1. Chapter 40
Nuclear Physics
Properties of Nuclei
15 • Calculate the binding energy and the binding energy per nucleon from
the masses given in Table 40-1 for (a) 12
C, (b) 56
Fe, and (c) 238
U.
Picture the Problem To find the binding energy of a nucleus we add the mass of
its neutrons to the mass of its protons and then subtract the mass of the nucleus
and multiply by c2
. To convert to MeV we multiply this result by 931.5 MeV/u.
The binding energy per nucleon is the ratio of the binding energy to the mass
number of the nucleus.
(a) For 12
C, Z = 6 and N = 6. Add the mass of the neutrons to that of the
protons:
u940098.12u6651.0086u825007.1666 np =×+×=+ mm
Subtract the mass of 12
C from this result:
( ) u9400.098u12u940098.1266 Cnp 12 =−=−+ mmm
Multiply the mass difference by c2
and convert to MeV:
( ) ( ) MeV2.92
u
MeV/5.931
u940098.0
2
22
b =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=Δ=
c
ccmE
and the binding energy per nucleon is MeV68.7
12
MeV2.92b
==
A
E
(b) For 56
Fe, Z = 26 and N = 30. Add the mass of the neutrons to that of the
protons:
u400463.56u6651.00803u825007.1263026 np =×+×=+ mm
Subtract the mass of 56
Fe from this result:
( ) u4580.528u942934.55u400463.563026 Cnp 12 =−=−+ mmm
52

2. Nuclear Physics 53
Multiply the mass difference by c2
and convert to MeV:
( ) ( ) MeV492
u
MeV/5.931
u4580.528
2
22
b =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=Δ=
c
ccmE
and the binding energy per nucleon is MeV79.8
56
MeV492b
==
A
E
(c) For 238
U, Z = 92 and N = 146. Add the mass of the neutrons to that of
the protons:
u990984.239u6651.008146u825007.19214692 np =×+×=+ mm
Subtract the mass of 238
U from this result:
( ) u207934.1u783050.238u990984.23914692 Unp 238 =−=−+ mmm
Multiply the mass difference by c2
and convert to MeV:
( ) ( ) MeV1802
u
MeV/5.931
u207934.1
2
22
b =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=Δ=
c
ccmE
and the binding energy per nucleon is MeV57.7
238
MeV1802b
==
A
E
19 •• The neutron, when isolated from an atomic nucleus, decays into a
proton, an electron, and an antineutrino as follows: n → 1
H +e−
+ν . The thermal
energy of a neutron is of the order of kT, where k is the Boltzmann constant. (a) In
both joules and electron volts, calculate the energy of a thermal neutron at 25ºC.
(b) What is the speed of this thermal neutron? (c) A beam of monoenergetic
thermal neutrons is produced at 25ºC and has an intensity I. After traveling
1350 km, the beam has an intensity of 1
2
I. Using this information, estimate the
half-life of the neutron. Express your answer in minutes.
Picture the Problem The speed of the neutrons can be found from their thermal
energy. The time taken to reduce the intensity of the beam by one-half, from I to
I/2, is the half-life of the neutron. Because the beam is monoenergetic, the
neutrons all travel at the same speed.

3. Chapter 4054
(a) The thermal energy of the neutron
is:
( )( )
meV7.25
J101.602
eV1
J1011.4
J1011.4
K27325J/K10381.1
19
21
21
23
thermal
=
×
××=
×=
+×=
=
−
−
−
−
kTE
(b) Equate and the kinetic
energy of the neutron to obtain:
thermalE 2
n2
1
thermal vmE = ⇒
n
thermal2
m
E
v =
Substitute numerical values and
evaluate v:
( ) km/s22.2
kg101.673
J104.112
27
21
=
×
×
= −
−
v
(c) Relate the half-life, ,21t to the
speed of the neutrons in the
beam:
v
x
t =21
Substitute numerical values and
evaluate :21t
min1.10
s60
min1
s086
km/s2.22
km1350
1/2
=
×==t
21 •• In 1920, 12 years before the discovery of the neutron, Ernest
Rutherford argued that proton–electron pairs might exist in the confines of the
nucleus in order to explain the mass number, A, being greater than the nuclear
charge, Z. He also used this argument to account for the source of beta particles in
radioactive decay. Rutherford’s scattering experiments in 1910 showed that the
nucleus had a diameter of approximately 10 fm. Using this nuclear diameter, the
uncertainty principle, and that beta particles have an energy range of 0.02 MeV to
3.40 MeV, show why the hypothetical electrons cannot be confined to a region
occupied by the nucleus.
Picture the Problem The Heisenberg uncertainty principle relates the uncertainty
in position, Δx, to the uncertainty in momentum, Δp, by ΔxΔp ≥
1
2
h.
Solve the Heisenberg equation for
Δp: x
p
Δ
≈Δ
2
h

