ISM chapter44 tủ tài liệu training

44
Nuclear Physics
CHAPTER OUTLINE
44.1

Some Properties of Nuclei

44.2

Nuclear Binding Energy

44.3

Nuclear Models

44.4

Radioactivity

44.5

The Decay Processes

44.6

Natural Radioactivity

44.7

Nuclear Reactions

44.8

Nuclear Magnetic Resonance and Magnetic Resonance Imaging

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

ANSWERS TO OBJECTIVE QUESTIONS
OQ44.1

Answer (b). The frequency increases linearly with the magnetic field
 
strength because the magnetic potential energy − µ • B is
proportional to the magnetic field strength.

OQ44.2

Answer (a). In the beta decay of

95
36

Kr, the emitted particles are an

electron, e, and an antineutrino, ν e . The emitted particles contain a
total charge of –e and zero nucleons. Thus, to conserve both charge
95
and nucleon number, the daughter nucleus must be 37
Rb, which
contains Z = 37 protons and A – Z = 95 – 37 = 58 neutrons. (Recall
that the electron and an antineutrino are produced by the decay on a

neutron into a proton.)
0
−1

1099
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1100

Nuclear Physics

OQ44.3

Answer (c). The emitted particle is not a nucleon because there is no
change in nucleon number, and conservation of charge requires
15 = 16 + Z → Z = –1, so the emitted particle is an electron. From
Equation 44.19, we see that 32
15 P decays by means of beta decay:
32
15

OQ44.4

P→

32
16

S + −10 e + ν e .

Answer (d). In a large sample, one half of the radioactive nuclei
initially present remain in the sample after one half-life has elapsed.
Hence, the fraction of the original number of radioactive nuclei
remaining after n half-lives have elapsed is (1/2)n = 1/2n. In this case
the number of half-lives that have elapsed is Δt T1 2 = 14 d 3.6 d ≈ 4.
Therefore, the approximate fraction of the original sample that
remains undecayed is 1/24 = 1/16.

OQ44.5

(i)

Answer (b). Since the samples are of the same radioactive
isotope, their half-lives are the same.

(ii) Answer (b). When prepared, sample G has twice the activity
(number of radioactive decays per second) of sample H. The
activity of a sample experiences exponential decay also;
therefore, after 5 half-lives, the activity of sample G is decreased
by a factor of 25, and after 5 half-lives the activity of sample H is
decreased by a factor of 25. So after 5 half-lives, the ratio of
activities is still 2:1.
OQ44.6

Answer (b). A gamma ray photon carries no nucleon number and no
charge, so there can be no change in these quantities.

OQ44.7

Answer (c). The nucleus 40
18 X contains A = 40 total nucleons, of which
Z = 18 are protons. The remaining A – Z = 40 – 18 = 22 are neutrons.

OQ44.8

Answer (b). Conservation of nucleon number requires 144 = 140 + A
→ A = 4, and conservation of charge requires 60 = 58 + Z → Z = 2.
The particle is 24 X = 24 He.

OQ44.9

Answer (d). The Q value for the reaction 94 Be + 24 He →
(using masses from Table 44.2)

(

12
6

C + 01 n is

)

Q = ( Δm) c 2 = m 9 Be + m 4 He − m 12 C − mn c 2
4

2

= [ 9.012 182 u + 4.002 603 u

6

−12.000 000 u − 1.008 665 u ] × ( 931.5 Mev u )

= 5.70 MeV
OQ44.10

(i)

Answer (a). The liquid drop model gives a simpler account of a
nuclear fission reaction, including the energy released and the
probable fission product nuclei.

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Chapter 44

1101

(ii) Answer (b). The shell model predicts magnetic moments by
necessarily describing the spin and orbital angular momentum
states of the nucleons.
(iii) Answer (b). Again, the shell model wins when it comes to
predicting the spectrum of an excited nucleus, as it allows only
quantized energy states, and thus only specific transitions.
OQ44.11

Answer (d). A free neutron can undergo beta decay into a proton

plus an electron and an antineutrino because its mass is greater than
the mass of a free proton. Energy conservation prevents a free proton
from decaying into a neutron plus a positron and a neutrino. (A
proton bound inside a nucleus can undergo beta decay into a neutron
if the final mass of the nucleus is less than that of the original
nucleus, as for example in the beta decay of sodium-22:
22
+
22
11 Na → e + ν + 10 Ne. )

OQ44.12

Answer (d). The reaction energy is the amount of energy released as
a result of a nuclear reaction. Equation 44.29 in the text implies that
the reaction energy is (initial mass – final mass) c2. The Q-value is
taken as positive for an exothermic reaction.

OQ44.13

Answer (c). To conserve nucleon number (mass number), it is
necessary that A + 4 = 234, or A = 230. Conservation of charge
(atomic number) demands that Z + 2 = 90, or Z = 88.

ANSWERS TO CONCEPTUAL QUESTIONS
CQ44.1

The alpha particle and the daughter nucleus carry equal amounts of
momentum in opposite directions. Since kinetic energy can be
p2

written as
, the small-mass alpha particle has much more of the
2m
decay energy than the recoiling nucleus.

CQ44.2

The statement is false. Both patterns show monotonic decrease over
time, but with very different shapes. For radioactive decay,
maximum activity occurs at time zero. Cohorts of people now living
will be dying most rapidly perhaps forty years from now. Everyone
now living will be dead within less than two centuries, while the
mathematical model of radioactive decay tails off exponentially
forever. A radioactive nucleus never gets old. It has constant
probability of decay however long it has existed.

CQ44.3

An alpha particle contains two protons and two neutrons. Because
the nuclei of heavy hydrogen (D and T) contain only one proton, they
cannot emit an alpha particle.

