CHAPTER 24 NUCLEAR REACTIONS AND
THEIR APPLICATIONS
FOLLOW–UP PROBLEMS
24.1A
Plan: Write a skeleton equation that shows fluorine-20 undergoing beta decay. Conserve mass and atomic number
by ensuring the superscripts and subscripts equal one another on both sides of the equation. Determine the identity
of the daughter nuclide by using the periodic table.
Solution:
Fluorine-20 has Z = 9. Its symbol is 209F. When it undergoes beta decay, a beta particle, 10β , and a daughter
nuclide, AZX , are produced.
20
A
0
9F  ZX + 1β
To conserve atomic number, Z must equal 10. Element is neon.
To conserve mass number, A must equal 20.
The identity of Z X is 20
10Ne.
A
The balanced equation is: 209F  20
10Ne +
24.1B
0
1β
A
0
Plan: Write a skeleton equation that shows an unknown nuclide, Z X , undergoing beta decay, 1β , to form
133
cesium–133, 55 Cs . Conserve mass and atomic number by ensuring the superscripts and subscripts equal one
another on both sides of the equation. Determine the identity of X by using the periodic table to identify the
element with atomic number equal to Z.
Solution:
The unknown nuclide yields cesium-133 and a  particle:
A
133
0
Z X  55 Cs + 1β
To conserve atomic number, Z must equal 54. Element is xenon.
To conserve mass number, A must equal 133.
A
133
The identity of Z X is 54 Xe .
133
133
0
The balanced equation is: 54 Xe  55 Cs + -1 β
Check: A = 133 = 133 + 0 and Z = 54 = 55 + (–1).
24.2A
Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in
the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number
greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the
ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons
and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and
or neutrons are related to stability whereas odd numbers are related to instability.
Solution:
a) 105B appears stable because its N/Z ratio (10  5)/5 = 1.00 is in the band of stability.
b) 58
23V appears unstable/radioactive because its N/Z ratio (58  23)/23 = 1.52 is too high and is above the band of
stability. Additionally, this nuclide has both odd N(35) and Z(23).
24.2B
Plan: Nuclear stability is found in nuclides with an N/Z ratio that falls within the band of stability. Nuclides with
an even N and Z, especially those nuclides that have magic numbers, are exceptionally stable. Examine the two
nuclides to see which of these criteria can explain the difference in stability.
Solution:
Phosphorus-31 has 16 neutrons and 15 protons, with an N/Z ratio of 1.07. Phosphorus-30 has 15 neutrons and 15
protons, with an N/Z ratio of 1.00. 31P has an even N while 30P has both an odd Z and an odd N. 31P has a slightly
higher N/Z ratio that is closer to the band of stability.
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24-1
24.3A
Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the
band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios
are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle
emission.
Solution:
a) Iron-61 has an N/Z ratio of (61 – 26)/26 = 1.35, which is too high for this region of the band. Iron-61 will
undergo – decay.
61
61
0
26 Fe  27 Co + 1β
new N/Z = (61 – 27)/27 = 1.26
Additionally, iron has an atomic mass of 55.85 amu. The A value of 61 is higher, suggesting beta decay.
b) Americium–241 has Z > 83, so it undergoes  decay.
241
237
4
95 Am  93 Np + 2 He
24.3B
Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the
band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios
are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle
emission.
Solution:
a) Titanium-40 has an N/Z ratio of (40 – 22)/22 = 0.81, which is too low for this region of the band. Titanium-40
will undergo positron decay or electron capture. Additionally, titanium’s atomic mass is 47.87 amu, which is
much higher than the A value of 40, also suggesting positron decay or electron capture.
b) Cobalt-65 has an N/Z ratio of (65 – 27)/27 = 1.40, which is too high for this region of the band. Cobalt-65 will
undergo beta decay. Additionally, cobalt’s atomic mass is 58.93 amu, which is much lower than the A value of
65, also suggesting beta decay.
24.4A
Plan: Specific activity of a radioactive sample is its decay rate per gram. Find the mass of the sample. Calculate
the specific activity by dividing the number of particles emitted per second (disintegrations per second = dps) by
the mass of the sample. Convert disintegrations per second to Ci by using the conversion factor between the two
units: 1 Ci = 3.70x1010 dps. Convert Ci to Bq by using the conversion factor between the two units:
1 Ci = 3.70x1010 Bq.
Solution:
a) Mass (g) of As = (3.4x10–8 mol As)
Specific activity (Ci/g) =
2.6x10–6 g
1.6x106 Ci
3.70x1010 dps
g
Ci
3.70x1010 Bq
= 1.5904x106 = 1.6x106 Ci/g
= 5.9200x1016 = 5.9x1016 Bq/g
Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation
A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number.
The decay rate is 9.97x1012 beta particles/h or more simply, 9.97x1012 nuclei/h.
Solution:
Decay rate = A = kN
k=
24.5A
= 2.5840x10–6 = 2.6x10–6 g As
1 mol As
1.53x1011 dps
1 Ci
b) Specific activity (Bq/g) =
24.4B
76 g As
A
N
=
9.97x1012 nuclei/h
6.50x10–2 mol (6.022x1023 nuclei/h)
= 2.5471x10–10 = 2.55x10–10 h–1
Plan: Use the half-life of 24Na to find k. Substitute the value of k, initial activity (A0), and time of decay (4 days)
into the integrated first-order rate equation to solve for activity at a later time (At).
Solution:
ln 2
ln 2
k=
=
= 0.0462098 h–1
t1/2
15 h
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24-2
ln At = ln A0 – kt
ln At = ln (2.5x109) – (0.0462098 h–1)(4.0 days)(24 h/day)
ln At = 17.203416
At = 2.960388x107 = 3.0x107 d/s
24.5B
Plan: Use the half-life of 59Fe to find k. Substitute the value of k and time of decay (17 days) into the integrated
first-order rate equation. Assume that A0 is 1.0 (100% of the original sample) and solve for At, the fraction of the
sample remaining after 17 days. Subtract the fraction remaining from the original sample (1.0) to calculate the
fraction that has decayed.
Solution:
ln 2
ln 2
k=
=
= 1.5576x10–2 days–1
44.5 days
t1/2
ln At = ln A0 – kt
ln At = ln (1.0) – (1.5576x10–2 days–1)(17 days)
ln At = –0.2648
At = 0.7674 = fraction of iron-59 remaining
Fraction of iron-59 decayed = 1.0 – 0.7674 = 0.2326 = 0.23 = fraction of iron-59 that has decayed
24.6A
Plan: The wood from the case came from a living organism, so A0 equals 15.3 d/min•g. Substitute the current
activity of the case (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of
carbon (5730 yr).
Solution:
ln 2
ln 2
k=
=
= 1.209680943x10–4 yr–1
5730 yr
t1/2
ln At = ln A0 – kt
ln [9.41 d/min•g] = ln [15.3 d/min•g] – (1.209680943x10–4 yr–1)(t)
–0.486079875 = – (1.209681x10–4 yr–1)(t)
t = 4018.2 = 4.02x103 years
24.6B
Plan: The woolen tunic came from a living organism, so A0 equals 15.3 d/min•g. Substitute the current activity of
the tunic (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of carbon
(5730 yr).
Solution:
ln 2
ln 2
k=
=
= 1.209680943x10–4 yr–1
5730 yr
t1/2
ln At = ln A0 – kt
ln [12.87 d/min•g] = ln [15.3 d/min•g] – (1.209680943x10–4 yr–1)(t)
–0.172953806 = – (1.209681x10–4 yr–1)(t)
t = 1429.7473 = 1430 years
24.7A
Plan: Nickel-58 has 28 protons and 30 neutrons in its nucleus. Calculate the change in mass (m) in one 58Ni
atom, convert to MeV and divide by 58 to obtain binding energy/nucleon.
Solution:
m = [(28 x mass H atom) + (30 x mass neutron)] – mass 58Ni atom
m = [(28 x 1.007825 amu) + (30 x 1.008665)] – 57.935346 amu
m = 0.543704 amu
931.5 MeV
Binding energy (MeV) = (0.543704 amu)
Binding Energy/nucleon =
56
506.460276 MeV
58 nucleons
1 amu
= 506.460276 MeV
= 8.7321 = 8.732 MeV/nucleon
The BE/nucleon of Fe is 8.790 MeV/nucleon. The energy per nucleon holding the 58Ni nucleus together is less
than that for 56Fe (8.732 < 8.790), so 58Ni is less stable than 56Fe.
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24-3
24.7B
Plan: Uranium-235 has 92 protons and 143 neutrons in its nucleus. Calculate the change in mass (m) in one 235U
atom, convert to MeV and divide by 235 to obtain binding energy/nucleon.
Solution:
m = [(92 x mass H atom) + (143 x mass neutron)] – mass 235U atom
m = [(92 x 1.007825 amu) + (143 x 1.008665)] – 235.043924 amu
m = 1.915071 amu
 931.5 MeV 
Binding energy (MeV) = 1.915071 amu  
 = 1783.888637 MeV
 1 amu 
1783.888637 MeV
= 7.591015 = 7.591 MeV/nucleon
235 nucleons
12
The BE/nucleon of C is 7.680 MeV/nucleon. The energy per nucleon holding the 235U nucleus together is less than that
for 12C (7.591 < 7.680), so 235U is less stable than 12C.
Binding Energy/nucleon =
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B24.1
In the s-process, a nucleus captures a neutron sometime over a long period of time. Then the nucleus emits a beta
particle to form another element. The stable isotopes of most heavy elements up to 209Bi form by the s-process.
The r-process very quickly forms less stable isotopes and those with A greater than 230 by multiple neutron
captures, followed by multiple beta decays.
B24.2
Plan: Find the change in mass of the reaction by subtracting the mass of the products from the
mass of the reactants and convert the change in mass to energy with the conversion factor
between amu and MeV. Convert the energy per atom to energy per mole by multiplying by
Avogadro’s number.
Solution:
m = mass of reactants – mass of products
= [(4)(1.007825)] – [4.00260 + (2)(5.48580x10–4)]
= 4.031300 – 4.003697 = 0.02760 amu /4He atom = 0.02760284 g/mol 4He
 0.02760284 amu 4 He   931.5 MeV 
Energy (MeV/atom) = 
 
 = 25.7120 = 25.71 MeV/atom

1 atom

  1 amu 
Convert atoms to moles using Avogadro’s number.
23
 25.7120 MeV   6.022x10 atoms 
25
25
Energy = 

