43.1: a)
Si
28
14
has 14 protons and 14 neutrons.
b)
Rb
85
37
has 37 protons and 48 neutrons.
c)
Tl
205
81
has 81 protons and 124 neutrons.
43.2: a) Using
,fm)2.1(
31
AR 
the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm.
b) Using
2
4 R

for each of the radii in part (a), the areas are 163
andfm353,fm
22
.fm633
2
c)
3

3
4
R

gives 195
,fm
3
624
3
fm
and 1499
.fm
3
d) The density is the same, since the volume and the mass are both proportional to A:
317
mkg103.2 
(see Example 43.1).
e) Dividing the result of part (d) by the mass of a nucleon, the number density is
.m1040.1fm14.0
3
443

43.3:
BμBμBμE
zzz
2)(







T533.0
T)J10051.5)(7928.2(2
Hz)1027.2(s)J1063.6(
2
so,But
27
734







B
hf
BhfE
z

43.4:
a) As in Example 43.2,
eV.1077.2)T30.2)(TeV1015245.3)(9130.1(2
78 
E
Since


SN and

are in opposite directions for a neutron, the antiparallel configuration is
lower energy. This result is smaller than but comparable to that found in the example for
protons.
b)
m 484λMHz,9.66 


f
c
h
E
f
43.5: a)

 SNBμ and.BU
z


point in the same direction for a proton. So if the
spin magnetic moment of the proton is parallel to the magnetic field,
,0

U
and if they
are antiparallel,
.0

U
So the parallel case has lower energy.
The frequency of an emitted photon has a transition of the protons between the two

states given by:
Hz.1002.7
)sJ1063.6(
)T65.1)(TJ10051.5)(7928.2(2
2
7
34
27












h
B
μ
h
EE
h
E
f
z


m.27.4
Hz1002.7
sm1000.3
λ
7
8





f
c
This is a radio wave.
b) For electrons, the negative charge means that the argument from part (a) leads to
the
2
1

s
m
state (antiparallel) having the lowest energy, since

SN and
point in
opposite directions. So an emitted photon in a transition from one state to the other has a
frequency
h
B
h

EE
h
E
f
z

2
2
1
2
1







But from Eq. (41.22),
m.1049.6
Hz1062.4
sm1000.3
λso
Hz1062.4
kg)1011.9(4
T)C)(1.651060.1)(00232.2(
4
)00232.2(
4
)00232.2(

2
)00232.2(
3
10
8
10
31
19













f
c
f
ππm
eB
f
m
e
S

m
e
μ
e
e
z
e
z

This is a microwave.
43.6: a)
%.0027.01066.2)eV10511.0()eV6.13(
56


b)
%.937.01037.9)MeV3.938()MeV795.8(
3


43.7: The binding energy of a deuteron is
eV.10224.2
6

The photon with this energy
has wavelength equal to
.m10576.5
eV)J10602.1)(eV10224.2(
s)m10998.2)(sJ10626.6(
λ

13
196
834







E
hc
43.8: a)
u,112.0)(7
NHn



mmm
which is 105 MeV, or 7.48 MeV per nucleon.
b) Similarly,
perMeV7.07orMeV,3.28u03038.0)(2
HenH
 mmm
nucleon,
slightly lower (compare to Fig. (43.2)).
43.9: a) For
B
11
5

the mass defect is:

)565
11
5enp
BM(mmmm 
u)MeVu)(931.5081815.0(energy bindingThe
u0.081815
u009305.11)u000549.0(5)u008665.1(6)u007277.1(5
B






E

MeV76.21

b) From Eq. (43.11):
A
ZA
C
A
ZZ
CACACE
2
4321B
)2()1(

3
1
3
2




and there is no
fifth term since Z is odd and A is even.

