CHAPTER 19 THE NUCLEUS: A CHEMIST'S VIEW · CHAPTER 19 THE NUCLEUS: A CHEMIST'S VIEW Questions 1....
Transcript of CHAPTER 19 THE NUCLEUS: A CHEMIST'S VIEW · CHAPTER 19 THE NUCLEUS: A CHEMIST'S VIEW Questions 1....
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
ΔE = Δmc2 = 0.09888 u
u
kg101.6605 27 (2.9979 × 10
8 m/s)
2 = 1.476 × 1011
J
nucleons12
J101.476
Nucleon
energyBinding 11
1.230 × 1012 J/nucleon
For 235
U (92e, 92p, and 143n):
Δm = 235.0439 92me [92(1.00782 me) + 143(1.00866)] = 1.9139 u
ΔE = Δmc2 = 1.9139 ×
u
kg101.66054 27 (2.99792 × 10
8 m/s)
2 = 2.8563 × 1010
J
750 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
nucleons235
J102.8563
Nucleon
energyBinding 10
1.2154 × 1012 J/nucleon
Because 56
Fe is the most stable known nucleus, the binding energy per nucleon for 56
Fe
(1.408 × 1012 J/nucleon) will be larger than that of
12C or
235U (see Figure 19.9 of the text).
46. For H21 : Mass defect = Δm = mass of H2
1 nucleus mass of proton mass of neutron. The
mass of the 2H nucleus will equal the atomic mass of
2H minus the mass of the electron in an
2H atom. From the text, the pertinent masses are me = 5.49 × 104
u, mp = 1.00728 u, and mn
= 1.00866 u.
Δm = 2.01410 u 0.000549 u (1.00728 u + 1.00866 u) = 2.39 × 103 u
ΔE = Δmc2 = 2.39 × 103
u × u
kg101.6605 27× (2.998 × 10
8 m/s)
2 = 3.57 × 1013
J
nucleons2
J1057.3
Nucleon
energyBinding 13
1.79 × 1013 J/nucleon
For H31 : Δm = 3.01605 0.000549 [1.00728 + 2(1.00866)] = 9.10 × 103
u
ΔE = 9.10 × 103 u ×
u
kg101.6605 27 × (2.998 × 10
8 m/s)
2 = 1.36 × 1012
J
nucleons3
J101.36
Nucleon
energyBinding 12
4.53 × 1013 J/nucleon
47. Let mLi = mass of 6Li nucleus; an
6Li nucleus has 3p and 3n.
−0.03434 u = mLi − (3mp + 3mn) = mLi − [3(1.00728 u) + 3(1.00866 u)]
mLi = 6.01348 u
Mass of 6Li atom = 6.01348 u + 3me = 6.01348 + 3(5.49 × 410 u) = 6.01513 u
(includes mass of 3 e)
48. Binding energy = nucleon
J10326.1 12 × 27 nucleons = 3.580 × 1110 J for each
27Mg nucleus
ΔE = Δmc2, Δm =
2c
EΔ =
28
11
)m/s109979.2(
J10580.3
= −3.983 2810 kg
Δm = −3.983 2810 kg × kg101.6605
u127
= −0.2399 u = mass defect
Let mMg = mass of 27
Mg nucleus; an 27
Mg nucleus has 12 p and 15 n.
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 751
−0.2399 u = mMg − (12mp + 15mn) = mMg − [12(1.00728 u) + 15(1.00866 u)]
mMg = 26.9764 u
Mass of 27
Mg atom = 26.9764 u + 12me, 26.9764 + 12(5.49 × 410 u) = 26.9830 u
(includes mass of 12 e)
49. 11 H + 1
1 H H21 +
01 e; Δm = (2.01410 u − me + me) − 2(1.00782 u − me)
Δm = 2.01410 − 2(1.00782) + 2(0.000549) = −4.4 × 410 u for two protons reacting
When 2 mol of protons undergoes fusion, Δm = −4.4 × 410 g.
