by rivers.

(d)

No. Uranium cannot be replenished by the radioactive decay of
other elements on Earth.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
P45.11

One kg of enriched uranium contains 3.40%
uranium-235 is

235
92

1153

U, so the mass of

m235 = 0.034 0(1 000 g) = 34.0 g
In terms of number of nuclei, this is equivalent to
⎛
⎞

1
N 235 = (34.0 g) ⎜
(6.02 × 1023  atoms/mol )
⎝ 235 g/mol ⎟⎠
= 8.71 × 1022  nuclei

If all these nuclei fission, the energy released is equal to

( 8.71 × 10

22

 nuclei ) ( 200 × 106  eV/nucleus )

× ( 1.602 × 10 –19  J/eV ) = 2.79 × 1012  J

Now, for the engine,

efficiency =

work output
heat input

or e =

PΔr cosθ
Qh

So the distance the ship can travel per kilogram of uranium fuel is
Δr =

Section 45.3
*P45.12

(a)

0.200 ( 2.79 × 1012  J )
eQh
=
= 5.58 × 106  m
5
P cos ( 0° )
1.00 × 10  N

Nuclear Reactors
With a specific gravity of 4.00, the density of soil is
ρ = 4.00 × 103 kg/m 3 . Thus, the mass of the top 1.00 m of soil is
2
⎡
1m ⎞ ⎤
2 ⎛
m = ρV = ( 4.00 × 10 kg/m ) ⎢( 1.00 m )( 43 560 ft ) ⎜
⎝ 3.281 ft ⎟⎠ ⎥⎦
⎣
= 1.62 × 107 kg
3

3

At a rate of 1 part per million, the mass of uranium in this soil is

mU =
(b)

m 1.62 × 107 kg
=
= 16.2 kg
106
106

Since 0.720% of naturally occurring uranium is
235
92

235
92

U, the mass of

U in the soil of part (a) is

m 235 U = ( 7.20 × 10−3 ) mU = ( 7.20 × 10−3 ) ( 16.2 kg )
92

= 0.117 kg = 117 g
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1154

P45.13

Applications of Nuclear Physics
In one minute there are N =

60.0 s
= 5.00 × 10 4 fissions.
−3
1.20 × 10 s

So the rate increases by a factor of ( 1.000 25 )
P45.14

(a)

4
⎛ 3V ⎞
For a sphere: V = π r 3 → r = ⎜
⎝ 4π ⎟⎠
3
4π r 2
A
3 ⎛ 36π ⎞
=
= =⎜
⎟
3
V ( 4 3)π r
r ⎝ V ⎠

(b)

50 000

= 2.68 × 105 .

13

, so

13

= 4.84V −1 3

For a cube: V = 3 →  = V 1 3 , so

A 62 6
= 3 = = 6V −1 3
V


(c)

⎛V⎞
For a parallelepiped: V = 2a 3 → a = ⎜ ⎟
⎝ 2⎠

13

, so

2
2
13
13
A ( 2a + 8a ) 5
⎛ 2⎞
⎛ 250 ⎞
−1 3
=
=
=
5
=
⎜⎝ ⎟⎠
⎜⎝
⎟⎠ = 6.30V
3
V
2a
a
V
V

(d) The answers show that the sphere has the smallest surface area
for a given volume and the brick has the greatest surface area of
the three. Therefore, The sphere has minimum leakage and the
parallelepiped has maximum leakage.
P45.15

Recall the radius of a nucleus of mass number A is r = aA1/3, where
a = 1.2 fm. The center to center distance of the nuclei of helium (A = 4)
and gold (A = 197) is the sum of their combined radii:
r = ( 1.2 fm ) ( 4 )

13

+ ( 1.2 fm ) ( 197 )

13

= 8.9 fm = 8.9 × 10−15 m

The electric potential energy is

ke q1q2
r
( 8.99 × 109 N ⋅ m2 / C2 ) ( 2 )(79)(1.60 × 10−19 C) e

U = qV =
=

8.9 × 10−15 m
= 2.6 × 107 eV = 26 MeV
P45.16

The power after three months is P = 10.0 MW = 1.00 × 107 J/s. If each
decay delivers 1.00 MeV = 1.60 × 10–13 J, then the number of decays/s

=

1.00 × 107 J/s
= 6.25 × 1019 Bq
−13
1.60 × 10 J

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
P45.17

(a)

1155

Do not think of the “reserve” as being held in reserve. We are
depleting it as fast as we choose. The remaining current balance
of irreplaceable 235U is 0.7% of the whole mass of uranium:
⎛ 103 kg ⎞ ⎛ 103 g ⎞
= 3.08 × 1010 g
( 0.007 00 )( 4.40 × 10 tons ) ⎜
⎟
⎜
⎟
⎝ 1 ton ⎠ ⎝ 1 kg ⎠
6

