741 CHAPTER 19 THE NUCLEUS: A CHEMIST'S VIEW Questions 1. Characteristic frequencies of energies emitted in a nuclear reaction suggest that discrete energy levels exist in the nucleus. The extra stability of certain numbers of nucleons and the predominance of nuclei with even numbers of nucleons suggest that the nuclear structure might be described by using quantum numbers. 2. No, coal-fired power plants also pose risks. A partial list of risks is: Coal Nuclear Air pollution Radiation exposure to workers Coal mine accidents Disposal of wastes Health risks to miners Meltdown (black lung disease) Terrorists Public fear 3. Beta-particle production has the net effect of turning a neutron into a proton. Radioactive nuclei having too many neutrons typically undergo -particle decay. Positron production has the net effect of turning a proton into a neutron. Nuclei having too many protons typically undergo positron decay. 4. a. Nothing; binding energy is related to thermodynamic stability, and is not related to kinetics. Binding energy indicates nothing about how fast or slow a specific nucleon decays. b. 56 Fe has the largest binding energy per nucleon, so it is the most stable nuclide. 56 Fe has the greatest mass loss per nucleon when the protons and neutrons are brought together to form the 56 Fe nucleus. The least stable nuclide shown, having the smallest binding energy per nucleon, is 2 H. c. Fusion refers to combining two light nuclei having relatively small binding energies per nucleon to form a heavier nucleus which has a larger binding energy per nucleon. The difference in binding energies per nucleon is related to the energy released in a fusion reaction. Nuclides to the left of 56 Fe can undergo fusion. Nuclides to the right of 56 Fe can undergo fission. In fission, a heavier nucleus having a relatively small binding energy per nucleon is split into two smaller nuclei having larger binding energy per nucleons. The difference in binding energies per nucleon is related to the energy released in a fission reaction.
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741
CHAPTER 19
THE NUCLEUS: A CHEMIST'S VIEW
Questions
1. Characteristic frequencies of energies emitted in a nuclear reaction suggest that discrete
energy levels exist in the nucleus. The extra stability of certain numbers of nucleons and the
predominance of nuclei with even numbers of nucleons suggest that the nuclear structure
might be described by using quantum numbers.
2. No, coal-fired power plants also pose risks. A partial list of risks is:
Coal Nuclear
Air pollution Radiation exposure to workers
Coal mine accidents Disposal of wastes
Health risks to miners Meltdown
(black lung disease) Terrorists
Public fear
3. Beta-particle production has the net effect of turning a neutron into a proton. Radioactive
nuclei having too many neutrons typically undergo -particle decay. Positron production has
the net effect of turning a proton into a neutron. Nuclei having too many protons typically
undergo positron decay.
4. a. Nothing; binding energy is related to thermodynamic stability, and is not related to
kinetics. Binding energy indicates nothing about how fast or slow a specific nucleon
decays.
b. 56
Fe has the largest binding energy per nucleon, so it is the most stable nuclide. 56
Fe has
the greatest mass loss per nucleon when the protons and neutrons are brought together to
form the 56
Fe nucleus. The least stable nuclide shown, having the smallest binding
energy per nucleon, is 2H.
c. Fusion refers to combining two light nuclei having relatively small binding energies per
nucleon to form a heavier nucleus which has a larger binding energy per nucleon. The
difference in binding energies per nucleon is related to the energy released in a fusion
reaction. Nuclides to the left of 56
Fe can undergo fusion.
Nuclides to the right of 56
Fe can undergo fission. In fission, a heavier nucleus having a
relatively small binding energy per nucleon is split into two smaller nuclei having larger
binding energy per nucleons. The difference in binding energies per nucleon is related to
the energy released in a fission reaction.
742 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
5. The transuranium elements are the elements having more protons than uranium. They are
synthesized by bombarding heavier nuclei with neutrons and positive ions in a particle
accelerator.
6. All radioactive decay follows first-order kinetics. A sample is analyzed for the 176
Lu and
176
Hf content, from which the first-order rate law can be applied to determine the age of the
sample. The reason 176
Lu decay is valuable for dating very old objects is the extremely long
half-life. Substances formed a long time ago that have short half-lives have virtually no nuclei
remaining. On the other hand, 176
Lu decay hasn’t even approached one half-life when dating
5-billion-year-old objects.