ΔE = Δmc2 = −4.4 × 710 kg × (3.00 × 10
8 m/s)
2 = −4.0 × 10
10 J
g01.1
mol1
protonsmol2
J100.4 10
= −2.0 × 1010
J/g of hydrogen nuclei
50. H21 + H3
1 He42 +
10 n; using atomic masses, the masses of the electrons cancel when
determining Δm for this nuclear reaction.
Δm = [4.00260 + 1.00866 − (2.01410 + 3.01605)] u = −1.889 × 210 u
For the production of 1 mol of He42 : Δm = −1.889 × 210 g = −1.889 × 510 kg
ΔE = Δmc2 = 1.889 × 10
-5 kg × (2.9979 × 10
8 m/s)
2 = 1.698 × 10
12 J/mol
For 1 nucleus of He42
nuclei100221.6
mol1
mol
J10698.123
12
= −2.820 × 1210 J/nucleus
Detection, Uses, and Health Effects of Radiation
51. The Geiger-Müller tube has a certain response time. After the gas in the tube ionizes to
produce a "count," some time must elapse for the gas to return to an electrically neutral state.
The response of the tube levels off because at high activities, radioactive particles are
entering the tube faster than the tube can respond to them.
52. Not all of the emitted radiation enters the Geiger-Müller tube. The fraction of radiation
entering the tube must be constant.
53. Water is produced in this reaction by removing an OH group from one substance and H from
the other substance. There are two ways to do this:
752 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
Because the water produced is not radioactive, methyl acetate forms by the first reaction in
which all the oxygen-18 ends up in methyl acetate.
54. The only product in the fast-equilibrium step is assumed to be N16
O18
O2, where N is the
central atom. However, this is a reversible reaction where N16
O18
O2 will decompose to NO
and O2. Because any two oxygen atoms can leave N16
O18
O2 to form O2, we would expect (at
equilibrium) one-third of the NO present in this fast equilibrium step to be N16
O and two-
thirds to be N18
O. In the second step (the slow step), the intermediate N16
O18
O2 reacts with
the scrambled NO to form the NO2 product, where N is the central atom in NO2. Any one of
the three oxygen atoms can be transferred from N16
O18
O2 to NO when the NO2 product is
formed. The distribution of 18
O in the product can best be determined by forming a
probability table.
N16
O (1/3) N18
O (2/3) 16
O (1/3) from N16
O18
O2 N16
O2 (1/9) N18
O16
O (2/9) 18
O (2/3) from N16
O18
O2 N16
O18
O (2/9) N18
O2 (4/9)
From the probability table, 1/9 of the NO2 is N16
O2, 4/9 of the NO2 is N18
O2, and 4/9 of the
NO2 is N16
O18
O (2/9 + 2/9 = 4/9). Note: N16
O18
O is the same as N18
O16
O. In addition,
N16
O18
O2 is not the only NO3 intermediate formed; N16
O218
O and N18
O3 can also form in the
fast-equilibrium first step. However, the distribution of 18
O in the NO2 product is the same as
calculated above, even when these other NO3 intermediates are considered.
55. 23592 U + 1
0 n 14458 Ce + 90
38 Sr + ? 10 n + ? 0
1 e; to balance the atomic number, we need 4
beta-particles, and to balance the mass number, we need 2 neutrons.
56. 23892 U + 1
0 n 23992 U 0
1 e + 23993 Np 0
1 e + 23994 Pu; plutonium-239 is the
fissionable material in breeder reactors.
57. Release of Sr is probably more harmful. Xe is chemically unreactive. Strontium is in the same
family as calcium and could be absorbed and concentrated in the body in a fashion similar to
Ca. This puts the radioactive Sr in the bones; red blood cells are produced in bone marrow.
Xe would not be readily incorporated into the body.