(b)

The number of moles of 235U in the reserve is

n=
(c)

m 3.08 × 1010 g
=
= 1.31 × 108 mole
M 235 g/mole

The number of moles found in part (b) corresponds to

⎛ 6.02 × 1023 atom ⎞ ⎛ 1 nucleus ⎞
N = nN A = ( 1.31 × 108 mole ) ⎜
⎟⎠ ⎜⎝ 1 atom ⎟⎠
⎝
1 mole
= 7.89 × 1031 nuclei
(d) We imagine each nucleus as fissioning, to release

(7.89 × 10
(e)

31

−13
⎛ 200 MeV ⎞ ⎛ 1.60 × 10 J ⎞
fissions ) ⎜
= 2.52 × 1021 J
⎝ 1 fission ⎟⎠ ⎜⎝ 1 MeV ⎟⎠

The definition of power is represented by
P = (energy converted)/ Δt , so we have

Δt =

energy
2.52 × 1021 J
1 yr
⎛
=
= ( 1.68 × 108 s ) ⎜
13
⎝ 3.156 × 107
P
1.5 × 10 J/s

⎞
⎟
s⎠

= 5.33 yr
(f)

P45.18

Fission is not sufficient to supply the entire world with energy at
a price of $130 or less per kilogram of uranium.

Assuming that the impossibility is not that he can have this control

over the process (which, as far as we know presently, is impossible),
let’s see what else might be wrong. The reaction can be written
1
0

141
94
1
n  +   235
92 U   →    57 La  +   35 Br  +  n ( 0 n )

where n is the number of neutrons released in the fission reaction. By
balancing the equation for electric charge and number of nucleons, we
find that n = 1. If one incoming neutron results in just one outgoing
neutron, the possibility of a chain reaction is not there, so this nuclear
reactor will not work.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1156
*P45.19

Applications of Nuclear Physics
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 number of fission events needed will be
N=

E
Eevent

=

8.64 × 1010 J
= 2.60 × 1021
−13
( 208 MeV )( 1.60 × 10 J/MeV )

and the mass of this number of 235U atoms is
⎛ N ⎞
⎛ 2.60 × 1021 atoms ⎞
m=⎜
Mmol = ⎜
( 235 g/mol )
⎝ 6.02 × 1023 atoms/mol ⎟⎠
⎝ N A ⎟⎠
= 1.01 g

P45.20

(a)

Since K = p2/2m, we have
⎛3
⎞
p = 2mK = 2m ⎜ kBT ⎟
⎝2

⎠
= 3(1.675 × 10−27 kg)(1.38 × 10−23 J/K) ( 300 K )
= 4.56 × 10−24 kg ⋅ m/s

(b)

The de Broglie wavelength of the particle is

h
6.626 × 10−34 J ⋅ s
λ= =
= 1.45 × 10−10 m = 0.145 nm
−24
p 4.56 × 10 kg ⋅ m/s
(c)

Section 45.4
P45.21

(a)

This size has the same order of magnitude as an atom’s outer
electron cloud, and is vastly larger than a nucleus.

Nuclear Fusion
Helium fusion proceeds according to
4
2

(b)

4

He + 2 He →

8
4

Be + γ

The beryllium produced by helium fusion fuses with another
alpha particle according to
8
4

Be + 24 He →

12
6

C +γ

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
(c)

1157

The total energy released in this pair of fusion reactions is

Q = ( Δm) c 2 = ⎡⎣ 2 m 4 He − m 8 B ⎤⎦ c 2 + ⎡⎣ m 8 B + m 4 He − m12 C ⎤⎦ c 2
= ⎡⎣ 3m 4 He − m12 C ⎤⎦ c 2

= ⎡⎣ 3 ( 4.002 602 u ) − 12.000 000 u ⎤⎦ ( 931.5 MeV u )
= 7.27 MeV
P45.22

From Equation 45.2, the energy released in the reaction
2
3
4
1
1 H + 1 H → 2 He + 0 n is 17.59 MeV per event. The total energy required
for the year is

E = ( 2 000 kWh month ) ( 12.0 months ) ( 3.60 × 106 J kWh )
= 8.64 × 1010 J
so the number of fusion events needed for the year is

N=

E
8.64 × 1010 J
=
Q ( 17.59 MeV event ) ( 1.602 × 10−13 J MeV )

= 3.07 × 1022 events
P45.23

The energy released in the reaction 11 H + 12 H → 32 He + γ is
Q = ( Δm) c 2 = ⎡ m1 H + m 2 H − m 3 He ⎤ c 2
⎣ 1
⎦
1
2
= [ 1.007 825 u + 2.014 102 u − 3.016 029 u ]( 931.5 MeV u )
= 5.49 MeV

P45.24

(a)