3
1
3
2
11
)4(5
)MeV(0.7100MeV)(11)80.17(11)MeV75.15(
B
 E

)11(
)1011(
)MeV69.23(
2



MeV.76.68
B

 E
So the percentage difference is
%62.0100
MeV76.21
MeV76.21MeV76.68


Eq. (43.11) has a greater percentage accuracy for
Ni.
62
43.10: a)






u62.929601u5)29(1.00782u5)34(1.008662934
CuHn
mmm
and
u
MeV
931.5(usingnucleon per MeV8.75orMeV,551iswhichu,0.592
).nucleons63
b) In Eq. (43.11), Z = 29 and N = 34, so the fifth term is zero. The predicted binding
energy is
3
1
3

2
)63(
)28)(29(
)MeV(0.7100MeV)(63)80.17()63)(MeV75.15(
B
E
MeV.556
(63)
(5)
MeV)69.23(
2


(The fifth term is zero since the number of neutrons is even while the number of protons
is odd, making the pairing term zero.)
This result differs from the binding energy found from the mass deficit by 0.86%,
a very good agreement comparable to that found in Example 43.4.
43.11:
Z is a magic number of the elements helium (Z = 2), oxygen (Z = 8), calcium (Z =
20), nickel (Z =28), tin (Z = 50) and lead (Z = 82). The elements are especially stable,
with large energy jumps to the next allowed energy level. The binding energy for these
elements is also large. The protons’ net magnetic moments are zero.
43.12: a)
u,1.9392146
UHn



mmm
which is

b)
MeV7.56c)orMeV,1080.1
3

per nucleon (using 931.5
u
MeV
and 238 nucleons).
43.13: a)
4by decreases2,by decreases:decay AZα

 UPb
235
92
239
94
b)
:same theremains1,by decreases:decay AZβ




MgNa
24
12
24
11
c)
:same theremains1,by decreases:decay AZ






NO
15
7
15
8
43.14: a)The energy released is the energy equivalent of
u,108.40
4
epn

 mmm
,b)keV.783or
pn
mm 
and the decay is not possible.
43.15:
Be)(He)(2
8
4
4
2
MMm 

u8.005305u)2(4.002603




u109.9
5
 m
43.16: a) A proton changes to a neutron, so the emitted particle is a positron
).(


b) The number of nucleons in the nucleus decreases by 4 and the number of protons
by 2, so the emitted particle is an alpha-particle. c) A neutron changes to a proton, so the
emitted particle is an electron
).(


43.17: If
Cdecay
14

is possible, then we are considering the decay
.NC
14
7
14
6



keV156
MeV0.156)uMeVu)(931.51086.1(So:u101.68
0.0005491u))7(0.000549u(14.003074u))6(0.000549u003242.14(

)N()C(
44
e
14
7
14
6





E
mMMm
43.18: a) As in the example, (0.000898 u)(931.5
MeV.0.836u)MeV

b)
MeV.0.700MeV0.014MeV0.122MeV0.836



43.19: a) If tritium is to be unstable with respect to


decay, then the mass of the
products of the decay must be less than the parent nucleus.
u3.014932u)582(0.000548u3.016029)He(
u3.015500u0.00054858u016049.3)H(
23

2
3
1




M
M

u,100.2)He()H(
5
e
3
2
3
1

 mMMm
so the decay is possible.
b) The energy of the products is just
keV.19MeV0.019u)MeV5.931)(u100.2(
5


E
43.20:
Note that Eq. 43.17 can be written as follows
.2
21

/
0
Tt
NN


The amount of elapsed time since the source was created is roughly 2.5 years.
Thus, we expect the current activity to be
Ci.3600Ci)25000(
5.271years
2.6years


N
The source is barely usable. Alternatively, we could calculate
1
21
s)0.132(year
)2(ln
λ


T
and use the Eq. 43.17 directly to obtain the same answer.
43.21: For
yT 5730,C
21
14



mindecays0.180;so2lnλ;
0
)2ln(
021
λ
0
21



AeAATeAA
Tt
t
a)
mindecays159,y1000  At
b)
mindecays43.0,y000,50  At
43.22: (a)
XSr
90
39
90
39



X has 39 protons and 90 protons plus neutrons, so it must be
Y.
90
(b) Use base 2 because we know the half life.