The chemical properties determine where a radioactive material may be concentrated in the
body or how easily it may be excreted. The length of time of exposure and what is exposed to
radiation significantly affects the health hazard. (See Exercise 58 for a specific example.)
18+ HO HCH3C OCH3
O
18+ H OCH3CH3C OH
O
H OHCH3CO CH3
O
H O CH3 ++CH3CO H
O
ii.
i.
18 18
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 753
58. (i) and (ii) mean that Pu is not a significant threat outside the body. Our skin is sufficient to
keep out the alpha-particles. If Pu gets inside the body, it is easily oxidized to Pu4+
(iv), which
is chemically similar to Fe3+
(iii). Thus Pu4+
will concentrate in tissues where Fe3+
is found.
One of these is the bone marrow, where red blood cells are produced. Once inside the body,
alpha-particles cause considerable damage.
Additional Exercises
59. The most abundant isotope is generally the most stable isotope. The periodic table predicts
that the most stable isotopes for exercises a-d are 39
K, 56
Fe, 23
Na, and 204
Tl. (Reference Table
19.2 of the text for potential decay processes.)
a. Unstable; 45
K has too many neutrons and will undergo beta-particle production.
b. Stable
c. Unstable; 20
Na has too few neutrons and will most likely undergo electron capture or
positron production. Alpha-particle production makes too severe of a change to be a
likely decay process for the relatively light 20
Na nuclei. Alpha-particle production
usually occurs for heavy nuclei.
d. Unstable; 194
Tl has too few neutrons and will undergo electron capture, positron
production, and/or alpha-particle production.
60. a. Cobalt is a component of vitamin B12. By monitoring the cobalt-57 decay, one can study
the pathway of vitamin B12 in the body.
b. Calcium is present in the bones in part as Ca3(PO4)2. Bone metabolism can be studied by
monitoring the calcium-47 decay as it is taken up in bones.
c. Iron is a component of hemoglobin found in red blood cells. By monitoring the iron-59
decay, one can study red blood cell processes.
61. N = 180 lb Cg14
Cmol1
Cg100
Cg106.1
bodyg100
Cg18
lb
g6.45314
141410
Cnuclei100.1Cmol
Cnuclei10022.6 1415
14
1423
Rate = kN; k = 112
1/2
s108.3s3600
h1
h24
d1
d365
yr1
yr5730
693.0
t
2ln
Rate = kN; k = decays/s3800)nucleiC100.1(s108.3 1415112
A typical 180 lb person produces 3800 beta particles each second.
754 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
62. t1/2 = 5730 yr; k = (ln 2)/t1/2; ln(N/N0) = −kt; 3.15
1.15ln =
yr5730
t)2(ln, t = 109 yr
No; from 14
C dating, the painting was produced during the early 1900s.
63. The third-life will be the time required for the number of nuclides to reach one-third of the
original value (N0/3).
0N
Nln = kt =
2/1t
t)6931.0(,
3
1ln = ,
yr4.31
t)6931.0( t = 49.8 yr
The third-life of this nuclide is 49.8 years.
64. ln(N/N0) = −kt; k = (ln 2)/t1/2 ; N = 0.001 × N0
0
0
N
N001.0ln = ,
yr100,24
t)2(ln ln(0.001) = −(2.88 × 105
)t, t = 2 × 105 yr = 200,000 yr
65.
0N
Nln = −kt = ,
yr3.12
t)2(ln
0
0
N
N17.0ln = −(5.64 × 210 )t, t = 31.4 yr
It takes 31.4 years for the tritium to decay to 17% of the original amount. Hence the watch
stopped fluorescing enough to be read in 1975 (1944 + 31.4).