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1102

Nuclear Physics

CQ44.4

In alpha decay, there are only two final particles, the alpha particle
and the daughter nucleus. There are also two conservation
principles, energy and momentum, that apply to the process. As a
result, the alpha particle must be ejected with a discrete energy to
satisfy both conservation principles. Beta decay, however, is a threeparticle decay involving the beta particle, the neutrino (or
antineutrino), and the daughter nucleus. As a result, the energy and
momentum can be shared in a variety of ways among the three
particles while still satisfying the two conservation principles. This
explains why the beta particle can have a continuous range of
energies.

CQ44.5

Carbon dating cannot generally be used to estimate the age of a rock,
because the rock was not alive to receive carbon, and hence
radioactive carbon-14, from the environment. Only the ages of
objects that were once alive can be estimated with carbon dating.

CQ44.6

The larger rest energy of the neutron means that a free proton in
space will not spontaneously decay into a neutron and a positron.
When the proton is in the nucleus, however, you must consider the
total rest energy of the nucleus. If it is energetically favorable for the
nucleus to have one fewer proton and one more neutron, then the
process of positron decay will occur to achieve this lower energy.

CQ44.7

I refers to nuclear spin quantum number.
(a)

(b)

5
5 3
⎛ 5⎞
Iz may have 2I + 1 = 2 ⎜ ⎟ + 1 = 6 values for I = , namely , ,
⎝ 2⎠
2
2 2
1
1
3
5
, − , − , and − .
2
2
2
2

For I = 3, there are 2I + 1 = 2(3) + 1 = 7 possible values for Iz.

CQ44.8

Extra neutrons are required to overcome the increasing electrostatic
repulsion of the protons. The neutrons participate in the net
attractive effect of the nuclear force, but feel no Coulomb repulsion.

CQ44.9

Nuclei with more nucleons than bismuth-209 are unstable because
the electrical repulsion forces among all of the protons is stronger
than the nuclear attractive force between nucleons.

CQ44.10

The nuclear force favors the formation of neutron-proton pairs, so a
stable nucleus cannot be too far away from having equal numbers of
protons and neutrons. This effect sets the upper boundary of the
zone of stability on the neutron-proton diagram. All of the protons
repel one another electrically, so a stable nucleus cannot have too
many protons. This effect sets the lower boundary of the zone of
stability.

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Chapter 44

1103

CQ44.11

Nucleus Y will be more unstable. The nucleus with the higher
binding energy requires more energy to be disassembled into its
constituent parts and has less available energy to release in a decay.

CQ44.12

After one half-life, one half the radioactive atoms have decayed.
After the second half-life, one half of the remaining atoms have
1 1 3
decayed. Therefore, + = of the original radioactive atoms have
2 4 4
decayed after two half-lives.

CQ44.13

Long-lived progenitors at the top of each of the three natural
radioactive series are the sources of our radium. As an example,
thorium-232 with a half-life of 14 Gyr produces radium-228 and
radium-224 at stages in its series of decays.

CQ44.14

Yes. The daughter nucleus can be left in its ground state or
sometimes in one of a set of excited states. If the energy carried by
the alpha particle is mysteriously low, the daughter nucleus can
quickly emit the missing energy in a gamma ray.

CQ44.15

The alpha particle does not make contact with the nucleus because of
electrostatic repulsion between the positively-charged nucleus and
the +2e alpha particle. To drive the alpha particle into the nucleus
would require extremely high kinetic energy.

CQ44.16

The samples would have started with more carbon-14 than we first
thought. We would increase our estimates of their ages.

CQ44.17

The photon and the neutrino are similar in that both particles have
zero charge and little or no mass. (The photon has zero mass, but
evidence suggests that neutrinos have a very small mass.) Both
particles are capable of transferring both energy and momentum.
They differ in that the photon has spin 1 and is involved in
electromagnetic interactions, while the neutrino has spin 21 , interacts
through the weak interaction, and is closely related to beta decay.

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1104

Nuclear Physics

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 44.1
P44.1

P44.2

Some Properties of Nuclei

The average nuclear radii are r = r0A1/3, where r0 = 1.2 × 10–15 m = 1.2 fm

and A is the mass number.
r = ( 1.2 fm ) ( 2 )

13

= 1.5 fm

(a)

For 12 H ,

(b)

For

60
27

(c)

For

197
79

Au ,

r = ( 1.2 fm ) ( 197 )

13

= 7.0 fm

(d) For

239
94

Pu ,

r = ( 1.2 fm ) ( 239 )

13

= 7.4 fm

(a)

r = ( 1.2 fm ) ( 60 )

13

Co ,

Approximate nuclear radii are given by r = r0A1/3. Thus, if a
nucleus of atomic number A has a radius approximately twothirds that of 230
88 Ra, we should have
r = r0 A1 3 =

or

(b)
(c)

= 4.7 fm

2
13
r0 ( 230 )
3

23
8
A = 3 ( 230 ) =
( 230) ≈ 68
3
27

One possible nucleus is

68
30

Zn .

Isotopes of other elements to the left and right of zinc in
the periodic table (from manganese to bromine) may
have the same mass number.

P44.3

(a)

The initial kinetic energy of the alpha particle must equal the
electrostatic potential energy at the distance of closest approach.

