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. a. Alpha particle ( 42 He) production is required to convert 214

84 Po into 210
82 Pb.
210
b. Beta particle ( −01 e) production is required to convert 82 Pb into 210
83 Bi.

202
c. Beta particle ( −01 e) production is required to convert 79 Au into 202
80 Hg.

3. Radiotracers generally have short half-lives. The radioactivity from the radiotracers is
monitored to study a specific area of the body. However, we don’t want long-lived
radioactivity in order to minimize potential damage to healthy tissue by the radioactivity.

4. 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

5. Beta-particle production has the net effect of turning a neutron into a proton. Radioactive
nuclei having too many neutrons typically undergo β-particle decay. Positron production has
the net effect of turning a proton into a neutron. Nuclei having too many protons typically
undergo positron decay.

6. 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.
56
b. Fe has the largest binding energy per nucleon, so it is the most stable nuclide. 56Fe has
the greatest mass loss per nucleon when the protons and neutrons are brought together to
form the 56Fe nucleus. The least stable nuclide shown, having the smallest binding
energy per nucleon, is 2H.

790
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 791

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 56Fe can undergo fusion.
Nuclides to the right of 56Fe 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.

7. 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.

8. All radioactive decay follows first-order kinetics. A sample is analyzed for the 176Lu and
176
Hf content, from which the first-order rate law can be applied to determine the age of the
sample. The reason 176Lu 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, 176Lu decay hasn’t even approached one half-life when dating
5-billion-year-old objects.

9. ∆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.
10. 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 235UF6 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 235U content in natural uranium.
11. 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.

12. Penetration power refers to the ability of radioactive decay particles or rays to penetrate
human tissue. Gamma radiation easily penetrates human tissue, beta particles can penetrate
about 1 cm, and alpha particles are stopped by the skin, so they are the least penetrating.

13. Somatic damage is immediate damage to the organism itself, resulting in sickness or death.
Genetic damage is damage to the genetic material which can be passed on to future
generations. The organism will not feel immediate consequences from genetic damage, but its
offspring may be damaged.

14. 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.
792 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

Exercises
Radioactive Decay and Nuclear Transformations

15. 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. 3
1H → 23 He + 0
−1 e b. 3 Li → 4 Be + −1 e
8 8 0

4 Be → 2 2 He
8 4

8
3 Li → 2 42 He + 0
−1 e

c. 7
4 Be + 0
−1 e → 73 Li d. 5 B → 4 Be
8 8
+ 0
+1 e

16. 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. 60
27 Co → 60
28 Ni + 0
−1 e b. 97
43Tc + 0
−1 e → 97
42 Mo

c. 99
43Tc → 99
44 Ru + 0
−1 e d. 239
94 Pu → 235
92 U + 42 He

17. 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. 238
92 U → 42 He + 234
90Th ; this is alpha-particle production.

b. 234
90 Th → 234
91 Pa + 0
−1 e ; this is β-particle production.

18. a. 51
24 Cr + 0
−1 e → 51
23V b. 131
53 I → 0
−1 e + 131
54 Xe c. 32
15 P → 0
−1 e + 32
16 S

19. a. 68 Ga
31 + 0
−1 e → 68
30 Zn b. 62 Cu
29 → 0
+1 e + 62 Ni
28

c. 212
87 Fr → 4
2 He + 208 At
85 d. 129 Sb
51 → 0
−1 e + 129
52 Te

20. a. 73 Ga
31 → 73
32 Ge + 0
−1 e b. 192
78 Pt → 188
76 Os + 4
2 He

c. 205 Bi
83 → 205 Pb
82 + 0
+1 e d. 241 Cm
96 + 0
−1 e → 241 Am
95

21. 235
92 U → 207
82 Pb + ? 42 He + ? 0
−1 e
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 793

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:

235
92 U → 207
82 Pb + 7 42 He + 4 0
−1 e

22. 242
96 Cm → 206 82 Pb + ? 2 He + ? −1 e; the change in mass number (242 − 206 = 36) is due
4 0

exclusively to the alpha-particles. A change in mass number of 36 requires 9 24 He particles

to be produced. The atomic number only changes by 96 − 82 = 14. The 9 alpha-particles
change the atomic number by 18, so 4 −01 e (4 beta-particles) are also produced in the decay
series of 242Cm to 206Pb.