We assume that the nuclei are stationary at closest approach, so
that the electrostatic potential energy equals the total energy E.
Then, from the isolated system model,

K f + U f = Ki + U i

→

Uf = E

then,

ke ( Z1e ) ( Z2 e )
=E
rmin

( 8.99 × 10
E=

9

N ⋅ m 2 C2 ) ( 1.60 × 10−19 C ) Z1Z2 ⎛
⎞
1 keV
−14
−16 ⎟
⎜
⎝ 1.60 × 10 J ⎠
1.00 × 10 m
2

= ( 144 keV ) Z1Z2
or E = 144Z1Z2 where E is in keV.
(b)

The energy is proportional to each atomic number.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1158

Applications of Nuclear Physics
(c)

Take Z1 = 1 and Z2 = 59 or vice versa. This choice minimizes the

product Z1 Z2. If extra cleverness is allowed, take Z1 = 0 and
Z2 = 60: use neutrons as the bombarding particles. A neutron is a
nucleon but not an atomic nucleus.

(d) For both the D-D and the D-T reactions, Z1 = Z2 = 1. Thus, the
minimum energy required in both cases is

⎛ 1 MeV ⎞
E = ( 2.30 × 10−14 J ) ⎜
⎝ 1.60 × 10−13 J ⎟⎠
= 144 keV for both, according to this model.
Section 45.4 in the text gives more accurate values for the critical
ignition temperatures, of about 52 keV for D-D fusion and 6 keV
for D-T fusion. The nuclei can fuse by tunneling. A triton moves
more slowly than a deuteron at a given temperature. Then D-T
collisions last longer than D-D collisions and have much greater
tunneling probabilities.
P45.25

(a)

The Q value for the D-T reaction is 17.59 MeV (from Equation
45.4). Specific energy content in fuel for D-T reaction (from Table
44.2, mass = 2.014 u + 3.016 u = 5.030 u):

(17.59 MeV ) (1.60 × 10−13
( 5.030 u ) (1.6605 × 10−27

J MeV )
kg u )

= 3.37 × 1014 J kg

The rate of fuel burning for the D-T reaction is then

rDT =

( 3.00 × 10

( 3.37 × 10

9

14

J s ) ( 3 600 s hr )

J kg ) ( 10−3 kg g )

= 32.1 g h burning of D and T
(b)

Using energy values from Equation 45.4, the specific energy
content in fuel for D-D reaction is:
Q=

1
( 3.27 + 4.03) = 3.65 MeV
2

From Table 44.2, the D-D mass is = 2(2.014 u) = 4.018 u. The
specific energy content in D-D fuel is

( 3.65 MeV ) (1.60 × 10−13 J MeV )
= 8.73 × 1013
−27
( 4.028 u ) (1.6605 × 10 kg u )

J kg

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45

1159

and the rate of fuel burning for the D-D reaction is

rDD =
P45.26

(a)

( 3.00 × 10

( 8.73 × 10

9

13

J s ) ( 3 600 s hr )

J kg ) ( 10−3 kg g )

= 124 g h

The radius of a nucleus with mass number A is r = aA1/3, where
a = 1.2 fm. The distance of closest approach is equal to the center
to center distance of the two nuclei:
13
13
rf = rD + rT = ( 1.20 × 10−15 m ) ⎡⎣( 2 ) + ( 3 ) ⎤⎦

= 3.24 × 10−15 m = 3.24 fm
(b)

At this distance, the electric potential energy is
9
2
2
−19
ke e 2 ( 8.99 × 10 N ⋅ m C ) ( 1.60 × 10 C )
Uf =
=
rf
3.24 × 10−15 m

2

= 7.10 × 10−14 J = 444 keV
(c)

Conserving momentum, mD vi = ( mD + mT ) v f or
⎛ mD ⎞
2
vf = ⎜
v
=
vi
i
5
⎝ mD + mT ⎟⎠

(d) To find the minimum initial kinetic energy of the deuteron, we
use Ki + Ui = Kf + Uf , where Ui = 0 because the deuteron starts
from very far away (infinity), and with the result from part (c),

Ki + 0 =

1
( mD + mT ) v 2f + U f
2
2

⎛ mD ⎞ 2
1
K i = ( mD + mT ) ⎜
vi + U f

2
⎝ mD + mT ⎟⎠
With some re-arrangement, we have
⎛ mD ⎞ ⎛ 1
⎛ mD ⎞
2⎞
Ki = ⎜
Ki + U f
⎜⎝ mD vi ⎟⎠ + U f = ⎜
⎟
⎝ mD + mT ⎠ 2
⎝ mD + mT ⎟⎠

or
⎛
mD ⎞
⎜⎝ 1 − m + m ⎟⎠ K i = U f
D
T

solving for the initial kinetic energy then gives
⎛ m + mT ⎞ 5
Ki = U f ⎜ D
= ( 444 keV ) = 740 keV
⎝ mT ⎟⎠ 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1160

