Chapter 29
Nuclear Physics
Quick Quizzes
1.
(c). At the end of the first half-life interval, half of the original sample has decayed and
half remains. During the second half-life interval, half of the remaining portion of the
sample decays. The total fraction of the sample that has decayed during the two half-lives
1 1 1 3
is +   = .
2 2 2 4
2.
(c). The half-life of a radioactive material is T1 2 =
( )
that material. Thus, if λA = 2λB , we have T1 2
A
ln 2
=
λ
, where λ is the decay constant for
ln 2
λA
=
( )
ln 2 1
= T1 2
2λB 2
B
=
1
(4 h) = 2 h .
2
3.
(a). Conservation of momentum requires the momenta of the two fragments be equal in
magnitude and oppositely directed. Thus, from KE = p 2 2m , the lighter alpha particle has
more kinetic energy that the more massive daughter nucleus.
4.
(a) and (b). Reactions (a) and (b) both conserve total charge and total mass number as
required. Reaction (c) violates conservation of mass number with the sum of the mass
numbers being 240 before reaction and being only 223 after reaction.
5.
(b). In an endothermic reaction, the threshold energy exceeds the magnitude of the Q
value by a factor of ( 1 + m M ) , where m is the mass of the incident particle, and M is the
mass of the target nucleus.
457
458
CHAPTER 29
Answers to Even Numbered Conceptual Questions
2.
Since the nucleus was at rest before decay, the total linear momentum will be zero both
before and after decay. This means that the alpha particle and the daughter nucleus must
recoil in opposite directions with equal magnitude momenta, mα vα = mD vD . The kinetic
energy of the alpha particle is then
 m2
 m 1
1
1
m
( KE )α = mα vα2 = mα  D2 vD2  = D  mD vD2  = D ( KE )D
2
2
 mα
 mα
 mα  2
Since the mass of the daughter nucleus will be much larger than that of the alpha particle,
the kinetic energy of the alpha particle, ( KE )α , is considerably larger than that of the
daughter nucleus, ( KE )D .
4.
An alpha particle is a doubly positive charged helium nucleus, is very massive and does
not penetrate very well. A beta particle is a singly negative charged electron and is very
light and only slightly more difficult to shield from. A gamma ray is a high energy photon,
or high frequency electromagnetic wave, and has high penetrating ability.
6.
Beta particles have greater penetrating ability than do alpha particles.
8.
The much larger mass of the alpha particle as compared to that of the beta particle ensures
that it will not deflect as much as does the beta, which has a mass about 7000 times
smaller.
10.
Consider the reaction
14
6
C →
14
7
N + −01e + ν . We have six positive charges before the event
on the carbon-14 nucleus. After the decay, we still have a net of six positive charges, as +7
from the nitrogen and −1 from the electron. Thus, in order to have conservation of charge,
the neutrino must be uncharged.
12.
We would have to revise our age values upward for ancient materials. That is, we would
conclude that the materials were older than we had thought because the greater cosmic
ray intensity would have left the samples with a larger percentage of carbon-14 when they
died, and a longer time would have been necessary for it to decay to the percentage found
at present.
14.
The amount of carbon-14 left in very old materials is extremely small, and detection
cannot be accomplished with a high degree of accuracy.
16.
(a) The nucleus loses two protons and two neutrons, which are ejected in the form of an
alpha particle. (b) In beta decay, the nucleus loses 1 neutron, but gains 1 proton, leaving
the total number of nucleons unchanged.
Nuclear Physics
Answers to Even Numbered Problems
2.
~10 28 protons , ~10 28 neutrons , ~10 28 electrons
4.
ρ n ρ a = 8.6 × 1013
6.
(a)
7.89 cm for
8.
(a)
1.9 × 10 −15 m
(b)
7.4 × 10 −15 m
10.
(a)
(c)
1.11 MeV nucleon
8.79 MeV nucleon
(b)
(d)
7.07 MeV nucleon
7.57 MeV nucleon
12.
Eb A = 8.765 MeV nucleon for 55 Mn ;
12
C , 8.21 cm for
13
C
(b)
7.89 cm
12
=
= 0.961
8.21 cm
13
Eb A = 8.786 MeV nucleon for 56 Fe ;
Eb A = 8.768 MeV nucleon for 59 Co
14.
7.93 MeV
16.
8.7 × 10 3 Bq
18.
(a)
20.
2.29 g
22.
1.72 × 10 4 yr
24.
1.66 × 10 3 yr
26.
12
6
8.06 d
C , 24 He ,
(b)
14
6
It is probably
131
53
I.
C
28.
235
92 U
a
231
90 Th
e231
91 Pa
a
227
89 Ac
a
223
87 Fr
a
223
88 Ra
e227
90 Th
a
219
85 At
a
219
86 Rn
e-
a
215
83 Bi
a
215
84 Po
e-
ea
211
82 Pb
e211
83 Bi
a
207
81 Tl
a
207
82 Pb
e211
84 Po
e-
459
460
CHAPTER 29
30.
4.28 MeV
32.
(a)
34.
1.67 × 10 4 yr
36.
9.96 × 10 3 yr
38.
4
2
66
28
Ni →
66
29
Cu + −01e +ν
(b)
186 keV
(b)
–2.79 MeV
He , 24 He
40.
17.3 MeV
42.
(a)
44.
22.0 MeV
46.
5.0 rad
48.
(a)
2.00 J kg
(b)
4.78 × 10 −4 °C
50.
(a)
2.5 × 10 −3 rem x-ray
(b)
≈ 38 times background levels
52.
(a)
10 h
(b)
3.2 m
54.
24 decays min
56.
fraction decayed = 0.003 5 ( 0.35% )
58.
(a) 2.7 fm
(b)
2
(d) 7.4 fm, 3.8 × 10 N , 18 MeV
60.
12 mg
62.
6 × 109 yr
64.
(a)
60.5 Bq L
(b)
40.6 d
66.
(a)
4.28 × 10 −12 J
(b)
1.19 × 10 57
10
5
B
1.5 × 10 2 N
(c)
2.6 MeV
(c)
107 billion years
Nuclear Physics
461
Problem Solutions
29.1
29.2
The average nuclear radii are r = r0 A1 3 , where r0 = 1.2 × 10 −15 m = 1.2 fm , and A is the
mass number.
r = ( 1.2 fm ) ( 2 )
= 1.5 fm
13
For
2
1
For
60
27
For
197
79
Au ,
r = ( 1.2 fm )( 197 )
= 7.0 fm
For
239
94
Pu ,
r = ( 1.2 fm )( 239 )
= 7.4 fm
H,
Co ,
r = ( 1.2 fm )( 60 )
13
= 4.7 fm
13
13
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 (say) 35 kg out of a total body
mass of 70 kg.
 1 nucleon 
28
28
N = 35 kg 
 ~10 protons and ~10 neutrons
-27
 1.67 × 10 kg 
The electron number is precisely equal to the proton number,
N e = ~10 28 electrons
29.3
4