23. a. 241 Am
95 → 4
2 He + 237
93 Np

b. 241
95 Am → 8 42 He + 4 −01 e + 209
83 Bi; the final product is 209
83 Bi.

c. 241
95 Am → 237
93 Np + α → 233
91 Pa + α→ 233
92 U +β → 229
90 Th + α → 225
88 Ra + α
↓
213
84 Po + β ← 213
83 Bi + α ← 217
85 At + α ← 221
87 Fr +α ← 225
89 Ac + β
↓
209
82 Pb + α → 209
83 Bi + β

The intermediate radionuclides are:

237 233 233 229 225 225 221 217 213 213 209
93 Np, 91 Pa, 92 U, 90 Th, 88 Ra, 89 Ac, 87 Fr, 85 At, 83 Bi, 84 Po, and 82 Pb

24. The complete decay series is:

232
90 Th → 228
88 Ra + 42 He → 228
89 Ac + 0
−1 e → 228
90 Th + 0
−1 e → 224
88 Ra + 42 He
↓
0
−1 e + 212
84 Po ← 0
−1 e + 212
83 Bi ← He + 4
2
212
82 Pb ← 4
2 He + 216
84 Po ← 220
86 Rn + 42 He
↓
208 4
82 Pb + 2 He

25. Reference Table 19.2 of the text for potential radioactive decay processes. 8B and 9B contain
too many protons or too few neutrons. Electron capture or 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.
12
B and 13B contain too many neutrons or too few protons. Beta-particle production lowers
the neutron to proton ratio, so we expect 12B and 13B to be β-emitters.
794 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

26. Reference Table 19.2 of the text for potential radioactive decay processes. 17F and 18F 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. 21F
contains too many neutrons or too few protons. Beta-particle production lowers the neutron-
to-proton ratio, so we expect 21F to be a beta-emitter.

27. a. 249
98 Cf + 18
8O → 263
106 Sg + 4 01 n b. 259
104 Rf ; 263
106 Sg → 42 He + 259
104 Rf

28. a. 240
95 Am + 4
2 He → 243
97 Bk + 1
0n b. 238
92 U + 12
6C → 244
98 Cf + 6 01 n

c. 249
98 Cf + 15
7N → 260
105 Db + 4 01 n d. 249
98 Cf + 10
5B → 257
103 Lr + 2 01 n

Kinetics of Radioactive Decay

29. For t1/2 = 12,000 yr:
ln 2 0.693 1 yr 1d 1h
k= = × × × = 1.8 × 10−12 s−1
t 1/2 12,000 yr 365 d 24 h 3600 s

Rate = kN = 1.8 × 10−12 s−1 × 6.02 × 1023 nuclei = 1.1 × 1012 decays/s

For t1/2 = 12 h:
ln 2 0.693 1h
k= = × = 1.6 × 10−5 s−1
t 1/2 12 h 3600 s

Rate = kN = 1.6 × 10−5 s−1 × 6.02 × 1023 nuclei = 9.6 × 1018 decays/s

For t1/2 = 12 s:
0.693
Rate = kN = × 6.02 × 1023 nuclei = 3.5 × 1022 decays/s
12 s

6.02 × 10 23 nuclei
32
1 mol K 3 PO 4 1 mol 32 P
30. a. 120 g K3PO4 × 32
× 32
×
213.3 mg K 3 PO 4 mol K 3 PO 4 mol
= 3.4 × 1023 P-32 nuclei
ln 2 0.69315 1h
k= = × = 1.35 × 10−5 s−1
t 1/2 14.3 h 3600 s
Ci
Rate = kN = 1.35 × 10−5 s−1 × 3.4 × 1023 nuclei × = 1.2 × 108 Ci
3.7 × 10 decays/s
10

ln 2 0.693 1 yr 1d 1h
b. k = = × × × = 9.2 × 10−13 s−1
t 1/2 24,000 yr 365 d 24 h 3600 s
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 795

Ci 1000 mCi
Rate = kN = 9.2 × 10−13 s−1 × 6.02 × 1023 nuclei × ×
3.7 × 10 decays/s
10
Ci

Rate = 1.5 × 104 mCi

31. All radioactive decay follows first-order kinetics where t1/2 = (ln 2)/k.
ln 2 0.693
t1/2 = = = 690 h
k 1.0 × 10 −3 h −1

ln 2 0.69315 1 yr 1d 1h
32. k= = × × × = 5.08 × 10 −11 s −1
t1 / 2 433 yr 365 d 24 h 3600 s
1 mol 6.022 × 10 23 nuclei
Rate = kN = 5.08 × 10 −11 s −1 × 5.00 g × ×
241 g mol
= 6.35 × 1011 decays/s

6.35 × 1011 alpha particles are emitted each second from a 5.00-g 241Am sample.

