803 CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW Radioactive Decay and Nuclear Transformations 1. a. Thermodynamic stability: the potential energy of a particular nucleus compared to the sum of the potential energies of its component protons and neutrons. b. Kinetic stability: the probability that a nucleus will undergo decomposition to form a different nucleus. c. Radioactive decay: a spontaneous decomposition of a nucleus to form a different nucleus. d. Beta-particle production: a decay process for radioactive nuclides where an electron is produced; the mass number remains constant and the atomic number changes. e. Alpha-particle production: a common mode of decay for heavy radioactive nuclides where a helium nucleus is produced, causing the atomic number and the mass number to change. f. Positron production: a mode of nuclear decay in which a particle is formed having the same mass as an electron but opposite in charge. g. Electron capture: a process in which one of the inner-orbital electrons in an atom is captured by the nucleus. h. Gamma-ray emissions; the production of high-energy photons (gamma rays) that fre- quently accompany nuclear decays and particle reactions. 2. Beta-particle production has the net effect of turning a neutron into a proton. Radioactive nuclei having too many neutrons typically undergo beta-particle decay. Positron production has the net effect of turning a proton into a neutron. Nuclei having too many protons typically undergo positron decay. 3. 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. Th He U 234 90 4 2 238 92 ; this is alpha-particle production. b. e Pa Th 0 1 234 91 234 90 ; this is -particle production.
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803
CHAPTER 20
THE NUCLEUS: A CHEMIST'S VIEW
Radioactive Decay and Nuclear Transformations
1. a. Thermodynamic stability: the potential energy of a particular nucleus compared to the
sum of the potential energies of its component protons and neutrons.
b. Kinetic stability: the probability that a nucleus will undergo decomposition to form a
different nucleus.
c. Radioactive decay: a spontaneous decomposition of a nucleus to form a different nucleus.
d. Beta-particle production: a decay process for radioactive nuclides where an electron is
produced; the mass number remains constant and the atomic number changes.
e. Alpha-particle production: a common mode of decay for heavy radioactive nuclides
where a helium nucleus is produced, causing the atomic number and the mass number
to change.
f. Positron production: a mode of nuclear decay in which a particle is formed having the
same mass as an electron but opposite in charge.
g. Electron capture: a process in which one of the inner-orbital electrons in an atom is
captured by the nucleus.
h. Gamma-ray emissions; the production of high-energy photons (gamma rays) that fre-
quently accompany nuclear decays and particle reactions.
2. Beta-particle production has the net effect of turning a neutron into a proton. Radioactive
nuclei having too many neutrons typically undergo beta-particle decay. Positron production
has the net effect of turning a proton into a neutron. Nuclei having too many protons typically
undergo positron decay.
3. 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.
804 CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW
4. a. 7331
Ga 7332
Ge + 01e b. 192
78Pt 188
76Os + 4
2He
c. 20583
Bi 20582
Pb + 01 e d. 241
96Cm + 0
1e 241
95Am
e. eNiCo 01
6028
6027 f. MoeTc 97
4201
9743
g. eRuTc 01
9944
9943 h. HeUPu 4
223592
23994
5. a. ZneGa 6830
01
6831 b. NieCu 62
2801
6229
c. AtHeFr 20885
42
21287 d. TeeSb 129
5201
12951
6. a. eHeH 01
32
31 b. eBeLi 0
184
83
eHe2Li
________________He2Be
01
42
83
42
84
c. LieBe 73
01
74
d. eBeB 01
84
85
7. 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. VeCr 5123
01
5124 b. XeeI 131
5401
13153
c. SeP 3216
01
3215
8. Fe5326 has too many protons. It will undergo either positron production, electron capture,
and/or alpha-particle production. Fe5926 has too many neutrons and will undergo beta-particle
production. (See Table 20.2 of the text.) The reactions are:
eCoFe;HeCrFe;MneFe;eMnFe 01
5927
5926
42
4924
5326
5325
01
5326
01
5325
5326
9. Reference Table 20.2 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
production 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 β-emitter.
10. a. NpHeAm 23793
42
24195
b. .Biisproductfinalthe;Bie4He8Am 20983
20983
01
42
24195
CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW 805
c. αRaαThβUαPaαNpAm 22588
22990
23392
23391
23793
24195
βAcαFrαAtαBiβPo 22589
22187
21785
21383
21384
βBiαPb 20983
20982
The intermediate radionuclides are:
Pband,Po,Bi,At,Fr,Ac,Ra,Th,U,Pa,Np 20982
21384
21383
21785
22187
22589
22588
22990
23392
23391
23793
11. ;e?He?PbBk 01
42
20782
24797 The change in mass number (247 - 207 = 40) is due ex-
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 e5 01 (five beta particles) are produced in the decay
series of 247
Bk to 207
Pb.