K i = Uf =
rmin

ke qQ
rmin

9
2
2
−19
k qQ ( 8.99 × 10 N ⋅ m C ) ( 2 ) ( 79 ) ( 1.60 × 10 C )
= e
=
Ki
( 0.500 MeV ) (1.60 × 10−13 J/MeV )

2

= 4.55 × 10−13 m = 455 × 10−15 m = 455 fm
(b)

Following the same logic as in part (a),

Ki =

1
k qQ
mα vi2 = e
2
rmin

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Chapter 44
–15

Now, for rmin = 300 fm = 300 × 10
velocity gives
vi =
=

1105

m, solving for the initial

2ke qQ
mα rmin
2 ( 8.99 × 109 N ⋅ m 2 / C2 ) ( 2 ) ( 79 ) ( 1.602 × 10−19 C )

(6.645 × 10

−27

2

kg ) ( 300 × 10−15 m )

vi = 6.05 × 106 m/s

P44.4

An iron nucleus (in hemoglobin) has a few more neutrons than
protons, but in a typical water molecule there are eight neutrons and
ten protons. So protons and neutrons are nearly equally numerous in
your body, each contributing mass (say) 35 kg:
(a)

⎛ 1 nucleon ⎞
35 kg ⎜
~ 1028 protons ,
−27
⎟
1.67
×
10
kg
⎝
⎠

(b)

and ~ 1028 neutrons .

(c)

The electron number is precisely equal to the proton number,

~ 1028 electrons .
P44.5

(a)

65
29

Cu has an A number of 65, so the radius of its nucleus is
r = r0 A1 3 = ( 1.2 fm ) ( 65 )

13

(b)

= 4.8 fm

The volume of the nucleus, assumed to be spherical in shape, is
3
4
4
4
V = π r 3 = π ⎡⎣ r03 A ⎤⎦ = π ⎡( 1.2 × 10−15 m ) ( 65 )⎤
⎦
3
3
3 ⎣

= 4.7 × 10−43 m 3
(c)

The density of the nucleus is

ρ=

3 ( 1.66 × 10−27 kg )
m
Am
3m
=
=
=
3
V 4 π ⎡ r 3 A ⎤ 4π r03 4π ( 1.2 × 10−15 m )
3 ⎣0 ⎦

= 2.3 × 1017 kg/m 3

P44.6

From ME = ρn V = ρn (4π r 3 3) , we find
⎛ 3 ME ⎞
r=⎜
⎝ 4 π ρn ⎟⎠

13

⎡ 3 ( 5.98 × 1024 kg ) ⎤
=⎢
17
3 ⎥
⎢⎣ 4 π ( 2.30 × 10 kg/m ) ⎥⎦

13

= 184 m

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1106
P44.7

Nuclear Physics
The number of neutrons in a star of two solar masses is
A=

2 ( 1.99 × 1030 kg )

1.67 × 10

−27

kg neutron

= 2.38 × 1057 neutrons

Therefore,
r = r0 A1 3 = ( 1.20 × 10−15 m ) ( 2.38 × 1057 )

13

= 1.6 × 10 4 m = 16 km

P44.8

(a)

The electric potential energy between two protons is

U = ke

q1q2
e2
= ke
r
r

⎡ ( 1.60 × 10−19 C )2 ⎤
⎥
= ( 8.99 × 10 N ⋅ m C ) ⎢
−15
⎢⎣ 4.00 × 10 m ⎥⎦
9

2

2

⎛
⎞ ⎛ 1 MeV ⎞
1 eV
×⎜
⎟
−19 ⎟ ⎜
⎝ 1.60 × 10 J ⎠ ⎝ 106 eV ⎠
= 0.360 MeV
(b)

P44.9

Figure P44.8 shows the highest point in the curve at about
4 MeV, a factor of ten higher than the value in (a).

By energy conservation,
1 2
mv = qΔV:
2

2mΔV = qr 2 B2

By Newton’s second law,
mv 2
= qvB:
r

2mΔV

qB2

r=

Comparing radii for particles with different masses but with the same
charge, we find that
r2
=
r1

2m2 ΔV qB2
2m1 ΔV qB2

m2

=

m1

For

12

C: m1 = 12 u and r1 = 7.89 cm

For

13

C:

r2
r2
=
=
r1 7.89 cm

m2
m1

=

13
12

→

8.21 cm

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Chapter 44
P44.10

1107

By energy conservation,
1 2
mv = qΔV:

2

2mΔV = qr 2 B2

By Newton’s second law,
mv 2
= qvB:
r

r=

2mΔV
qB2

Comparing radii for particles with different masses but with the same
charge, we find that
2m2 ΔV qB2

r2
=
r1

*P44.11

(a)

2m1 ΔV qB

2

=

m2
m1

→

r2 =

m2
r1
m1

The magnitude of the maximum Coulomb force is given by

Fmax =
=

ke q1q2
2
rmin

( 8.99 × 10

9

2
N ⋅ m 2 C2 ) ⎡( 2 )( 6 )( 1.60 × 10−19 C ) ⎤
⎣
⎦

2
−14
(1.00 × 10 m )

= 27.6 N
(b)

From Newton’s second law,

amax =
(c)

Fmax
27.6 N
=
= 4.16 × 1027 m/s 2
−27
mα
6.64 × 10 kg

The potential energy of the system at the time of the maximum
force is
U max =

ke q1q2
rmin

⎧ ( 8.99 × 109 N ⋅ m 2 C2 ) ⎡( 2 )( 6 )( 1.60 × 10−19 C )2 ⎤ ⎫
⎪
⎣

⎦⎪
=⎨
⎬
−14
1.00 × 10 m )
(
⎪
⎪
⎩
⎭
⎛
⎞ ⎛ 1 MeV ⎞
1 eV
×⎜
⎟
−19 ⎟ ⎜
⎝ 1.60 × 10 J ⎠ ⎝ 106 eV ⎠
= 1.73 MeV

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1108
P44.12

Nuclear Physics
We obtain the alpha particle’s momentum from

1
1 ( mv )

Eα = 7.70 MeV = mv 2 =
2
2 m
(a)

2

→

mv = 2mEα

The de Broglie wavelength of the alpha particle is (mass from
Table 44.1)

λ=
=

h
=
mα vα

h
2mα Eα
6.626 × 10−34 J ⋅ s

2 ( 6.64 × 10−27 kg ) ( 7.70 × 106 eV ) ( 1.60 × 10−19 J eV )

= 5.18 × 10−15 m = 5.18 fm

(b)

P44.13

Since λ is much less than the distance of closest approach, the
alpha particle may be considered a particle.