From ME = ρ n V = ρ n  π r 3  , we find
3

 3 ME 
r =

 4 π ρn 
13
 3 ( 5.98 × 10 24 kg ) 

=
17
3
 4 π ( 2.3 × 10 kg m ) 
13
= 1.8 × 10 2 m
462
CHAPTER 29
29.4
The mass of the hydrogen atom is approximately equal to that of the proton,
1.67 × 10 −27 kg . If the radius of the atom is r = 0.53 × 10 −10 m , then
ρa =
3 ( 1.67 × 10 −27 kg )
m
m
=
=
= 2.7 × 10 3 kg m 3
V ( 4 3 ) π r 3 4 π ( 0.53 × 10 −10 m )3
The ratio of the nuclear density to this atomic density is
ρ n 2.3 × 1017 kg m 3
=
= 8.6 × 1013
3
3
ρ a 2.7 × 10 kg m
29.5
(a)
Fmax
2
−19
9
2
2 

k e q1 q2 ( 8.99 × 10 N ⋅ m C ) ( 2 )( 6 ) ( 1.60 × 10 C ) 
= 2 =
= 27.6 N
2
rmin
(1.00 × 10−14 m )
(b) amax =
(c)
Fmax
27.6 N
=
= 4.16 × 10 27 m s 2
mα
6.64 × 10 -27 kg
PEmax =
k e q1 q2
= Fmax ⋅ rmin = ( 27.6 N ) ( 1.00 × 10 −14 m )
rmin
 1 MeV 
= 2.76 × 10 −13 J 
 = 1.73 MeV
−13
 1.60 × 10 J 
Nuclear Physics
29.6
(a) From conservation of energy, ∆KE = − ∆PE , or
1
mv 2 = q ( ∆V ) . Also, the centripetal
2
acceleration is supplied by the magnetic force,
so
mv 2
= qvB , or v = qBr m
r
The energy equation then yields r = 2m ( ∆V ) qB2
For
12
C , m=12u
2 12 ( 1.66 × 10 −27 kg )  ( 1 000 V )
r=
= 7.89 cm
(1.60 × 10−19 C ) ( 0.200 T )2
And
For
13
C , m=13u
r=
and
(b)
2 13 ( 1.66 × 10 −27 kg )  ( 1 000 V )
= 8.21 cm
(1.60 × 10−19 C ) ( 0.200 T )2
2m1 ( ∆V ) qB2
r1
=
=
r2
2m2 ( ∆V ) qB2
m1
m2
r12 7.89 cm
=
= 0.961 and
r13 8.21 cm
29.7
12u
= 0.961 , so they do agree.
13u
(a) At the point of closest approach, PE f = KEi , so
or v =
=
k e ( 2 e )( 79 e )
rmin
=
1
mα v 2
2
2 k e ( 2 e )( 79 e )
mα rmin
316 ( 8.99 × 10 9 N ⋅ m 2 C 2 )( 1.60 × 10 −19 C )
( 6.64 × 10−27 kg ) ( 3.2 × 10−14 m )
463
2
= 1.9 × 107 m s
464
CHAPTER 29
(b) KEi =
=
29.8
1
mα v 2
2
2
1
1 MeV
6.64 × 10 −27 kg )( 1.85 × 107 m s ) 
(
× 10 −13
2
1.60

Using r = r0 A1 3 , with r0 = 1.2 × 10 −15 m = 1.2 fm , gives:
(a) For 24 He, A = 4 , and
238
92
(b) For
29.9