Applications of Nuclear Physics
(e)

The nuclei can fuse possibly by tunneling through the potential
energy barrier.
3

P45.27

(a)

⎛ 1 609 m ⎞
V = ( 317 × 10 mi ) ⎜
= 1.32 × 1018 m 3
⎝ 1 mi ⎟⎠
6

3

From the periodic table, H has atomic mass 1.007 9 and O has
atomic mass 15.999 4, so water has atomic mass 18.015 2.

mwater = ρV = ( 103 kg m 3 ) ( 1.32 × 1018 m 3 ) = 1.32 × 1021 kg
⎛ MH 2 ⎞
⎛ 2.016 ⎞
21
mH2 = ⎜
⎟⎠ ( 1.32 × 10 kg )
⎟ mH2 O = ⎜⎝
18.015

⎝ MH 2 O ⎠
= 1.48 × 1020 kg

mDeuterium = ( 0.030 0% ) mH2 = ( 0.030 0 × 10−2 ) ( 1.48 × 1020 kg )
= 4.43 × 1016 kg
The number of deuterium nuclei in this mass is
N=

4.43 × 1016 kg
mDeuterium
=
= 1.33 × 10 43
mDeuteron ( 2.014 u ) ( 1.66 × 10−27 kg u )

Since two deuterium nuclei are used per fusion, 21 H + 21 H → 24 He,
N
the number of events is
= 6.63 × 10 42.
2
The energy released per event is

Q = ⎡⎣ M 2 H + M 2 H − M 4 He ⎤⎦ c 2

= [ 2 ( 2.014 102 ) − 4.002 603 ] u ( 931.5 MeV u )
= 23.8 MeV

The total energy available is then

⎛ 1.60 × 10−13 J ⎞
⎛ N⎞

42
E = ⎜ ⎟ Q = ( 6.63 × 10 ) ( 23.8 MeV ) ⎜
⎝ 2⎠
⎝ 1 MeV ⎟⎠
= 2.53 × 1031 J
(b)

The time this energy could possibly meet world requirements is

1 yr
2.53 × 1031 J
⎛
⎞
Δt = =
= ( 1.69 × 1016 s ) ⎜
7
13
⎝ 3.16 × 10 s ⎟⎠
P 100 ( 1.50 × 10 J s )
E

= 5.34 × 108 yr

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
P45.28

(a)

1161

Including both ions and electrons, the number of particles in the
plasma is N = 2nV, where n is the ion density and V is the volume
of the container. Application of Equation 21.6 gives the total
energy as

E=

3
NkBT = 3nVkBT
2

6
3
⎡
3 ⎛ 10 cm ⎞ ⎤
    = 3 ( 2.00 × 10 cm ) ⎢( 50.0 m ) ⎜
⎥
⎝ 1 m 3 ⎟⎠ ⎦
⎣
13

−3

× ( 1.38 × 10−23 J K ) ( 4.00 × 108 K )

7
E = 1.66 × 10 J

(b)

The specific heat of water is c = 4 186 J/kg . °C, and the energy
required to raise the temperature of one kilogram of water from
27.0°C to 100°C is given by Equation 20.4:

Q = mcΔT = ( 1.00 kg ) ( 4 186 J/kg ⋅°C )( 100°C − 27.0°C )
= 3.06 × 105 J
From Table 20.2, the heat of vaporization of water is
Lv = 2.26 × 106 J kg , so that a total of
E1 kg = 3.06 × 105 J + 2.26 × 106 J = 2.57 × 106 J

is required to boil away each kilogram of water initially at 27.0°C.
The mass of water that could be boiled away is therefore
E
1.66 × 107 J
m=
=
= 6.45 kg
E1 kg 2.57 × 106 J kg

P45.29

(a) Taking m ≈ 2mp for deuterons, we have
1
3
mv 2 = kBT
2
2

The root-mean-square speed is

vrms

3 ( 1.38 × 10 –23  J/K ) ( 4.00 × 108  K )
3kBT
=
=
2mp
2 ( 1.67 × 10 –27  kg )
= 2.23 × 106  m/s

(b)

The confinement time in the absence of confinement measures is

Δt =

x
0.100 m
=
 10 –7 s
6
v 2.23 × 10  m/s

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1162

P45.30

Applications of Nuclear Physics
(a) By adding 1 + 6 = 7 and 1 + 12 = 13, we have 11 H +

12
6

C→

13
7

N + γ so

13

nucleus A is N.
(b)

Now 13 – 0 = 13 and 7 – 1 = 6, so the positron decay is
13
13
N → 136 C + 01 e + ν and nucleus B is C.
7

(c)

