Chemistry 2nd Edition by
Paul Flowers
Complete Chapter Solutions Manual
are included (Ch 1 to 21)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
Chapter 1 Essential Ideas
Chapter 2 Atoms, Molecules, and Ions
Chapter 3 Composition of Substances and Solutions
Chapter 4 Stoichiometry of Chemical Reactions
Chapter 5 Thermochemistry
Chapter 6 Electronic Structure and Periodic Properties of Elements
Chapter 7 Chemical Bonding and Molecular Geometry
Chapter 8 Advanced Theories of Covalent Bonding
Chapter 9 Gases
Chapter 10 Liquids and Solids
Chapter 11 Solutions and Colloids
Chapter 12 Kinetics
Chapter 13 Fundamental Equilibrium Concepts
Chapter 14 Acid-Base Equilibria
Chapter 15 Equilibria of Other Reaction Classes
Chapter 16 Thermodynamics
Chapter 17 Electrochemistry
Chapter 18 Representative Metals, Metalloids, and Nonmetals
Chapter 19 Transition Metals and Coordination Chemistry
Chapter 20 Organic Chemistry
Chapter 21 Nuclear Chemistry
,OpenStax Chemistry 2e
21.1: Nuclear Structure and Stability
Chemistry 2e
21: Nuclear Chemistry
21.1: Nuclear Structure and Stability
1. Write the following isotopes in hyphenated form (e.g., “carbon-14”)
(a) 24
11 Na
29
(b) 13 Al
73
(c) 36 Kr
194
(d) Ir 77
Solution
(a) sodium-24; (b) aluminum-29; (c) krypton-73; (d) iridium-194
2. Write the following isotopes in nuclide notation (e.g., “ 146 C ”)
(a) oxygen-14
(b) copper-70
(c) tantalum-175
(d) francium-217
Solution
(a) 148 O ; (b) 70 175 217
29 Cu ; (c) 73Ta ; (d) 87 Fr
3. For the following isotopes that have missing information, fill in the missing information to
complete the notation
(a) 34
14 X
36
(b) X P
57
(c) X Mn
121
(d) X 56
Solution
(a) 34 36 57 121
14 Si ; (b) 15 P ; (c) 25 Mn ; (d) 56 Ba
4. For each of the isotopes in Exercise 1, determine the numbers of protons, neutrons, and
electrons in a neutral atom of the isotope.
Solution
(a) sodium-24: 11 protons, 13 neutrons, 11 electrons; (b) aluminum-29: 13 protons, 16 neutrons,
13 electrons; (c) krypton-73: 36 protons, 37 neutrons, 36 electrons; (d) iridium-194: 77 protons,
117 neutrons, 77 electrons
5. Write the nuclide notation, including charge if applicable, for atoms with the following
characteristics:
(a) 25 protons, 20 neutrons, 24 electrons
(b) 45 protons, 24 neutrons, 43 electrons
(c) 53 protons, 89 neutrons, 54 electrons
(d) 97 protons, 146 neutrons, 97 electrons
Solution
142 −1
(a) 45
25 Mn
+1
; (b) 69 +2 243
45 Rh ; (c) 53 I ; (d) 97 Bk
6. Calculate the density of the 24
12 Mg nucleus in g/mL, assuming that it has the typical nuclear
diameter of 1 10 cm and is spherical in shape.
–13
Solution
Page 1 of 3
, OpenStax Chemistry 2e
21.1: Nuclear Structure and Stability
24.31 amu 1.6605 10−24 g amu −1 4.037 10−23 g
= −40
= 8 1016 g cm −3
4
(3.1416)(1 10−13 cm2 )3 5.236 10 cm 3
3
7. What are the two principal differences between nuclear reactions and ordinary chemical
changes?
Solution
Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange
atoms. Nuclear reactions involve much larger energies than chemical reactions and have
measureable mass changes.
8. The mass of the atom 23 11 Na is 22.9898 amu.
(a) Calculate its binding energy per atom in millions of electron volts.
(b) Calculate its binding energy per nucleon.
Solution
(a) The binding energy per atom is calculated from the mass defect, the difference between the
actual mass of the nuclide and its component parts. First, determine the theoretical mass of 23
11 Na ,
which contains 11 protons, 11 electrons, and 12 neutrons:
protons: 11 1.0073 amu = 11.0803 amu
electrons: 11 0.00055 amu = 0.00605 amu
neutrons: 11 1.0087 amu = 12.1044 amu
theoretical mass: 23.1908 amu
mass defect = 23.1908 amu – 22.9898 amu = 0.2010 amu
To use the Einstein conversion, the mass must be expressed in kilograms:
1.6605 × 10−27 kg
mass defect = 0.2010 amu = 3.3376 10–28 kg
1 amu
E = mc2 = (3.3376 10–28 kg)(2.9979 108 m/s)2
= 2.9996 10–11 kg m2/s2 = 2.9996 10–11 J per nucleon;
In terms of megaelectron volts, use the conversion factor:
1 MeV = 1.602189 10–13 J, which gives:
1 MeV
2.996 10–11 J/nucleus = 187.2 MeV per nucleus;
1.602189 × 10 −13 J
1 nucleus
(b) 187.2 MeV/nucleus = 8.140 MeV per nucleon]
23 nucleons
9. Which of the following nuclei lie within the band of stability shown in Figure 21.2?
(a) chlorine-37
(b) calcium-40
(c) 204Bi
(d) 56Fe
(e) 206Pb
(f) 211Pb
(g) 222Rn
(h) carbon-14
Solution
(a), (b), (c), (d), and (e)