The volume of each of the golf balls is
V=

4 4 4
3
π r = π ( 0.021 5 m ) = 4.16 × 10−5 m 3
3
3

We take the nuclear density from Example 44.2. Then, the mass of a
golf-ball sized nuclear matter is

m = ρV = ( 2.3 × 1017 kg/m 3 ) ( 4.16 × 10−5 m 3 ) = 9.6 × 1012 kg
and the gravitational force between two such balls is

9.6 × 1012 kg )
m1m2
−11
2
2 (
F = G 2 = ( 6.67 × 10 N ⋅ m / kg )
r
(1.00 m )2

2

F = 6.1 × 1015 N toward each other.
P44.14

(a)

Let V represent the volume of the tank. The number of molecules
present is

N = nN A =

(1.013 × 105 N/m2 )V (6.022 × 1023 )
PV
=
RT ( 8.315 J/mol ⋅ K )( 273 K )

= ( 2.69 × 1025 m −3 ) V

The volume of one molecule is
3

−10
⎛4
⎞ 8π ⎛ 1.00 × 10 m ⎞
2 ⎜ πr3 ⎟ =
= 1.047 × 10−30 m 3
⎜
⎟
⎝3
⎠

3 ⎝
2
⎠

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Chapter 44

1109

The volume of all the molecules is

( 2.69 × 10

25

m −3 ) V ( 1.047 × 10−30 m 3 ) = 2.82 × 10 –5 V

So the fraction of the volume occupied by the hydrogen
molecules is 2.82 × 10 –5. An atom is precisely one half of a
molecule.
(b)

The fraction occupied by the nucleus is found from
4 3
3
πr
nuclear volume
⎛ r ⎞

= 3
=⎜
⎟
atomic volume 4 π (d / 2)3 ⎝ d / 2 ⎠
3
3

⎛ 1.20 × 10−15 m ⎞
=⎜
= 1.38 × 10−14
⎝ 0.500 × 10−10 m ⎟⎠

In linear dimension, the nucleus is small inside the atom in the
way a fat strawberry is small inside the width of the Grand
Canyon. In terms of volume, the nucleus is really small.

Section 44.2
P44.15

Nuclear Binding Energy

Using Equation 44.2, the binding energy per nucleon is
A
Eb ⎡⎣ ZM ( H ) + Nmn − M ( Z X ) ⎤⎦ ⎛ 931.5 MeV ⎞
=
⎜⎝
⎟⎠
A
A
u

Using atomic masses as given in Table 44.2,
(a)

For 21 H:

Eb 1( 1.007 825 u ) + 1( 1.008 665 u ) − 2.014 102 u
=
A
2
⎛ 0.002 388 u ⎞ ⎛ 931.5 MeV ⎞
=⎜
⎟⎠ ⎜⎝
⎟⎠ = 1.11 MeV
⎝
2
u
(b)

For 24 He:

Eb 2 ( 1.007 825 u ) + 2 ( 1.008 665 u ) − 4.002 603 u
=
A
4
⎛ 0.030 377 u ⎞ ⎛ 931.5 MeV ⎞
=⎜
⎟⎠ ⎜⎝
⎟⎠ = 7.07 MeV
⎝

4
u

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1110

Nuclear Physics
(c)

For

56
26

Fe:

Eb 26 ( 1.007 825 u ) + 30 ( 1.008 665 u ) − 55.934 942 u
=
A
56
⎛ 0.528 458 u ⎞ ⎛ 931.5 MeV ⎞
=⎜
⎟⎠ ⎜⎝
⎟⎠ = 8.79 MeV
⎝
56
u
(d) For

238
92

U:

Eb 92 ( 1.007 825 u ) + 146 ( 1.008 665 u ) − 238.050 783 u
=
A
238
⎛ 1.934 207 u ⎞ ⎛ 931.5 MeV ⎞
=⎜
⎟⎠ ⎜⎝
⎟⎠ = 7.57 MeV
⎝
238
u
P44.16

We use Equation 44.2,

Eb ( MeV ) = ⎡⎣ ZM ( H ) + Nmn − M ( AZ X ) ⎤⎦ ( 931.494 MeV/u )
Then, for

23
11

Na ,

23

⎡
⎤
Eb ( 23
11 Na ) = ⎣ 11M ( H ) + 12mn − M ( 11 Na ) ⎦ ( 931.494 MeV/u )

= ⎡⎣11( 1.007 825 u ) + 12 ( 1.008 665 u ) − 22.989 769 u ⎤⎦
× ( 931.494 MeV/u )
= 186.565 MeV
Eb 186.565 MeV
=
= 8.11 MeV
A
23

and
For

23
12

Mg ,
23
Eb = Eb ( 12
Mg )

23
= ⎡⎣12M ( H ) + 11mn − M ( 12
Mg ) ⎤⎦ ( 931.494 MeV/u )

= ⎡⎣12 ( 1.007 825 u ) + 11( 1.008 665 u ) − 22.994 124 u ⎤⎦

× ( 931.494 MeV/u )
= 181.726 MeV
and

Eb 181.726 MeV
=
= 7.90 MeV
A
23

The difference is
23
23
ΔEb Eb ( 11 Na ) − Eb ( 12 Mg )
=
A
A
= 8.11 MeV − 7.90 MeV = 0.210 MeV
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Chapter 44

1111

The binding energy per nucleon is greater for 23
11 Na by 0.210 MeV .
23
There is less proton repulsion in 11 Na; it is the more stable nucleus.
P44.17

From Equation 44.2, the binding energy of a nucleus is

Eb ( MeV ) = ⎡⎣ ZM ( H ) + Nmn − M ( AZ X ) ⎤⎦ ( 931.494 MeV/u )
For

15
8

O:

Eb = [ 8 ( 1.007 825 u ) + 7 ( 1.008 665 u ) − 15.003 065 u ]

× ( 931.494 MeV/u ) = 111.96 MeV

For

15
7

N:

Eb = [ 7 ( 1.007 825 u ) + 8 ( 1.008 665 u ) − 15.000 109 u ]

× ( 931.494 MeV/u ) = 115.49 MeV

Therefore, the binding energy is greater for
P44.18

N by 3.54 MeV.