yr190
2log
01.0log)yr28(
2log
01.0log
201.0
2
21
00
0
21
21






T
t
AA
AA
Tt
Tt
43.23: a)
HeeH
3
2
0
1

3
1


b)
210
λ
0
)2ln(λand100.0, TNNeNN
t



;100.0
21
)2ln( Tt
e

 );100.0(ln)2ln(
21
 Tt
y9.40
2ln
)100.0ln(
21



T
t

43.24: a)
)s1070.3)(10500(Ci500
1106 


dt
dN
sdec1085.1
7
dtdN
s106.69
d)s(86,40012
2ln2ln
λ
λ
2ln
7
21
21


dT
T
nuclei1077.2
s
10
69
.
6
s/dec1085.1

λ
λ
13
17
7





dtdN
NN
dt
dN
The mass of this
131
many
Ba nuclei is
)nucleuskg1066.1131(nuclei1077.2
2713 
m

ng0.6g100.6kg100.6
912


(b)
t
eAA
λ

0


days108
s400,86
d1
s1029.9
s1069.6
)5001(ln
λ
)5001(ln
λ)500/1(ln
)Ci500(Ci1
6
17
λ

















t
t
e
μμ
t
43.25:
2/1
/)2ln(
0
λ
0
Tt
t
eAeAA




)(ln
)2ln(
0
2
1
AA
T
t

days80.2

)83183091(ln
)days00.4)(2ln(
)(ln
)2ln(
0
2
1

AA
t
T
43.26:
N
dt
dN
λ
111
7
s1036.1λ
yr1
s1015.3
yr1620
2ln2ln
λ
2
1













T
atoms10665.2
g226
atoms10022.6
g1
25
23











N
Bq1062.3
s
dec

1062.3)s1036.1)(10665.2(
λ
10
1011125



N
dt
dN
Convert to Ci:
Ci98.0
Bq1070.3
Ci1
Bq1062.3
10
10











43.27:
Find the total number of carbon atoms in the sample.

y5730s10807.1λ2)ln(
)s(10836.3
)(
λ;λ
sdecays3.00mindecays180
atoms14-carbon1082.7)10016.6()103.1(soatoms,10016.6
mol
kg10(12.011mol)atoms10022.6()kg100.12(
;
11
2/1
112
11231223
tot
3233
AAtot











T
N
tN

NtN
tN
N
MNmnNN
Mmn
43.28:
a) Solving Eq. (43.19) for
λ
,

.s1017.4
)hrsec(3600)dayhrs24)(yeardays365()y27.5(
2ln2ln
λ
19
21


T
b)
.1061.3
)g1066.1()60(
g1060.3
u
17
24
5







A
m
N
c)
Bq,1051.1λ
9
 N
dt
dN
which is d) 0.0408 Ci. The same calculation for radium,
with larger
A
and longer half-life (lower
λ
) gives
.Ci1057.3
)226()yrs600.1(
)60()yrs27.5(
Ci0408.0
λλ
5
Ra1/2Ra
Co1/2Co
CoCoRaRA





















AT
AT
NN
43.29: a)
sdecays1056.7Bq1056.7
)0(
1111

dt
dN
and

.s1075.3

min)s60(min)8.30(
693.0693.0
λ
14
21


T
so
nuclei.102.02
s103.75
sdecays107.56)0(
λ
1
)0(
15
14
11





dt
dN
N
b) The number of nuclei left after one half-life is
15
1001.1
2

)0(

N
nuclei, and the
activity is half:
.sdecays1078.3
11

dt
dN
c) After three half lives (92.4 minutes) there is an eighth of the original amount
14
1053.2 N
nuclei, and an eighth of the activity:
.sdecays1045.9
10







dt
dN
43.30: The activity of the sample is
,kgBq102
kg)(0.500min)sec(60
mindecays3070


while the
activity of atmospheric carbon is 255
kgBq
(see Example 43.9). The age of the sample
is then
.y7573
y1021.1
)255102(ln
λ
)255102(ln
4