 = 1.54838x10 = 1.548x10 MeV/mol

atom
1
mol



B24.3
The simultaneous fusion of three nuclei is a termolecular process. Termolecular processes have a very low
probability of occurring. The bimolecular fusion of 8Be with 4He is more likely.
B24.4
Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the
right side must be equal.
Solution:
210
210
0
83 Bi  84 Po + 1
210
206
4
84 Po  82 Pb + 2 
206
1
209
82 Pb + 3 0 n  82 Pb
209
210
0
82 Pb  83 Bi + 1
210
84 Po is Nuclide A
206
82 Pb is Nuclide B
209
82 Pb is Nuclide C
210
83 Bi is Nuclide D
END–OF–CHAPTER PROBLEMS
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24-4
24.1
a) Chemical reactions are accompanied by relatively small changes in energy while nuclear reactions are
accompanied by relatively large changes in energy.
b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction.
c) Both chemical and nuclear reaction rates increase with higher reactant concentrations.
d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield
increases in a chemical reaction. The presence of more radioactive reagent results in more decay product, so a
higher reactant concentration increases the yield in a nuclear reaction.
24.2
a) The percentage of sulfur atoms that are sulfur-32 is 95.02%, the same as the relative abundance of 32S.
b) The atomic mass is larger than the isotopic mass of 32S. Sulfur-32 is the lightest isotope, as stated in the
problem, so the other 5% of sulfur atoms are heavier than 31.972070 amu. The average mass of all the sulfur
atoms will therefore be greater than the mass of a sulfur-32 atom.
24.3
a) She found that the intensity of emitted radiation is directly proportional to the concentration of the element in
the various samples, not to the nature of the compound in which the element occurs.
b) She found that certain uranium minerals were more radioactive than pure uranium, which implied that they
contained traces of one or more as yet unknown, highly radioactive elements. Pitchblende is the principal ore of
uranium.
24.4
Plan: Radioactive decay that produces a different element requires a change in atomic number (Z, number of
protons).
Solution:
A
ZX
A = mass number (protons + neutrons)
Z = number of protons (positive charge)
X = symbol for the particle
N = A – Z (number of neutrons)
a) Alpha decay produces an atom of a different element, i.e., a daughter with two less protons and two less
neutrons.
A
ZX
 AZ42Y + 42 He
A
ZX
 Z A1Y + 01
2 fewer protons, 2 fewer neutrons
b) Beta decay produces an atom of a different element, i.e., a daughter with one more proton and one less neutron.
A neutron is converted to a proton and  particle in this type of decay.
1 more proton, 1 less neutron
c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged.
A
Z X*
 ZA X + 00 
( ZA X * = energy rich state), no change in number of protons or neutrons.
d) Positron emission produces an atom of a different element, i.e., a daughter with one less proton and one more
neutron. A proton is converted into a neutron and positron in this type of decay.
A
ZX
 Z A1Y + 01
1 less proton, 1 more neutron
e) Electron capture produces an atom of a different element, i.e., a daughter with one less proton and one more
neutron. The net result of electron capture is the same as positron emission, but the two processes are different.
A
0
A
Z X + 1e  Z 1Y
1 less proton, 1 more neutron
A different element is produced in all cases except (c).
24.5
The key factor that determines the stability of a nuclide is the ratio of the number of neutrons to the number of
protons, the N/Z ratio. If the N/Z ratio is either too high or not high enough, the nuclide is unstable and decays.
3
2 He
2
2 He
24.6
N/Z = 1/2
N/Z = 0/2, thus it is more unstable.
A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert
protons to neutrons. The conversion of neutrons to protons occurs by beta decay:
1
1
0
0 n  1 p + 1
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24-5
The conversion of protons to neutrons occurs by either positron decay:
1
1
0
1 p  0 n + 1
or electron capture:
1
0
1
1 p + 1e  0 n
Neutron-rich nuclides, with a high N/Z, undergo  decay. Neutron-poor nuclides, with a low N/Z, undergo
positron decay or electron capture.
24.7
Both positron emission and electron capture increase the number of neutrons and decrease the number of protons.
The products of both processes are the same. Positron emission is more common than electron capture among
lighter nuclei; electron capture becomes increasingly common as nuclear charge increases. For Z < 20, +
emission is more common; for Z > 80, electron capture is more common.
24.8
Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the
right side must be equal.
Solution:
234
4
230
92 U  2 He + 90Th
232
0
232
b) 93 Np + 1e  92 U
12
0
12
c) 7 N  1 + 6 C
a)
24.9
26
0
Mass: 234 = 4 + 230;
Charge: 92 = 2 + 90
Mass: 232 + 0 = 232;
Charge: 93 + (–1) = 92
Mass: 12 = 0 + 12;
Charge: 7 = 1 + 6
26
a) 11 Na  1 + 12 Mg
223
0
223
87 Fr  1 + 88 Ra
212
4
208
c) 83 Bi  2  + 81Tl
b)
24.10
Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the
right side must be equal.
Solution:
a) The process converts a neutron to a proton, so the mass number is the same, but the atomic number increases by
one.
27
0
27
12 Mg  1 + 13 Al
Mass: 27 = 0 + 27;
Charge: 12 = –1 + 13
b) Positron emission decreases atomic number by one, but not mass number.
23
0
23
12 Mg  1 + 11 Na
Mass: 23 = 0 + 23;
Charge: 12 = 1 + 11
c) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant,
but atomic number decreases by one.
103
0
103
46 Pd + 1e  45 Rh
24.11
32
0
Mass: 103 + 0 = 103;
Charge: 46 + (–1) = 45
32
a) 14 Si  1 + 15 P
218
4
214
84 Po  2  + 82 Pb
110
0
110
c) 49 In + 1e  48 Cd
b)
24.12
Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the
right side must be equal.
Solution:
a) In other words, an unknown nuclide decays to give Ti-48 and a positron.
48
48
0
23V  22Ti + 1
Mass: 48 = 48 + 0;
Charge: 23 = 22 + 1
b) In other words, an unknown nuclide captures an electron to form Ag-107.
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24-6
107
0
107
48 Cd + 1e  47 Ag
Mass: 107 + 0 = 107;
Charge: 48 + (–1) = 47
c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle.
210
206
4
86 Rn  84 Po + 2 He
Mass: 210 = 206 + 4;
Charge: 86 = 84 + 2
24.13
241
241
0
94 Pu  95 Am + 1
228
228
0
b) 88 Ra  89 Ac + 1
207
203
4
c) 85 At  83 Bi + 2 
24.14
Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and
the right side must be equal.
Solution:
a) In other words, an unknown nuclide captures an electron to form Ir-186.
a)
186
78 Pt
0
186
+ 1e  77 Ir
Mass: 186 + 0 = 186;
Charge: 78 + (–1) = 77
b) In other words, an unknown nuclide decays to give Fr-221 and an alpha particle.
225
221
4
89 Ac  87 Fr + 2 He
Mass: 225 = 221 + 4;
Charge: 89 = 87 + 2
c) In other words, an unknown nuclide decays to give I-129 and a beta particle.
129
129
0
52Te  53 I + 1
24.15
52
52
Mass: 129 = 129 + 0;
Charge: 52 = 53 + (–1)
0
a) 26 Fe  25 Mn + 1
219
215
4
86 Rn  84 Po + 2 
81
0
81
c) 37 Rb + 1e  36 Kr
b)
24.16
Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in
the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number
greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the
ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons
and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and
or neutrons are related to stability whereas odd numbers are related to instability.
Solution:
20
a) 8 O appears stable because its Z (8) value is a magic number, but its N/Z ratio (20  8)/8 = 1.50 is too high and
this nuclide is above the band of stability;
20
8 O is unstable.
59
b) 27 Co might look unstable because its Z value is an odd number, but its N/Z ratio (59  27)/27 = 1.19 is in the
59
band of stability, so 27 Co appears stable.
9
c) 3 Li appears unstable because its N/Z ratio (9  3)/3 = 2.00 is too high and is above the band of stability.
24.17
24.18
146
60 Nd
114
b) 48 Cd
88
c) 42 Mo
a)
N/Z = 86/60 = 1.4
Stable, N/Z ok
N/Z = 66/48 = 1.4
Stable, N/Z ok
N/Z = 46/42 = 1.1
Unstable, N/Z too small for this region of the band
Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in
the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number
greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the
ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons
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24-7
and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and
or neutrons are related to stability whereas odd numbers are related to instability.
Solution:
a) For the element iodine Z = 53. For iodine-127, N = 127  53 = 74. The N/Z ratio for 127I is 74/53 = 1.4. Of the
examples of stable nuclides given in the book, 107Ag has the closest atomic number to iodine. The N/Z ratio for
107
Ag is 1.3. Thus, it is likely that iodine with six additional protons is stable with an N/Z ratio of 1.4.
b) Tin is element number 50 (Z = 50). The N/Z ratio for 106Sn is (106  50)/50 = 1.1. The nuclide 106Sn is unstable
with an N/Z ratio that is too low.
c) For 68As, Z = 33 and N = 68  33 = 35 and N/Z = 1.1. The ratio is within the range of stability, but the nuclide is
most likely unstable because there is an odd number of both protons and neutrons.
24.19
24.20
48
a) 19 K N/Z = 29/19 = 1.5
Unstable, N/Z too large for this region of the band
79
b) 35 Br
33
c) 18 Ar
N/Z = 44/35 = 1.3
Stable, N/Z okay
N/Z = 14/18 = 0.78
Unstable, N/Z too small
Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while
a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo 
decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, +
emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of
decay for a heavy, unstable nucleus (Z > 83).
Solution:
238
92 U: Nuclides with Z > 83 decay through  decay.
48
b) The N/Z ratio for 24 Cr is (48 – 24)/24 = 1.00. This number is below the band of stability because N is too
a)
low and Z is too high. To become more stable, the nucleus decays by converting a proton to a neutron, which is
positron decay. Alternatively, a nucleus can capture an electron and convert a proton into a neutron through
electron capture.
50
c) The N/Z ratio for 25 Mn is (50 – 25)/25 = 1.00. This number is below the band of stability, so the nuclide
undergoes positron decay or electron capture.
24.21
a) 47 Ag
111
beta decay N/Z = 1.4 which is too high
41
b) 17 Cl
110
c) 44 Ru
beta decay N/Z = 1.4 which is too high
beta decay N/Z = 1.5 which is too high
24.22
Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while
a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo 
decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, +
emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of
decay for a heavy, unstable nucleus (Z > 83).
Solution:
a) For carbon-15, N/Z = 9/6 = 1.5, so the nuclide is neutron-rich. To decrease the number of neutrons and increase
the number of protons, carbon-15 decays by beta decay.
b) The N/Z ratio for 120Xe is 66/54 = 1.2. Around atomic number 50, the ratio for stable nuclides is larger than 1.2,
so 120Xe is proton-rich. To decrease the number of protons and increase the number of neutrons, the xenon-120
nucleus either undergoes positron emission or electron capture.
c) Thorium-224 has an N/Z ratio of 134/90 = 1.5. All nuclides of elements above atomic number 83 are unstable
and decay to decrease the number of both protons and neutrons. Alpha decay by thorium-224 is the most likely
mode of decay.
24.23
a)
106
49 In
positron decay or electron capture N/Z = 1.2
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24-8
141
24.24
b) 63 Eu
positron decay or electron capture N/Z = 1.2
241
c) 95 Am
alpha decay
N/Z = 1.5
Plan: Stability results from a favorable N/Z ratio, even numbers of N and/or Z, and the occurrence of magic
numbers.
Solution:
52
The N/Z ratio of 24 Cr is (52  24)/24 = 1.17, which is within the band of stability. The fact that Z is even does not
52
account for the variation in stability because all isotopes of chromium have the same Z. However, 24 Cr has 28
neutrons, so N is both an even number and a magic number for this isotope only.
40
20 Ca
24.25
N/Z = 20/20 = 1.0
It lies in the band of stability, and N and Z are both even and magic.
24.26
237
4
233
93 Np  2  + 91 Pa
233
0
233
91 Pa  1 + 92 U
233
4
229
92 U  2  + 90Th
229
4
225
90Th  2  + 88 Ra
24.27
Alpha emission produces helium ions which readily pick up electrons to form stable helium atoms.
24.28
The equation for the nuclear reaction is 92 U  82 Pb + __ 1 + __ 2 He
To determine the coefficients, notice that the beta particles will not impact the mass number. Subtracting the mass
number for lead from the mass number for uranium will give the total mass number for the alpha particles
released, 235  207 = 28. Each alpha particle is a helium nucleus with mass number 4. The number of helium
atoms is determined by dividing the total mass number change by 4, 28/4 = 7 helium atoms or seven alpha
particles. The equation is now
235
207
0
4
235
207
0
4
92 U  82 Pb + __ 1 + 7 2 He
To find the number of beta particles released, examine the difference in number of protons (atomic number)
between the reactant and products. Uranium, the reactant, has 92 protons. The atomic number in the products, lead
atom and 7 helium nuclei, total 96. To balance the atomic numbers, four electrons (beta particles) must be emitted
to give the total atomic number for the products as 96  4 = 92, the same as the reactant. In summary, seven alpha
particles and four beta particles are emitted in the decay of uranium-235 to lead-207.
235
207
0
4
92 U  82 Pb + 1 + 7 2 He
24.29
a) In a scintillation counter, radioactive emissions are detected by their ability to excite atoms and cause them to
emit light.
b) In a Geiger-Müller counter, radioactive emissions produce ionization of a gas that conducts a current to a
recording device.
24.30
Since the decay rate depends only on the number of radioactive nuclei, radioactive decay is a first-order process.
24.31
No, it is not valid to conclude that t1/2 equals 1 min because the number of nuclei is so small (six nuclei). Decay
rate is an average rate and is only meaningful when the sample is macroscopic and contains a large number of
nuclei, as in the second case. Because the second sample contains 6x1012 nuclei, the conclusion that
t1/2 = 1 min is valid.
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24-9
24.32
High-energy neutrons in cosmic rays enter the upper atmosphere and keep the amount of 14C nearly constant
through bombardment of ordinary 14N atoms. This 14 C is absorbed by living organisms, so its proportion stays
relatively constant there also.
14
1
14
1
7 N + 0 n  6 C + 1H
24.33
Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing
the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert
disintegrations per second to Ci by using the conversion factor between the two units.
Solution:
1 Ci = 3.70x1010 dps
 1.56x106 dps   1 mg  