7. E = mc2; the key difference is the mass change when going from reactants to products. In
chemical reactions, the mass change is indiscernible. In nuclear processes, the mass change is
discernible. It is the conversion of this discernible mass change into energy that results in the
huge energies associated with nuclear processes.
8. Effusion is the passage of a gas through a tiny orifice into an evacuated container. Graham’s
law of effusion says that the effusion of a gas in inversely proportional to the square root of
the mass of its particle. The key to effusion, and to the gaseous diffusion process, is that they
are both directly related to the velocity of the gas molecules, which is inversely related to the
molar mass. The lighter 235
UF6 gas molecules have a faster average velocity than the heavier 238
UF6 gas molecules. The difference in average velocity is used in the gaseous diffusion
process to enrich the 235
U content in natural uranium.
9. The temperatures of fusion reactions are so high that all physical containers would be
destroyed. At these high temperatures, most of the electrons are stripped from the atoms. A
plasma of gaseous ions is formed that can be controlled by magnetic fields.
10. The linear model postulates that damage from radiation is proportional to the dose, even at
low levels of exposure. Thus any exposure is dangerous. The threshold model, on the other
hand, assumes that no significant damage occurs below a certain exposure, called the
threshold exposure. A recent study supported the linear model.
Exercises
Radioactive Decay and Nuclear Transformations
11. All nuclear reactions must be charge balanced and mass balanced. To charge balance,
balance the sum of the atomic numbers on each side of the reaction, and to mass balance,
balance the sum of the mass numbers on each side of the reaction.
a. eHeH 01
32
31 b. eBeLi 0
184
83
eHe2Li
________________He2Be
01
42
83
42
84
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 743
c. LieBe 73
01
74
d. eBeB 01
84
85
12. All nuclear reactions must be charge balanced and mass balanced. To charge balance,
balance the sum of the atomic numbers on each side of the reaction, and to mass balance,
balance the sum of the mass numbers on each side of the reaction.
a. eNiCo 01
6028
6027 b. MoeTc 97
4201
9743
c. eRuTc 01
9944
9943 d. HeUPu 4
223592
23994
13. All nuclear reactions must be charge balanced and mass balanced. To charge balance,
balance the sum of the atomic numbers on each side of the reaction, and to mass balance,
balance the sum of the mass numbers on each side of the reaction.
a. ThHeU 23490
42
23892 ; this is alpha-particle production.
b. ePaTh 01
23491
23490 ; this is -particle production.
14. a. VeCr 5123
01
5124 b. XeeI 131
5401
13153 c. SeP 32
1601
3215
15. a. 6831 Ga + 0
1e 68
30 Zn b. 6229 Cu 0
1 e + 6228 Ni
c. 21287 Fr 4
2 He + 20885
At d. 12951
Sb 01e + 129
52Te
16. a. 7331 Ga 73
32 Ge + 01e b. 192
78Pt 188
76Os + 4
2 He
c. 20583
Bi 20582
Pb + 01 e d. 241
96Cm + 0
1e 241
95Am
17. 23592 U 207
82 Pb + ? 42 He + ? 0
1e
From the two possible decay processes, only alpha-particle decay changes the mass number.
So the mass number change of 28 from 235 to 207 must be done in the decay series by seven
alpha particles. The atomic number change of 10 from 92 to 82 is due to both alpha-particle
production and beta-particle production. However, because we know that seven alpha-par-
ticles are in the complete decay process, we must have four beta-particle decays in order to
balance the atomic number. The complete decay series is summarized as:
23592 U 207
82 Pb + 7 42 He + 4 0
1 e
18. 24797 Bk 207
82 Pb + ? 42 He + 0
1 e; the change in mass number (247 207 = 40) is due
exclusively to the alpha-particles. A change in mass number of 40 requires 10 He42 particles
to be produced. The atomic number only changes by 97 82 = 15. The 10 alpha-particles
change the atomic number by 20, so 5 01 e (5 beta-particles) are produced in the decay series
of 247
Bk to 207
Pb.