 = 7.1 MeV
J
For
93
41
r = ( 1.2 fm )( 4 )
= 1.9 fm = 1.9 × 10 −15 m
13
He, A = 238 , and r = ( 1.2 fm )( 238 )
13
= 7.4 fm = 7.4 × 10 −15 m
Nb ,
∆m = 41 mH + 52 mn − mNb
= 41( 1.007 825u ) + 52 ( 1.008 665u ) − ( 92.906 376 8 u ) = 0.865 028 u
Thus,
For
197
79
Eb ( ∆m ) c 2 ( 0.865 028 u )( 931.5 MeV u )
=
=
= 8.66 MeV nucleon
A
A
93
Au ,
∆m = 79 ( 1.007 825u ) + 118 ( 1.008 665u ) − ( 196.966 543u ) = 1.674 102u
and
29.10
Eb ( ∆m ) c 2 ( 1.674 102u )( 931.5 MeV u )
=
=
= 7.92 MeV nucleon
A
A
197
(a) For 12 H ,
∆m = 1( 1.007 825u ) + 1( 1.008 665u ) − ( 2.014 102u ) = 0.002 388 u
and
Eb ( ∆m ) c 2 ( 0.002 388 u )( 931.5 MeV u )
=
=
= 1.11 MeV nucleon
A
A
2
Nuclear Physics
(b) For 24 He ,
∆m = 2 ( 1.007 825u ) + 2 ( 1.008 665u ) − ( 4.002 602u ) = 0.030 378 u
and
(c) For
Eb ( ∆m ) c 2 ( 0.030 378 u )( 931.5 MeV u )
=
=
= 7.07 MeV nucleon
A
A
4
56
26
Fe ,
∆m = 26 ( 1.007 825u ) + 30 ( 1.008 665u ) − ( 55.934 940 ) = 0.528 460 u
Eb ( ∆m ) c 2 ( 0.528 460 u )( 931.5 MeV u )
and
=
=
= 8.79 MeV nucleon
A
A
56
(d) For
238
92
U,
∆m = 92 ( 1.007 825u ) + 146 ( 1.008 665u ) − ( 238.050 784 ) = 1.934 206 u
and
29.11
For
15
8
Eb ( ∆m ) c 2 ( 1.934 206 u )( 931.5 MeV u )
=
=
= 7.57 MeV nucleon
A
A
238
O,
∆m = 8 ( 1.007 825u ) + 7 ( 1.008 665u ) − ( 15.003 065 ) = 0.120 190 u
and
For
Eb
15
7
15
0
= ( ∆m ) c 2 = ( 0.120 190 u )( 931.5 MeV u ) = 111.957 MeV
N,
∆m = 7 ( 1.007 825u ) + 8 ( 1.008 665u ) − ( 15.000 108 ) = 0.123 987 u
and
Eb
Therefore,
15
N
= ( ∆m ) c 2 = ( 0.123 987 u )( 931.5 MeV u ) = 115.494 MeV
∆Eb = Eb
15
N
− Eb
15
0
= 3.54 MeV
465
466
29.12
CHAPTER 29
∆m = Z mH + ( A − Z ) mn − m and Eb A = ∆m ( 931.5 MeV u ) A
55
25
Mn
( in u )
∆m
( in u )
( in MeV )
Z
(A – Z)
25
30
54.938 048
0.517 527
8.765
Nucleus
m
Eb A
56
26
Fe
26
30
55.934 940
0.528 460
8.786
59
27
Co
27
32
58.933 198
0.555 357
8.768
Therefore, 56
26 Fe has a greater binding energy per nucleon than its neighbors. This gives
us finer detail than is shown in Figure 29.4.
29.13
For
23
11
Na ,
∆m = 11 ( 1.007 825u ) + 12 ( 1.008 665u ) − ( 22.989 770 u ) = 0.200 285u
Eb ( ∆m ) c 2 ( 0.200 285u )( 931.5 MeV u )
=
=
= 8.111 MeV nucleon
A
A
23
and
For
23
12
Mg ,
∆m = 12 ( 1.007 825u ) + 11 ( 1.008 665u ) − ( 22.994 127 u ) = 0.195 088 u
so
Eb ( ∆m ) c 2 ( 0.195 088 u )( 931.5 MeV u )
=
=
= 7.901 MeV nucleon
A
A
23
The binding energy per nucleon is
greater for
23
11
Na by 0.210 MeV nucleon
23
This is attributable to less proton repulsion in 11
Na
29.14
42
43
The sum of the mass of 20
Ca plus the mass of a neutron exceeds the mass of 20
Ca . This
difference in mass must represent the mass equivalence of the energy spent removing a
43
42
Ca to produce 20
Ca plus a free neutron. Thus,
the last neutron from 20
(
)
E = m 42 Ca + mn − m 43 Ca c 2 = ( 41.958 622 u + 1.008 665 u − 42.958 770 u ) c 2
or
20
20
E = ( 0.008 517 u )( 931.5 MeV u ) = 7.93 MeV
Nuclear Physics
29.15
The decay constant is λ =
ln 2
, so the activity is
T1 2
3.0 × 1016 ) ln 2
(
N ln 2
R = λN =
=
= 1.7 × 1010 decays s
4
T1 2
( 14 d ) ( 8.64 × 10 s d )
or
29.16


1 Ci
R = ( 1.7 × 1010 decays s ) 
 = 0.46 Ci
10
 3.7 × 10 decays s 
The activity is R = R0 e − λt where λ =
R = R0 e
29.17
(
− t ln 2 T1 2
)
ln 2
. Thus,
T1 2
= ( 1.1 × 10 4 Bq ) e
−
( 2.0 h ) ln 2
6.05 h
= 8.7 × 10 3 Bq
(a) The decay constant is
λ=
ln 2
ln 2
=
= 9.98 × 10 −7 s −1
T1 2 ( 8.04 d ) ( 8.64 × 10 4 s d )
(b) R = λ N , so the required number of nuclei is
N=
29.18
R
λ
=
0.50 × 10 −6 Ci  3.7 × 1010 decays s 
10
 = 1.9 × 10 nuclei
-7
−1 
9.98 × 10 s 
1 Ci