1 Ci
= 2.55528x10–2 = 2.56x10–2 Ci/g
Specific activity (Ci/g) = 
 1.65 mg   103 g   3.70x1010 dps 




24.34
24.35
24.36
24.37
  4.13x108 d   1 h  
 
  3600 s   
h

1 Ci



–6
–6
Specific activity (Ci/g) = 

 = 1.1925x10 = 1.2x10 Ci/g

10
2.6 g
3.70x10
dps







Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing
the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert
disintegrations per second to Bq by using the conversion factor between the two units.
Solution:
A becquerel is a disintegration per second (dps).
  7.4x104 d   1 min  
 
 

  min   60 s    1 Bq 
8
8
Specific activity (Bq/g) = 

 = 1.43745x10 = 1.4x10 Bq/g
 106 g    1 dps 
 8.58 g 

 1 g  


 

  3.77x107 d   1 min  
 
 

min

 60 s    1 Bq 

Specific activity (Bq/g) = 
  1 dps  = 587.2274 = 587 Bq/g
 103 g 



1.07 kg 



1
kg




Plan: The decay constant is the rate constant for the first-order reaction.
Solution:
N
Decay rate = 
= kN
t
1 atom

= k(1x1012 atom)
day
k = 1x1012 d1
24.38
N
= kN
t
 (2.8x1012 atom/1.0 yr) = k(1 atom)
Decay rate = 
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24-10
k = 2.8x1012 yr1
24.39
Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation
A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number.
The decay rate is 1.39x105 atoms/yr or more simply, 1.39x105 yr–1 (the disintegrations are assumed).
Solution:
N
Decay rate = A = –
= kN
t

 1.00x1012 mol   6.022x1023 atoms 
1.39x105 atoms
= k
 


1.00 yr
1 mol



1.39x105 atom/yr = k(6.022x1011 atom)
k = (1.39x105 atom/yr)/6.022x1011 atom
k = 2.30820x10–7 = 2.31x10–7 yr–1
24.40
24.41
N
= kN
t
– (–1.07x1015 atom/1.00 h) = k[(6.40x10–9 mol)(6.022x1023 atom/mol)]
(1.07x1015 atom/1.00 h) = k (3.85408x1015 atom)
k = [(1.07x1015 atom/1.00 h)]/(3.85408x1015 atom)
k = 0.2776 = 0.278 h–1
Decay rate = A = –
Plan: Radioactive decay is a first-order process, so the integrated rate law is ln Nt = ln N0 – kt
First find the value of k from the half-life and use the integrated rate law to find Nt. The time unit in
the time and the k value must agree.
Solution:
t1/2 = 1.01 yr
t = 3.75x103 h
ln 2
ln 2
t1/2 =
or k =
t1/2
k
ln 2
= 0.686284 yr–1
1.01 yr
ln Nt = ln N0 – kt
k=


 1 d   1 yr 
ln Nt = ln [2.00 mg] – (0.686284 yr–1) 3.75x103 h 


 24 h   365 d 
ln Nt = 0.399361
Nt = e0.399361
Nt = 1.49087 = 1.49 mg
24.42
t1/2 = 1.60x103 yr
t=?h
ln 2
ln 2
k=
=
= = 0.000433217 yr–1
t1/2 1.60x103 yr
ln [0.185 g] = ln [2.50 g] – (0.000433217 yr–1)(t)
t = 6010.129 yr
 365 d   24 h 
7
7
t =  6010.129 yr  

 = 5.264873x10 = 5.26x10 h
1
yr
1
d



24.43
Plan: Lead-206 is a stable daughter of 238U. Since all of the 206Pb came from 238U, the starting amount of 238U was
(270 mol + 110 mol) = 380 mol = N0. The amount of 238U at time t (current) is 270 mol = Nt. Find k from the
first-order rate expression for half-life, and then substitute the values into the integrated rate law and solve
for t.
Solution:
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24-11
ln 2
ln 2
or k =
t1/2
k
ln 2
k=
= 1.540327x10–10 yr–1
4.5x109 yr
t1/2 =
ln Nt = ln N0 – kt
ln
or
ln
N0
= kt
Nt
380 mol
= (1.540327x10–10 yr–1)(t)
270 mol
0.3417492937 = (1.540327x10–10 yr–1)(t)
t = 2.21868x109 = 2.2x109 yr
24.44
The ratio (0.735) equals Nt/N0 so N0/Nt = 1.360544218
ln 2
ln 2
k=
=
= = 1.2096809x10–4 yr–1
t1/2
5730 yr
N0
= kt
Nt
ln 1.360544218 = (1.2096809x10–4 yr–1)(t)
0.30788478 = (1.2096809x10–4 yr–1)(t)
t = 2.54517x103 = 2.54x103 yr
ln
24.45
Plan: The specific activity of the potassium-40 is the decay rate per mL of milk. Use the conversion factor
1 Ci = 3.70x1010 disintegrations per second (dps) to find the disintegrations per mL per s; convert the time
unit to min and change the volume to 8 oz.
Solution:
 6 x1011 mCi   103 Ci   3.70x1010 dps   60 s   1000 mL   1 qt   1 cup 
Activity = 
 
 
 



 8 oz 

mL
1 Ci

  1 mCi  
  1 min   1.057 qt   4 cups   8 oz 
= 31.50426 = 30 dpm
24.46
Plutonium-239 (t1/2 = 2.41x104 yr)
Time = 7(t1/2) = 7(2.41x104 yr) = 1.6870x105 = 1.69x105 yr
24.47
Plan: Both Nt and N0 are given: the number of nuclei present currently, Nt, is found from the moles of 232Th. Each
fission track represents one nucleus that disintegrated, so the number of nuclei disintegrated is added to the
number of nuclei currently present to determine the initial number of nuclei, N0. The rate constant, k, is calculated
from the half-life. All values are substituted into the first-order decay equation to find t.
Solution:
ln 2
ln 2
t1/2 =
or k =
t1/2
k
ln 2
k=
=4.95105129x10–11 yr–1
10
1.4x10 yr
 6.022x1023 Th atoms 
9
Nt = 3.1x10 15 mol Th 
 = 1.86682x10 atoms Th

1 mol Th


9
4
N0 =1.86682x10 atoms + 9.5x10 atoms = 1.866915x109 atoms
N
ln 0 = kt
or
ln Nt = ln N0 – kt
Nt


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24-12
1.866915x109 atoms
= (4.95105129x10–11 yr–1)(t)
1.86682x109 atoms
5.088738x10–5 = (4.95105129x10–11 yr–1)(t)
t = 1.027809x106 = 1.0x106 yr
ln
24.48
The mole relationship between 40K and 40Ar is 1:1. Thus, 1.14 mmol 40Ar = 1.14 mmol 40K decayed.
ln 2
ln 2
k=
=
= 5.5451774x10–10 yr–1
t1/2
1.25x109 yr
ln
ln
N0
= kt
Nt
1.38  1.14 mmol
= (5.5451774x10–10 yr–1)(t)
1.38 mmol
0.6021754 = (5.5451774x10–10 yr–1)(t)
t = 1.08594x109 = 1.09x109 yr
27
13 Al
4
30
1
24.49
+ 2 He  15 P + 0 n
They experimentally confirmed the existence of neutrons, and were the first to produce an artificial radioisotope.
24.50
Both gamma radiation and neutron beams have no charge, so neither is deflected by electric or magnetic fields.
Neutron beams differ from gamma radiation in that a neutron has mass approximately equal to that of a proton.
Researchers observed that a neutron beam could induce the emission of protons from a substance. Gamma rays do
not cause such emissions.
24.51
A proton, for example, exits the first tube just when it becomes positive and the next tube becomes negative.
Pushed by the first tube and pulled by the second, the proton accelerates across the gap between them.
24.52
Protons are repelled from the target nuclei due to the interaction of like (positive) charges. Higher energy is
required to overcome the repulsion.
24.53
Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the
right side must be equal. In the shorthand notation, the nuclide to the left of the parentheses is the reactant while
the nuclide written to the right of the parentheses is the product. The first particle inside the parentheses is the
projectile particle while the second substance in the parentheses is the ejected particle.
Solution:
a) An alpha particle is a reactant with 10B and a neutron is one product. The mass number for the reactants is
10 + 4 = 14. So, the missing product must have a mass number of 14 – 1 = 13. The total atomic number for the
reactants is 5 + 2 = 7, so the atomic number for the missing product is 7.
10
4
1
13
5 B + 2 He  0 n + 7 N
b) A deuteron (2H) is a reactant with 28Si and 29P is one product. For the reactants, the mass number is 28 + 2 = 30
and the atomic number is 14 + 1 = 15. The given product has mass number 29 and atomic number 15, so the
missing product particle has mass number 1 and atomic number 0. The particle is thus a neutron.
28
2
1
29
14 Si + 1H  0 n + 15 P
c) The products are two neutrons and 244Cf with a total mass number of 2 + 244 = 246, and an atomic number of
98. The given reactant particle is an alpha particle with mass number 4 and atomic number 2. The missing reactant
must have mass number of 246 – 4 = 242 and atomic number 98 – 2 = 96. Element 96 is Cm.
242
4
1
244
96 Cm + 2 He  2 0 n + 98 Cf
24.54
31
1
1
29
a) 15 P +   1H + 0 n + 14 Si
31
P (, p, n) 29Si
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24-13
b)
252
98 Cf
10
1
257
c)
238
4
1
239
92 U + 2 He  3 0 n + 94 Pu
+ 5 B  5 0 n + 103 Lr
252
Cf (10B, 5n) 257Lr
238
U (, 3n) 239Pu
24.55
249
249
98 Cf
249
98 Cf
24.56
12
257
1
6 C  104 Rf + 4 0 n
15
260
1
+ 7 N  105 Db + 4 0 n
18
263
1
+ 8 O  106 Sg + 4 0 n
a) 98 Cf +
b) 249Cf (12C, 4n) 257Rf
249
Cf (15N, 4n) 260Db
249
Cf (18O, 4n) 263Sg
Gamma radiation has no mass or charge while alpha particles are massive and highly charged. These differences
account for the different effect on matter that these two types of radiation have. Alpha particles interact with
matter more strongly than gamma particles due to their mass and charge. Therefore alpha particles penetrate
matter very little. Gamma rays interact very little with matter due to the lack of mass and charge. Therefore
gamma rays penetrate matter more extensively.
24.57
In the process of ionization, collision of matter with radiation dislodges an electron. The free electron and the
positive ion that result are referred to as an ion-pair.
24.58
Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to
radiation than an adult’s slowly dividing cells.
24.59
The hydroxyl free radical forms more free radicals which go on to attack and change surrounding biomolecules,
whose bonding and structure are delicately connected with their function. These changes are irreversible, as
opposed to the reversible changes produced by OH–. 
24.60
Plan: The rad is the amount of radiation energy absorbed in J per body mass in kg: 1 rad = 0.01 J/kg. Change the
mass unit from pounds to kilograms. The conversion factor between rad and gray is 1 rad = 0.01 Gy.
Solution:
 3.3x10 7 J   2.205 lb  