Similarly, we have 11 H + 136 C →

14
7

N + γ and nucleus C is N.
14

(d) The hydrogen nuclei keep piling on like rugby players after a
tackle. We have 11 H + 147 N → 158 O + γ and nucleus D is 15O.
15
8

O→

15
7

N + 01 e + ν , so nucleus E is N.
15

(e)

Now

(f)

We calculate 15 + 1 – 4 = 12 and 7 + 1 – 2 = 6 to identify
12
1
H + 157 N → 126 C + 24 He and nucleus F is C.
1

(g)

The original carbon-12 nucleus is returned. One carbon nucleus
can participate in the fusions of colossal numbers of hydrogen
nuclei, four after four. Carbon is a catalyst.
The two positrons immediately annihilate with electrons
according to 01 e + −10 e → 2γ . The overall reaction, obtained by
adding all eight reactions, can be represented as
1
1

H+

12
6

C + 11 H + 11 H + 11 H + 2 −10 e → 24 He +

12
6

C + 7γ + 2ν

This simplifies to 4 ( 11 H ) + 2 −10 e → 24 He + 2ν . The net reaction is
identical to the net reaction in the proton–proton cycle which
predominates in the Sun. In energy terms the reaction can be

(

)

considered as 4 11 H atom → 24 He atom + 26.7 MeV, where the Q
value of energy output was computed in Chapter 39, Problem 67
and again in Problem 59 in this chapter.
P45.31

(a)

Lawson’s criterion for the D-T reaction is nτ ≥ 1014 s cm 3 . For a
confinement time of τ = 1.00 s, this requires a minimum ion
density of n = 1014 cm −3 .

(b)

At the ignition temperature of T = 4.5 × 107 K and the ion density
found above, the plasma pressure is

P = 2nkBT
⎡
⎛ 106 cm 3 ⎞ ⎤
−23
7
= 2 ⎢( 1014 cm −3 ) ⎜
3
⎟⎠ ⎥ ( 1.38 × 10 J K ) ( 4.5 × 10 K )
⎝
1
m
⎣

⎦
 = 1.2 × 105 J m 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
(c)

1163

The required magnetic energy density is then

B2
uB =
≥ 10P = 10 ( 1.2 × 105 J m 3 ) = 1.2 × 106 J m 3
2 µ0
which requires a magnetic field of magnitude

B ≥ 2 µ0 ( 10P ) = 2 ( 4π × 10−7 N A 2 ) ( 1.24 × 106 J m 3 )
= 1.8 T
This is a very strong field.

Section 45.5
P45.32

(a)

Radiation Damage
The number of x-ray images made per year is (assuming a 2-week
vacation)

n = ( 8 x-ray d ) ( 5 d wk ) ( 50 wk yr ) = 2.0 × 103 x-ray yr

The average dose per photograph is

5.0 rem yr
= 2.5 × 10−3 rem x-ray = 2.5 mrem x-ray
3
2.0 × 10 x-ray yr
(b)

The technician receives low-level background radiation at a rate
of 0.13 rem/yr. The ration dose of 5.0 rem/yr received as a result
of the job to background is

5.0 rem yr
= 38
0.13 rem yr
The technician’s occupational exposure is high compared to
background radiation—it is 38 times 0.13 rem/yr.
P45.33

(a)

I = I 0 e − µ x , so x = −

1 ⎛ I⎞
–1
ln ⎜ ⎟ , with µ = 1.59 cm .
µ ⎝ I0 ⎠

I0
,
2
1
⎛ 1⎞
x=−
ln ⎜ ⎟ = 0.436 cm
−1
⎝ 2⎠
1.59 cm

When the intensity I =

(b)

I0
,
1.00 × 10 4
1
1
⎛
⎞
x=−
ln ⎜
= 5.79 cm
−1
4⎟
⎝
1.59 cm
1.00 × 10 ⎠

When I =

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1164

P45.34

Applications of Nuclear Physics

(a)

(b)
P45.35

I = I 0 e − µ x , so x = −

1 ⎛ I⎞
.
ln
µ ⎜⎝ I 0 ⎟⎠

When intensity I =

1 ⎛ I⎞
1 ⎛ 1 ⎞ ln ( 2 )
I0
, x = − ln ⎜ ⎟ = − ln ⎜ ⎟ =

.
2
µ ⎝ I0 ⎠
µ ⎝ 2⎠
µ

When intensity I = f I0, x = −

ln f
1 ⎛ I⎞
1
ln ⎜ ⎟ = − ln ( f ) = −
.
µ ⎝ I0 ⎠
µ
µ

The source delivers 100 mrad of 2.00-MeV γ -rays/h at a 1.00-m
distance. The RBE for these γ -rays is 1.0 (from Table 45.1).
(a)

From Equation 45.6,
dose in rem = dose in rad × RBE
1.00 rem = dose in rad × 1.0   

or,

dose in rad = 1.00 rad = ( 100 × 10−3 rad/h ) Δt

which gives Δt = 10.0 h.