(a) From R = R0 e − λt , where λ =
0.842 = e − λ ( 2.00 d )
Thus,
T1 2 =
ln 2
λ
or
=
ln 2
, and R = 0.842R0 when t = 2.00 d , we find
T1 2
λ =−
ln ( 0.842 )
= 8.60 × 10 −2 d −1
2.00 d
ln 2
= 8.06 d
8.60 × 10 −2 d −1
(b) Comparing the computed half-life to the measured half-lives of the radioactive
materials in Appendix B shows that this is probably
131
53
I or Iodine-131
467
468
CHAPTER 29
29.19
Recall that the activity of a radioactive sample is directly proportional to the number of
radioactive nuclei present, and hence, to the mass of the radioactive material present.
Thus,
R
N
m 0.25 × 10 −3 g
=
=
=
= 0.25
1.0 × 10 −3 g
R0 N 0 m0
0.25 = e − λ ( 2.0 h ) and
From R = R0 e − λt , we obtain
Then, the half-life is
29.20
T1 2 =
when t = 2.0 h
ln 2
λ
=
λ =−
ln ( 0.25 )
= 0.693 h −1
2.0 h
ln 2
= 1.0 h
0.693 h −1
Recall that the activity of a radioactive sample is directly proportional to the number of
radioactive nuclei present and hence, to the mass of the radioactive material present.
R
N
m
=
=
R0 N 0 m0
and
R = R0 e − λt becomes
The decay constant is
λ=
ln 2
ln 2
=
= 0.181 d −1
T1 2 3.83 d
Thus,
m = m0 e − λt
If m0 = 3.00 g and the elapsed time is t = 1.50 d , the mass of radioactive material
remaining is
m = m0 e − λt = ( 3.00 g ) e
29.21
(
)
− 0.181 d −1 ( 1.50 d )
= 2.29 g
From R = R0 e − λt , with R = ( 1.00 × 10 −3 ) R0 , we find e − λt =
and
t=−
ln ( R R0 )
λ
R
R0
 ln ( R R0 ) 
= − T1 2 

 ln 2 
 ln ( 1.00 × 10 −3 ) 
 = 4.31 × 10 3 yr
= − ( 432 yr ) 
ln
2


29.22
Using R = R0 e − λt , with R R0 = 0.125 , gives λ t = − ln ( R R0 )
or
t=−
ln ( R R0 )
λ
 ln ( R R0 ) 
 ln ( 0.125 ) 
4
= − T1 2 
 = − ( 5 730 yr ) 
 = 1.72 × 10 yr
ln
2
 ln 2 


Nuclear Physics
29.23
(a) The initial activity is R0 = 10.0 mCi , and at t = 4.00 h , R = 8.00 mCi . Then, from
R = R0 e − λt , the decay constant is
λ =−
ln ( R R0 )
t
=−
ln ( 0.800 )
= 5.58 × 10 −2 h −1
4.00 h
ln 2
and the half-life is T1 2 =
(b) N 0 =
λ
−3
10
−2
The number of
N0 =
ln 2
= 12.4 h
5.58 × 10 −2 h −1
(10.0 × 10 Ci )( 3.70 × 10 s 1 Ci ) =
=
( 5.58 × 10 h ) (1 h 3 600 s )
R = R0 e − λt = ( 10.0 mCi ) e
(c)
29.24
R0
λ
=
90
38
−1
−1
(
)
− 5.58×10 −2 h −1 ( 30 h )
2.39 × 1013 nuclei
= 1.9 mCi
Sr nuclei initially present is
5.00 kg
total mass
=
= 3.35 × 10 25
mass per nucleus ( 89.907 7 u ) ( 1.66 × 10 −27 kg u )
The half-life of
90
38
Sr is T1 2 = 29.1 yr (Appendix B), so the initial activity is
R0 = λ N 0 =
( 3.35 × 1025 ) ln 2
N 0 ln 2
=
= 2.53 × 1016 counts s
7
T1 2
29.1
yr
3.156
×
10
s
yr
(
)(
)
The time when the activity will be R = 10.0 counts min is
t=−
ln ( R R0 )
λ
ln R R
( ) (ln 2 )
= − T1 2
0
 10.0 min −1  1 min  
ln 