We find the mass difference, ΔM = ZmH + Nmn − M, and then the
ΔM ( 931.5 )
E
binding energy per nucleon, b =
, in units of MeV. The
A
A
results are tabulated below.

Nuclei

Z

N

M in u

ΔM in u

Eb
in MeV
A

55

Mn

25

30

54.938 050

0.517 5

8.765

56

Fe

26

30

55.934 942

0.528 46

8.790

59

Co

27

32

58.933 200

0.555 35

8.768

∴ 56 Fe has a greater
P44.19

15
7

(a)

The isobar with the highest neutron-to-proton ratio is
ratio is

(b)

Eb
than its neighbors.
A

139
57

139
55

Cs ; the

N A − Z 139 − 55 84
=
=
=
= 1.53
Z
Z
55
55

La is stable, so has the largest binding energy per nucleon

(8.378 MeV).

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1112

Nuclear Physics
(c)

The isobars are close in Figure 44.6, the plot of binding energy per
nucleon versus mass number, and there is not much detail, so we
may assume they have about the same binding energy, or missing
mass. However, neutrons have more mass than protons, so the
isobar with more neutrons (thus, fewer protons) should be more
massive:

P44.20

(a)

139
55

Cs .

The radius of the 40Ca nucleus is,

R = r0 A1 3 = ( 1.20 × 10−15 m ) ( 40 )

13

= 4.10 × 10−15 m

The energy required to overcome electrostatic repulsion is
9
2
2
−19
3keQ 2 3 ( 8.99 × 10 N ⋅ m /C ) ⎡⎣ 20 ( 1.602 × 10 C ) ⎤⎦
U=
=
5R
5 ( 4.10 × 10−15 m )

2

= 1.35 × 10−11 J = 84.2 MeV
(b)

The binding energy of 40
20 Ca (Z = 20, N = A – Z = 20) is (using
Equation 44.2 and masses from Table 44.2),

Eb = ⎡⎣ 20 ( 1.007 825 u ) + 20 ( 1.008 665 u ) − 39.962 591 u ⎤⎦
× ( 931.5 MeV/u )
= 342 MeV
(c)

The nuclear force is so strong that the binding energy greatly
exceeds the minimum energy needed to overcome electrostatic
repulsion.

P44.21

Removal of a neutron from

43
20

Ca would result in the residual nucleus,

Ca . If the required separation energy is ΔEn , the overall process can
be described by
42
20

42
mass ( 43
20 Ca ) + ΔEn = mass ( 20 Ca ) + mass ( n )
43
ΔEn = mass ( 42
20 Ca ) + mass ( n ) − mass ( 20 Ca )

From Table 44.2,
ΔEn = ( 41.958 618 u + 1.008 665 u − 42.958 767 u )

× ( 931.5 MeV/u )

= 7.93 MeV

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Chapter 44

Section 44.3
P44.22

1113

Nuclear Models

The curve of binding energy shows that a heavy nucleus of mass
number A = 200 has binding energy about
⎛
MeV ⎞

⎜⎝ 7.8 nucleon ⎟⎠ (200 nucleons) ≈ 1.56 GeV

Thus, it is less stable than its potential fission products, two
middleweight nuclei of A = 100, together having binding energy

2(8.7 MeV/nucleon)(100 nucleons) ≈ 1.74 GeV
Fission then releases about

1.74 GeV − 1.56 GeV ~ 200 MeV

ANS. FIG. P44.22
P44.23

(a)

In Equation 44.3, the first or “Volume” term is,
E1 = C1A = (15.7 MeV)(56) = 879 MeV
The second, or “Surface” term is,
23
E2 = −C2 A 2 3 = − ( 17.8 MeV ) ( 56 ) = −260 MeV
The third, or “Coulomb” term is,

E3 = −C3

Z ( Z − 1)
( 26) ( 25)
= − ( 0.71 MeV )
13
A
( 56)1 3

= −121 MeV
and the fourth, or “Asymmetry” term is,

E4 = C4

( A − 2Z )2
A

= − ( 23.6 MeV )

( 56 − 52 )2
56

= − 6.74 MeV

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1114

Nuclear Physics
The binding energy is then

Eb = C1 A − C2 A

23

Z ( Z − 1)
( A − 2Z )

− C3
− C4
13
A
A

2

= 879 MeV − 260 MeV − 121 MeV − 6.74 MeV = 491 MeV
(b)

P44.24

(a)

(b)

The percentages for each of the terms is as follows

term1 :

E1
E
= 179% ; term 2 : 2 = −53.0% ;
Eb
Eb

term 3:

E3

E
= −24.6% ; term 4: 4 = −1.37%
Eb
Eb

Nucleons on the surface have fewer neighbors with which to
interact. The surface term is negative to reduce the estimate
from the volume term, which assumes that all nucleons have
the same number of neighbors.
The volume to surface ratio for a sphere of radius r is
3
Volume ( 4 3 ) π r
1
=
= r
2
Area
4π r
3

The volume to surface ratio for a cube of side length L is
Volume L3
1
= 2 = L
Area
6L
6

The sphere has a larger ratio to its characteristic length, so it
would represent a larger binding energy and be more plausible

for a nuclear shape.