4. Nuclear Physics 55
Assuming that electrons are
contained within the nucleus, the
uncertainty in their momenta must
be:
( )
m/skg1025.5
m10102
sJ1005.1
21
15
34
⋅×=
×
⋅×
≈Δ
−
−
−
p
The kinetic energy of the electron is
given by:
pcK =
Substitute numerical values and evaluate K for a nuclear electron:
( )( )
MeV88.9
J101.602
eV1
J1058.1m/s10998.2m/skg1025.5 19
12821
=
×
××=×⋅×= −
−−
K
This result contradicts experimental observations that show that the energy of
electrons in unstable atoms is of the order of 1 to 1000 eV. Hence the premise on
which it is based (that electrons are contained within the nucleus) must be false.
Radioactivity
29 •• Plutonium is very toxic to the human body. Once it enters the body it
collects primarily in the bones, although it also can be found in other organs. Red
blood cells are synthesized within the marrow of the bones. The isotope 239
Pu is
an alpha emitter that has a half-life of 24 360 years. Because alpha particles are an
ionizing radiation, the blood-making ability of the marrow is, in time, destroyed
by the presence of 239
Pu. In addition, many kinds of cancers will also develop in
the surrounding tissues because of the ionizing effects of the alpha particles. (a) If
a person accidentally ingested 2.0 μg of 239
Pu and all of it is absorbed by the
bones of the person, how many alpha particles are produced per second within the
body of the person? (b) When, in years, will the activity be 1000 alpha particles
per second?
Picture the Problem Each 239
Pu nucleus emits an alpha particle whose activity,
A, depends on the decay constant of 239
Pu and on the number N of nuclei present
in the ingested 239
Pu. We can find the decay constant from the half-life and the
number of nuclei present from the mass ingested and the atomic mass of 239
Pu.
Finally, we can use the dependence of the activity on time to find the time at
which the activity be 1000 alpha particles per second.
(a) The activity of the nuclei present
in the ingested 239
Pu is given by:
NA λ= (1)

5. Chapter 4056
Find the constant for the decay of
239
Pu:
( )
( )( )
113
2/1
s1002.9
Ms/y56.31y24360
693.02ln
−−
×=
==
t
λ
Express the number of nuclei N
present in the mass mPu of 239
Pu
ingested:
Pu
Pu
A M
m
N
N
= ⇒
Pu
A
Pu
M
N
mN =
where MPu is the atomic mass of 239
Pu.
Substitute numerical values and
evaluate N: ( )
nuclei1004.5
g/mol239
nuclei/mol1002.6
g0.2
15
23
×=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
= μN
Substitute numerical values in
equation (1) and evaluate A:
( )( )
/s1055.4
1004.5s1002.9
3
15113
α
α
×=
××= −−
A
(b) The activity varies with time
according to:
t
eAA λ−
= 0 ⇒
λ−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
= 0
ln
A
A
t
Substitute numerical values and
evaluate t:
( )
y1032.5
y
Ms56.31
s1002.9
s/1055.4
s/1000
ln
4
113
3
×=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×−
⎟
⎠
⎞
⎜
⎝
⎛
×
=
−−
α
α
t
31 • The fissile material 239
Pu is an alpha emitter. Write the reaction that
describes 239
Pu undergoing alpha decay. Given that 239
Pu, 235
U, and an alpha
particle have respective masses of 239.052 156 u, 235.043 923 u, and
4.002 603 u, use the relations appearing in Problem 32 to calculate the kinetic
energies of the alpha particle and the recoiling daughter nucleus.
Picture the Problem We can write the equation of the decay process by using the
fact that the post-decay sum of the Z and A numbers must equal the pre-decay
values of the parent nucleus. The Q value in the equations from Problem 32 is
given by Q = −(Δm)c2
.
239
Pu undergoes alpha decay
according to:
Q++→ α4
2
235
92
239
94 UPu