2.53 × 1016 s −1  60 s  

= − ( 29.1 yr )
= 1.66 × 10 3 yr
ln 2
29.25
212
83
95
36
Bi →
Kr →
144
60
208
81
95
37
Tl + 24 He
Rb + −01e
Nd → 24 He + 140
58 Ce
469
470
29.26
CHAPTER 29
12
5
B →
234
90
29.27
12
6
C + −01e
4
Th → 230
88 Ra + 2 He
14
6
C → 147 N + −01e
40
20
40
Ca → +01e + 19
K
98
44
Ru → 24 He +
144
60
Nd →
4
2
94
42
Mo
He + 140
58 Ce
29.28
235
92 U
a
231
90 Th
ea
231
91 Pa
227
89 Ac
a
e227
90 Th
223
87 Fr
a
ea
223
88 Ra
219
85 At
a
ea
219
86 Rn
215
83 Bi
ea
215
84 Po
a
211
82 Pb
e211
83 Bi
a
e211
84 Po
29.29
207
81 Tl
ea
207
82 Pb
56
The more massive 56
27 Co decays into the less massive 26 Fe . To conserve charge, the
charge of the emitted particle must be +1e. Since the parent and the daughter have the
same mass number, the emitted particle must have essentially zero mass. Thus, the
decay must be positron emission or e+ decay . The decay equation is
56
27
Co →
56
26
Fe + e+ + ν
Nuclear Physics
29.30
The energy released in the decay
238
92
U → 24 He +
234
90
471
Th is
Q = ( ∆m ) c 2 =  m 238 U − ( m 4 He + m 234 Th )  c 2
=  238.050 784 u − ( 4.002 602 u + 234.043 583 u )  ( 931.5 MeV u )
= 4.28 MeV
29.31
The Q value of a decay, Q = ( ∆m ) c 2 , is the amount of energy released in the decay. Here,
∆m is the difference between the mass of the original nucleus and the total mass of the
decay products. If Q > 0 , the decay may occur spontaneously.
40
40
Ca → e+ + 19
K , the masses of the electrons do not automatically
(a) For the decay 20
cancel. Thus, we add 20 electrons to each side of the decay to yield neutral atoms
and obtain
(
40
20
Then,
40
Ca + 20e− ) → e+ + ( 19
K + 19e− ) + e−
(
Q = m 40 Ca
20
atom
− m 40 K
19
atom
144
60
49
Caatom → e+ + 40 K atom + e−
)
− 2me c 2 =  39.962 591 u − 39.964 000 u − 2 ( 0.000 549 u )  c 2
or Q = ( −0.002 507 u ) c 2 < 0
(b) In the decay
or
so the decay cannot occur spontaneously
Nd →24 He + 140
58 Ce , we may add 60 electrons to each side forming all
neutral atoms and use masses from Appendix B to find
(
)
Q = m144 Nd − m 4 He − m140 Ce c 2 = ( 143.910 082 u − 4.002 602 u − 139.905 434 u ) c 2
60
2
58
or Q = ( +0.002 046 u ) c 2 > 0
29.32
(a)
66
28
Ni →
66
29
Cu + −10 e +ν
so the decay can occur spontaneously
472
CHAPTER 29
(b) Because of the mass differences, neglect the kinetic energy of the recoiling daughter
nucleus in comparison to that of the other decay products. Then, the maximum
kinetic energy of the beta particle occurs when the neutrino is given zero energy.
That maximum is
KEmax = ( m 66 Ni − m 66 Cu ) c 2 = ( 65.929 1 u − 65.928 9 u )( 931.5 MeV u )
= 0.186 MeV = 186 keV
29.33
In the decay 13 H → 23 He + -01e + ν , the anti-neutrino is massless. Adding 1 electron to
each side of the decay gives
(
3
1
H + e − ) → ( 23 He + 2 e− ) + ν , or 13 H atom → 23 Heatom + ν .
Therefore, using neutral atomic masses from Appendix B, the energy released is
E = ( ∆m ) c 2 = ( m 3 H − m 3 He ) c 2 = ( 3.016 049 u − 3.016 029 u )( 931.5 MeV u )
= 0.018 6 MeV = 18.6 keV
29.34
The initial activity of the 1.00-kg carbon sample would have been
 15.0 counts min 
4
-1
R0 = ( 1.00 × 10 3 g ) 
 = 1.50 × 10 min
1.00
g


From R = R0 e − λt , and T1 2 = 5 730 yr for
t=−
ln ( R R0 )
λ
= − T1 2
14
C (Appendix B), the age of the sample is
ln ( R R0 )
ln 2
 2.00 × 10 3 min −1 
ln 

1.50 × 10 4 min −1 

= − ( 5 730 yr )
= 1.67 × 10 4 yr
ln 2
29.35
From R = R0 e − λt , and T1 2 = 5 730 y for
t=−
ln ( R R0 )
λ
= − T1 2
ln ( R R0 )
ln 2
14
C (Appendix B), the age of the sample is
= − ( 5 730 yr )
ln ( 0.600 )
= 4.22 × 10 3 yr
ln 2
Nuclear Physics
29.36
The total number of carbon nuclei in the sample is
 21.0 × 10 −3 g 
23
21
N =
 ( 6.02 × 10 nuclei mol ) = 1.05 × 10
 12.0 g mol 
Of these, one in every 7.70 × 1011 was originally 14 C . Hence, the initial number of 14 C
nuclei present was N 0 = 1.05 × 10 21 7.70 × 1011 = 1.37 × 10 9 , and the initial activity was
9
N 0 ln 2 ( 1.37 × 10 ) ln 2  1 yr   7 d 
3
-1
R0 = λ N 0 =
=
 = 3.17 × 10 week


T1 2
( 5 730 yr )  365.24 d   1 week 
The current activity, accounting for counter efficiency, is
R=
count rate 837 week -1
=
= 951 week -1
efficiency
0.880
Therefore, from R = R0 e − λt , the age of the sample is
t=−
ln ( R R0 )
λ
= − T1 2
ln ( R R0 )
ln 2


951 week −1
ln 
−1 
3
3.17 × 10 week 
= − ( 5 730 yr ) 
= 9.96 × 10 3 yr
ln 2
29.37
(a)
27
13
Al + 24 He → 10 n +
30
15
P
(b) Q = ( ∆m ) c 2 = ( m 27 Al + m 4 He − mn − m 30 P ) c 2
Q = [ 26.981 538 u + 4.002 602u − 1.008 665 u − 29.978 310 u ] ( 931.5 MeV u )
= − 2.64 MeV
4
2
29.38
7
3
He + 147 N → 11H + 187 O
Li + 11H → 24 He + 24 He
473
474
CHAPTER 29
29.39
(a)
29.40
21
10
24
Ne + 24 He → 12
Mn + 10 n
U + 10 n → 90
38 Sr +
(b)
235
92
144
54
(c)
2 11H → 12 H + e+ + ν
Xe + 2 10 n
Q = ( ∆m ) c 2 = ( m1 H + m 7 Li − 2 m 4 He ) c 2
= 1.007 825u + 7.016 003u − 2 ( 4.002 602 u )  ( 931.5 MeV u ) = 17.3 MeV
29.41
29.42
(a)
10
5
B + 24 He → 11H +
13
6
C
(b)
13
6
C + 11H → 24 He +
10
5
B
(a)
7
3
Li + 24 He →
10
5
B + 10 n
(b) Q = ( ∆m ) c 2 = ( m7 Li + m 4 He − m10 B − mn ) c 2
= [7.016 003u + 4.002 602u − 10.012 936 u − 1.008 665 u ] ( 931.5 MeV u )
= − 2.79 MeV
29.43
(a) Requiring that both charge and the number of nucleons (atomic mass number) be
conserved, the reaction is found to be
197
79
0
Au + 10 n →198
80 Hg + −1 e + ν
Note that the antineutrino has been included to conserve electron-lepton number
which will be discussed in the next chapter.
Nuclear Physics
475
(b) We add 79 electrons to both sides of the reaction equation given above to produce
neutral atoms so we may use mass values from Appendix B. This gives
197
79
Au atom + 10 n →198
80 Hg atom + ν
and, remembering that the neutrino is massless, the Q value is found to be
(
)
Q = ( ∆m ) c 2 = m197 Au + mn − m198 Hg c 2 = ( 196.966 543 u + 1.008 665 u − 197.966 750 u ) c 2
79
80
or Q = ( +0.008 458 u )( 931.5 MeV u ) = 7.88 MeV
The kinetic energy carried away by the daughter nucleus is negligible. Thus, the
energy released may be split in any manner between the electron and neutrino, with
the maximum kinetic energy of the electron being 7.88 MeV
29.44
The energy released in the reaction 10 n + 24 He → 12 H + 13 H is
Q = ( ∆m ) c 2 = ( mn + m 4 He − m 2 H − m 3 H ) c 2
= [1.008 665 u + 4.002 602 u − 2.014 102 u − 3.016 049 u ] ( 931.5 MeV u )
= − 17.6 MeV
The threshold energy of the incident neutron is then