1 rad
a) Dose (rad) = 
= 5.39x10–7 = 5.4x10–7 rad
 135 lb   1 kg   1x102 J /kg 




0.01
gy


–9
–9
b) Gray (rad) = 5.39x107 rad 
 = 5.39x10 = 5.4x10 Gy
 1 rad 

24.61

 1 rad 
a) Dose (rad) = (8.92x10–4 Gy) 
 = 0.0892 rad
 0.01 Gy 
 0.01 J/kg 
–3
–3
b) Energy (J) = (0.0892 rad) 
 3.6 kg  = 3.2112x10 = 3.2x10 J
1
rad


24.62
Plan: Multiply the number of particles by the energy of one particle to obtain the total energy absorbed.
Convert the energy to dose in grays with the conversion factor 1 rad = 0.01 J/kg = 0.01 Gy. To find the millirems,
convert grays to rads and multiply rads by RBE to find rems. Convert rems to mrems. Convert the dose to
sieverts with the conversion factor 1 rem = 0.01 Sv.
Solution:



a) Energy (J) absorbed = 6.0x105  8.74x1014 J/ = 5.244x10–8 J
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24-14


5.244x108 J  1 rad   0.01 Gy 
–10
–10
Dose (Gy) =

 = 7.4914x10 = 7.5x10 Gy


J
70. kg
 0.01 kg   1 rad 


 1 mrem 
 1 rad 
–5
–5
b) rems = rads x RBE = 7.4914x1010 Gy 
 = 7.4914x10 = 7.5x10 mrem
 1.0   3
0.01
Gy


 10 rem 


 10 3 rem   0.01 Sv 
= 7.4914x10–10 = 7.5x10–10 Sv
Sv = 7.4914x10 5 mrem 
 1 mrem   1 rem 





1.77x10   2.20x10
a) Dose =
24.63
24.64

J /   103 g   1 rad 

 = 1.46943 = 1.47 rad

 

265 g
 1 kg   0.01 J kg 


b) Dose = (1.46943 rad)(0.01 Gy/1 rad) = 1.46943x10–2 = 1.47x10–2 Gy
c) Dose = (1.46943 rad)(0.75 rem/rad)(0.01 Sv/rem) = 1.10207x10–2 = 1.10x10–2 Sv


 2.50 pCi   1x1012 Ci   3.70x1010 dps 
 3600 s   8.25x1013 J   1 rad 
Dose = 
65
h

 
 

 
 
1 Ci
 95 kg   1 pCi  
 1 h   disint.   0.01 J kg 



10
13
= 1.8796974x10–8 = 1.9x10–8 rad
Dose = (1.8796974x10–8 rad)(0.01 Gy/1 rad) = 1.8796974x10–10 = 1.9x10–10 Gy
24.65
Use the time and disintegrations per second (Bq) to find the number of 60Co atoms that disintegrate, which equals
the number of  particles emitted. The dose in rads is calculated as energy absorbed per body mass.


 475 Bq   103 g   1 dps   5.05x1014 J 
 60 s   1 rad 
Dose = 
24.0
min












 1.858 g   1 kg   1 Bq   1 disint. 
 1 min   0.01 J kg 


= 1.8591x10–3 = 1.86x10–3 rad
24.66
A healthy thyroid gland incorporates dietary I – into I-containing hormones at a known rate. To assess thyroid
function, the patient drinks a solution containing a trace amount of Na131I, and a scanning monitor follows the
uptake of 131I into the thyroid. Technetium-99 is often used for imaging the heart, lungs, and liver.
24.67
NAA does not destroy the sample while chemical analysis does. Neutrons bombard a non-radioactive sample,
“activating” or energizing individual atoms within the sample to create radioisotopes. The radioisotopes decay
back to their original state (thus, the sample is not destroyed) by emitting radiation that is different for each
isotope.
24.68
In positron-emission tomography (PET), the isotope emits positrons, each of which annihilates a nearby electron.
In the process, two  photons are emitted simultaneously, 180° apart from each other. Detectors locate the sites
and the image is analyzed by computer.
24.69
The concentration of 59 Fe in the steel sample and the volume of oil would be needed.
24.70
The oxygen in formaldehyde comes from methanol because the oxygen isotope in the methanol reactant appears
in the formaldehyde product. The oxygen isotope in the chromic acid reactant appears in the water product, not
the formaldehyde product. The isotope traces the oxygen in methanol to the oxygen in formaldehyde.
24.71
The mass change in a chemical reaction was considered too minute to be significant and too small to measure with
even the most sophisticated equipment.
24.72
When a nucleus forms from its nucleons, there is a decrease in mass called the mass defect. This decrease in mass
is due to mass being converted to energy to hold the nucleus together. This energy is called the binding energy.
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-15
24.73
Energy is released when a nuclide forms from nucleons. The nuclear binding energy is the amount of energy
holding the nucleus together. Energy is absorbed to break the nucleus into nucleons and is released when
nucleons “come together.”
24.74
The binding energy per nucleon is the average amount of energy per each component (proton and neutron) part of
the nuclide. The binding energies per nucleon are helpful in comparing the stabilities of different combinations
and to provide information on the potential processes a nuclide can undergo to become more stable. The binding
energy per nucleon varies considerably. The greater the binding energy per nucleon, the more strongly the
nucleons are held together and the more stable the nuclide.
24.75
Plan: The conversion factors are: 1 MeV = 106 eV and 1 eV = 1.602x1019 J.
Solution:
 106 eV 
a) Energy (eV) = 0.01861 MeV 
= 1.861x104 eV
 1 MeV 


 106 eV   1.602x10 19 J 
15
15
b) Energy (J) =  0.01861 MeV  
 = 2.981322x10 = 2.981x10 J
 1 MeV  
1 eV



24.76


1 eV
 1000 J  

= 9.8002x106 = 9.80x106 eV
a) Energy (eV) = 1.57x10 15 kJ 

19 
1
kJ
1.602x10
J



 1 MeV 
6
b) Energy (MeV) = 9.8002x10 eV  6
 9.8002 = 9.80 MeV
 10 eV 

24.77

Plan: Convert moles of 239Pu to atoms of 239Pu using Avogadro’s number. Multiply the number of atoms by the
energy per atom (nucleus) and convert the MeV to J using the conversion 1 eV = 1.602x10–19 J.
Solution:
 6.022x1023 atoms 
23
Number of atoms = 1.5 mol 239 Pu 
 = 9.033x10 atoms

mol




 5.243 MeV   106 eV   1.602x10 19 J 
11
11
Energy (J) = 9.033x1023 atoms 
 
 = 7.587075x10 = 7.6x10 J
 
1
atom
1
MeV
1
eV





24.78
24.79


  103 J  

  1 MeV  
8.11x105 kJ
1 eV
1 mol 49 Cr
Energy (MeV) = 




 3.2x10 3 mol 49 Cr   1 kJ   1.602x1019 J   106 eV   6.022x1023 nuclei 






= 2.6270 = 2.6 MeV
Plan: Oxygen-16 has eight protons and eight neutrons. First find the Δm for the nucleus by subtracting the given
mass of one oxygen atom from the sum of the masses of eight 1H atoms and eight neutrons. Use the conversion
factor 1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total
number of nucleons (protons and neutrons) in the oxygen nuclide to obtain binding energy per nucleon. Convert
Δm of one oxygen atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy
per mole of oxygen, use the relationship E = mc2. m must be converted to units of kg/mol.
Solution:
Mass of 8 1H atoms = 8 x 1.007825 = 8.062600 amu
Mass of 8 neutrons = 8 x 1.008665 = 8.069320 amu
Total mass =16.131920 amu
m = 16.131920  15.994915 = 0.137005 amu/16O = 0.137005 g/mol 16O
 0.137005 amu 16 O   931.5 MeV 
a) Binding energy (MeV/nucleon) = 
 
 = 7.976259844 = 7.976 MeV/nucleon
 16 nucleons

  1 amu 
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-16
 0.137005 amu 16 O   931.5 MeV 
b) Binding energy (MeV/atom) = 
 
 = 127.6201575 = 127.6 MeV/atom

1 atom

  1 amu 
c) E = mc2


2
 0.137005 g 16 O   1 kg 
1 J   1 kJ 
8
Binding energy (kJ/mol) = 
  3  2.99792x10 m/ s 
 3 
2
mol
 kg•m 2   10 J 

  10 g 
s 

= 1.23133577x1010 = 1.23134x1010 kJ/mol

24.80

m is calculated from the mass of 82 protons (1H) and 124 neutrons vs. the mass of the lead nuclide.
Mass of 82 1H atoms = 82 x 1.007825 = 82.641650 amu
Mass of 124 neutrons = 124 x 1.008665 = 125.074460 amu
Total mass = 207.716110 amu
m = 207.716110  205.974440 = 1.741670 amu/206Pb = 1.741670 g/mol 206Pb
 1.741670 amu 206 Pb   931.5 MeV 
a) Binding energy (MeV/nucleon) = 
 