33. This sample goes through 2 half-lives in 10.0 minutes:

1.00 × 1020 t1/2 5.00 × 1019 t1/2 2.50 × 1019 nuclides

So each half-life is 10.0/2 = 5.00 minutes.

34. 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:

100% 50% 25% 12.5%

t1/2 t1/2 t1/2

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 × 105 yr

35. 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 47Ca2+ to be ordered.

ln 2  N  − (0.693) t  5.0 μg Ca 2+  − 0.693(2.0 d )

k= ; ln  = − kt = , ln = = − 0.31
 
t1 / 2  N0  t1 / 2  m0  4. 5 d
5.0
= e−0.31 = 0.73, m0 = 6.8 µg of 47Ca2+ needed initially
m0
107.0 μg 47
CaCO 3
6.8 µg 47Ca2+ × = 15 µg 47CaCO3 should be ordered at the minimum.
47.0 μg 47
Ca 2+
796 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

ln 2 0.6931 1d 1h
36. a. k = = × × = 6.27 × 10 −7 s −1
t1 / 2 12.8 d 24 h 3600 s

 1 mol 6.022 × 10 23 nuclei 

b. Rate = kN = 6.27 × 10 −7 s −1 ×  28.0 × 10 −3 g × × 

 64.0 g mol 
Rate = 1.65 × 1014 decays/s

c. 25% of the 64Cu 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.

ln 2  N  − (0.6931) 72.0 yr  N 
37. t = 72.0 yr; k = ; ln  = −kt = = −1.73,   = e−1.73 = 0.177
t1/ 2  N0  28.9 yr  0
N

17.7% of the 90Sr remains as of July 16, 2017.

38. Assuming 2 significant figures in 1/100:

ln(N/N0) = −kt; N = (0.010)N0; t1/2 = (ln 2)/k

− (ln 2) t − (0.693) t
ln(0.010) = = , t = 53 days
t1 / 2 8.0 d

ln 2  N  − (ln 2)t  N  − (0.693)(48.0 h )

39. k= ; ln  = − kt = ; ln  = = 5.5

t1/2  N 0  t 1/2  N 0  6.0 h

N
= e −5.5 = 0.0041; the fraction of 99Tc that remains is 0.0041, or 0.41%.
N0
32
32.0 mg P ln 2
40. 175 mg Na332PO4 × 32
= 33.9 mg 32P; k =
165.0 mg Na 3 PO 4 t1/ 2
 N  − (0.6931) t  m  − 0.6931(35.0 d )
ln   = − kt = , ln  = ; carrying extra sig. figs.:
 N0  t1/ 2  33.9 mg  14.3 d

ln(m) = −1.696 + 3.523 = 1.827, m = e1.827 = 6.22 mg 32P remains

 24 h 60 min 
− 0.693 3.0 d × × 
 N  − (ln 2) t  1.0 g   d h 
41. ln  = −kt = , ln  =
 N0  t1 / 2  m0  1.0 × 103 min

 1.0 g  1.0
ln  = −3.0, = e −3.0 , m0 = 20. g 82Br needed
 0 
m m 0

1 mol 82 Br 1 mol Na 82 Br 105.0 g Na 82 Br

20. g 82Br × × × = 26 g Na82Br
82.0 g mol 82 Br mol Na 82 Br
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 797

42. Assuming the current year is 2017, t = 71 yr.

 N  − (0.693)t  N  − 0.693(71 yr) 0.10 decay events
ln  = −kt = , ln  = , N=
 N0  t 1/2  5.5  12.3 yr min • 100. g water

ln 2  N  − (0.693)t  N  − 0.693(15,000 yr)

43. k= ; ln  = −kt = , ln  = = −1.8
t1/ 2  N0  t 1/2  13.6  5730 yr

N
= e −1.8 = 0.17, N = 13.6 × 0.17 = 2.3 counts per minute per g of C
13.6
If we had 10. mg C, we would see:
1g 2.3 counts 0.023 counts
10. mg × × =
1000 mg min g min
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 14C activity. Thus 14C dating
is not practical for very small samples.