m
KEmin =  1 + n

m 4 He

29.45
(a)
18
8
O + 11H → 189 F +
1
0


1.008 665 u 
 Q =  1 +
 − 17.6 MeV = 22.0 MeV
4.002
602u



n
(b) The energy released in this reaction is
Q = ( ∆m ) c 2 = ( m18 O + m1 H − m18 F − mn ) c 2
= 17.999 160 u + 1.007 825u − m18 F − 1.008 665 u  ( 931.5 MeV u )
Since we know that Q = − 2.453 MeV , we find that
m18 F = 18.000 953 u
476
CHAPTER 29
29.46
For equivalent doses, it is necessary that
( heavy ion dose in rad ) × RBEheavy ions = ( x-ray dose in rad ) × RBEx-rays
or
29.47
ion dose in rad =
( x-ray dose in rad ) × RBEx-rays (100 rad )(1.0 )
=
=
RBEheavy ions
20
5.0 rad
For each rad of radiation, 10 − 2 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 = ( 25.0 rad )  10 − 2
 ( 75.0 kg ) = 18.8 J
rad 

29.48
(a) Each rad of radiation delivers 10 − 2 J of energy to each kilogram of absorbing
material. Thus, the energy delivered per unit mass with this dose is
J kg 
E

= ( 200 rad )  10 − 2
 = 2.00 J kg
rad 
m

(b) From E = Q = mc ( ∆T ) , the expected temperature rise with this dosage is
∆T =
29.49
E m
2.00 J kg
=
= 4.78 × 10 − 4 °C
c
4 186 J kg ⋅ °C
The rate of delivering energy to each kilogram of absorbing material is
J kg
E m
 − 2 J kg 

 = ( 10 rad s )  10
 = 0.10
rad 
s
 ∆t 

The total energy needed per unit mass is

J 
5
E m = c ( ∆T ) =  4 186
 ( 50°C ) = 2.1 × 10 J kg
kg ⋅ °C 

so the required time will be
∆t =
energy needed 2.1 × 10 5 J kg
=
= 2.1 × 106
delivery rate
0.10 J kg ⋅ s
1d


s
 = 24 d
4
 8.64 × 10 s 
Nuclear Physics
29.50
477
(a) The number of x-rays taken per year is
production = ( 8 x-ray d )( 5 d week )( 50 weeks yr ) = 2.0 × 10 3 x-ray yr
so the exposure per x-ray taken is
exposure rate =
exposure
5.0 rem yr
=
= 2.5 × 10 −3 rem x-ray
3
production 2.0 × 10 x-ray yr
(b) The exposure due to background radiation is 0.13 rem yr . Thus, the work-related
exposure of 5.0 rem yr is
5.0 rem yr
≈ 38 times background levels
0.13 rem yr
29.51
(a) From N =
R
λ
=
R0 e − λ t
λ
 T1 2 R0
=
 ln 2
 − t ln 2 T1 2
, the number of decays occurring during the
e

10-day period is
 T1 2 R0 
− t ln 2 T1 2
∆N = N 0 − N = 
 1− e
 ln 2 
(
)
 ( 14.3 d ) ( 1.31 × 106 decay s )  8.64 × 10 4 s 
− ( 10.0 d ) ln 2 14.3 d
=

 1− e
ln
2
1
d



(
= 8.97 × 1011 decays , and one electron is emitted per decay.
(b) The total energy deposited is found to be

 1.60 × 10 −16 J 
keV 
11
E =  700
 ( 8.97 × 10 decays ) 
 = 0.100 J
decay 
1 keV



(c) The total absorbed dose (measured in rad) is given by
dose =
=
energy deposited per unit mass
energy deposition per rad
( 0.100 J
0.100 kg )
= 100 rad
 -2 J kg 
 10

rad 


)