 = 7.8755612 = 7.876 MeV/nucleon

206 nucleons

  1 amu 
 1.741670 amu 206 Pb   931.5 MeV 
b) Binding energy (MeV/atom) = 
 
 = 1622.3656 = 1622 MeV/atom

1 atom

  1 amu 



  1kJ 
2
 1.741670 g 206 Pb   1 kg 
1J
8
c) Binding energy (kJ/mol) = 
  3  2.99792 x10 m / s 
 3 
2

mol
 kg•m 2   10 J 

  10 g 
s 

= 1.5653301x1011 = 1.56533x1011 kJ/mol

24.81

Plan: Cobalt-59 has 27 protons and 32 neutrons. First find the Δm for the nucleus by subtracting the given mass of
one cobalt atom from the sum of the masses of 27 1H atoms and 32 neutrons. Use the conversion factor
1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total number
of nucleons (protons and neutrons) in the cobalt nuclide to obtain binding energy per nucleon. Convert Δm of
one cobalt atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy per
mole of cobalt, use the relationship E = mc2. m must be converted to units of kg/mol.
Solution:
Mass of 27 1H atoms = 27 x 1.007825 = 27.211275 amu
Mass of 32 neutrons = 32 x 1.008665 = 32.27728 amu
Total mass = 59.488555 amu
m = 59.488555 – 58.933198 = 0.555357 amu/59Co = 0.555357 g/mol 59Co
 0.555357 amu 59 Co   931.5 MeV 
a) Binding energy (MeV/nucleon) = 
 
 = 8.768051619 = 8.768 MeV/nucleon

59 nucleons

  1 amu 
 0.555357 amu 59 Co   931.5 MeV 
b) Binding energy (MeV/atom) = 
 
 = 517.3150 = 517.3 MeV/atom

1 atom

  1 amu 
c) Use E = mc2


2
 0.555357 g 59 Co   1 kg 
1 J   1 kJ 
8
Binding energy (kJ/mol) = 
  3  2.99792x10 m/s 
 3 
2
mol
 kg•m 2   10 J 

  10 g 
s 

= 4.9912845x1010 = 4.99128x1010 kJ/mol

24.82

m is calculated from the mass of 53 protons (1H) and 78 neutrons vs. the mass of the iodine nuclide.
Mass of 53 1H atoms = 53 x 1.007825 = 53.414725 amu
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-17
Mass of 78 neutrons = 78 x 1.008665 = 78.675870 amu
Total mass = 132.090595 amu
m = 132.090595 – 130.906114 = 1.184481 amu/131I = 1.184481 g/mol 131I
 1.184481 amu 131 I   931.5 MeV 
= 8.422473676 = 8.422 MeV/nucleon
a) Binding energy (MeV/nucleon) = 
 131 nucleons   1 amu 



 1.184481 amu 131 I   931.5 MeV 
b) Binding energy (MeV/atom) = 
 
 = 1103.34405 = 1103 MeV/atom

1 atom

  1 amu 
c) E = mc2


2
 1.184481 g 131 I   1 kg 
1 J   1 kJ 
8
Binding energy (kJ/mol) = 
  3  2.99792x10 m/s 
 3 
2
mol
 kg•m 2   10 J 

  10 g 
s 

= 1.06455518x1011 = 1.06456x1011 kJ/mol

24.83
80
0

80
a) 35 Br  1 + 36 Kr (reaction 1)
80
0
80
35 Br + 1e  34 Se (reaction 2)
b) Reaction 1:
m = 79.918528 – 79.916380 = 0.002148 amu
Reaction 2:
m = 79.918528 – 79.916520 = 0.002008 amu
Since E = (m)c2, the greater mass change (reaction 1) will release more energy.
24.84
The minimum number of neutrons from each fission event that must be absorbed by the nuclei to sustain the chain
reaction is one. In reality, due to neutrons lost from the fissionable material, two to three neutrons are generally
needed to continue a self-sustaining chain reaction.
24.85
In both radioactive decay and fission, radioactive particles are emitted, but the process leading to the emission is
different. Radioactive decay is a spontaneous process in which unstable nuclei emit radioactive particles and
energy. Fission occurs as the result of high-energy bombardment of nuclei with small particles that cause the
nuclides to break into smaller nuclides, radioactive particles, and energy.
In a chain reaction, all fission events are not the same. The collision between the small particle emitted in the
fission and the large nucleus can lead to splitting of the large nuclei in a number of ways to produce several
different products.
24.86
Enriched fissionable fuel is needed in the fuel rods to ensure a sustained chain reaction. Naturally occurring
235
U is only present in a concentration of 0.7%. This is consistently extracted and separated until its concentration
is between 3-4%.
24.87
a) Control rods are movable rods of cadmium or boron which are efficient neutron absorbers. In doing so, they
regulate the flux of neutrons to keep the reaction chain self-sustaining which prevents the core from overheating.
b) The moderator is the substance flowing around the fuel and control rods that slows the neutrons, making them
better at causing fission.
c) The reflector is usually a beryllium alloy around the fuel-rod assembly that provides a surface for neutrons that
leave the assembly to collide with and therefore, return to the fuel rods.
24.88
The water serves to slow the neutrons so that they are better able to cause a fission reaction. Heavy water
2
1
( 1 H2O or D2O) is a better moderator because it does not absorb neutrons as well as light water ( 1 H2O ) does, so
more neutrons are available to initiate the fission process. However, D2O does not occur naturally in great
abundance, so production of D2O adds to the cost of a heavy water reactor. In addition, if heavy water does
3
absorb a neutron, it becomes tritiated, i.e., it contains the isotope tritium, 1H , which is radioactive.
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24-18
24.89
The advantages of fusion over fission are the simpler starting materials (deuterium and tritium), and no long-lived
toxic radionuclide by-products.
24.90
Virtually all the elements heavier than helium, up to and including iron, are produced by nuclear fusion reactions
in successively deeper and hotter layers of massive stars. Iron is the point at which fusion reactions cease to be
energy producers. Elements heavier than iron are produced by a variety of processes, primarily during a
supernova event, which distribute the Sun’s ash into the cosmos to form next generation suns and planets. Thus,
the high cosmic and Earth abundance of iron is consistent with it being the most stable of all nuclei.
24.91
Mass of reactants: 3.01605 + 2.0140 = 5.03005 amu
Mass of products: 4.00260 + 1.008665 = 5.011265 amu
m = mass of reactants – mass of products = 5.03005 – 5.011265 = 0.018785 amu = 0.018785 g/mol
E = mc2


2
1 J   1 kJ 
 0.018785 g   1 kg 
8
Energy (kJ/mol) = 
 3 
  3  2.99792x10 m/s 
2
mol

  10 g 
 kg•m 2   10 J 
s 

= 1.6883064x109 = 1.69x109 kJ/mol


24.92
243
4
239
95 Am  2 He + 93 Np
239
0
239
93 Np  1 + 94 Pu
239
4
235
94 Pu  2 He + 92 U
239
239
235
93 Np , 94 Pu , and 92 U were present as products in the decay of Am-243.
24.93
Plan: Use the masses given in the problem to calculate the mass change (reactant – products) for the reaction.
The conversion factor between amu and kg is 1 amu = 1.66054x10–27 kg. Use the relationship E = mc2 to
convert the mass change to energy.
Solution:
243
239
4
a) 96 Cm  94 Pu + 2 He
m (amu) = 243.0614 amu  (4.0026 + 239.0522) amu = 0.0066 amu
 1.661x10 24 g   1 kg 
–29
–29
m (kg) =  0.0066 amu  
  3  = 1.09626x10 = 1.1x10 kg

1
amu

  10 g 


1
J
–13
–13
b) E = mc2 = 1.09626x10 kg 2.99792x10 m/s 
 = 9.85266x10 = 9.9x10 J
2
 kg•m 2 
s 


13
23
 9.85266x10
J   6.022x10 reactions   1 kJ 
8
8
c) E released = 
 
  3  = 5.93317x10 = 5.9x10 kJ/mol

reaction
mol


  10 J 
This is approximately one million times larger than a typical heat of reaction.

24.94
29

8


2
a) First, determine the amount of activity released by the 239Pu for the duration spent in the body (16 h) using the
relationship A = kN. The rate constant is derived from the half-life and N is calculated using the molar mass and
Avogadro’s number.

ln 2 
  1 day   1 h 
ln 2
1 yr
–13 –1
k=
= 



 = 9.113904x10 s
4
365.25
day
24
h
3600
s
t1/2



 2.41x10 yr  
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-19
 10 6 g   1 mol Pu   6.022x1023 atoms Pu 
15
N = 1.00 g Pu  
 = 2.5196653x10 atoms Pu
 1 g   239 g Pu  
1
mol
Pu




 1 disint. 
3
A = kN = (9.113904 x 10–13 s–1)( 2.5196653x1015 atoms Pu) 
 = 2.2963988x10 d/s
1
atom
Pu


Each disintegration releases 5.15 MeV, so d/s can be converted to MeV. Convert MeV to J (using
1.602x10–13 J = 1 MeV) and J to rad (using 0.01 J/kg = 1 rad).


 2.2963988x103 d/s   5.15 MeV   1.602x10 13 J   1 rad   3600 s 
Energy = 
 
 
 16 h 
 

J 
85 kg
MeV

  disint.  
  0.01 kg   1 h 


= 1.2838686x10–4 = 1.28x10–4 rads
b) Since 0.01 Gy = 1 rad, the worker receives:
Dose = (1.2838686x10–4 rad)(0.01 Gy/rad) = 1.2838686x10–6 = 1.28x10–6 Gy
24.95
Plan: Determine k for 14C using the half-life (5730 yr). Determine the mass of carbon in 4.58 g of CaCO3.
Divide the given activity of the C in d/min by the mass of carbon to obtain the activity in d/min•g; this is At and is
compared to the activity of a living organism (A0 = 15.3 d/min•g) in the integrated rate law, solving for t.
Solution:
ln 2
ln 2
k=
=
= 1.2096809x10–4 yr–1
t1/2
5730 yr
 1 mol CaCO3  1 mol C  12.01 g C 
Mass (g) of C = 4.58 g CaCO3 


 = 0.5495634 g C
 100.09 g CaCO3  1 mol CaCO3   1 mol C 
3.2 d/min
At =
= 5.8228 d/min•g
0.5495634 g
Using the integrated rate law:
A 
ln  t  = –kt
A0 = 15.3 d/min•g (the ratio of 12 C:14 C in living organisms)
A
 0
 5.8228 d/min•g 
–4
–1
ln 
 = – (1.2096809x10 yr )(t)
 15.3 d/min•g 
t = 7986.17 = 8.0x103 yr
24.96
Find the rate constant from the rate of decay and the initial number of atoms. Use rate constant to calculate halflife.
Initial number of atoms:
 106 g   1 mol RaCl 2   1 mol Ra   6.022x1023 Ra atoms 
Ra atoms =  5.4 g RaCl 2  

 1 g   297 g RaCl   1 mol RaCl  
1 mol Ra
2 
2 



= 1.09490909x1016 Ra atoms
A
A = kN or k =
N

  1d s 
1.5x105 Bq

 = 1.36997675x10–11 s–1
k= 
 1.09490909x1016 Ra atoms   Bq 



ln 2
ln 2
t1/2 =
=
= 5.05955x1010 = 5.1x1010 s
k
1.36997675x1011 s1
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24-20
24.97
ln 2
ln 2
=
= 1.2096809x10–4 yr–1
t1/2
5730 yr
k=
 1 mol 14 C   6.022x1023 atoms 14 C 
14
14
Number of atoms = 10 8 g 14 C 
 = 4.3014x10 atoms C
14
 14 g 14 C  
1
mol
C