Section 44.4
*P44.25

Radioactivity

We use Equation 44.7 for the exponential decay rate of the sample,
R = R0 e − λ t , where

λ=

ln 2
= 0.026 7 h −1
26.0 h

Since we require a 90% decrease in activity,

R
= 0.100 = e − λ t
R0

→

ln ( 0.100 ) = − λ t

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Chapter 44

1115

then,
t=

P44.26

(a)

2.30
= 86.4 h
0.026 7/h

From R = R0 e − λ t , the decay constant is
1 ⎛ R ⎞ ⎛ 1 ⎞ ⎛ 10.0 mCi ⎞
−2
−1
λ = ln ⎜ 0 ⎟ = ⎜
⎟⎠ ln ⎜⎝
⎟⎠ = 5.58 × 10 h
⎝
⎠
⎝
t
R
4.00 h
8.00 mCi
= 1.55 × 10−5 s −1

(b)

The half-life is
T1 2 =

(c)

ln 2
= 12.4 h
λ

The number of original atoms can be found if we convert the
initial activity from curies into becquerels (decays per second):
1 Ci ≡ 3.70 × l010 Bq.

R0 = 10.0 mCi = ( 10.0 × 10−3 Ci ) ( 3.70 × 1010 Bq/Ci )
= 3.70 × 108 Bq
Since R0 = λ N 0 , the original number of nuclei is

R0 3.70 × 108  decays/s
N0 =
=
= 2.39 × 1013  atoms
–5
λ
1.55 × 10  s
(d) The decay rate after thirty hours is
R = R0 e − λ t = ( 10.0 mCi ) exp ⎡⎣( −5.58 × 10−2 h −1 ) ( 30.0 h ) ⎤⎦
= 1.88 mCi

P44.27

The decay law is

dN/dt = – λ N
Then, the decay constant is

λ=−

⎛
⎞ ⎛ −6.00 × 1011 nuclei ⎞
1 ⎛ dN ⎞
1
=
−
⎜
⎟
⎟⎠
⎜⎝ 1.00 × 1015 nuclei ⎟⎠ ⎜⎝
N ⎝ dt ⎠
s

= 6.00 × 10 –4 s –1
and the half-life is
T1/2 =

ln 2
= 1.16 × 103 s
λ

(This is 19.3 minutes.)
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1116
P44.28

Nuclear Physics
According to Equation 44.7, the time dependence of the decay rate is
R = R0 e − λΔ t . From this equation we can derive a relation between the
change in decay rate over the time interval Δt to the decay constant.
We start with R = R0 e − λΔ t . Then, rearranging and taking the natural log
of both sides gives
e − λ Δt =

or

R
R0

⎛ R⎞
ln e − λΔ t = ln ⎜ ⎟
⎝ R0 ⎠

(

→

)

⎛ R⎞
⎛R ⎞
− λ Δt = ln ⎜ ⎟ = − ln ⎜ 0 ⎟
⎝ R⎠
⎝ R0 ⎠

Solving,

λ=

1 ⎛ R0 ⎞
ln ⎜ ⎟
Δt ⎝ R ⎠

Now, because λ =

ln 2
, we can relate the time interval Δt to the halfT1/2

life:

λ=

1 ⎛ R0 ⎞
ln ⎜ ⎟
Δt ⎝ R ⎠

→

ln 2

1
⎛R ⎞
=
ln ⎜ 0 ⎟
T1/2 ( ln 2 ) Δ t ⎝ R ⎠
1
1
⎛R ⎞
=
ln ⎜ 0 ⎟
T1/2 ( ln 2 ) Δ t ⎝ R ⎠
T1/2 =

P44.29

( ln 2 ) Δ t

ln ( R0 R )

The number of nuclei that decay during the interval will be

(

ΔN = N 1 − N 2 = N 0 e − λ t1 − e − λ t2

)

First we find the decay constant λ:

λ=

ln 2 0.693
=
= 0.010 7 h −1 = 2.97 × 10−6 s −1
T1 2 64.8 h

Now we find N0:

N0 =

R0
( 40.0 µCi ) ⎛ 3.70 × 104 s −1 ⎞
=
⎟⎠
λ 2.97 × 10−6 s −1 ⎜⎝
µCi

= 4.98 × 1011 nuclei

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Chapter 44

1117

Substituting in these values,

N 1 − N 2 = ( 4.98 × 1011 ) ⎡⎣ e − ( ln 2 64.8 h )(10.0 h ) − e − ( ln 2 64.8 h )(12.0 h ) ⎤⎦
N 1 − N 2 = 9.47 × 109 nuclei

P44.30

The number of nuclei that decay during the interval will be

(

N 1 − N 2 = N 0 e − λ t1 − e − λ t2

)

We wish to write this expression in terms of the half-life T1/2 and the
initial decay rate R0. First, from the definition of λ, we have

λ=

ln 2
ln 2 ( −t T1 2 )
−t T
→ e−λ t = e
= 2 12
T1 2

Now we find N0:

N0 =

R0 R0T1 2
=
λ
ln 2

Substituting in these expressions, we find that
N1 − N 2 =

P44.31

(a)

R0T1 2
ln 2

(e

− λ t1

)

− e − λ t2 =

R0T1 2
ln 2

(2

−t1 T1 2

−2

−t2 T1 2

)

The decay constant is

λ=
=

ln 2
ln 2
=
= 0.086 2 d −1
T1 2 8.04 d
0.086 2 ⎛ 1 d ⎞
−3
−1
⎜
⎟ = 3.59 × 10 h
d ⎝ 24 h ⎠

9.98 × 10−7 ⎛ 1 h ⎞
−7 −1
=
⎜⎝
⎟⎠ = 9.98 × 10 s
h
3600 s
(b)