t
43.31: a)
)ys10156.3()y1028.1(
693.0693.0
79
21


T

.s1072.1
117 

In
Kofg1063.1

406
m
there are
nuclei.1045.2
)kg1066.1(40
kg1063.1
16
27
9






N
So
.sdecays0.421nuclei)10(2.45)s1072.1(λ
16117


N
dt
dN
b)
.Ci1014.1
CiBq1070.3
Bq421.0
11
10





dt
dN
43.32:
.Ci113.0Ci1013.1Bq1017.4
s86,400
decays10360
73
6
μ


43.33: a)
.s1091.4
)ys10156.3()y1047.4(
693.0693.0
λ
118
79
21




T
b)
decays/s.1044.4

Cis
decays
1070.3)Ci1020.1(Ci1020.1
51055










.
s1091.4
sdecays1044.41
But
118
5




dt
dN
NN
dt
dN



NmN u)238(nuclei1004.9
22

kg.0357.0)10kg)(9.041066.1)(238(
2227


m
c) Each decay emits one alpha particle. In 60.0 g of uranium there are
523118
23
27
107.46nuclei)1052.1()s1091.4(λ
nuclei1052.1
)kg1066.1(238
kg0600.0






N
dt
dN
N
alpha particles emitted each second.
43.34:
(a) rem = rad


RBE
200 = x(10)
x = 20 rad
(b) 1 rad deposits 0.010
kgJ
, so 20 rad deposit
kgJ0.20
. This radiation
affects 25 g
(0.025 kg) of tissue, so the total energy is

mJ5.0J105.0)kgJ(0.20kg)(0.025
3


(c) Since RBE = 1 for

-rays, so rem = rad. Therefore
20 rad = 20 rem
43.35: 1 rad =
2
10

Gy, so 1 Gy = 100 rad and the dose was 500 rad
rem=(rad) (RBE) = (500 rad) (4.0) = 2000 rem
kgJ5.0so,kgJ1Gy1 
43.36: a)
rem.540)Svrem(100Sv5.4


b)
The RBE of 1 gives an absorbed dose of
540 rad. c) The absorbed dose is 5.4 Gy, so the total energy absorbed is

kg)(65Gy)(5.4
J.351
The energy required to raise the temperature of 65
C010.0kg
o
is
(4190kg)(65
kJ.3C)(0.01K)kgJ 
43.37:
a) We need to know how many decays per second occur.
.s1079.1
)ys10156.3()y3.12(
693.0693.0
λ
19
7
21




T
The number of tritium atoms is
19
10
s101.79

)CiBq10(3.70Ci)(0.35
λ
1
)0(




dt
dN
N
18
102540.7)0(  N
nuclei. The number of remaining nuclei after one week is just
18)s3600()24()7()s1079.1(18λ
107.2462week)1()1025.7()0()week1(
19



NeeNN
t
nuclei
decays.107.8week)1()0(
15
 NNN
So the energy absorbed is
J.6.24)eVJ10(1.60eV)5000()108.7(
1915
total



γ
ENE
So the absorbed dose
is
rad.5.12kgJ125.0
kg)(50
J)(6.24

Since RBE = 1, then the equivalent dose is
12.5 rem.
b) In the decay, antinentrinos are also emitted. These are not absorbed by the body,
and so some of the energy of the decay is lost (about 12
keV
).
43.38:
a) From Table (43.3), the absorbed dose is 0.0900 rad. b)
The energy absorbed
is
J;101.35kg)(0.150)kgJ10(9.00
44 

each proton has energy
J,101.282
13

so
the number absorbed is
c)1005.1

9

The RBE for alpha particles is twice that for
protons, so only half as many,
,1027.5
8

would be absorbed.
43.39: a)
m
10
2.00
)sm10(3.00s)J10(6.63)10(6.50
11
83410
total







Nhc
NEE
J.106.46
4
total

 E

b) The absorbed dose is the energy divided by tissue mass:
`
rad.0.108
kgJ
rad100
)kgJ10(1.08
kg0.600
J106.46
dose
3
4














The rem dose for x rays (RBE = 1) is just 0.108 rem.
43.40:
α
117106
108.41s)10(3.156)CiBq10(3.7Ci)10(0.72 


particles. The
absorbed dose is
rad.108Gy1.08
kg)(0.50
)eVJ10(1.602eV)10(4.0)10(8.41
19611



The equivalent dose is (20) (108 rad) = 2160 rem.
43.41: a)
 241and7492SoHe.XBeH
4
2
9
4
2
1
ZAA
A
Z
Li.isXso,3
7
3
Z
b)
He)(Li)(Be)(H)(
4
2