A = kN = [(1.2096809x10–4 yr–1)(4.3014x1014 atoms 14C)](1 disintegration/1 atom) = 5.2033x1010 dpyr


13
 5.2033x1010 dpyr 
 0.156 MeV   1.602x10 J   1 rad 
Dose = 
= 2.001x10–3 = 10–3 rad
yr  




  1 MeV


  0.01 J 
65
kg
disint.






kg 


24.98

Plan: Determine how many grams of AgCl are dissolved in 1 mL of solution. The activity of the radioactive Ag+
indicates how much AgCl dissolved, given a starting sample with a specific activity (175 nCi/g). Convert g/mL to
mol/L (molar solubility) using the molar mass of AgCl.
Solution:
 1.25x102 Bq   1 dps  
  1 nCi   1 g AgCl 
1 Ci
–6
Concentration = 
 
  9  

 = 1.93050x10 g AgCl/mL
10

mL
1
Bq
175
nCi
3.70
x10
dps
10
Ci






 1.93050x10 6 g AgCl   1 mol AgCl   1 mL 
–5
–5
Molarity = 
 
  3  = 1.34623x10 = 1.35x10 M AgCl

mL
143.4
g
AgCl
10
L






24.99
a) The process shown is fission in which a neutron bombards a large nucleus, splitting that nucleus
into two nuclei of intermediate mass.
1
b) 0 n +
235
144
90
1
92 U  55 Cs + 37 Rb + 2 0 n
144
c) 55 Cs , with 55 protons and 89 neutrons, has a n/p ratio of 1.6. This ratio places this isotope above the band of
stability and decay by beta particle emission is expected.
24.100 Plan: Determine the value of k from the half-life. Then determine the fraction from the integrated rate law.
Solution:
ln 2
ln 2
k=
=
= 9.90210x10–10 yr–1
t1/2
7.0x108 yr
ln
N0
= kt = (9.90210x10–10 yr–1)(2.8x109 yr) = 2.772588
Nt
N0
= 15.99998844
Nt
Nt
= 0.062500 = 6.2x10–2
N0
24.101 Determine the value of k from the half-life. Then determine the age from the integrated rate law.
ln 2
ln 2
k=
=
= 1.540327x10–10 yr–1
9
t1/2
4.5x10 yr
N0
= kt
Nt
6 9
ln
=(1.540327x10-10 yr-1 )(t)
9
0.5108256 = (1.540327x10–10 yr–1)(t)
t = 3.316345x109 = 3.3x109 yr
ln
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24-21
24.102 Plan: Find the rate constant, k, using any two data pairs (the greater the time between the data points, the
greater the reliability of the calculation). Calculate t1/2 using k. Once k is known, use the integrated rate law to
find the percentage lost after 2 h. The percentage of isotope remaining is the fraction remaining after 2.0 h
(Nt where t = 2.0 h) divided by the initial amount (N0), i.e., fraction remaining is Nt/N0. Solve the first-order rate
expression for Nt/N0, and then subtract from 100% to get fraction lost.
Solution:
N
a) ln t = –kt
N0
 495 photons/s 
ln 
 = –k(20 h)
 5000 photons/s 
–2.312635 = –k(20 h)
k = 0.11563 h–1
ln 2
ln 2
t1/2 =
=
= 5.9945 = 5.99 h (Assuming the times are exact, and the emissions have three
k
0.11563 h 1
significant figures.)
b) ln
ln
Nt
= –kt
N0
Nt
= – (0.11563 h–1)(2.0 h) = –0.23126
N0
Nt
Nt
= 0.793533
x 100% = 79.3533%
N0
N0
The fraction lost upon preparation is 100% – 79.3533% = 20.6467% = 21%.
ln 2
ln 2
=
= 5.5451774x10–10 yr–1
9
t1/2
1.25x10 yr
A = kN where A = dps N = number of atoms
Number of atoms = (1.0 mol 40K)(6.022x1023 atoms/mol) = 6.022x1023 atoms 40K
24.103 k =
 5.5451774x10 10 

  1 day   1 h   1 disint. 
1 yr
7
23
A= 
 6.022x10 atoms 



 = 1.05816x10 dps

yr
365.25
day
24
h
3600
s
1
atom







Dose (Ci) = (1.05816x107 dps)(1 Ci/3.70x1010 dps) = 2.85990x10–4 = 2.9x10–4 Ci
Dose = (1.05816x107 dps)(1 Bq/1 dps) = 1.05816x107 = 1.1x107 Bq


24.104 Plan: Use the given relationship for the fraction remaining after time t, where t = 10.0 yr, 10.0x103 yr, and
10.0x104 yr.
Solution:
a) Fraction remaining after 10.0 yr =
t
t1
10.0
2
=  1 
2
10.0x103
3
1
yr =  
2
4
1
yr =  
2
b) Fraction remaining after 10.0x10
c) Fraction remaining after 10.0x10

1
2
10.0x104
5730
5730
5730
= 0.998791 = 0.999
= 0.298292 = 0.298
= 5.5772795x10–6 = 5.58x10–6
d) Radiocarbon dating is more reliable for b) because a significant quantity of 14C has decayed and a significant
quantity remains. Therefore, a change in the amount of 14C would be noticeable. For the fraction in a), very little
14
C has decayed and for c) very little 14C remains. In either case, it will be more difficult to measure the change so
the error will be relatively large.
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24-22
24.105
210
0
210
86 Rn + 1e  85 At
Mass = (2.368 MeV)(1 amu/931.5 MeV) = 0.0025421363 amu
Mass 210At = mass 210Rn + electron mass – mass equivalent of energy emitted.
= (209.989669 + 0.000549 – 0.0025421363) amu = 209.9876759 = 209.98768 amu
24.106 Plan: At one half-life, the fraction of sample is 0.500. Find n for which (0.900)n = 0.500.
Solution:
(0.900)n = 0.500
n ln (0.900) = ln (0.500)
n = (ln 0.500)/(ln 0.900) = 6.578813 = 6.58 h
24.107 a)  decay by vanadium-52 produces chromium-52.
51
1
52
52
0
23V + 0 n  23V  24 Cr + 1
51
52
23V (n,) 24 Cr
b) Positron emission by copper-64 produces nickel-64.
63
1
64
29 Cu + 0 n  29 Cu
63
+ 64
29 Cu (n, ) 28 Ni
64
0
 28 Ni + 1
c)  decay by aluminum-28 produces silicon-28.
27
1
28
28
0
13 Al + 0 n  13 Al  14 Si + 1
27
28
13 Al (n,) 14 Si
24.108 Determine k for 90Sr.
ln 2
ln 2
k=
=
= 0.023902 yr–1
t1/2
29 yr
a) ln
ln
Nt
= –kt
N0
Nt
= – (0.023902 yr–1)(10 yr) = – 0.23902
0.0500 g
Nt
= 0.787399
0.0500 g
(0.787399)(0.0500 g) = 0.03936995 = 0.039 g 90Sr
N
b) ln t = –kt
N0
100  99.9%
= – (0.023902 yr–1)(t)
100%
t = 289.003 = 3x102 yr (The calculation 100 – 99.9 limits the answer to one significant figure.)
ln
24.109 a)
12
4
16
6 C + 2 He  8 O
13
 931.5 MeV   1.602x10
b) 7.7x10 2 amu 

1 MeV
 1 amu  


J   1 kJ 
–14
–14
  3  = 1.1490425x10 = 1.1x10 kJ
10
J



24.110 Plan: The production rate of radon gas (volume/hour) is also the decay rate of 226Ra. The decay rate, or activity, is
proportional to the number of radioactive nuclei decaying, or the number of atoms in 1.000 g of 226Ra, using the
relationship A = kN. Calculate the number of atoms in the sample, and find k from the half-life. Convert the
activity in units of nuclei/time (also disintegrations per unit time) to volume/time using the ideal gas law.
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24-23
Solution:
226
4
222
88 Ra  2 He + 86 Rn
ln 2
ln 2
=
= 4.33879178x10–4 yr–1(1 yr/8766 h) = 4.94510515x10–8 h–1
t1/2
1599 yr
The mass of 226Ra is 226.025402 amu/atom or 226.025402 g/mol.
k=

  6.022x1023 Ra atoms 
1 mol Ra
21
N = 1.000 g Ra  
 = 2.6643023x10 Ra atoms
 
226.025402
g
Ra
1
mol
Ra



–8 –1
21
A = kN = (4.94510515x10 h )(2.6643023x10 Ra atoms) = 1.3175255x1014 Ra atoms/h
This result means that 1.318x1014 226Ra nuclei are decaying into 222Rn nuclei every hour. Convert atoms of 222Rn
into volume of gas using the ideal gas law.
 1.3175255x1014 Ra atoms   1 atom Rn  
1 mol Rn

Moles of Rn/h = 
 



23

h

  1 atom Ra   6.022x10 Rn atoms 
= 2.1878537x10–10 mol Rn/h
L•atm 

2.1878537x1010 mol Rn/h  0.08206
 273.15 K 
nRT
mol•K 

V=
=
1 atm
P
–9
–9
= 4.904006x10 = 4.904x10 L/h
Therefore, radon gas is produced at a rate of 4.904x10–9 L/h. Note: Activity could have been calculated as decay
in moles/time, removing Avogadro’s number as a multiplication and division factor in the calculation.


24.111 Determine k:
ln 2
ln 2
k=
=
= 0.0216608 s–1
t1/2
32 s
ln
Nt
= –kt
N0
90%
= – (0.0216608 s–1)(t)
100%
t = 4.86411 = 4.9 s
ln
133
The N/Z ratio for 140Cs is too high.
79
It has an even number of neutrons compared with 78Br.
c) 12 Mg
24
The N/Z ratio equals 1.
14
d) 7 N
The N/Z ratio equals 1.
24.112 a) 55 Cs
b) 35 Br
24.113 Plan: Determine k from the half-life and then use the integrated rate law, solving for time.
Solution:
ln 2
ln 2
k=
=
= 0.0239016 yr–1
t1/2
29 yr
ln
Nt
= –kt
N0
 1.0x104 particles 
= – (0.0239016 yr–1)(t)
ln 
 7.0x104 particles 


–1.945910 = – (0.0239016 yr–1)(t)
t = 81.413378 = 81 yr
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24-24
6
6
12
6 C (dilithium)
6
12
12
b) m = 2(mass 3 Li ) – mass 6 C = 2(6.015121 amu) – 12.000000 amu = 0.030242 amu/atom 6 C (dilithium)
24.114 a) 3 Li + 3 Li 
m = (0.030242 amu/atom)(1.66054x10–27 kg/amu) =5.02180507x10–29 kg/atom