Thus a person would have to stand there 10.0 hours to receive
1.00 rem from a 100-mrad/h source.
(b)

If the γ -radiation is emitted isotropically, the dosage rate falls off
1
as 2 .
r
Thus a dosage 10.0 mrad/h would be received at a distance
r = 10.0 m = 3.16 m .

*P45.36

For each gray (GY) or radiation, 1 J of energy is delivered to each
kilogram of absorbing material. Thus, the total energy delivered in this
whole body dose to a 75.0-kg person is
⎛ J/kg ⎞
E = ( 0.250 Gy ) ⎜ 1
(75.0 kg ) = 18.8 J
⎝ Gy ⎟⎠

P45.37

By definition, one rad increases the energy of one kilogram of the
absorbing material by 1.00 × 10-2 J. The energy starts as energy carried
by electromagnetic radiation, and turns entirely into internal energy.
The 1 000 rad or 10.0 gray = 10.0 Gy will then put 10.0 J/kg into the
body, to raise its temperature by the same amount as 10.0 J/kg of
energy input by heat from a higher-temperature energy source. In
Q = mcΔT we have Q/m = 10.0   J/kg and

ΔT =

⎛
⎞
Q1
1
= ( 10.0 J/kg ) ⎜
= 2.39 × 10−3 °C
⎟
mc
⎝ 4 186 J/kg ⋅°C ⎠

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
P45.38

1165

Assume all the energy from the x-ray machine is absorbed by the
water and that no energy leaves the cup of water by heat or thermal
radiation. The energy input to the cup and the temperature of the
water are related by

TER = mcΔT
Because the power input P is equal to TER /ΔT, we have
PΔt = mcΔT    →    Δt = 

mcΔT

P

where we have solved for the time interval required to raise the
temperature of the water. We note that the temperature of the water
will increase until it is 100°C, after which the latent heat of
vaporization of Lv = 2.26 × 106 J kg would have to be added to boil the
water. For the purposes of this problem, we limit ourselves to
increasing the temperature of the water to 100°C. Substituting
numerical values gives

Δt = 

m ( 4 186 J/kg ⋅ °C ) ( 50.0°C )

(10.0 rad/s ) (1 × 10–2  J/kg ) m

 = 2.09 × 106  s = 24.2 d

Therefore, it would take over 24 days just to increase the water’s
temperature to 100°C, and much longer to boil it, and this technique
will not work for a 20-minute coffee break!
P45.39

The number of nuclei in the original sample is

N0 =

mass present
5.00 kg
=

mass of nucleus ( 89.907 7 u ) ( 1.66 × 10 –27  kg/u )

= 3.35 × 1025  nuclei
The decay constant is

λ=

ln 2
0.693
=
= 2.38 × 10 –2  yr –1 = 4.53 × 10 –8  min –1
T1/2 29.1 yr

The original activity is

R0 = λ N 0 = ( 4.53 × 10 –8  min –1 ) ( 3.35 × 1025  nuclei )
= 1.52 × 1018  decays/min
The law of decay then gives us

10.0 decays/min
R
=
= 6.59  × 10 –18 =  e – λt
R0 1.52 × 1018  decays/min

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1166

Applications of Nuclear Physics
and the time interval is
t=

P45.40

− ln ( R / R0 ) − ln ( 6.59 × 10 –18 )
=
= 1.66 × 103 yr
–2
–1
λ
2.38 × 10  yr

If half of the 0.140-MeV gamma rays are absorbed by the patient, the
total energy absorbed is
E=

( 0.140 MeV ) ⎡⎛ 1.00 × 10−8 g ⎞ ⎛ 6.02 × 1023 nuclei ⎞ ⎤
⎢⎜
⎟⎜
⎣⎝ 98.9 g mol ⎠ ⎝

2

1 mol

= ( 4.26 × 1012 MeV ) ( 1.60 × 10−13 J MeV ) = 0.682 J

Thus, the dose received is Dose =

P45.41

⎟⎠ ⎥
⎦

0.682 J ⎛ 1 rad ⎞
= 1.14 rad
60.0 kg ⎜⎝ 10−2 J kg ⎟⎠

The decay constant is λ = ln 2 T1 2 = ln 2 17.0 d . The number of nuclei
remaining after 30.0 days is

⎡⎛ − ln 2 ⎞
⎤
N = N 0 e − λT = N 0 exp ⎢⎜
30.0 d ⎥ = 0.294N 0
⎟
⎣⎝ 17.0 d ⎠
⎦
The number decayed is N0 – N = N0 (1 – 0.294) = 0.706N0.
Then the energy release is

⎛ 1.60 × 10−19 J ⎞
2.12 J = ( 0.706N 0 ) ( 21.0 × 103 eV ) ⎜
⎟⎠
1 eV
⎝
N0 =
(a)

