Solutions Manual for
Chemistry The Molecular
Nature of Matter and Change
3rd Canadian Edition By
Martin Silberberg (All
Chapters 1-25, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse 25-1.
,CHAPTER 25 NUCLEAR REACTIONS AND
THEIR APPLICATIONS
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B25.1 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.
B25.2 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 u 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)]u – [4.00260 + (2)(5.48580x10–4)]u
= (4.031300 – 4.003697)u = 0.02760 u /4He atom = 0.02760284 g/mol 4He
0.02760284 u 4 He 931.5 MeV
Energy (MeV/atom) = = 25.7120 MeV/atom = 25.71 MeV/atom
1 atom 1u
Convert atoms to moles using Avogadro’s number.
25.7120 MeV 6.022x10 atoms
23
Energy = = 1.54838x1025 MeV/atom = 1.548x1025 MeV/mol
atom 1 mol
B25.3 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.
B25.4 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:
83 Bi → 84 Po + −1
210 210 0 210
84 Po is nuclide A
210
84 Po → 82 Pb + 2
206 4 206
82 Pb is nuclide B
206
82 Pb + 3 01 n → 209
82 Pb
209
82 Pb is nuclide C
209
82 Pb → 83 Bi + −01
210 210
83 Bi is nuclide D
END–OF–CHAPTER PROBLEMS
25.1 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.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page
25-1
Instructor’s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.
,25.2 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 u. The average mass of all the sulfur atoms
will therefore be greater than the mass of a sulfur-32 atom.
25.3 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.
25.4 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.
Z X → Z − 2Y + 2 He
A A− 4 4
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.
Z X → Z +1Y + −1
A A 0
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.
Z X → Z −1Y + +1
A A 0
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.
Z X + −1 e → Z −1Y
A 0 A
1 less proton, 1 more neutron
A different element is produced in all cases except (c).
25.5 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 N/Z = 1/2
2
2 He N/Z = 0/2, thus it is more unstable.
25.6 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:
0 n → 1 p + −1
1 1 0
The conversion of protons to neutrons occurs by either positron decay:
1 p → 0 n + 1
1 1 0
or electron capture:
1 p + −1 e → 0 n
1 0 1
Neutron-rich nuclides, with a high N/Z, undergo decay. Neutron-poor nuclides, with a low N/Z, undergo
positron decay or electron capture.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page
25-2
Instructor’s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.
, 25.7 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.
25.8 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:
a) The process involves the loss of an particle from the nucleus. For each particle emited, the mass number
decreases by four and the atomic number decreases by two.
92 U → 2 He + 90Th
234 4 230
Mass: 234 = 4 + 230; Charge: 92 = 2 + 90
b) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant
93 Np + −1 e → 92 U
232 0 232
Mass: 232 + 0 = 232; Charge: 93 + (–1) = 92
c) Positron emission decreases atomic number by one, but not mass number.
7 N → 1 + 6 C
12 0 12
Mass: 12 = 0 + 12; Charge: 7 = 1 + 6
25.9 a) 26
11 Na → −1
0
+ 26
12 Mg
b) 223
87 Fr → −1
0
+ 223
88 Ra
c) 212
83 Bi → 2
4
+ 208
81Tl
25.10 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.
12 Mg → −1 + 13 Al Charge: 12 = –1 + 13
27 0 27
Mass: 27 = 0 + 27;
b) Positron emission decreases atomic number by one, but not mass number.
12 Mg → 1 + 11 Na
23 0 23
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.
46 Pd + −1 e → 45 Rh
103 0 103
Mass: 103 + 0 = 103; Charge: 46 + (–1) = 45
14 Si → −1
32 0 32
25.11 a) +
15 P
84 Po → 2
218 4
b) + 214
82 Pb
c) 110 0
49 In + −1 e → 48 Cd
110
25.12 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.
23V → 22Ti + 1
48 48 0
Mass: 48 = 48 + 0; Charge: 23 = 22 + 1
b) In other words, an unknown nuclide captures an electron to form Ag-107.
48 Cd + −1 e → 47 Ag
107 0 107
Mass: 107 + 0 = 107; Charge: 48 + (–1) = 47
c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle.
86 Rn → 84 Po + 2 He
210 206 4
Mass: 210 = 206 + 4; Charge: 86 = 84 + 2
25.13 a) 241
94 Pu → 241
95 Am + −1
0
b) 228
88 Ra → 228
89 Ac + −1
0
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page
25-3
Instructor’s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.