10. Which of the following nuclei lie within the band of stability shown in Figure 21.2?
Page 2 of 3
Paul Flowers
Complete Chapter Solutions Manual
are included (Ch 1 to 21)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
Chapter 1 Essential Ideas
Chapter 2 Atoms, Molecules, and Ions
Chapter 3 Composition of Substances and Solutions
Chapter 4 Stoichiometry of Chemical Reactions
Chapter 5 Thermochemistry
Chapter 6 Electronic Structure and Periodic Properties of Elements
Chapter 7 Chemical Bonding and Molecular Geometry
Chapter 8 Advanced Theories of Covalent Bonding
Chapter 9 Gases
Chapter 10 Liquids and Solids
Chapter 11 Solutions and Colloids
Chapter 12 Kinetics
Chapter 13 Fundamental Equilibrium Concepts
Chapter 14 Acid-Base Equilibria
Chapter 15 Equilibria of Other Reaction Classes
Chapter 16 Thermodynamics
Chapter 17 Electrochemistry
Chapter 18 Representative Metals, Metalloids, and Nonmetals
Chapter 19 Transition Metals and Coordination Chemistry
Chapter 20 Organic Chemistry
Chapter 21 Nuclear Chemistry
,OpenStax Chemistry 2e
21.1: Nuclear Structure and Stability
Chemistry 2e
21: Nuclear Chemistry
21.1: Nuclear Structure and Stability
1. Write the following isotopes in hyphenated form (e.g., “carbon-14”)
(a) 24
11 Na
29
(b) 13 Al
73
(c) 36 Kr
194
(d) Ir 77
Solution
(a) sodium-24; (b) aluminum-29; (c) krypton-73; (d) iridium-194
2. Write the following isotopes in nuclide notation (e.g., “ 146 C ”)
(a) oxygen-14
(b) copper-70
(c) tantalum-175
(d) francium-217
Solution
(a) 148 O ; (b) 70 175 217
29 Cu ; (c) 73Ta ; (d) 87 Fr
3. For the following isotopes that have missing information, fill in the missing information to
complete the notation
(a) 34
14 X
36
(b) X P
57
(c) X Mn
121
(d) X 56
Solution
(a) 34 36 57 121
14 Si ; (b) 15 P ; (c) 25 Mn ; (d) 56 Ba
4. For each of the isotopes in Exercise 1, determine the numbers of protons, neutrons, and
electrons in a neutral atom of the isotope.
Solution
(a) sodium-24: 11 protons, 13 neutrons, 11 electrons; (b) aluminum-29: 13 protons, 16 neutrons,
13 electrons; (c) krypton-73: 36 protons, 37 neutrons, 36 electrons; (d) iridium-194: 77 protons,
117 neutrons, 77 electrons
5. Write the nuclide notation, including charge if applicable, for atoms with the following
characteristics:
(a) 25 protons, 20 neutrons, 24 electrons
(b) 45 protons, 24 neutrons, 43 electrons
(c) 53 protons, 89 neutrons, 54 electrons
(d) 97 protons, 146 neutrons, 97 electrons
Solution
142 −1
(a) 45
25 Mn
+1
; (b) 69 +2 243
45 Rh ; (c) 53 I ; (d) 97 Bk
6. Calculate the density of the 24
12 Mg nucleus in g/mL, assuming that it has the typical nuclear
diameter of 1 10 cm and is spherical in shape.
–13
Solution
Page 1 of 3
, OpenStax Chemistry 2e
21.1: Nuclear Structure and Stability
24.31 amu 1.6605 10−24 g amu −1 4.037 10−23 g
= −40
= 8 1016 g cm −3
4
(3.1416)(1 10−13 cm2 )3 5.236 10 cm 3
3
7. What are the two principal differences between nuclear reactions and ordinary chemical
changes?
Solution
Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange
atoms. Nuclear reactions involve much larger energies than chemical reactions and have
measureable mass changes.
8. The mass of the atom 23 11 Na is 22.9898 amu.
(a) Calculate its binding energy per atom in millions of electron volts.
(b) Calculate its binding energy per nucleon.
Solution
(a) The binding energy per atom is calculated from the mass defect, the difference between the
actual mass of the nuclide and its component parts. First, determine the theoretical mass of 23
11 Na ,
which contains 11 protons, 11 electrons, and 12 neutrons:
protons: 11 1.0073 amu = 11.0803 amu
electrons: 11 0.00055 amu = 0.00605 amu
neutrons: 11 1.0087 amu = 12.1044 amu
theoretical mass: 23.1908 amu
mass defect = 23.1908 amu – 22.9898 amu = 0.2010 amu
To use the Einstein conversion, the mass must be expressed in kilograms:
1.6605 × 10−27 kg
mass defect = 0.2010 amu = 3.3376 10–28 kg
1 amu
E = mc2 = (3.3376 10–28 kg)(2.9979 108 m/s)2
= 2.9996 10–11 kg m2/s2 = 2.9996 10–11 J per nucleon;
In terms of megaelectron volts, use the conversion factor:
1 MeV = 1.602189 10–13 J, which gives:
1 MeV
2.996 10–11 J/nucleus = 187.2 MeV per nucleus;
1.602189 × 10 −13 J
1 nucleus
(b) 187.2 MeV/nucleus = 8.140 MeV per nucleon]
23 nucleons
9. Which of the following nuclei lie within the band of stability shown in Figure 21.2?
(a) chlorine-37
(b) calcium-40
(c) 204Bi
(d) 56Fe
(e) 206Pb
(f) 211Pb
(g) 222Rn
(h) carbon-14
Solution
(a), (b), (c), (d), and (e)
10. Which of the following nuclei lie within the band of stability shown in Figure 21.2?
Page 2 of 3