7
3
9
4
2
1
MMMMm 

u.
10
678
.
7
u
002603
.
4
u
016003
.
7
u
012182
.
9
u
014102
.
2
3







So
MeV.7.152)uMeV(931.5u)10(7.678)(
32


cmE
c)
The threshold energy is taken to be the potential energy of the two reactants when
they just “touch.” So we need to know their radii:
m102.5(9)m)102.1(
m101.5(2)m)102.1(
153115
9
153115
2
Be
H




r
r
So the centers’ separation is

m100.4
15
r
Thus
J013.2
)m100.4(4
C)1060.1(4
4
1
13
15
0
219
BeH
0







πεr
qq
πε
U
MeV.1.4eV101.4
6
 U
43.42:

,u1097.1
2
HHeHHe
1
1
4
2
2
1
3
2

 mmmm
so the energy released is 18.4 MeV.
43.43: a) As in Ex.
,3571and,610142),a41.43(










ZZAA
Li.Xso
6
3


b) As in Ex. (43.41b), using
,Li)(,u003074.14)N(,u014102.2)H(
6
3
14
7
2
1
 MMM
u,010882.0,u012937.10)B(u,6.01521
10
5
 mMand
so energy is absorbed in the
reaction.
MeV.10.14u)MeV5.931(u)010882.0(  Q
c) From Eq. (43.24):
K
m
M
M
K


cm
so
MeV11.6MeV)14.10(
u0.14
u01.2u0.14

cm











K
M
mM
K
43.44:


1.93)molmolecules10(6.023)eVJ10(1.602eV)10(200
23196
,molJ10
13
which is far higher than typical heats of combustion.
43.45: The mass defect is
)U(U)(
*236
92n
235
92

MmMm 
u007025.0
u
045562
.
236
u
008665
.
1
u
043923
.
235






m
So the internal excitation of the nucleus is:
MeV544.6
)uMeV(931.5u)(0.007025)(
2

 cmQ
43 .46: a)
10.174and5023









AZ
b) The nuclide is a boron nucleus,
and
u,1000.3
3
BnLiHe

 mmmm
and so 2.79 MeV of energy is absorbed.
43.47:
The energy liberated will be
MeV.586.1
)vMeV(931.5
v)7.016929v4.002603v(3.016029Be)()He(He)(
7
4
4
2
3
2




MMM
43.48: a)
XZZAA
A
Z
2.so14124.so2824.XMgSi
24
12
28
14


is an

particle.
b)
)vMeV(931.5v)27.976927v4.002603v429850.23(
2
 mcKE
γ

MeV9.984

43.49: Nuclei:




24
2

)2(4
2
HeYX
ZA
Z
zA
Z
Add the mass of
Z
electrons to each side and we find:



Y)(X)(
4
2
A
Z
A
Z
MMm
He),(
4
2
M
where now we have the mass of the neutral atoms. So as long as the mass of
the original neutral atom is greater than the sum of the neutral products masses, the
decay can happen.
43.50:
Denote the reaction as

.YX
1


 e
A
Z
A
Z
The mass defect is related to the change in the neutral atomic masses by
),(])1([][
YXeeYeX
mmmmZmZmm 
where
X
m
and
Y
m
are the masses as tabulated in, for instance, Table (43.2).
43.51:





)1(
1
YX
Z

A
Z
zA
Z
Adding (Z –1) electron to both sides yields






YX
1
A
Z
A
Z
So in terms of masses:





  
 
 
 
.2YX
YX
YX

e
1
e
1
e
e
1
mMM
mMmM
mMMm
A
Z
A
Z
A
Z
A
Z
A
Z
A
Z







So the decay will occur as long as the original neutral mass is greater than the sum

of the
neutral product mass and two electron masses.
43.52: Denote the reaction as
Y.X
1
A
Z
A
Z
e



The mass defect is related to the change in the neutral atomic masses by
),(])1([][
YXeYeeX
mmmZmmZmm 
where
X
m
and
Y
m
are the masses as tabulated in, for instance, Table (43.2).
43.53: a) Only the heavier one
Al)(
25
13
can decay into the lighter one
Mg).(