744 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
19. a. 24195
Am 42 He + 237
93Np
b. 24195 Am 8
42 He + 4
01 e +
20983 Bi; the final product is 209
83 Bi.
c. 24195 Am 237
93 Np + 23391Pa + 233
92 U + 22990 Th + 225
88 Ra +
21384 Po + 213
83Bi + 21785 At + 221
87 Fr + 22589 Ac +
20982 Pb + 209
83 Bi +
The intermediate radionuclides are:
23793 Np, 233
91 Pa, 23392 U, 229
90 Th, 22588 Ra, 225
89 Ac, 22187 Fr, 217
85 At, 21383 Bi, 213
84 Po, and 20982 Pb
20. The complete decay series is:
23290 Th 228
88 Ra + 42 He 228
89 Ac + 01e 228
90 Th + 01e 224
88 Ra + 42 He
01e + 212
84 Po 01e + 212
83 Bi 42 He + 212
82 Pb 42 He + 216
84 Po 22086 Rn + 4
2 He
20882 Pb + 4
2 He
21. 5326 Fe has too many protons. It will undergo either positron production, electron capture,
and/or alpha-particle production. 5926 Fe has too many neutrons and will undergo beta-particle
production. (See Table 19.2 of the text.)
22. Reference Table 19.2 of the text for potential radioactive decay processes. 17
F and 18
F contain
too many protons or too few neutrons. Electron capture and positron production are both
possible decay mechanisms that increase the neutron to proton ratio. Alpha-particle produc-
tion also increases the neutron-to-proton ratio, but it is not likely for these light nuclei. 21
F
contains too many neutrons or too few protons. Beta-particle production lowers the neutron-
to-proton ratio, so we expect 21
F to be a beta-emitter.
23. a. 24998 Cf + 18
8 O 263106Sg + 4 1
0 n b. 259104 Rf ; 263
106Sg 42 He + 259
104 Rf
24. a. 24095 Am + 4
2 He 24397 Bk + 1
0 n b. 23892 U + 12
6 C 24498 Cf + 6 1
0 n
c. 24998 Cf + 15
7 N 260105 Db + 4 1
0 n d. 24998 Cf + 10
5 B 257103 Lr + 2 1
0 n
Kinetics of Radioactive Decay
25. All radioactive decay follows first-order kinetics where t1/2 = (ln 2)/k.
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 745
t1/2 = 13 h100.1
693.0
k
2ln
= 690 h
26. k = s3600
h1
h24
d1
d365
yr1
yr433
69315.0
t
2ln
2/1
= 5.08 × 111 s10
Rate = kN = 5.08 × 111 s10 × 5.00 g mol
nuclei10022.6
g241
mol1 23
= 6.35 × 1011
decays/s
6.35 × 1011
alpha particles are emitted each second from a 5.00-g 241
Am sample.
27. Kr-81 is most stable because it has the longest half-life. Kr-73 is hottest (least stable); it
decays most rapidly because it has the shortest half-life.
12.5% of each isotope will remain after 3 half-lives:
t1/2
100% 50%t1/2
25%t1/2
12.5%
For Kr-73: t = 3(27 s) = 81 s; for Kr-74: t = 3(11.5 min) = 34.5 min
For Kr-76: t = 3(14.8 h) = 44.4 h; for Kr-81: t = 3(2.1 × 105 yr) = 6.3 × 10
5 yr
28. a. k = s3600
h1
h24
d1
d8.12
6931.0
t
2ln
2/1
= 6.27 × 17 s10
b. Rate = kN = 6.27 × 17 s10
mol
nuclei10022.6
g0.64
mol1g100.28
233
Rate = 1.65 × 1014
decays/s
c. 25% of the 64
Cu will remain after 2 half-lives (100% decays to 50% after one half-life,
which decays to 25% after a second half-life). Hence 2(12.8 days) = 25.6 days is the time
frame for the experiment.
29. Units for N and N0 are usually number of nuclei but can also be grams if the units are
identical for both N and N0. In this problem, m0 = the initial mass of 47
Ca2+
to be ordered.
31.0d5.4
)d0.2(693.0
m
Caμg0.5ln,
t
t)693.0(kt
N
Nln;
t
2lnk
0
2
2/102/1
0m
0.5 = e
−0.31 = 0.73, m0 = 6.8 µg of
47Ca
2+ needed initially
6.8 µg 47
Ca2+
× 247
347
Cagμ0.47
CaCOgμ0.107 = 15 µg
47CaCO3 should be ordered at the minimum.