2
1
J
29
8
–12
E = mc2 = 5.02180507x10 kg/atom 2.99792x10 m/s 
 = 4.5133595x10 J/atom
2
kg•m


s2 




 4.5133595 x10 12 J  


1 atom
1 amu
E= 
 


27

atom

  12.000000 amu   1.66054x10 kg 
= 2.2650059x1014 = 2.2650x1014 J/kg dilithium
1
4
0
c) 4 1H  2 He + 2 1
2 positrons are released.
12
d) For dilithium ( 6 C ):
Mass = (5.02180507x10–29 kg/atom)(1 atom/12.000000 amu) (1 amu/1.66054x10–27 kg 12C)
= 2.5201667x10–3 = 2.5202x10–3 kg/kg 12C
4
For 2 He (The mass of a positron is the same as the mass of an electron.)
1
4
0
m = 4 (mass 1H ) – [mass 2 He + 2 mass 1e ]
= 4(1.007825 amu) – [4.00260 amu + 2(5.48580x10–4 amu)] = 0.02760 amu/atom
= (0.02760 amu/atom)(1.66054x10–27 kg/amu) = 4.58309x10–29 kg/atom
Mass = (4.58309x10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054x10–27 kg 4He)
= 6.895517x10–3 = 6.896x10–3 kg/kg 4He
2
3
4
1
6
1
4
3
e) 1H + 1H  2 He + 0 n
m = [2.0140 amu + 3.01605 amu] – [4.00260 amu + 1.008665 amu]
= 5.0300 amu – 5.01126 amu = 0.0188 amu
= (0.0188 amu/atom)(1.66054x10–27 kg/amu) = 3.1218152x10–29 kg/atom
Mass = (3.1218152x10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054x10–27 kg 4He)
= 4.696947x10–3 = 4.70x10–3 kg/kg 4He
f) 3 Li + 0 n  2 He + 1H
 1 mol 3 H   3.01605 g 3 H   1 mol 6 Li   103 g 6 Li   1 kg 3 H 
3
Mass 1H / kg 6Li = 
 1 mol 6 Li   1 mol 3 H   6.015121 g 6 Li   1 kg 6 Li   103 g 3 H 






= 0.5014113598 = 0.501411 kg 3H/kg 6Li
 103 g   1 mol 3 H   6.022x1023 atom 3 H   3.121852x1029 kg 
Mass = (0.5014113598 kg 3H) 
 

 1 kg   3.01605 g 3 H  
atom
mol 3 H





= 3.1254222x10–3 = 3.125x10–3 kg
Change in mass for dilithium reaction:
Mass = (5.02180507x10–29 kg/atom 12C)(1 atom 12C/2 atoms 6Li)(6.022x1023 atoms 6Li/mol)
(1 mol 6Li/6.015121 g 6Li)(103 g/kg 6Li) = 2.513774x10–3 = 2.514x10–3 kg
The change in mass for the dilithium reaction is slightly less than that for the fusion of tritium with
deuterium.
24.115 Plan: Convert pCi to Bq using the conversion factors 1 Ci = 3.70x1010 Bq and 1 pCi = 10–12 Ci. For part b),
use the first-order integrated rate law to find the activity at the later time (t = 9.5 days). You will first need to
calculate k from the half-life expression. For part c), solve for the time at which Nt = the EPA recommended
level.
Solution:
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24-25
12
10
 4.0 pCi   10 Ci   3.70x10 Bq 
a) Activity (Bq/L) = 

 
 = 0.148 = 0.15 Bq/L

1 Ci
 L   1 pCi  

The safe level is 0.15 Bq/L.
ln 2
ln 2
b) k =
=
= 0.181452 d–1
t1/2
3.82 d
ln
Nt
= – kt
N0
ln
Nt
= – (0.181452 d–1)(9.5 d) = –1.723794
41.5 pCi/L
Nt
= 0.17838806
41.5 pCi/L
Nt = 7.403104 pCi/L
12
10
 7.403104 pCi   10 Ci   3.70 x10 Bq 
Activity (Bq/L) = 

 
 = 0.2739148 = 0.27 Bq/L

L
1 Ci

  1 pCi  

c) The desired activity is 0.15 Bq/L, however, the room air currently contains 0.27 Bq/L.
N
ln t = – kt
N0
0.15 Bq/L
= – (0.181452 d–1)(t)
0.2739148 Bq/L
– 0.602182 = – (0.181452 d–1)(t)
t = 3.318685 = 3.3 d
It takes 3.3 d more to reach the recommended EPA level. A total of 12.8 (3.3 + 9.5) d is required to reach
recommended levels when the room was initially measured at 41.5 pCi/L.
ln
24.116 k =
ln 2
ln 2
=
= 0.05653729 yr–1
12.26 yr
t1/2
ln
Nt
= – kt
N0
ln
Nt
= – (0.05653729 yr–1)(5.50 yr) = –0.310955097
N0
Nt
= 0.732746777
N0
fraction lost = 1 – 0.732746777 = 0.267253222 = 0.267
24.117
239
0
239
239
0
4
235
4
231
92 U  1 + 93 Np  94 Pu + 1  2  + 92 U  2  + 90Th
235
This could begin the 92 U decay series.
24.118 Plan: Convert mCi to Ci to disintegrations per second; multiply the dps by the energy of each disintegration in
MeV and convert to energy in J. Recall that 1 rad = 0.01 J/kg. The mass of the child must be converted from
lb to kg.
Solution:
 103 Ci   3.70x1010 dps   5.59 MeV   1.602x1013 J 
–5
Energy (J/s) = 1.0 mCi  
 
 = 3.3134166x10 J/s
 
 1 mCi  
1
Ci
1
disint.
1
MeV





Convert the mass of the child to kg:
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24-26
 54 lb  
1 kg 
 = 24.4897960 kg
 2.205 lb 
Time for 1.0 mrad to be absorbed:
 103 rad   0.01 J /kg   24.4897960 kg 
= 7.3911008 = 7.4 s
Time (s) = 1.0 mrad  
 1 mrad   1 rad   3.3134166x105 J /s 




24.119 k =
ln
ln 2
ln 2
=
= 1.2096809x10–4 yr–1
5730 yr
t1/2
Nt
= – kt
N0
 12.9 d /min • g 
–4
–1
ln 
 = – (1.2096809x10 yr )(t)
15.3
d/min
•
g


– 0.170625517 = – (1.2096809x10–4 yr–1)(t)
t = 1410.500216 = 1.41x103 yr
The earthquake occurred 1410 years ago.
24.120 Assuming our atmosphere present today was not fully developed and the fact that much of the cosmic ionizing
radiation is absorbed by the atmosphere, organisms would have been exposed to more cosmic radiation a billion
years ago. In addition, the number of radioactive nuclei such as uranium would have been larger. This would also
have the effect of exposing organisms to greater ionizing radiation from Earth.
24.121 Because the 1941 wine has a little over twice as much tritium in it, just over one half-life has passed between the
two wines. Therefore, the older wine was produced before 1929 (1941– 12.26) but not much earlier than that. To
find the number of years back in time, use the first-order rate expression, where N0 = 2.32 N, Nt = N and t = years
transpired between the manufacture date and 1941.
ln 2
ln 2
k=
=
= 0.05653729 yr–1
12.26 yr
t1/2
ln
Nt
= – kt
N0
N
= – (0.05653729 yr–1)(t)
2.32 N
– 0.841567 = – (0.05653729 yr–1)(t)
t = 14.92857 = 14.9 yr
The wine was produced in (1941 – 15) = 1926.
ln
24.122 If 99% of Pu-239 decays, 1% remains.
ln 2
ln 2
k=
=
= 2.87612938x10-5 yr–1
t1/2
2.41x104 yr
ln
Nt
= – kt
N0
1%
= – (2.87612938x10–5 yr–1)(t)
100%
–4.60517 = – (2.8761293x10–5 yr–1)(t)
t = 1.6011693x105 = 2x105 yr
(The 1% limits the answer to one significant figure.)
ln
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-27
24.123 k =
ln
ln 2
ln 2
=
= 1.2096809x10–4 yr–1
5730 yr
t1/2
Nt
= – kt
N0
0.61 pCi/g
= – (1.2096809x10–4 yr–1)(t)
6.89 pCi/g
–2.4243674 = – (1.2096809x10–4 yr–1(t)
t = 2.0041x104 = 2.0x104 yr
ln
24.124 Plan: Determine the change in mass for the reaction by subtracting the masses of the products from the masses of
the reactants. Use conversion factors to convert the mass change in amu to energy in eV and then to J.
Solution:
m = mass of reactants – mass of products
= (14.003074 + 1.008665) – (14.003241 + 1.007825) = 0.000673 amu
6
 931.5 MeV   10 eV 
5
5
Energy (eV) =  0.000673 amu  

 = 6.268995x10 = 6.27x10 eV

1
amu
1
MeV



 1.602x1019 J   1 kJ   6.022x1023 
Energy (kJ/mol) = 6.268995x105 eV 
  3  


1 eV

  10 J   1 mol 
= 6.0478524x107 = 6.05x107 kJ/mol


24.125 Calculate m.
Mass of 3 1H atoms = 3 x 1.007825 = 3.023475 amu
Mass of 4 neutrons = 4 x 1.008665 = 4.034660 amu
Total mass = 7.058135 amu
m = 7.058135 – 7.016003 = 0.042132 amu/7Li = 0.042132 g/mol 7Li
 0.042132 amu 7 Li   931.5 MeV   106 eV 
7
7
Binding energy = 
 
 = 3.9245958x10 = 3.925 10 eV/nucleus
 

1
atom
1
amu
1
MeV




Convert MeV to kJ.


  1 kJ 
2
 0.042132 g 7 Li   1 kg 
1J
8
Binding energy = 
  3  2.99792x10 m/s 
 3 
2

mol
 kg•m 2   10 J 

  10 g 
s 

= 3.7866237x109 = 3.7866x109 kJ/mol

24.126
146
0
64 Gd + 1e


146
146
0
4
142
63 Eu  62 Sm + 1  2 He + 60 Nd
3 R 
24.127 a) Kinetic energy = 1/2 mv2 = 
T
2  NA 


 3  8.314 J/mol•K  1.00x10 K
1
Energy =  
= 2.07090668x10–17 = 2.07x10–17 J/atom 1H
23
6.022x10 atom/mol
2
b) A kilogram of 1H will annihilate a kilogram of anti-H; thus, two kilograms will be converted to energy:
Energy = mc2 = (2.00 kg)(2.99792x108 m/s)2(J/(kg•m2/s2) = 1.7975048x1017 J
6
 1.7975048x1017 J   1 kg   1.0078 g H  


1 mol H
1 H atom
Number of atoms = 
  3  


17 
23

1 kg H

  10 g   1 mol H   6.022x10 atoms H   2.0709066x10 J 
7
= 1.4525903x10 = 1.45x107 H atoms
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24-28
1
4
0
c) 4 1H  2 He + 2 1
(Positrons have the same mass as electrons.)
4
m = [4(1.007825 amu)] – [4.00260 amu + 2(0.000549 amu)] = 0.027602 amu / 2 He

 0.027602 g   1 kg   1 mol He  
1 mol H
7
m = 
 1.4525903x10 H atoms
 3 

23
mol
He
4
mol
H

  10 g  
  6.022x10 H atoms 
= 1.6644967x10–22 kg
Energy = (m)c2 = (1.6644967x10–22 kg) (2.99792x108 m/s)2(J/(kg•m2/s2)
= 1.4959705x10–5 = 1.4960x10–5 J
d) Calculate the energy generated in part b):
 1.7975048x1017 J   1 kg   1.0078 g H  

1 mol H
–10
Energy = 
  3  
 = 3.0081789x10 J

23

1
kg
H
1
mol
H
  6.022x10 atoms H 

  10 g  
–5
Energy increase = (1.4959705x10 – 3.0081789x10–10 ) J = 1.4959404x10–5 = 1.4959x10–5 J

1
3

0
e) 3 1H  2 He + 11
3
3
m = [3(1.007825 amu)] – [3.01603 amu + 0.000549 amu] = 0.006896 amu/ 2 He = 0.006896 g/mol 2 He

 0.006896 g   1 kg   1 mol He  
1 mol H
7
m = 
 1.4525903x10 H atoms
 3 

23
mol
He
3
mol
H
10
g
6.022x10
H
atoms





= 5.54470426x10–23 kg
Energy = (m)c2 = (5.54470426x10–23 kg)(2.99792x108 m/s)2(J/(kg•m2/s2) = 4.9833164x10–6 = 4.983x10–6 J
No, the Chief Engineer should advise the Captain to keep the current technology.