6. Nuclear Physics 57
The Q value for the decay is given by:
( ) ( )[ ] ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−=
u
MeV/931.5 2
2
UPu
c
cmmmQ α
Substitute numerical values and evaluate Q:
( ) ( )[ ]
MeV24.5
u
931.5MeV/
u4.002603u235.043923u239.052156
2
2
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−=
c
cQ
From Problem 32, the kinetic energy
of the alpha particle is given by: Q
A
A
K ⎟
⎠
⎞
⎜
⎝
⎛ −
=
4
α
Substitute numerical values and
evaluate Kα: ( )
MeV5.15
MeV5.24
239
4239
=
⎟
⎠
⎞
⎜
⎝
⎛ −
=αK
From Problem 32, the kinetic energy
of the 235
U is given by: A
Q
K
4
U =
Substitute numerical values and
evaluate :U235K
( ) keV2.89
235
MeV24.54
U235 ==K
35 •• Radiation has been used for a long time in medical therapy to control
the development and growth of cancer cells. Cobalt-60, a gamma emitter that
emits photons that have energies of 1.17 MeV and 1.33 MeV, is used to irradiate
and destroy deep-rooted cancers. Small needles made of 60
Co of a specified
activity are encased in gold and used as body implants in tumors for time periods
that are related to tumor size, tumor cell reproductive rate, and the activity of the
needle. (a) A 1.00 μg sample of 60
Co, that has a half-life of 5.27 y, that is used to
irradiate a small internal tumor with gamma rays, is prepared in the cyclotron of a
medical center. Determine the activity of the sample in curies. (b) What will the
activity of the sample be 1.75 y from now?
Picture the Problem We can use NR λ=0 to find the initial activity of the
sample and to find the activity of the sample after 1.75 y.t
eRR λ−
= o

7. Chapter 4058
(a) The initial activity of the sample
is the product of the decay constant λ
for 60
Co and the number of atoms N
of 60
Co initially present in the
sample:
NR λ=0
Express N in terms of the mass m
of the sample, the molar mass M
of 60
Co, and Avogadro’s number
NA:
AN
M
m
N =
Substituting for N yields:
M
mN
R A
0
λ
=
The decay constant is given by: ( )
1/2
2ln
t
=λ
Substitute for λ to obtain: ( )
Mt
mN
R
1/2
A
0
2ln
=
Substitute numerical values and evaluate R0:
( )( )
mCi13.1
s103.7
Ci1
s10183.4
mol
g
60
y
Ms31.56
y27.5
mol
nuclei
106.022g101.002ln
110
17
236
0
=
×
××=
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
⎟
⎠
⎞
⎜
⎝
⎛
××
= −
−
−
R
(b) The activity varies with time
according to:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
== 5.27y
0.693
oo
t
t
eReRR λ
Evaluate R at t = 1.75 y:
( )
mCi898.0
mCi1.13 5.27y
1.75y0.693
=
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
−
eR
Nuclear Reactions
43 •• (a) Use the values 14.003 242 u and 14.003 074 u for the atomic
masses of and 14
, respectively, to calculate the Q value (in MeV) for the β-
decay reaction
14
C N
6
14
C → 7
14
N + e−
+ ve . (b) Explain why you should not add the mass
of the electron to that of atomic 14
for the calculation in (a).N

8. Nuclear Physics 59
Picture the Problem We can use ( ) 2
cmQ Δ−= to find the Q values for this
reaction.
(a) The masses of the atoms are: u242003.14C14 =m
u074003.14N14 =m
Calculate the increase in mass:
u168000.0
u242003.14u074003.14
if
−=
−=
−=Δ mmm
Calculate the Q value: ( )
( )
MeV156.0
u
MeV/
5.931u168000.0
2
2
2
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−=
Δ−=
c
c
cmQ
(b) The masses given are for atoms, not nuclei, so for nuclear masses the masses
are too large by the atomic number multiplied by the mass of an electron. For the
given nuclear reaction, the mass of the carbon atom is too large by and the
mass of the nitrogen atom is too large by . Subtracting from both sides
of the reaction equation leaves an extra electron mass on the right. Not including
the mass of the beta particle (electron) is mathematically equivalent to explicitly
subtracting from the right side of the equation.
e6m
e7m e6m
e1m
Fission and Fusion
45 • Assuming an average energy of 200 MeV per fission, calculate the
number of fissions per second needed for a 500-MW reactor.
Picture the Problem The power output of the reactor is the product of the
number of fissions per second and energy liberated per fission.
Express the required number N of
fissions per second in terms of the
power output P and the energy
released per fission Eper fission:
fissionperE
P
N =