66. Δm = −2(5.486 × 10-4
u) = −1.097 × 310 u
ΔE = Δmc2 = −1.097 × 310 u ×
u
kg101.6605 27 × (2.9979 × 10
8 m/s)
2
= −1.637 × 1310 J
Ephoton = 1/2(1.637 × 1310 J) = 8.185 × 1410 J = hc/λ
λ = E
hc =
J108.185
m/s102.9979sJ106.626114
834
= 2.427 × 1210 m = 2.427 × 310 nm
67. 20,000 ton TNT × Umol
Ug235
J102
Umol1
TNTton
J104235
235
13
2359
= 940 g 235
U 900 g 235
U
This assumes that all of the 235
U undergoes fission.
68. In order to sustain a nuclear chain reaction, the neutrons produced by the fission must be
contained within the fissionable material so that they can go on to cause other fissions. The
fissionable material must be closely packed together to ensure that neutrons are not lost to the
outside. The critical mass is the mass of material in which exactly one neutron from each
fission event causes another fission event so that the process sustains itself. A supercritical
situation occurs when more than one neutron from each fission event causes another fission
event. In this case, the process rapidly escalates and the heat build up causes a violent
explosion.
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 755
69. Mass of nucleus = atomic mass – mass of electron = 2.01410 u – 0.000549 u = 2.01355 u
urms =
1/2
M
RT3
=
1/27
g)kg/1000g(12.01355
K)10mol)(4J/K3(8.3145
= 7 × 10
5 m/s
KEavg =
u
kg101.66u2.01355
2
1mu
2
127
2 (7 × 105 m/s)
2 = 8 × 1610 J/nuclei
We could have used KEave = (3/2)RT to determine the same average kinetic energy.
70. HmassHmass;HnH2nH 11
11
11
10
11
10
11 1.00728 u = mass of proton = mp
Δm = 3mp + mn (mp + mn) = 2mp = 2(1.00728) = 2.01456 u
ΔE = Δmc2 = 2.01456 amu ×
amu
kg1066056.1 27 × (2.997925 × 10
8 m/s)
2
ΔE = 3.00660 × 10−10 J of energy is absorbed per nuclei, or 1.81062 × 10
14 J/mol nuclei.
The source of energy is the kinetic energy of the proton and the neutron in the particle
accelerator.
71. All evolved oxygen in O2 comes from water and not from carbon dioxide.
72. Sr-90 is an alkaline earth metal having chemical properties similar to calcium. Sr-90 can
collect in bones, replacing some of the calcium. Once embedded inside the human body, beta-
particles can do significant damage. Rn-222 is a noble gas, so one would expect Rn to be
unreactive and pass through the body quickly; it does. The problem with Rn-222 is the rate at
which it produces alpha-particles. With a short half-life, the few moments that Rn-222 is in
the lungs, a significant number of decay events can occur; each decay event produces an
alpha-particle that is very effective at causing ionization and can produce a dense trail of
damage.
ChemWork Problems
The answers to the problems 73-78 (or a variation to these problems) are found in OWL. These
problems are also assignable in OWL.
Challenge Problems
79. k =
02/1 N
Nln;
t
2ln
1/2t
(0.693)tkt
For 238
U: 50.0eN
N,693.0
yr105.4
)yr105.4)(693.0(
N
Nln 693.0
09
9
0
756 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
For 235
U: 012.0eN
N,39.4
yr101.7
)yr105.4)(693.0(
N
Nln 39.4
08
9
0
If we have a current sample of 10,000 uranium nuclei, 9928 nuclei of 238
U and 72 nuclei of 235
U are present. Now let’s calculate the initial number of nuclei that must have been present
4.5 × 109 years ago to produce these 10,000 uranium nuclei.
For 238
U: nucleiU100.250.0
nuclei9928
50.0
NN,50.0
N
N 23840
0
For 235
U: nucleiU100.6012.0
nuclei72
012.0
NN 2353
0
So 4.5 billion years ago, the 10,000-nuclei sample of uranium was composed of 2.0 × 104
238U nuclei and 6.0 × 10
3
235U nuclei. The percent composition 4.5 billion years ago would
have been:
100nuclei total)102.010(6.0
nucleiU102.043
2384
= 77%
238U and 23%
235U
80. Total activity injected = 86.5 × 103
Ci
Activity withdrawn OHmL
Ci101.8
OHmL2.0
Ci103.6
2
6
2
6
Assuming no significant decay occurs, then the total volume of water in the body multiplied
by 1.8 × 106
Ci/mL must equal the total activity injected.