 N  − (0.6931)t  1.2  − (0.6931)t

44. ln  = −kt = , ln  = , t = 2.0 × 104 yr
 0
N t 1/2  13 . 6  5730 yr

45. Assuming 1.000 g 238U present in a sample, then 0.688 g 206Pb is present. Because 1 mol 206Pb
is produced per mol 238U decayed:
1 mol Pb 1 mol U 238 g U
238
U decayed = 0.688 g Pb × × × = 0.795 g 238U
206 g Pb mol Pb mol U
Original mass 238U present = 1.000 g + 0.795 g = 1.795 g 238U
 N  − (ln 2)t  1.000 g  − 0.693(t )
ln  = −kt = , ln  = , t = 3.8 × 109 yr
 0
N t 1/ 2  1 . 795 g  4 . 5 × 10 9
yr

40 40
46. a. The decay of K is not the sole source of Ca.
40 40 40
b. Decay of K is the sole source of Ar and no Ar is lost over the years.
0.95 g 40 Ar
c. = current mass ratio
1.00 g 40 K
40 40 40 40
0.95 g of K decayed to Ar. 0.95 g of K is only 10.7% of the total K that decayed,
or:
40
0.107(m) = 0.95 g, m = 8.9 g = total mass of K that decayed
40
Mass of K when the rock was formed was 1.00 g + 8.9 g = 9.9 g.
 1.00 g 40 K  − (ln 2) t − (0.6931) t
ln  = −kt =
 = , t = 4.2 × 109 years old
×
40 9
 9 . 9 g K  t 1/ 2 1.27 10 yr
798 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

d. If some 40Ar escaped, then the measured ratio of 40Ar/40K 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

ΔE 3.9 × 10 23 kg m 2 /s 2
47. ΔE = Δmc2, Δm = = = 4.3 × 106 kg
c 2
(3.00 × 10 m/s)
8 2

The sun loses 4.3 × 106 kg of mass each second. Note: 1 J = 1 kg m2/s2

1.8 × 1014 kJ 1000 J 3600 s 24 h

48. × × × = 1.6 × 1022 J/day
s kJ h day
ΔE 1.6 × 10 22 J
ΔE = Δmc2, Δm = = = 1.8 × 105 kg
c2 (3.00 × 108 m/s) 2
1.8 × 105 kg of solar material provides 1 day of solar energy to the earth.

1 kJ 1g 1 kg
1.6 × 1022 J × × × = 5.0 × 1014 kg
1000 J 32 kJ 1000 g
5.0 × 1014 kg of coal is needed to provide the same amount of energy as 1 day of solar energy.

49. 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 m2/s2.

232
For 94 Pu (94 e, 94 p, 138 n):

mass of 232Pu nucleus = 3.85285 × 10 −22 g − mass of 94 electrons

mass of 232Pu nucleus = 3.85285 × 10 −22 g − 94(9.10939 × 10 −28 ) g = 3.85199 × 10 −22 g

Δm = 3.85199 × 10 −22 g − (mass of 94 protons + mass of 138 neutrons)

Δm = 3.85199 × 10 −22 g − [94(1.67262 × 10 −24 ) + 138(1.67493 × 10 −24 )] g

= −3.168 × 10 −24 g

For 1 mol of nuclei: Δm = −3.168 × 10 −24 g/nuclei × 6.0221 × 1023 nuclei/mol

= −1.908 g/mol

ΔE = Δmc2 = (−1.908 × 10 −3 kg/mol)(2.9979 × 108 m/s)2 = −1.715 × 1014 J/mol

231
For 91 Pa (91 e, 91 p, 140 n):

mass of 231Pa nucleus = 3.83616 × 10 −22 g − 91(9.10939 × 10-28) g = 3.83533 × 10 −22 g

CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 799

Δm = 3.83533 × 10 −22 g − [91(1.67262 × 10 −24 ) + 140(1.67493 × 10 −24 )] g

= −3.166 × 10 −24 g
2
− 3.166 × 10 − 27 kg 6.0221 × 10 23 nuclei  2.9979 × 10 8 m 
ΔE = Δmc = 2
× ×  

nuclei mol  s 
= −1.714 × 1014 J/mol

50. 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 × 10−4 u.