Chemistry The Molecular
Nature of Matter and Change
3rd Canadian Edition By
Martin Silberberg (All
Chapters 1-25, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse 25-1.
,CHAPTER 25 NUCLEAR REACTIONS AND
THEIR APPLICATIONS
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B25.1 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.
B25.2 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 u 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)]u – [4.00260 + (2)(5.48580x10–4)]u
= (4.031300 – 4.003697)u = 0.02760 u /4He atom = 0.02760284 g/mol 4He
0.02760284 u 4 He 931.5 MeV
Energy (MeV/atom) = = 25.7120 MeV/atom = 25.71 MeV/atom
1 atom 1u
Convert atoms to moles using Avogadro’s number.
25.7120 MeV 6.022x10 atoms
23
Energy = = 1.54838x1025 MeV/atom = 1.548x1025 MeV/mol
atom 1 mol
B25.3 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.
B25.4 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:
83 Bi → 84 Po + −1
210 210 0 210
84 Po is nuclide A
210
84 Po → 82 Pb + 2
206 4 206
82 Pb is nuclide B
206
82 Pb + 3 01 n → 209
82 Pb
209
82 Pb is nuclide C
209
82 Pb → 83 Bi + −01
210 210
83 Bi is nuclide D
END–OF–CHAPTER PROBLEMS
25.1 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.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page
25-1
Instructor’s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.
,25.2 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 u. The average mass of all the sulfur atoms
will therefore be greater than the mass of a sulfur-32 atom.
25.3 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.
25.4 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.
Z X → Z − 2Y + 2 He
A A− 4 4
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.
Z X → Z +1Y + −1
A A 0
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.
Z X → Z −1Y + +1
A A 0
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.
Z X + −1 e → Z −1Y
A 0 A
1 less proton, 1 more neutron
A different element is produced in all cases except (c).
25.5 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 N/Z = 1/2
2
2 He N/Z = 0/2, thus it is more unstable.
25.6 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:
0 n → 1 p + −1
1 1 0
The conversion of protons to neutrons occurs by either positron decay:
1 p → 0 n + 1
1 1 0
or electron capture:
1 p + −1 e → 0 n
1 0 1
Neutron-rich nuclides, with a high N/Z, undergo decay. Neutron-poor nuclides, with a low N/Z, undergo
positron decay or electron capture.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page
25-2
Instructor’s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.
, 25.7 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.
25.8 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:
a) The process involves the loss of an particle from the nucleus. For each particle emited, the mass number
decreases by four and the atomic number decreases by two.
92 U → 2 He + 90Th
234 4 230
Mass: 234 = 4 + 230; Charge: 92 = 2 + 90
b) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant
93 Np + −1 e → 92 U
232 0 232
Mass: 232 + 0 = 232; Charge: 93 + (–1) = 92
c) Positron emission decreases atomic number by one, but not mass number.
7 N → 1 + 6 C
12 0 12
Mass: 12 = 0 + 12; Charge: 7 = 1 + 6
25.9 a) 26
11 Na → −1
0
+ 26
12 Mg
b) 223
87 Fr → −1
0
+ 223
88 Ra
c) 212
83 Bi → 2
4
+ 208
81Tl
25.10 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.
12 Mg → −1 + 13 Al Charge: 12 = –1 + 13
27 0 27
Mass: 27 = 0 + 27;
b) Positron emission decreases atomic number by one, but not mass number.
12 Mg → 1 + 11 Na
23 0 23
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.
46 Pd + −1 e → 45 Rh
103 0 103
Mass: 103 + 0 = 103; Charge: 46 + (–1) = 45
14 Si → −1
32 0 32
25.11 a) +
15 P
84 Po → 2
218 4
b) + 214
82 Pb
c) 110 0
49 In + −1 e → 48 Cd
110
25.12 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.
23V → 22Ti + 1
48 48 0
Mass: 48 = 48 + 0; Charge: 23 = 22 + 1
b) In other words, an unknown nuclide captures an electron to form Ag-107.
48 Cd + −1 e → 47 Ag
107 0 107
Mass: 107 + 0 = 107; Charge: 48 + (–1) = 47
c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle.
86 Rn → 84 Po + 2 He
210 206 4
Mass: 210 = 206 + 4; Charge: 86 = 84 + 2
25.13 a) 241
94 Pu → 241
95 Am + −1
0
b) 228
88 Ra → 228
89 Ac + −1
0
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page
25-3
Instructor’s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.