From R = λN, the number of radioactive nuclei in a 6.40 mCi of 131 I is
N=

(c)

R 6.40 × 10−3 Ci ⎛ 3.70 × 1010 s −1 ⎞
=
= 2.37 × 1014 nuclei
−7
−1 ⎜
⎟
λ 9.98 × 10 s ⎝
Ci
⎠

From Equation 44.7, R = λN, the decay rate R also undergoes
exponential decay; thus, after one half-life, the rate drops from R0
to R0/2. The number of half-lives that have elapsed after 40.2 d is
n = t/T1/2 = 40.2 d/8.04 d = 5, so the remaining activity of the
sample is
R=

R0 R0 6.40 mCi
=
=
= 0.200 mCi
2n 25
32

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1118
P44.32

Nuclear Physics
(a)

From Equation 44.6, the fraction remaining at t = 5.00 yr will be

N
−t ln 2 T1 2
= e − λt = e
= e − ( 5.00 yr ) ln 2 (12.33 yr ) = 0.755
N0
(b)

At t = 10.0 yr,

N
−t ln 2 T1 2
− 10.0 yr ) ln 2 ( 12.33 yr )
= e − λt = e
=e (
= 0.570
N0
(c)

At t = 123.3 yr,

N
−t ln 2 T1 2

− 123.3 yr ) ln 2 ( 12.33 yr )
= e − λt = e
=e (
= e −10ln 2 = 9.766 × 10−4
N0
(d)

No. The decay model depends on large numbers of nuclei.
After some long but finite time, only one undecayed nucleus
will remain. It is likely that the decay of this final nucleus
will occur before infinite time.

P44.33

The number remaining after time

T1/2 ln 2
is
=
2
2λ

N = N 0 e − λt = N 0 e − λ ( ln 2/2 λ ) = N 0 ( e − ln 2 )

1/2

⎛ 1⎞
= N0 ⎜ ⎟
⎝ 2⎠

1/2

=

N0
2

The number decaying in this first half of the first half-life is

ΔN first half = N 0 −
=

2
2

(

⎛
N0
1 ⎞
2⎞
⎛
= ⎜1−
N0 = ⎜ 1 −
N0
⎟
⎝
2 ⎟⎠
2
2⎠

⎝

)

2 − 1 N0

The number remaining after time T1/2 is

N0
, so the number decaying
2

in the second half of the first half-life is

ΔN second half =
=

⎛ 2 1⎞
N0 N0 ⎛ 1 1 ⎞
−
=⎜
− ⎟ N0 = ⎜
− N0
2 ⎝ 2 2⎠
2
⎝ 2 2 ⎟⎠
1
2

(

)

2 − 1 N0

The ratio required is then

(

)

2
2 − 1 N0
ΔN first half =
2
=
= 2 = 1.41
1
ΔN second half =
2 − 1 N0
2

(

)

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Chapter 44

P44.34

(a)

1119

dN 2
= rate of change of N2
dt

= rate of production of N2 – rate of decay of N2
= rate of decay of N1 – rate of decay of N2

= λ1 N 1 − λ2 N 2
(b)

From the trial solution,

N 2 (t ) =
∴

N 10 λ1 − λ2 t
e
− e − λ1t
λ1 − λ2

(

)

dN 2
N λ
= 10 1 − λ2 e − λ2 t + λ1e − λ1t
dt
λ1 − λ2

(

)

[1]

(

)

dN 2
N λ
+ λ2 N 2 = 10 1 − λ2 e − λ2 t + λ1e − λ1t
dt
λ1 − λ2
N λ
+ 10 1 λ2 e − λ2 t − λ2 e − λ1t
λ1 − λ2
N λ
= 10 1 ( λ1 − λ2 ) e − λ1t = λ1 N 1
λ1 − λ2

(

So
(c)

)

dN 2
= λ1 N 1 − λ2 N 2 as required.
dt

The functions plotted in ANS. FIG. P44.34(c) are

Po nuclei: N 1 ( t ) = 1 000e − ( ln 2 3.10 min ) t
Pb nuclei: N 2 ( t ) = 1 130.8 ⎡⎣ e − ( ln 2 26.8 min ) t − e − ( ln 2 3.10 min ) t ⎤⎦

ANS. FIG. P44.34(c)
(d) From the graph, tm ≈ 10.9 min

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1120

Nuclear Physics

(e)

From equation [1],

dN 2

= 0 if
dt

λ2 e − λ2 t = λ1e − λ1t
∴ e( λ1 − λ2 )t =

Thus, t = tm =
(f)

λ1
λ2
ln ( λ1 λ2 )

λ1 − λ2

.

With λ1 = ln 2 ( 3.10 min ) , λ2 = ln 2 ( 26.8 min ) , this formula gives

tm =

ln ( λ1 λ2 )

λ1 − λ2

⎡ ln 2 ( 3.10 min ) ⎤
⎛ 26.8 min ⎞
ln ⎢
ln ⎜
⎥

ln
2
26.8
min
(
)⎦
⎝ 3.10 min ⎟⎠
⎣
=
=
ln 2 ⎞
1
1
⎛ ln 2
⎛
⎞
−
−
⎜⎝
⎟⎠ ln 2 ⎜⎝
⎟
3.10 min 26.8 min
3.10 min 26.8 min ⎠
= 10.9 min
This result is in agreement with the result of part (d).

Section 44.5
P44.35

The Decay Processes

Atomic masses are given in Table 44.2.
(a)

For this e+ decay,

Q = ( MX − MY − 2me ) c 2

= ⎡⎣ 39.962 591 u − 39.963 999 u − 2 ( 0.000 549 u ) ⎤⎦

× ( 931.5 MeV/u )

Q = −2.33 MeV
Since Q < 0, the decay cannot occur spontaneously.