746 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
30. a. 0.0100 Ci × Ci
s/decays107.3 10 = 3.7 × 10
8 decays/s; k =
2/1t
2ln
Rate = kN,
s3600
h1
h87.2
6931.0
s
decays107.3 8
× N, N = 5.5 × 1012
atoms of 38
S
5.5 × 1012
atoms 38
S × Smol
SONamol1
atoms1002.6
Smol138
4
38
2
23
38
= 9.1 × 1210 mol Na238
SO4
9.1 × 1210 mol Na238
SO4 × 4
38
2
4
38
2
SONamol
SONag0.148 = 1.3 × 910 g = 1.3 ng Na2
38SO4
b. 99.99% decays, 0.01% left;
100
01.0ln = −kt =
h87.2
t)6931.0( , t = 38.1 hours 40 hours
31. t = 68.0 yr; k = 2/1t
2ln;
0N
Nln = −kt =
yr28.9
yr068 )69310( .. = −1.63,
0N
N = e
−1.63 = 0.196
19.6% of the 90
Sr remains as of July 16, 2013.
32. Assuming 2 significant figures in 1/100:
ln(N/N0) = −kt; N = (0.010)N0; t1/2 = (ln 2)/k
ln(0.010) = d0.8
t)693.0(
t
t)2(ln
2/1
, t = 53 days
33. k = ;t
2)t(lnkt
N
Nln;
t
2ln
1/202/1
h0.6
)h0.48)(693.0(
N
Nln
0
= 5.5
5.5
0
eN
N 0.0041; the fraction of
99Tc that remains is 0.0041, or 0.41%.
34. 175 mg Na332
PO4 4
32
3
32
PONamg0.165
Pmg0.32 = 33.9 mg
32P;
2/1t
2lnk
d3.14
)d0.35(6931.0
mg9.33
mln,
t
t)6931.0(kt
N
Nln
2/10
; carrying extra sig. figs.:
ln(m) = 1.696 + 3.523 = 1.827, m = e1.827
= 6.22 mg 32
P remains
35.
0N
Nln = −kt =
2/1t
t)2(ln,
0m
g0.1ln =
min100.1
h
min60
d
h24d0.3693.0
3
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 747
0m
g0.1ln = −3.0,
0.3
0
em
0.1 , m0 = 20. g 82
Br needed
20. g 82
Br BrNamol
BrNag0.105
Brmol
BrNamol1
g0.82
Brmol182
82
82
8282
= 26 g Na82
Br
36. Assuming the current year is 2013, t = 67 yr.
0N
Nln = −kt =
2/1t
t)693.0(,
5.5
Nln = ,
yr12.3
yr)0.693(67 N =
waterg100.min
eventsdecay0.13
37. k = 2/1t
2ln;
0N
Nln = −kt =
yr5730
)yr000,15(693.0
6.13
Nln,
t
t)693.0(
2/1
= −1.8
6.13
N= 8.1e = 0.17, N = 13.6 × 0.17 = 2.3 counts per minute per g of C
If we had 10. mg C, we would see:
10. mg × min
counts023.0
gmin
counts3.2
mg1000
g1
It would take roughly 40 min to see a single disintegration. This is too long to wait, and the
background radiation would probably be much greater than the 14
C activity. Thus 14
C dating
is not practical for very small samples.
38.
0N
Nln = −kt =
2/1t
t)6931.0(,
6.13
2.1ln =
yr5730
t)6931.0(, t = 2.0 × 10
4 yr
39. Assuming 1.000 g 238
U present in a sample, then 0.688 g 206
Pb is present. Because 1 mol 206
Pb
is produced per mol 238
U decayed:
238
U decayed = 0.688 g Pb × Umol
Ug238
Pbmol
Umol1
Pbg206
Pbmol1 = 0.795 g
238U
Original mass 238
U present = 1.000 g + 0.795 g = 1.795 g 238
U
0N
Nln = −kt =
yr105.4
)t(693.0
g795.1
g000.1ln,
t
t)2(ln9
2/1
, t = 3.8 × 10
9 yr
40. a. The decay of 40
K is not the sole source of 40
Ca.
b. Decay of 40
K is the sole source of 40
Ar and no 40
Ar is lost over the years.
c. Kg00.1
Arg95.040
40
= current mass ratio
0.95 g of 40
K decayed to 40
Ar. 0.95 g of 40
K is only 10.7% of the total 40
K that decayed,
or:
748 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
0.107(m) = 0.95 g, m = 8.9 g = total mass of
40K that decayed
Mass of 40
K when the rock was formed was 1.00 g + 8.9 g = 9.9 g.