2.12 J
= 8.94 × 1014
−15
2.37 × 10 J

The initial activity is
R0 = λ N 0 =

(b)

⎛ 1d ⎞
ln 2
8.94 × 1014 ) ⎜
= 4.22 × 108 Bq
(
17.0 d
⎝ 86 400 s ⎟⎠

We find the total mass contained in the seeds from

original sample mass = m = N 0 mone atom
⎡
⎛ 1.66 × 10−27 kg ⎞ ⎤
= 8.94 × 10 ⎢( 103 u ) ⎜
⎟⎠ ⎥
⎝
1u
⎣
⎦
14

Then,

m = 1.53 × 10−10 kg = 1.53 × 10−7 g = 153 ng

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45
P45.42

1167

The nuclei initially absorbed are (mass from Table 44.2)
⎛ 6.02 × 1023 nuclei mol ⎞
N 0 = ( 1.00 × 10−9 g ) ⎜
= 6.70 × 1012
⎟
89.9
g
mol
⎝
⎠

The number of decays in time t is

(

)

(

ΔN = N 0 − N = N 0 1 − e − λ t = N 0 1 − e

−( ln 2 )t T1 2

)

At the end of 1 year,

⎧⎪
⎡⎛ − ln 2 ⎞
⎤ ⎫⎪
ΔN = N 0 − N = ( 6.70 × 1012 ) ⎨1 − exp ⎢⎜
1.00
yr
⎥⎬
⎟
⎣⎝ 29.1 yr ⎠
⎦ ⎭⎪
⎩⎪
= 1.58 × 1011
The energy deposited is

E = ( 1.58 × 1011 ) ( 1.10 MeV ) ( 1.60 × 10−13 J MeV ) = 0.027 7 J
Thus, the dose received is
⎛ 0.027 7 J ⎞
Dose = ⎜
= 3.96 × 10−4 J kg = 0.039 6 rad
⎟

70.0
kg
⎝
⎠

Section 45.6
P45.43

(a)

Uses of Radiation
With I ( x ) =

1
I 0 , I ( x ) = I 0 e − µ x becomes
2

1
I 0 = I 0 e −0.72 x mm
2
2 = e +0.72 x mm → ln 2 = 0.72 x mm → x =

( ln 2 ) mm
0.72

= 0.963 mm

(b)

The intensity reaching the detector through x1 = 0.800 mm of steel

is I1 = I 0 e − µ x1 . That transmitted by thickness x2 = 0.700 mm is
I 2 = I 0 e − µ x2 . The fractional change is

I 2 − I1 I 0 e − µ x2 − I 0 e − µ x1
=
= e µ (x1 − x2 ) − 1 = e(0.720/mm)(0.100 mm ) − 1
− µ x1
I1
I0e
= e 0.0720 − 1 = +0.074 7 = 7.47%
As the thickness decreases, the intensity increases by 7.47%.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1168
P45.44

Applications of Nuclear Physics
(a)

Starting with N = 0 radioactive atoms at t = 0, the rate of increase
is (production – decay)

dN
= R − λN
dt

dN = ( R − λ N ) dt.

so

The variables are separable.
N

∫
0

so

t
dN
= ∫ dt:
R−λN 0

⎛ R − λN ⎞
ln ⎜
= −λ t
⎝ R ⎟⎠

Therefore

P45.45

−

1−

λ
N = e−λ t
R

1 ⎛ R− λN⎞
ln ⎜
⎟ =t
λ ⎝ R ⎠

and
→

⎛ R− λN⎞
−λ t
⎜⎝
⎟⎠ = e .
R
N=

R
1 − e−λ t .
λ

(

)

R
.
λ

(b)

The maximum number of radioactive nuclei would be

(a)

The number of photons is

(b)

Natural copper is 69.17% 63Cu and 30.83% 65Cu. Thus, if the
sample contains NCu copper atoms, the number of atoms of each
isotope is N63 = 0.691 7 NCu and N65 = 0.308 3 NCu. Therefore,

10 4 MeV
= 9.62 × 103. Since only 50%
1.04 MeV
of the photons are detected, the number of 65Cu nuclei decaying is
twice this value, or 1.92 × 104. In two half-lives, three-fourths of
3
the original nuclei decay, so N 0 = 1.92 × 10 4 and N0 = 2.56 × 104.
4
65
This is 1% of the Cu, so the number of 65Cu is 2.56 × 106 ~ 106 .