9. Chapter 4060
Substitute numerical values and
evaluate N:
119
19
8
s1056.1
MeV200
J101.602
eV1
s
J
1000.5
MeV200
MW500
−
−
×=
×
××
=
=N
47 •• Consider the following fission reaction:
92 . The masses of the neutron,235
U +n → 42
95
Mo+ 57
139
La +2n + Q 235
U, 95
Mo, and
139
La are 1.008 665 u, 235.043 923 u, 94.905 842 u, and 138.906 348 u,
respectively. Calculate the Q value, in MeV, for this fission reaction.
Picture the Problem We can use Q = −(Δm)c2
, where Δm = mf − mi, to
calculate the Q value.
The Q value, in MeV, is given by:
( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Δ−=
u
MeV/5.931 2
2 c
cmQ
Calculate the change in mass Δm:
( )
( )
u068223.0
u6651.008u923235.043
u6651.0082u348138.906u842905.94
if
−=
+−
++=
−=Δ mmm
Substitute for Δm and evaluate Q:
( ) MeV208
u
MeV/931.5
u068223.0
2
2
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−=
c
cQ
General Problems
53 • The counting rate from a radioactive source is 6400 counts/s. The half-
life of the source is 10 s. Make a plot of the counting rate as a function of time for
times up to 1 min. What is the decay constant for this source?
Picture the Problem We can use the given information regarding the half-life of
the source to find its decay constant. We can then plot a graph of the counting rate
as a function of time.

10. Nuclear Physics 61
The decay constant is related to the
half-life of the source:
( ) ( )
1
1
21
s069.0
s0693.0
s10
2ln2ln
−
−
=
===
t
λ
The activity of the source is given
by:
( ) ( )tt
eeRR
1
s0693.0
0 Bq6400
−
−−
== λ
The following graph of ( ) ( )t
eR
1
s0693.0
Bq6400
−
−
= was plotted using a spreadsheet
program.
0
1000
2000
3000
4000
5000
6000
7000
0 10 20 30 40 50 60
t (s)
R(Bq)
57 •• Show that the 109
Ag nucleus is stable and does not undergo alpha
decay, . The mass of the47
109
Ag → 2
4
He + 45
105
Rh + Q 109
Ag nucleus is 108.904 756
u, and the products of the decay are 4.002 603 u and 104.905 250 u, respectively.
Picture the Problem We can show that 109
Ag is stable against alpha decay by
demonstrating that its Q value is negative.
The Q value, in MeV, for this reaction is:
( )[ ] ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−=
u
MeV/
5.931
2
2
AgRh
c
cmmmQ α
Substitute numerical values and evaluate Q:
( )[ ]
MeV88.2
u
MeV/
5.931u756904.108u250905.104u603002.4
2
2
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−=
c
cQ

11. Chapter 4062
Remarks: Alpha decay occurs spontaneously and the Q value will equal the
sum of the kinetic energies of the alpha particle and the recoiling daughter
nucleus, . Kinetic energy cannot be negative; hence, alpha decay
cannot occur unless the mass of the parent nucleus is greater than the sum of
the masses of the alpha particle and daughter nucleus, . Alpha
decay cannot take place unless the total rest mass decreases.
DKKQ α +=
DP mmm α +>
69 ••• Assume that a neutron decays into a proton and an electron without the
emission of a neutrino. The energy shared by the proton and the electron is then
0.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the
momentum of the proton must be equal and opposite the momentum of the
electron. This determines the ratio of the energies of the two particles, but because
the electron is relativistic, the exact calculation of these relative energies is
somewhat challenging. (a) Assume that the kinetic energy of the electron is
0.782 MeV and calculate the momentum p of the electron in units of MeV/c.
Hint: Use (Equation 39-27). (b) Using your result from Part
(a), calculate the kinetic energy p
( )22222
mccpE +=
2
/2mp of the proton. (c) Because the total energy
of the electron and the proton is 0.782 MeV, the calculation in Part (b) gives a
correction to the assumption that the energy of the electron is 0.782 MeV. What
percentage of 0.782 MeV is this correction?
Picture the Problem The momentum of the electron is related to its total energy
through and its total relativistic energy E is the sum of its kinetic
and rest energies.
2
0
222
EcpE +=
(a) Relate the total energy of the
electron to its momentum and rest
energy:
2
0
222
EcpE += (1)
The total relativistic energy E of the
electron is the sum of its kinetic
energy and its rest energy:
0EKE +=
Substitute for E in equation (1) to
obtain:
( ) 2
0
222
0 EcpEK +=+
Solving for p gives: ( )
c
EKK
p 02+
=