478
CHAPTER 29
29.52
(a) The dose (in rem) received in time ∆t is given by
dose = ( dose in rad ) × RBE
rad  
rem 


=  100 × 10 − 3
 ∆t  × ( 1.00 ) =  0.100
 ∆t
h  
h 


If this dose is to be 1.0 rem, the required time is
∆t =
1.0 rem
= 10 h
0.100 rem h
(b) Assuming the radiation is emitted uniformly in all directions, the intensity of the
radiation is given by I = I 0 4π r 2
Therefore,
I 0 4π r 2
( 1.0 m )
Ir
=
=
2
I1 I 0 4π ( 1.0 m )
r2
and r = ( 1.0 m )
29.53
100 mrad h
I1
= ( 1.0 m )
= 3.2 m
10 mrad h
Ir
From R = R0 e − λt , the elapsed time is
t=−
29.54
2
ln ( R R0 )
λ
= − T1 2
ln ( R R0 )
ln 2
= − ( 14.0 d )
ln ( 20.0 mCi 200 mCi )
ln 2
= 46.5 d
Assuming that the carbon-14 activity of a living organism 20 000 years ago was the same
as today, the original activity of the sample was
R0 = ( 15.0 decays min ⋅ g )( 18 g ) = 2.7 × 10 2 decays min
The decay constant for carbon-14 is
λ=
ln 2
ln 2
=
= 1.21 × 10 −4 yr −1
T1 2 5 730 yr
so the expected current activity is
R = R0 e − λt = ( 2.7 × 10 2 decay min ) e
(
)
− 1.21×10 −4 yr −1 ( 20 000 yr )
= 24 decays min
Nuclear Physics
29.55
The energy released in the reaction 12 H + 12 H → 23 He + 10 n is
Q = ( ∆m ) c 2 = ( 2 m 2 H − m 3 He − mn ) c 2
=  2 ( 2.014 102u ) − 3.016 029 u − 1.008 665 u  ( 931.5 MeV u )
= + 3.27 MeV
Since Q > 0 , no threshold energy is required
29.56
The speed of a neutron having a kinetic energy of
KE = ( 0.040 eV ) ( 1.60 × 10 −19 J eV ) = 6.4 × 10 −21 J
is
v=
2 ( 6.4 × 10 −21 J )
2 ( KE )
=
= 2.8 × 10 3 m s
1.675 × 10 −27 kg
mn
and the time required to travel 10.0 km is
t=
∆x 10.0 × 10 3 m
=
= 3.6 s
2.8 × 10 2 m s
v
The decay constant for free neutrons is
λ=
ln 2
ln 2  1 min 
−4
−1
=

 = 9.6 × 10 s
T1 2 12 min  60 s 
Since the number remaining after time t is N = N 0 e − λt , the fraction having decayed in
this time is
fraction decayed =
29.57
(a)
N0 =
N0 − N
− ( 9.6×10 −4 s−1 ) ( 3.6 s )
= 1 − e − λt = 1 − e
= 0.003 5 or 0.35%
N0
mass of sample
1.00 kg
=
= 2.52 × 10 24
mass per atom ( 239.05 u ) ( 1.66 × 10 −27 kg u )
2.52 × 10 24 ) ln 2
(
N 0 ln 2
=
= 2.29 × 1012 Bq
(b) R0 = λ N 0 =
7
T1 2
( 24 120 yr ) ( 3.156 × 10 s yr )
479
480
CHAPTER 29
(c) From R = R0 e − λt , the required time is
t=−
=−
29.58
(a)
ln ( R R0 )
λ
=−
T1 2 ln ( R R0 )
ln 2
( 24 120 yr ) ln ( 0.100
2.29 × 1012 )
ln 2
r = r0 A1 3 = ( 1.2 × 10 −15 m ) ( 12 )
13
= 1.07 × 10 6 yr
= 2.7 × 10 −15 m = 2.7 fm
(b) With Z = 6 ,
F=
k e e ( Z − 1) e 
r2
=
5 ( 8.99 × 10 9 N ⋅ m 2 C 2 )( 1.60 × 10 −19 C )
( 2.7 × 10
-15
m)
2
2
or F = 1.5 × 10 2 N
(c) The work done is the increase in the electrical potential energy, or
W = PE r − PE r =∞ =
k e e ( Z − 1) e 
r
−0=
5 ( 8.99 × 10 9 N ⋅ m 2 C 22 )( 1.60 × 10 −19 C )
2.7 × 10 −15 m
 1 MeV 
= 4.2 × 10 −13 J 
 = 2.6 MeV
-13
 1.60 × 10 J 
(d) Repeating the previous calculations for
r = r0 ( 238 )
13
and
W=
= 7.4 fm , F =
k e e ( 91) e 
r
238
92
k e e ( 91) e 
r2
= 18 MeV
U (with Z = 92 and A = 238 ) gives
= 3.8 × 10 2 N
2
Nuclear Physics
29.59
(a) If we assume all the 87 Sr nuclei came from the decay of
number of 87 Rb nuclei was
87
Rb nuclei, the original
N 0 = 1.82 × 1010 + 1.07 × 109 = 1.93 × 1010
Then, from N = N 0 e − λt , the elapsed time is
ln ( N N 0 )
t=−
λ
=−
T1 2 ln ( N N 0 )
ln 2
 1.82 × 1010 
10
4.8
10
yr
ln
×
(
)  1.93 × 1010 

 = 4.0 × 109 yr
=−
ln 2
(b) It could be no older . It could be younger if some 87 Sr were initially present
29.60
From R = λ N , the number of
N=
R
λ
=
60
Co nuclei present in a 10 Ci source is
( )
R T1 2
ln 2
decay s  ( 5.2 yr )  3.156 × 107 s 