25
12
b)
XZA
A
Z
 1,0X)Mg()Al(
25
12
25
13
is a positron




decay
or
XZA




1,0MgXAl)(
25
12
A
Z
25
13

is an electron


electron capture
c) Using the nuclear masses, we calculate the mass defect for


decay:

u10495.3
)u00054858.0(2u985837.24u990429.24
)12)Mg(()13)Al((
3
ee
25
12e
25
13




mmMmMm

.MeV255.3)uVMe5.931()u10495.3()(
32


cmQ
For electron capture:


u104.592
u985837.24u990429.24Mg)()Al(
3
25
12
25
13



MMm
MeV.277.4u)VMe(931.5u)10592.4()(
32


cmQ
43.54: a)
MeV.41.5oru,1081.5
3
HePbPo
4
2
206
82
210
84


Qmmm

The energy of the
alpha particle is
)210206(
times this, or
MeV30.5
(see Example 43.5)
b)
,0u1035.5
3
HBiPo
1
1
209
83
210
84


mmm
so the decay is not possible.
c)
0,u1022.8
3
n
PoPo
209
84
210
84



mmm
so the decay is not possible.
d)
,
PoAt
210
84
210
85
mm

so the decay is not possible (see Problem (43.50)).
e)
,2
Po
e
Bi
210
84
210
83
mmm


so the decay is not possible (see Problem (43.51)).
43.55: Using Eq: (43.12):

H
24

11
2
BnH
11Na)( MMcENmZMM
A
Z
%.022.0100
23.990963
23.985823.990963
error%
u9858.23
uMeV931.5
MeV)(198.31
u).00866513(1u)511(1.00782Na)(
MeV.31.198(24)MeV)39(
24
))11(224(
)MeV69.23(
(24)
(10)(11)
MeV7100.0((24)MeV)80.17()24(MeV75.15But.13
24
11
34
2
31
32
B
2
Bn









M
EcEm
If the binding energy term is neglected,
u24.1987Na)(
24
11
M
and so the percentage error
would be
%.87.0100
990963
.
23
990963.231987.24


43.56: The

-particle will have
230
226
of the mass energy (see Example 45.5)


MeV.4.69oru10032.5)(
230
226
3
RaTh



mmm
43.57:



u197.966752u968225.197)Hg()Au(HgAu
198
80
198
79
198
80
198
79
MMm

MeV.372.1uMeV(931.5u)10
473.1()( wasavailableenergy totaltheAndu.101.473
3
23





QcmQ
The emitted photon has energy
MeV,412.0
so the emitted electron must have kinetic
energy equal to
MeV.0.960MeV0.412MeV372.1


43.58: (See Problem (43.51))
u.1003.12
3
e
BC
11
5
11
6

 mmm
Decay is energetically
possible.
43.59:



CN
13

6
13
7
As in Problem 43.51,

β
decay has a mass defect in terms of
neutral atoms of

u101.287
u)582(0.00054813.003355u005739.13
2C)()N(
3
e
13
6
13
7



 mMMm
Therefore the decay is possible because the initial mass is greater than the final mass.
43.60:
a) A least-squares fit to log of the activity vs. time gives a slope of
,hr5995.0λ
1

for a half-life of
hr.16.1

λ
2ln

b) The initial activity is
so,λ
0
N

.1020.1
s)3600hr1()hr(0.5995
Bq)1000.2(
8
1
4
0



N
c)
.1081.1
6λ
0

 t
eN
43.61: The activity
 )0(λ
)0(
so)(

λ
)(
but
)(
)(
N
dt
dN
tN
dt
tdN
dt
tdN
tA
.λ
00
AN 
Taking the derivative of
ttt
e
dt
dN
eN
dt
tdN
eNtN
λλ
0
λ
0