24.128 Multiply the equation E = hc/ by Avogadro’s number (N) to get the energy change per mole.
6.022x10 /mol 6.626x10 J•s 2.99792x10 m / s  1pm  = 1.3702442x10 J/mol
E = Nhc/ =
34
23
8
 12 
 10 m 
Determine the mass through a series of conversions from the chapter and the inside back cover:
 1.3702442x1010 J  
  1 amu   1.66054x1027 kg 
1 Mev
Mass (kg) = 




 

13


mol
1 amu

  1.602x10 J   931.5 MeV  

= 1.52476x10–7 = 1.52x10–7 kg/mol
8.73pm 
24.129 k =
ln
10
ln 2
ln 2
=
= 0.131526979 yr–1
5.27 yr
t1/2
Nt
= – kt
N0
70.%
= – (0.131526979 yr–1)(t)
100%
– 0.3566749439 = – (0.131526979 yr–1)(t)
t = 2.71180 yr
The source must be replaced after the time calculated above. Two years would be March 1, 2009. Then add
0.7 yr which is (365.25 x 0.7) = 256 d (actually the significant figures limit this to 3x102 d). The final date is
November 12, 2009.
ln
24.130 Plan: The difference in the energies of the two  particles gives the energy of the ray released to get from
excited state I to the ground state. Use E = hc/ to determine the wavelength. The energy of the gamma ray
must be converted from MeV to J. The energy of the 2%  particle is equal to the highest energy  particle minus
the energy of the two  rays (the rays are from excited state II to excited state I, and from excited state I to the
ground state).
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24-29
Solution:
a) Energy of the  ray = (4.816 – 4.773) MeV = 0.043 MeV




6.626x1034 J•s 2.99792x108 m/s  1 MeV 
hc
=
= 2.8836x10–11 = 2.9x10–11 m

13 
E
 0.043 MeV 
1.602x10
J


b) (4.816 – 0.043 – 0.060) MeV = 4.713 MeV
=
24.131 a) Student 2 is correct. The fraction of uranium-238 remaining is equal to the amount of uranium-238 remaining
divided by the initial amount. The initial amount is equal to the sum of the amount of uranium-238 remaining plus
the amount of lead-206 that resulted from the decay of uranium-238.
b) Amount U-238 = 2(Pb-206)
2x
x
238
U
2x
2x
2
Fraction U-238 = 238 92 206
=
=
=
3x
2x  x
3
92 U  82 Pb
ln 2
ln 2
=
= 1.540327x10–10 yr–1
t1/2
4.5x109 yr
N
ln t = – kt
N0
k=
2 
ln  3  = – (1.540327x 0–10 yr–1)(t)
 1 


– 0.405465108 = – (1.540327x10–10 yr–1)(t)
t = 2.63233x109 = 2.6x109 years old
24.132
232
228
4
90Th  88 Ra + 2 
228
228
0
88 Ra  89 Ac + 1
228
228
0
89 Ac  90Th + 1
228
224
4
90Th  88 Ra + 2 
224
220
4
88 Ra  86 Rn + 2 
220
216
4
86 Rn  84 Po + 2 
216
212
4
84 Po  82 Pb + 2 
212
212
0
82 Pb  83 Bi + 1
212
212
0
83 Bi  84 Po + 1
212
208
4
84 Po  82 Pb + 2 
24.133 The age of Egyptian mummies is on the order of a few thousand years. An isotope with a half-life of a few
thousand years would be the best choice. Carbon-14, with a half-life of 5730 yr, would be the best choice.
24.134 Plan: Multiply each of the half-lives by 20 (the number of half-lives is considered to be exact).
Solution:
a) 242Cm
20(163 d) = 3.26x103 d
214
b) Po
20(1.6x10–4 s) = 3.2x10–3 s
232
20(1.39x1010 yr) = 2.78x1011 yr
c) Th
24.135 If the blade was made in 100 AD it is about 1.9x103 years old.
ln 2
ln 2
k=
=
= 1.2096809x10–4 yr–1
5730 yr
t1/2
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24-30
At
= – kt
A0
Handle:
 10.1 d/min•g 
–4
–1
ln 
 = – (1.2096809x10 yr )(t)
15.3
d/min•g


– 0.415317404 = – (1.2096809x10–4 yr–1)(t)
t = 3.43328x103 = 3.43x103 yr
Inlaid:
 13.8 d/min•g 
–4
–1
ln 
 = – (1.2096809x10 yr )(t)
15.3
d/min•g


– 0.103184236 = – (1.2096809x10–4 yr–1)(t)
t = 8.52987x102 = 8.53x102 yr
Ribbon:
 12.1 d/min•g 
–4
–1
ln 
 = – (1.2096809x10 yr )(t)
15.3
d/min•g


– 0.2346473758 = – (1.2096809x10–4 yr–1)(t)
t = 1.939746x103 = 1.94x103 yr
Sheath:
 15.0 d/min•g 
–4
–1
ln 
 = – (1.2096809x10 yr )(t)
15.3
d/min•g


– 0.0198026273 = – (1.2096809x10–4 yr–1)(t)
t = 1.63701x102 = 1.64x102 yr
The ribbon is nearest in age to the blade.
24.136 a) When 1.00 kg of antimatter annihilates 1.00 kg of matter, the change in mass is:
m = 0 – 2.00 kg = –2.00 kg. The energy released is calculated from E = mc2.
E = (–2.00 kg)(2.99792x108 m/s)2(J/(kg•m2/s2) = –1.7975049x1017 = –1.80x1017 J
The negative value indicates the energy is released.
b) Assuming that four hydrogen atoms fuse to form the two protons and two neutrons in one helium atom and
release two positrons, the energy released can be calculated from the binding energy of helium-4.
1
4
0
4 1H  2 He + 2 1
m = [4(1.007825 amu)] – [4.00260 amu + 2(0.000549 amu)]
4
= 0.02760 amu per 2 He formed
ln
4
 0.02760 amu   1 He 
2
Total m = 1.00x105 H atoms 

  4 H atoms  = 6.90x10 amu
4
He





 931.5 MeV   1.602x1016 kJ 
–10
Energy = 6.90 x 102 amu 
 = 1.02966x10 kJ per antiH collision
 
1 MeV
 1 amu  



 103 g   1 mol antiH   6.022x1023 antiH 
= 5.9742x1026 antiH
AntiH atoms = 1.00 kg  
 1 kg   1.008 g antiH   1 mol antiH 




 1.02966x10 10 kJ 
16
16
26
Energy released = 
 5.9742x10 antiH = 6.15139x10 = 6.15x10 kJ

antiH


c) From the above calculations, the procedure in part b) with excess hydrogen produces more energy per kilogram
of antihydrogen.


24.137 Plan: Einstein’s equation is E = mc2, which is modified to E = mc2 to reflect a mass difference. The speed of
light, c, is 2.99792x108 m/s. The mass of exactly one amu is 1.66054x10–27 kg (inside back cover of text). When
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24-31
the quantities are multiplied together, the unit will be kg•m2/s2, which is also the unit of joules. Convert J to MeV
using the conversion factor 1.602x10–13 J = 1 MeV.
Solution:
E = (m)c2



  1 MeV
2
 1.66054x1027 kg  
J

8
E = 1amu  
  2.99792x10 m/s 

13 
2
1 amu

 kg•m 2   1.602x10 J 

 
s 

= 9.3159448x102 = 9.316x102 MeV


24.138 The original amount of uranium is the current amount plus the amount converted to lead. We need to determine
the amount of decayed uranium from the lead in the sample:
 1 mol 206 Pb   1 mol 238 U   238 g 238 U 
= 0.0265728 g 238U
Mass (g) of 238U = 0.023 g 206 Pb 
 206 g 206 Pb   1 mol 206 Pb   1 mol 238 U 




Original mass of 238U = (0.065 + 0.0265728) g 238U = 0.0915728 g 238U
ln 2
ln 2
k=
=
= 1.540327x10–10 yr–1
t1/2
4.5x109 yr

ln

Nt
= – kt
N0
 0.065 g 
–10
–1
ln 
 = – (1.540327x10 yr )(t)
0.0915728
g


– 0.3427470145 = – (1.540327x10–10 yr–1)(t)
t = 2.2251575x109 = 2.2x109 years old
242
238
4
94 Pu  92 U* + 2 
238
238
92 U*  92 U + 
24.139 a)
b) Determine the MeV of the  ray by using E = hc/ to determine the energy.
6.626x10
34


J•s 2.99792x108 m/s  1 nm   1 MeV

= 0.044975 MeV
 9  
13 

 0.02757 nm 
 10 m   1.602x10 J 
Adding the two energies together gives the energy required to go directly from plutonium-242 to the lowest
energy uranium-238:
Total energy = (4.853 + 0.044975) MeV = 4.897975 = 4.898 MeV
E=
24.140
hc
=
249
18
263
1
98 Cf + 8 O  106 Sg + 4 0 n
263
259
4
106 Sg  104 Rf + 2 
259
255
4
104 Rf  102 No + 2 
255
251
4
102 No  100 Fm + 2 
24.141 Plan: The rate of formation of plutonium-239 depends on the rate of decay of neptunium-239 with a half-life of
2.35 d. Calculate k from the half-life equation and use the integrated rate law to find the time necessary to react
90% of the neptunium-239 (10% left).
Solution:
ln 2
ln 2
k=
=
= 0.294956 d–1
t1/2
2.35 d
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-32
ln
Nt
= – kt
N0


 1.00 kg  10.%

100%  = – (0.294956 d–1)(t)

ln


1.00 kg




– 2.3025851 = – (0.294956 d–1)(t)
t = 7.806538 = 7.81 d
24.142 a) Half of the atoms do not remain after each half-life. Decay is a random process. On average half of the atoms
remain after one half-life, but individual simulations may vary.
b) Increasing the number of atoms is more realistic. The number of remaining atoms varies less. Also, a real
radioactive sample consists of many atoms.
24.143 a) Ideally 128 atoms remain after the first half-life, 64 atoms remain after the second, 32 atoms remain after the
third, 16 atoms remain after the fourth, and 8 remain after the fifth.
b) To make the simulation more realistic, make the size of the sample (# of atoms) much larger.
9
10
7
24.144 a) Nucleus 1 is 4 Be ; Nucleus 2 is 4 Be ; Nucleus 3 is 4 Be
b) The n/p ratio for the stable Nucleus 1 = 5/4 = 1.25; the n/p ratio for Nucleus 2 = 6/4 = 1.5. Nucleus 2 has an
n/p ratio that is too high and the most likely mode of decay is beta particle emission. The n/p ratio of
Nucleus 3 = 3/4 = 0.75. This ratio is too low and the expected modes of decay are electron capture and/or
positron emission.
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
24-33