V × OHmL
Ci108.1
2
6 = 8.65 × 10
2 Ci, V = 4.8 × 10
4 mL H2O
Assuming a density of 1.0 g/mL for water, the mass percent of water in this 150-lb person is:
100lb150
g453.6
lb1
mL
OHg1.0OHmL104.8 2
24
= 71%
81. Assuming that the radionuclide is long-lived enough that no significant decay occurs during
the time of the experiment, the total counts of radioactivity injected are:
0.10 mL × mL
cpm100.5 3 = 5.0 × 10
2 cpm
Assuming that the total activity is uniformly distributed only in the rat’s blood, the blood
volume is:
V × mL
cpm48= 5.0 × 10
2 cpm, V = 10.4 mL = 10. mL
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 757
82. a. From Table 18.1: 2 H2O + 2 e → H2 + 2 OH
E° = − 0.83 V
oZr
oOH
ocell EEE
2 = − 0.83 V + 2.36 V = 1.53 V
Yes, the reduction of H2O to H2 by Zr is spontaneous at standard conditions because
ocellE > 0.
b. (2 H2O + 2 e → H2 + 2 OH
) × 2
Zr + 4 OH → ZrO2•H2O + H2O + 4 e
3 H2O(l) + Zr(s) → 2 H2(g) + ZrO2•H2O(s)
c. ΔG° = −nFE° = −(4 mol e)(96,485 C/mol e
)(1.53 J/C) = −5.90 × 10
5 J = −590. kJ
E = E° − n
0591.0log Q; at equilibrium, E = 0 and Q = K.
E° = n
0591.0log K, log K =
0591.0
)53.1(4 = 104, K 10
104
d. 1.00 × 103 kg Zr ×
Zrmol
Hmol2
Zrg22.91
Zrmol1
kg
g1000 2 = 2.19 × 104 mol H2
2.19 × 104 mol H2 ×
2
2
Hmol
Hg016.2 = 4.42 × 10
4 g H2
V = atm1.0
K)K)(1273atm/molL082060mol)(10(2.19
P
nRT4
. = 2.3 × 10
6 L H2
e. Probably yes; less radioactivity overall was released by venting the H2 than what would
have been released if the H2 had exploded inside the reactor (as happened at Chernobyl).
Neither alternative is pleasant, but venting the radioactive hydrogen is the less unpleasant
of the two alternatives.
83. a. 12
C; it takes part in the first step of the reaction but is regenerated in the last step. 12
C is
not consumed, so it is not a reactant.
b. 13
N, 13
C, 14
N, 15
O, and 15
N are the intermediates.
c. 4 11 H → He4
2 + 2 e01 ; Δm = 4.00260 u − 2 me + 2 me − [4(1.00782 u me)]
Δm = 4.00260 − 4(1.00782) + 4(0.000549) = −0.02648 u for four protons reacting
For 4 mol of protons, Δm = −0.02648 g, and ΔE for the reaction is:
ΔE = Δmc2 = − 2.648 × 510 kg × (2.9979 × 10
8 m/s)
2 = −2.380 × 10
12 J
For 1 mol of protons reacting: Hmol4
J10380.21
12 = −5.950 × 10
11 J/mol
1H
758 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW
84. a. 23892 U 222
86 Rn + ? 42 He + ? e0
1 ; to account for the mass number change, four alpha-
particles are needed. To balance the number of protons, two beta-particles are needed.
222
86 Rn 42 He + 218
84 Po; polonium-218 is produced when 222
Rn decays.
b. Alpha-particles cause significant ionization damage when inside a living organism.