Kg9.9
Kg00.1ln
40
40
= −kt =yr1027.1
t)6931.0(
t
t)2(ln9
2/1
, t = 4.2 × 10
9 years old
d. If some 40
Ar escaped, then the measured ratio of 40
Ar/40
K is less than it should be. We
would calculate the age of the rock to be less than it actually is.
Energy Changes in Nuclear Reactions
41. ΔE = Δmc2, Δm =
28
2223
2 m/s)10(3.00
/smkg103.9
c
E
= 4.3 × 10
6 kg
The sun loses 4.3 × 106 kg of mass each second. Note: 1 J = 1 kg m
2/s
2
42. day
h24
h
s3600
kJ
J1000
s
kJ108.1 14
= 1.6 × 1022
J/day
ΔE = Δmc2, Δm
28
22
2 m/s)10(3.00
J101.6
c
EΔ 1.8 × 10
5 kg of solar material provides
1 day of solar energy to the earth
1.6 × 1022
J g1000
kg1
kJ32
g1
J1000
kJ1 = 5.0 × 10
14 kg of coal is needed to provide the
same amount of energy
43. We need to determine the mass defect Δm between the mass of the nucleus and the mass of
the individual parts that make up the nucleus. Once Δm is known, we can then calculate ΔE
(the binding energy) using E = mc2. Note: 1 J = 1 kg m
2/s
2.
For 23294 Pu (94 e, 94 p, 138 n):
mass of 232
Pu nucleus = 3.85285 × 2210 g − mass of 94 electrons
mass of 232
Pu nucleus = 3.85285 × 2210 g − 94(9.10939 × 2810 ) g = 3.85199 × 2210 g
Δm = 3.85199 × 2210 g − (mass of 94 protons + mass of 138 neutrons)
Δm = 3.85199 × 2210 g − [94(1.67262 × 2410 ) + 138(1.67493 × 2410 )] g
= −3.168 × 2410 g
For 1 mol of nuclei: Δm = −3.168 × 2410 g/nuclei × 6.0221 × 1023
nuclei/mol
= −1.908 g/mol
ΔE = Δmc2 = (−1.908 × 310 kg/mol)(2.9979 × 10
8 m/s)
2 = −1.715 × 10
14 J/mol
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 749
For 23191Pa (91 e, 91 p, 140 n):
mass of 231
Pa nucleus = 3.83616 × 2210 g − 91(9.10939 × 10-28
) g = 3.83533 × 2210 g
Δm = 3.83533 × 2210 g − [91(1.67262 × 2410 ) + 140(1.67493 × 2410 )] g
= −3.166 × 2410 g
ΔE = Δmc2 =
282327
s
m109979.2
mol
nuclei100221.6
nuclei
kg10166.3
= 1.714 × 1014
J/mol
44. From the text, the mass of a proton = 1.00728 u, the mass of a neutron = 1.00866 u, and the
mass of an electron = 5.486 × 104 u.
Mass of Fe5626 nucleus = mass of atom mass of electrons = 55.9349 26(0.0005486)
= 55.9206 u
;Fen30H26 5626
10
11 Δm = 55.9206 u [26(1.00728) + 30(1.00866)] u = 0.5285 u
ΔE = Δmc2 = 0.5285 u
u
kg101.6605 27 (2.9979 × 10
8 m/s)
2 = 7.887 × 1011 J
nucleons56
J10887.7
Nucleon
energyBinding 11
1.408 × 1012 J/nucleon
45. Let me = mass of electron; for 12
C (6e, 6p, and 6n): Mass defect = Δm = [mass of 12
C
nucleus] [mass of 6 protons + mass of 6 neutrons]. Note: Atomic masses given include the
mass of the electrons.
Δm = 12.00000 u 6me [6(1.00782 me) + 6(1.00866)]; mass of electrons cancel.
Δm = 12.00000 [6(1.00782) + 6(1.00866)] = 0.09888 u