N 63 0.691 7
=
N 65 0.308 3
or

⎛ 0.6917 ⎞
⎛ 0.6917 ⎞

N 63 = ⎜
N =
2.56 × 106 ) = 5.75 × 106
⎝ 0.3083 ⎟⎠ 65 ⎜⎝ 0.3083 ⎟⎠ (

The total mass of copper present is then

mCu = ( 62.93 u ) N 63 + ( 64.93 u ) N 65

mCu = ⎡⎣( 62.93 u ) ( 5.75 × 106 ) + ( 64.93 u )( 2.56 × 106 ) ⎤⎦

× ( 1.66 × 10−24 g u )

= 8.77 × 10−16 g ~ 10−15 g

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 45

1169

Additional Problems
P45.46

(a)

(

)

The energy released by the 11 H + 115 B → 3 24 He reaction is
Q = ⎡ M 1 H + M 11 B − 3M 4 He ⎤ c 2
5
2
⎣ 1
⎦

Q = ⎡⎣1.007 825 u+11.009 305 u − 3 ( 4.002 603 u ) ⎤⎦

× ( 931.5 MeV u )

= 8.68 MeV

The particles must have enough kinetic energy to overcome their
mutual electrostatic repulsion so that they can get close enough
to fuse.

(b)

P45.47

From momentum conservation, we have


0 = mLi v Li + mα v α or mLi vLi = mα vα
Thus,

( m v ) ⎛ m2 ⎞
1

1 ( mLi vLi )
K Li = mLi vLi2 =
= α α = ⎜ α ⎟ vα2
2
2 mLi
2mLi
⎝ 2mLi ⎠
2

2

⎡ ( 4.002 6 u ) 2 ⎤ ⎛
2
6
K Li = ⎢
⎥ ⎜⎝ 9.25 × 10 m s ⎞⎟⎠
⎢⎣ 2 ( 7.016 0 u ) ⎥⎦

= ( 1.14 u )( 1.66 × 10− 27 kg/u ) ⎛⎝⎜ 9.25 × 106 m s ⎞⎠⎟

2

K Li = 1.62 × 10−13 J = 1.01 MeV
1
2
( ΔPmax ) .
ρ v (ω smax ) , and from Equation 17.10, I =
2
2 ρv
Substituting the second expression for I into the first and solving

for smax gives
2

P45.48

(a)

We have I =

smax

1 ⎛ 2I ⎞
= ⎜ ⎟
ω ⎝ ρv ⎠

12

1
=
ω

⎡ 2 ( ΔPmax )2 ⎤
⎢
⎥
⎢⎣ ρ v 2 ρ v ⎥⎦

12

=

ΔPmax
ωρ v

Solving for ΔPmax and assuming smax ~ 2.5 m,
ΔPmax = ωρ vsmax = ( 1 s −1 ) ( 1.20 kg/m 3 ) ( 343 m/s )( 2.5 m )
 103 Pa

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


1170

Applications of Nuclear Physics
(b)

The change in volume is given by

ΔV = 4π r 2 Δ r = 4π ( 14.0 × 103 m ) ( 2.5 m )
2

9
3
= 1.23 × 108 m 3 ~ 6 × 10 m

(c)

The energy carried by the blast wave is

W = ( ΔPmax ) ( ΔV ) = ( 103 Pa ) ( 6 × 109 m 3 ) = 6 × 1012 J
(d) Since the blast wave carries only 10% of the bomb’s energy,

1
6 × 1016 J = ( yield ) , and the bomb yield is then
10
14
yield = 6 × 1013 J ~10 J

(e)

The yield in terms of tons of TNT is

6 × 1013 J
= 1.42 × 10 4 ton TNT ~ 10 4 ton TNT
4.2 × 109 J ton TNT
*P45.49

The Japanese call it the original child bomb.
(a)

Suppose each 235 U fission releases 208 MeV of energy. Then, the
number of nuclei that must have undergone fission is

total release
5 × 1013 J
N=
=
energy per nuclei ( 208 MeV ) ( 1.60 × 10−13 J MeV ) .
= 1.5 × 1024 nuclei
(b)
P45.50

(a)

⎛
⎞
1.5 × 1024 nuclei
mass = ⎜
( 235 g mol ) ≈ 0.6 kg
23
⎝ 6.02 × 10 nuclei mol ⎟⎠
Subtracting the background counts, the decay counts are
N1 = 372 – 5(15) = 297 in the first 5.00 min interval and
N2 = 337 – 5(15) = 262 in the second. The midpoints of the time
intervals are separated by T = 5.00 min. We use R = R0 e − λ t , taking
t = T and identifying R0 = N1/T = 297/5 min and R = N2/T =
262/5 min. We have then

N 2 ⎛ N1 ⎞ − λ T
=⎜
⎟e
T ⎝ T ⎠

262
⎛ 297 ⎞ − ( ln 2 T1 2 )( 5.00 min )
=⎜
⎟e
5 min ⎝ 5 min ⎠

or

which gives

e

(

− ln 2 T1

2

)T = N 2

N1

or

e

(

− ln 2 T1

2

)( 5.00 min ) = 262
297

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.