12. a. nBkHeAm 10
24397
42
24095 b. n6CfCU 1
024498
126
23892
c. n4DbNCf 10
260105
157
24998 d. n2LrBCf 1
0257103
105
24998
13. a. n4SgOCf 10
263106
188
24998 b. RfHeSg;Rf 259
10442
263106
259104
14. The most abundant isotope is generally the most stable isotope. The periodic table predicts
that the most stable isotopes for parts a-d are 39
K, 56
Fe, 23
Na, and 204
Tl. (Reference Table 20.2
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.
Kinetics of Radioactive Decay
15. k = s3600
h1
h24
d1
d365
yr1
yr433
69315.0
t
2ln
2/1
= 5.08 × 111 s10
806 CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW
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.
16. Kr-81 is most stable because it has the longest half-life, whereas Kr-73 is hottest (least stable)
since it has the shortest half-life.
12.5% of each isotope will remain after 3 half-lives:
For Kr73: t = 3(27 s) = 81 seconds
For Kr74: t = 3(11.5 min) = 34.5 minutes
For Kr76: t = 3(14.8 h) = 44.4 hours
For Kr81: t = 3(2.1 × 105 yr) = 6.3 × 10
5 years
17. 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
18. ktN
Nln
0
; k = (ln 2)/t1/2 ; N = 0.001 × N0
,yr100,24
t)2(ln
N
N001.0ln
0
0
ln(0.001) = (2.88 × 105
)t, t = 200,000 years
19.
0
0
0 N
N17.0ln,
yr3.12
t)2(lnkt
N
Nln (5.64 × 102
)t, t = 31.4 years
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).
20. a. 0.0100 Ci × Ci
decays/s107.3 10 = 3.7 × 10
8 decays/s; k =
2/1t
2ln
100% 50% 25% 12.5t1/2
t1/2
t1/2
CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW 807
Rate = kN, s
decays107.3 8 =
s3600
h1
h87.2
6931.0 × N, N = 5.5 × 10
12 atoms of
38S
5.5 × 1012
atoms 38
S × Smol
SONamol1
atoms1002.6
Smol138
4
38
2
23
38
= 9.1 × 1012 mol Na2
38SO4
9.1 × 1012 mol Na2
38SO4
4
38
2
4
38
2
SONamol
SONag0.148 = 1.3 × 109
g = 1.3 ng Na238
SO4
b. 99.99% decays, 0.01% left; lnh87.2
t)6931.0(kt
100
01.0
, t = 38.1 hours 40 hours
21. t = 67.0 yr; k = 2/1t
2ln;
0N
Nln = kt =
yr28.9
yr.0(0.6931)67 = 1.61,
0N
N =
1.61e = 0.200
20.0% of the 90
Sr remains as of July 16, 2012.
22. 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
23. a. s3600
h1
h24
d1
d8.12
6931.0
t
2lnk
2/1
= 6.27 × 107 s
1
b. Rate = kN = 6.27 × 107 s
1
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.
24. 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
808 CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW
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.
25. Plants take in CO2 during the photosynthesis process, which incorporates carbon, including 14
C, into its molecules. As long as the plant is alive, the 14
C/12
C ratio in the plant will equal
the ratio in the atmosphere. When the plant dies, 14
C is not replenished because 14
C decays by
beta-particle production. By measuring the 14
C activity today in the artifact and comparing
this to the assumed 14
C activity when the plant died to make the artifact, an age can be
determined for the artifact. The assumptions are that the 14
C level in the atmosphere is
constant or that the 14
C level at the time the plant died can be calculated. A constant 14
C level
is a pure assumption, and accounting for variation is complicated. Another problem is that
some of the material must be destroyed to determine the 14
C level.
26. 238
U has a half-life of 4.5 × 109 years. In order to be useful, we need a significant number of
decay events by 238
U to have occurred. With the extremely long half-life of 238
U, the period of
time required for a significant number of decay events is on the order of 108 years. This is the
time frame of when the earth was formed. 238
U is not useful for aging 10,000-year-old objects
or less because a measurable quantity of decay events has not occurred in 10,000 years or
less. 14
C is good at dating these objects because 14
C has a half-life on the order of 103 years.
14C is not useful for dating ancient objects because of the relatively short half-life; no
discernable amount of 14
C will remain after 108 years.
27. 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
yr105.4
t)693.0(
g795.1
g000.1ln,
t
t)2(lnkt
N
Nln
92/10
, t = 3.8 × 10
9 years
28. 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:
(0.107)m = 0.95 g, m = 8.9 g = total mass of 40
K that decayed
CHAPTER 20 THE NUCLEUS: A CHEMIST'S VIEW 809
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
yr1027.1
t)6931.0(
t
t)2(lnkt
92/1
, t = 4.2 × 10
9 years
d. If some 40
Ar escaped, then the measured ratio of 40
Ar/40
K would be less than it should
be. We would calculate the age of the rocks to be less than it actually is.