Because the half-life of 222
Rn is relatively short, a significant number of alpha-particles
will be produced when 222
Rn is present (even for a short period of time) in the lungs.
c. 22286 Rn 4
2 He + 21884 Po; 218
84 Po 42 He + 214
82 Pb; polonium-218 is produced when
radon-222 decays. 218
Po is a more potent alpha-particle producer since it has a much
shorter half-life than 222
Rn. In addition, 218
Po is a solid, so it can get trapped in the lung
tissue once it is produced. Once trapped, the alpha-particles produced from polonium-
218 (with its very short half-life) can cause significant ionization damage.
d. Rate = kN; rate = Ci
sec/decays107.3
pCi
Ci101
L
pCi0.4 1012
= 0.15 decays/s•L
k = s3600
h1
h24
d1
d82.3
6391.0
t
2ln
2/1
= 2.10 × 16 s10
N = 16 s102.10
decays/s0.15
K
rate L
= 7.1 × 10
4 222
Rn atoms/L
atoms1002.6
Rnmol1
L
atomsRn101.723
2222224
= 1.2 ×
1910 mol 222
Rn/L
85. Moles of I =
counts105.0
minImol1
min
counts3311
= 6.6 × 1110 mol I
[I] =
L150.0
Imol106.6 11 = 4.4 × 1010 mol/L
Hg2I2(s) Hg22+
(aq) + 2 I(aq) Ksp = [Hg2
2+][I
]
2
Initial s = solubility (mol/L) 0 0
Equil. s 2s
From the problem, 2s = 4.4 × 1010 mol/L, s = 2.2 × 1010 mol/L.
Ksp = (s)(2s)2 = (2.2 × 1010 )(4.4 × 1010 )
2 = 4.3 × 2910
86. 21 H +
21 H 4
2 He; Q for 21 H = 1.6 ×
1910 C; mass of deuterium = 2 u.
E = r
)Q(Qm/CJ109.0 2129
=
m102
C)10(1.6m/CJ109.015
21929
= 1 × 1310 J per alpha particle
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 759
KE = 1/2 mv2; 1 × 1310 J = 1/2 (2 u × 1.66 × 2710 kg/u)v
2, v = 8 × 10
6 m/s
From the kinetic molecular theory discussed in Chapter 5:
urms = ,M
RT32/1
where M = molar mass in kilograms = 2 × 310 kg/mol for deuterium
8 × 106 m/s =
1/2
3 kg102
mol)(T)J/K3(8.3145
, T = 5 × 10
9 K
Integrative Problems
87. 24997 Bk + 22
10 Ne 267107 Bh + ?; this equation is charge balanced, but it is not mass balanced.
The products are off by 4 mass units. The only possibility to account for the 4 mass units is
to have 4 neutrons produced. The balanced equation is:
24997 Bk + 22
10 Ne 267107 Bh + 4 1
0 n
0N
Nln = −kt =
2/1t
t)6931.0(,
199
11ln =
s0.15
t)6931.0(, t = 62.7 s
Bh: [Rn]7s25f
146d
5 is the expected electron configuration.
88. 5826 Fe + 2 1
0 n 6027 Co + ?; in order to balance the equation, the missing particle has no
mass and a charge of 1−; this is an electron.
An atom of 6027 Co has 27 e, 27 p, and 33 n. The mass defect of the
60Co nucleus is:
m = (59.9338 – 27me) – [27(1.00782 – me) + 33(1.00866)] = − 0.5631 u
E = mc2 = − 0.5631 u ×
u
kg101.6605 27 × (2.9979 × 10
8 m/s)
2 = − 8.403 × 1110 J
Nucleon
energyBinding =
nucleons60
J10403.8 11 = 1.401 × 1210 J/nucleon
The emitted particle was an electron, which has a mass of 9.109 × 3110 kg. The deBroglie
wavelength is:
mv
hλ =
m/s)102.998(0.90kg109.109
sJ106.626831
34
= 2.7 × 1210 m