19
= ( 10 Ci )  3.7 × 1010

 = 8.8 × 10

Ci  ln2 
1 yr


The number that was present 30 months (2.5 years) ago must have been
N0 =
N
t ⋅ln 2 T1 2
2.5 yr ) ln 2 5.2 yr
=Ne
= ( 8.8 × 1019 ) e (
= 1.2 × 10 20
− λt
e
and the initial mass of
60
Co was
m = N 0 matom = ( 1.2 × 10 20 ) ( 59.93 u ) ( 1.66 × 10 −27 kg u ) 
= 1.2 × 10 −5 kg = 12 mg
481
482
CHAPTER 29
29.61
The total activity of the working solution at t = 0 was
( R0 )total = ( 2.5
mCi mL ) ( 10 mL ) = 25 mCi
Therefore, the initial activity of the 5.0-mL sample which will be drawn from the 250-mL
working solution was
( R0 )sample = ( R0 )total 
5.0 mL 
 5.0 mL 
−4
 = ( 25 mCi ) 
 = 0.50 mCi = 5.0 × 10 Ci
 250 mL 
 250 mL 
With a half-life of 14.96 h for
R = R0 e − λt = R0 e
− t ln 2 T1 2
24
Na (Appendix B), the activity of the sample after 48 h is
= ( 5.0 × 10 − 4 Ci ) e − ( 48 h ) ln 2 (14.96 h )
= 5.4 × 10 − 5 Ci = 54 µ Ci
29.62
The decay constants for the two uranium isotopes are
and
(
λ1 =
ln 2
ln 2
=
= 9.9 × 10 −10 yr −1 for
9
T1 2 0.70 × 10 yr
λ2 =
ln 2
ln 2
=
= 1.55 × 10 −10 yr −1 for
T1 2 4.47 × 109 yr
(
235
92
U
238
92
)
U
)
If there were N 0 nuclei of each isotope present at t = 0 , the ratio of the number
remaining would be
ln ( N 235 N 238 )
N 235 N 0 e − λ1t
=
= e ( λ2 − λ1 ) t , and the elapsed time is t =
− λ2 t
λ2 − λ1
N 238 N 0 e
With a measured value of 0.007 for this ratio, the estimated age is
t=
ln ( 0.007 )
= 6 × 109 yr
( 1.55 − 9.9 ) × 10−10 yr −1
Nuclear Physics
29.63
483
The original activity per unit area is
5.0 × 106 Ci  1 km 
−4
2
2
2

 = 5.0 × 10 Ci m = 5.0 × 10 µ Ci m
1.0 × 10 4 km 2  10 3 m 
2
R0 =
From R = R0 e − λt , the required time is
t=−
=−
29.64
ln ( R R0 )
λ
=−
T1 2 ln ( R R0 )
ln 2
( 28.7 yr ) ln ( 2.0 µ Ci
5.0 × 10 2 µ Ci )
ln 2
(a) The mass of a single
40
= 2.3 × 10 2 yr
K atom is
m = ( 39.964 u ) ( 1.66 × 10 −27 kg u ) = 6.63 × 10 −26 kg = 6.63 × 10 −23 g
Therefore, the number of
N=
40
K nuclei in a liter of milk is
total mass of 40 K present ( 2.00 g L )( 0.011 7 100 )
=
= 3.53 × 1018 L
mass per atom
6.63 × 10 −23 g
and the activity due to potassium is
3.53 × 1018 L ) ln 2
(
N ln 2
=
= 60.5 Bq L
R = λN =
T1 2
(1.28 × 109 yr )( 3.156 × 107 s yr )
(b) Using R = R0 e − λt , the time required for the
potassium is given by
t=−
ln ( R R0 )
λ
=−
T1 2 ln ( R R0 )
ln 2
=−
131
I activity to decrease to the level of the
( 8.04 d ) ln ( 60.5 2 000 )
ln 2
= 40.6 d
484
CHAPTER 29
29.65
The total activity due to
R = R0 e − λt = R0 e
59
Fe at the end of the 1 000-h test will be
− t ⋅ln 2 T1 2
= ( 20.0 µ Ci ) e
(
)
−  10 3 h ⋅ln 2 ( 45.1 d )( 24 h d ) 


= 10.5 µ Ci
The total activity in the oil at the end of the test is
 800 min -1 
Roil = 
 ( 6.5 L )
L


 1 min  1 µ Ci 
= 5.2 × 10 3 min -1 
= 2.3 × 10 −3 µ Ci

4
-1 
60
s
3.7
10
s
×



Therefore, the fraction of the iron that was worn away during the test is
fraction =
Roil 2.3 × 10 −3 µ Ci
=
= 2.2 × 10 −4
10.5 µ Ci
R
This represents a mass of
mworn away = ( fraction ) ⋅ ( total mass of iron )
= ( 2.2 × 10 −4 ) ( 0.20 kg ) = 4.4 × 10 −5 kg
so the wear rate was
29.66
4.4 × 10 −5 kg
= 4.4 × 10 −8 kg h
1 000 h
(a) Adding 2 electrons to each side of the reaction
4 11H + 2e− → 24 He + 2ν + γ
gives 4 ( 11 H + e − ) → ( 24 He + 2 e− ) + 2ν + γ , or
4 11Hatom →24 Heatom + 2ν + γ
Since the neutrinos and the photon are massless, the available energy is
Q = ( ∆m ) c 2 =  4 m1 H − m 4 He  c 2
=  4 ( 1.007 825 u ) − 4.002 602 u  ( 931.5 MeV u )
= 26.7 MeV ( 1.60 × 10 −13 J MeV ) = 4.28 × 10 −12 J
Nuclear Physics
(b) N =
mSun 1.99 × 10 30 kg
=
= 1.19 × 10 57
m1 H 1.67 × 10 -27 kg
(c) With 4 hydrogen nuclei consumed per reaction, the total energy available is
 1.19 × 10 57 
N
−12
45
E =  Q = 
 ( 4.28 × 10 J ) = 1.27 × 10 J
4
 4


At a power output of P = 3.76 × 10 26 W , this supply would last for
t=
1.27 × 10 45 J
E
=
= 3.39 × 1018
P 3.76 × 10 26 J s
1 yr


11
s
 = 1.07 × 10 yr
7
 3.156 × 10 s 
= 107 × 109 yr = 107 billion years
485
486
CHAPTER 29