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Chapter 44
(b)

1121

For this alpha decay,

Q = ( MX − MY − 2me ) c 2

= [ 97.905 287 u − 4.002 603 u − 93.905 088 u ]

× ( 931.5 MeV/u )

Q = −2.24 MeV
Since Q < 0, the decay cannot occur spontaneously.
(c)

For this alpha decay,
Q = ( MX − MY − 2me ) c 2

= [ 143.910 083 u − 4.002 603 u − 139.905 434 u ]

× ( 931.5 MeV/u )

Q = 1.91 MeV

Since Q > 0, the decay can occur spontaneously.
P44.36

(a)

3
1

The reaction is

H → 23 He + e− + ν .

Adding one electron, the reaction becomes
3
1

H nucleus + e− → 23 He nucleus + 2e− + ν

Ignoring the slight difference in ionization energies, we have
3
1

(b)

H atom → 23 He atom + ν

The total energy released is the Q value:
Q = ( MH-3 − MHe-3 ) c 2

Q = ( 3.016 049 u − 3.016 029 u ) ( 931.5 MeV u )
= 0.018 6 MeV = 18.6 keV

P44.37

From Equation 44.21, carbon-14 undergoes beta decay:
14
6

C→

14
7

N + e− + ν

Adding six electrons to each side, this is the same as

14
6

C atom →

14
7

N atom + ν

The Q value is
Q = ( MC-14 − MN-14 − mν ) c 2
= [14.003 242 u − 14.003 074 u − 0 ]( 931.5 MeV/u )
= 0.156 MeV
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1122
P44.38

Nuclear Physics
Total Z and A are conserved.
(a)

A gamma ray has zero charge and it contains no protons or
neutrons. So for a gamma ray Z = 0 and A = 0. Keeping the total
values of Z and A for the system conserved requires Z = 28 and A
= 65 for X. With this atomic number it must be nickel, and the
65
28

nucleus must be in an excited state, so X is
(b)

(c)

An alpha particle, α = 24 He, has Z = 2 and A = 4. Total initial Z is
84, and total initial A is 215, so for X we require
Z = 84 = ZX + 2

→

ZX = 82

A = 215 = AX + 4

→

AX = 211,

→ Pb, and
→ X is

211
82

Pb .

A positron, e+ = 01 e , has charge the same as a nucleus with Z = 1.
A neutrino, 00ν , has no charge. Neither contains any protons or

neutrons. So X must have by conservation Z = 26 + 1 + 0 = 27; so,
X is Co. And A = 55 + 0 + 0 = 55: X is

P44.39

Ni∗ .

55
27

Co .

Atomic masses are given in Table 44.2. We calculate the energy
released by the reaction, its Q-value, as
Q = ( MU-238 − MTh-234 − MHe-4 ) c 2

Q = ( 238.050 783 − 234.043 596 − 4.002 603 ) u ( 931.5 MeV u )
= 4.27 MeV
P44.40

(a)

The decay constant is λ = ln2/10 h = 0.0693/h. The number of
parent nuclei is given by N P = N P, 0 e − λt = 1.00 × 106 e −0.0693t , where
t is in hours.

(

)

The number of daughter nuclei is equal to the number of missing
parent nuclei,

(

)(

)

N d = N P, 0 − N P, 0 e − λt = 1.00 × 106 1 − e −0.0693t , where t is in hours.
(b)

The number of daughter nuclei starts from zero at t = 0. The
number of stable product nuclei always increases with time and
asymptotically approaches 1.00 × 106 as t increases without limit.

(c)

The minimum number of daughter nuclei is zero at t = 0. The
maximum number of daughter nuclei asymptotically approaches
1.00 × 106 as t increases without limit.

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Chapter 44

1123

(d) The rate of change is

dN d
= ( 1.00 × 106 ) (0 + 0.0693 e −0.0693t ) = 6.93 × 10 4 e −0.0693t
dt
dN d
is in decays per hour and t is in hours. The rate of
dt
change has its maximum value, 6.93 × 104 h–1, at t = 0, after which
the rate decreases more and more, approaching zero as t
increases without limit.

where

P44.41

(a)

The reaction for one particle is e− + p → n + ν .

(b)

For nuclei,

15
8

O + e− →

15
7

N + ν.

Add seven electrons to both sides to obtain
15
8

O atom →

15
7

N atom + ν

From Table 44.2 of atomic masses,

Q = ( 15.003 065 u − 15.000 109 u ) ( 931.5 MeV u )
= 2.75 MeV
P44.42

(a)

The number of carbon atoms in the sample is
⎛ 0.021 0 g ⎞ ⎛ 6.02 × 1023 atoms ⎞
NC = ⎜
= 1.05 × 1021
⎜
⎟
⎟
⎝
⎠

12.0
g
mol
mol
⎝
⎠

(b)

1 in 7.70 × 1011 carbon atoms is a 14C atom. Then,

1

( N 0 )C-14 = 1.05 × 1021 ⎛⎜⎝ 7.70 × 1011 ⎞⎟⎠ =
(c)

1.37 × 109

The decay constant for 14C is

λC-14 =

ln 2
1 yr
⎛
⎞
= 1.21 × 10−4 yr −1 ⎜
7
⎝ 3.16 × 10 s ⎟⎠
5 730 yr

= 3.83 × 10−12 s −1
(d) We use R = λ N = λ N 0 e − λ t . At t = 0,
⎡ 7 ( 86 400 s ) ⎤
R0 = λ N 0 = ( 3.83 × 10−12 s −1 ) ( 1.37 × 109 ) ⎢
⎥
⎣ 1 week ⎦
= 3.17 × 103 decays week

(e)

At time t,

R=

837
= 951 decays week .
0.880

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