12. Nuclear Physics 63
Substitute numerical values and evaluate p:
( )( )
c
c
p MeV/19.1
MeV511.02MeV782.0MeV782.0
=
×+
=
(b) Because pp = −pe:
p
2
p
p
2m
p
K =
Substitute numerical values (see
Table 7-1 for the rest energy of a
proton) and evaluate Kp:
( )
( ) eV752
MeV/28.9382
MeV/188.1
2
2
p ==
c
c
K
(c) The percent correction is:
%0962.0
MeV0.782
eV752p
==
K
K
73 ••• Frequently, the daughter nucleus of a radioactive parent nucleus is
itself radioactive. Suppose the parent nucleus, designated by P, has a decay
constant λP; while the daughter nucleus, designated by D, has a decay constant λD.
The number of daughter nuclei ND are then given by the solution to the
differential equation
dND/dt = λPNP – λDND
where NP is the number of parent nuclei. (a) Justify this differential equation.
(b) Show that the solution for this equation is
N D t( )=
N P0λP
λD
− λP
e−λP t
− e−λDt
( )
where NP0 is the number of parent nuclei present at t = 0 when there are no
daughter nuclei. (c) Show that the expression for ND in Part (b) gives ND (t) > 0
whether λP > λD or λD > λP. (d) Make a plot of NP (t) and ND (t) as a function of
time when τD = 3τP, where τD and τP are the mean lifetimes of the daughter and
parent nuclei, respectively.
Picture the Problem We can differentiate ( ) ( )tt
ee
N
tN DP
PD
PP0
D
λλ
λλ
λ −−
−
−
= with
respect to t to show that it is the solution to the differential equation
dND/dt = λPNP − λDND.

13. Chapter 4064
(a) The rate of change of ND is the rate of generation of D nuclei minus the rate of
decay of D nuclei. The generation rate is equal to the decay rate of P nuclei,
which equals λPNP. The decay rate of D nuclei is λDND.
(b) We’re given that:
DDPP
D
NN
dt
dN
λλ −= (1)
( ) ( )tt
ee
N
tN DP
PD
PP0
D
λλ
λλ
λ −−
−
−
= (2)
t
eNN P
P0P
λ−
= (3)
Differentiate equation (2) with respect to t to obtain:
( )[ ] ( )[ ] [ ]tttt
ee
N
ee
dt
dN
tN
dt
d DPDP
DP
PD
PP0
PD
PP0
D
λλλλ
λλ
λλ
λ
λλ
λ −−−−
+−
−
=−
−
=
Substitute this derivative in equation (1) to obtain:
[ ] ( )⎥
⎦
⎤
⎢
⎣
⎡
−
−
−=+−
−
−−−−− ttttt
ee
N
eNee
N DPPDP
PD
PP0
DP0PDP
PD
PP0 λλλλλ
λλ
λ
λλλλ
λλ
λ
Multiply both sides by
PD
PD
λλ
λλ −
and simplify to obtain:
[ ] ( )
[ ]tt
tt
tttt
ttttt
ee
N
eNe
N
eNeNe
N
eN
eeNeNee
N
DP
DP
DPPP
DPPDP
DP
D
P0
P0
D
PP0
P0P0
D
PP0
P0
P0P0
D
PD
DP
D
P0
λλ
λλ
λλλλ
λλλλλ
λλ
λ
λ
λ
λ
λ
λ
λλ
λλ
λ
−−
−−
−−−−
−−−−−
+−=
+−=
+−−=
−−
−
=+−
which is an identity and confirms that equation (2) is the solution to equation (1).
(c) If λP > λD the denominator and expression in parentheses are both negative for
t > 0. If λP < λD the denominator and expression in parentheses are both positive
for t > 0.

14. Nuclear Physics 65
(d) The following graph was plotted using a spreadsheet program.
0.00
0 5 10 15 20
Number of parent (P) nuclei
Number of daughter ( D) nuclei
τ τ τ τp p p p
NPO
N
N
N
NPO
PO
PO
PO
0.20
0.40
0.60
0.80
1.00
Time