)0(
λ
)(
)(


.)(or
λ
0
t
eAtA


43.62: From Eq.43.17











21
T
2)ln(
0
λ

0
λ
0
but)(
t
tt
eNeNeNtN
 
.where
2
1
)(So.
21
0
)
2
1
(ln
0
2)(ln
0
21
21
T
t
nNtNeNeN
n
T
t
T

t

































Recall
.and,)(),ln(ln
ln
xeeexxa
xaaxaxa

43.63:
119
710
21
s1062.4
)ys10156.3()y1075.4(
693.0693.0




T
87
87
0
y)s10156.3()y106.4()s1062.4(
00
870087
0694.1
79119
8787

8787
NNeNN
eNNeNN
tt






But we also know that




85
85
87
8785
87
3856.0
)2783.01(
2783.0
2783.0
N
N
N
NN
N
.2920.0

))3856.0(0694.11(
)3856.0(0694.1
So.3856.0
8785
87
85
00
0
0




NN
N
N
So the original percentage of
850
87
85
(29%.isRb NN 
since it doesn’t decay.)
43.64: a)


Gy0.0682kg)0.70()eVJ10(1.602MeV)1077.4()1025.6(
19612
6.82 rad. b) (20)(6.82 rad)=136 rem
 Bq1017.1
)2ln(

u
λ(c)
9
21
TA
m
N
d)mCi.31.6
s,1034.5
Bq1017.1
1025.6
3
9
12



about an hour and a half. Note that this time is
so small in comparison with the half-life that the decrease in a
ctivity of the source may be
neglected.
43.65: a)
sdecays109.6Ci)sdecays10Ci(3.70106.2
6104


dt
dN
so in one
second there is an energy delivered of

s.J1062.9
)eVJ10(1.60eV)10(1.25s)00.1()s106.9(
2
1
2
1
7
19616











γ
Et
dt
dN
E
b) Absorbed dose
kg500.0
sJ106.9
7



m
E

rad.109.1
skgJ
rad
100skgJ109.1
46 











c) Equivalent dose
rem.101.3rad)109.1(7.0
44 

d)
days.17s105.1
srem101.3
rem200
6
4




43.66:
a) After 4.0 min = 240 s, the ratio of the number of nuclei is
.1242
2
2
2.122
1
9.26
1
)240(
9.26240
2.122240










b) After 15.0 min = 900 s, the ratio is
.1015.7
7

43.67:
t

e
N
N
λ
0
21.0



y13000
693
.
0
5730
)21.0ln(
λ
)21.0ln(

y
t
43.68:
The activity of the sample will have decreased by a factor of
;210097.1
60s)min1(min)counts5.8(
Ci)Bq10(3.70Ci)102.4(
06.206
106




this corresponds to 20.06 half-lifes, and the elapsed time is 40.1 h. Note the retention of
extra figures in the exponent to avoid roundoff error. To the given two figures the time is
40 h.
43.69:
For deuterium:
a)
 
J1061.7
(2)m)102.1(24
C)1060.1(
4
1
14
3115
0
2192
0








r
e
U

MeV0.48


b)
n
3
2
2
1
He)(H)(2 mMMm 
MeV270.3)uMeV(931.5u)1051.3()(
u103.51
u
1.008665
u
3.016029
u)
014102
.
2
(
2
32
3







cmE


J105.231
13

c) A mole of deuterium has
23
10022.6 
molecules, so the energy per mole is
J.103.150J)10231.5()10022.6(
111323


This is over a million times more than the
heat of combustion.
43.70: a)
u,1030.1
2
HNO
1
1
15
7
16
8

 mmm
so the proton separation energy is
u,1068.1b)MeV.1.12
2
n

OO
15
8
16
8

 mmm
so the neutron separation energy is
MeV.7.15
c) It takes less energy to remove a proton.
43.71: Mass of
K
40
atoms in 1.00 kg is
kg.102.52kg)102.1()101.2(
743 

Number of atoms
.10793.3
)ukg10u(1.66140
kg1052.2
18
27
7







N
.ydecays10054.2
1028.1
)10793.3()693.0(
λ
9
9
18




y
N
dt
dN
So in 50 years the energy absorbed is:

y)
decay10(2.054)y50(decay)MeV50.0(
9
E
J.108.22MeV1014.5
310 

So the absorbed dose is


rad)J100(J)1022.8(
3

82.0
rad and since the RBE = 1.0, the equivalent dose is 0.82 rem.
43.72: In terms of the number
N
of cesium atoms that decay in one week and the mass
kg,0.1

m
the equivalent dose is
.10535.1
J)10(2.283
Sv)(3.5kg)0.1(
soJ),10283.2(
MeV))(0.51(1.5)MeV)66.0()1((
)E(RBE)ERBE)((Sv5.3
13
13
13
ee








N
m
N

m
N
m
N
γγ
The number
0
N
of atoms present is related to
so,by
λ
0
t
NeNN 
.10536.1)10
535.1(sec1030.7
)yrsec10(3.156yr)07.30(
693.0
λ
13days)sec10(8.64days)7()sec1030.7(13
λ
0
110
7
4110








e
NeN
t
43.73: a)
M
m
m
vv
cm



v
Mm
M
Mm
m
vvv
m













M
m
vm
v
m




2
2
2
2
2
2
2
2
22
)(2
1
)(2
1
)(2
1
2
1
2

1
v
Mm
m
Mm
mM
Mm
M
v
Mm
Mm
v
Mm
mM
vMvmK
Mm























cm
KK
Mm
M
Kmv
Mm
M













2
2

1
b) For an endoenergetic reaction


0
cm
 QQK
at threshold. Putting this into part
(a) gives


Q
M
mM
KK
m
M
M
Q
thth





43.74:



K

mM
M
K


, where

K
is the energy that the

-particle would have if the
nucleus were infinitely massive. Then,



MMKMMM
OsOs


u94821.181MeV76.2
182
186
2
c
43.75:







n
94
38
140
54
235
92
SrXeU mMMMm 
     
MeV.185uMeV931.5u0.1983
u0.1983
u1.008665u93.915360u139.921636u230439.235
2






cmE
43.76:
a) A least-squares fit of the log of the activity vs.
time for the times later than 4.0 hr
gives a fit with correlation


6
1021



and decay constant of
1
hr361.0

, corresponding
to a half-life of 1.92 hr. Extrapolating th
is back to time 0 gives a contribution to the rate of
about 2500/s for this longer-lived species. A least-squares fit of the log of the activity vs.
time for times earlier than 2.0 hr gives a fit with correlation = 0.994, indicating the
presence of only two species.
b) By trial and error, the data is fit by a decay rate modeled by








hr361.0hr733.1
Bq2500Bq5000
tt
eeR


.
This would correspond to half-lives of 0.400 hr and 1.92 hr.
c) In this model, there are
7

1004.1 
of the shorter-lived species and
7
1049.2 
of the
longer-lived species.
d) After 5.0 hr, there would be
3
1080.1 
of the shorter-lived species and
6
1010.4 
of the longer-lived species.
43.77: (a) There are two processes occurring: the creation of
I
128
by the neutron
irradiation, and they decay of the newly produced
I
128
. So
KNK
dt
dN
whereλ
is the
rate of production by the neutron irradiation. Then
 





N t
dt
N
K
Nd
0 0
.
λ




 
 
 
.
λ
1
λlnλln
λλln
λ
0
t
N
eK
tN
tKNK
tNK








b) The activity of the sample is








sdecays105.11λ
6λt
eKtN



















t
e
min25
693.0
1
. So the activity is




t
e
02772.06
1sdecays105.1


, with t in minutes. So
the activity









dt
Nd
at various times is:
Bq;105.1
min)180(
Bq;103.1min)75(
Bq;101.1
min)50(
Bq;105.7min)25(
Bq;106.3
min)10(
Bq;101.4min)1(
66
65
54






















dt
tNd
t
dt
Nd
dt
tNd
t
dt
Nd
dt
tNd
t
dt
Nd
c)
 
atoms102.3
02772.0
)60()105.1(
λ

9
6
max



K
N
.
d) The maximum activity is at saturation, when the rate being produced equals
that decaying and so it equals
.sdecays105.1
6

43.78:
The activity of the original iron, after 1000 hours of operation, would be
Bq108306.12)CiBq103.7(Ci)104.9(
5)dhr24d45(hr)1000(106


.
The activity of the oil is 84 Bq, or
4
105886.4


of the total iron activity, and this must
be the fraction of the mass worn, or mass of
g1059.4
2


. The rate at which the piston
rings lost their mass is then
hrg1059.4
5

.