Mass of 56
26 Fe nucleus = mass of atom − mass of electrons = 55.9349 − 26(0.0005486)
= 55.9206 u

26 11H + 30 10 n → 56
26 Fe; Δm = 55.9206 u − [26(1.00728) + 30(1.00866)] u = −0.5285 u
1.6605 × 10 −27 kg
ΔE = Δmc2 = −0.5285 u × × (2.9979 × 108 m/s)2 = −7.887 × 10−11 J
u
Binding energy 7.887 × 10 −11 J
= = 1.408 × 10−12 J/nucleon
Nucleon 56 nucleons

51. Let me = mass of electron; for 12C (6e, 6p, and 6n): Mass defect = Δm = [mass of 12C
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

1.6605 × 10 −27 kg
ΔE = Δmc2 = −0.09888 u × × (2.9979 × 108 m/s)2 = −1.476 × 10−11 J
u
Binding energy 1.476 × 10 −11 J
= = 1.230 × 10−12 J/nucleon
Nucleon 12 nucleons
For 235U (92e, 92p, and 143n):

Δm = 235.0439 − 92me − [92(1.00782 − me) + 143(1.00866)] = −1.9139 u

1.66054 × 10 −27 kg
ΔE = Δmc2 = −1.9139 × × (2.99792 × 108 m/s)2 = −2.8563 × 10−10 J
u

Binding energy 2.8563 × 10 −10 J

= = 1.2154 × 10−12 J/nucleon
Nucleon 235 nucleons

Because 56Fe is the most stable known nucleus, the binding energy per nucleon for 56Fe
(1.408 × 10−12 J/nucleon) will be larger than that of 12C or 235U (see Figure 19.9 of the text).
800 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

52. For 21 H : Mass defect = Δm = mass of 21 H 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
2
H atom. From the text, the pertinent masses are me = 5.49 × 10−4 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 × 10−3 u

1.6605 × 10 −27 kg
ΔE = Δmc2 = −2.39 × 10−3 u × × (2.998 × 108 m/s)2 = −3.57 × 10−13 J
u
Binding energy 3.57 × 10 −13 J
= = 1.79 × 10−13 J/nucleon
Nucleon 2 nucleons

For 13 H : Δm = 3.01605 − 0.000549 − [1.00728 + 2(1.00866)] = −9.10 × 10−3 u

1.6605 × 10 −27 kg
ΔE = −9.10 × 10−3 u × × (2.998 × 108 m/s)2 = −1.36 × 10−12 J
u
Binding energy 1.36 × 10 −12 J
= = 4.53 × 10−13 J/nucleon
Nucleon 3 nucleons

ΔE − 3.086 × 1012 J
53. ΔE = Δmc2, Δm = = = −3.434 × 10 −5 kg
c 2
(2.9979 × 10 m/s)
8 2

The mass defect for 1 mol of 6Li is −3.434 × 10 −5 kg = −0.03434 g. The mass defect for one
6
Li nuclei is −0.03434 u.

Let mLi = mass of 6Li nucleus; an 6Li nucleus has 3p and 3n.

Mass defect = −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 × 10 −4 u) = 6.01513 u

(includes mass of 3 e−)

1.326 × 10 −12 J
54. Binding energy = × 27 nucleons = 3.580 × 10 −11 J for each 27Mg nucleus
nucleon
ΔE − 3.580 × 10 −11 J
ΔE = Δmc2, Δm = = = −3.983 10 −28 kg
c2 (2.9979 × 108 m/s) 2
1u
Δm = −3.983 10 −28 kg × = −0.2399 u = mass defect
1.6605 × 10 − 27 kg
27
Let mMg = mass of Mg nucleus; an 27Mg nucleus has 12 p and 15 n.

Mass defect = −0.2399 u = mMg − (12mp + 15mn) = mMg − [12(1.00728 u) + 15(1.00866 u)]
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 801

mMg = 26.9764 u

Mass of 27Mg atom = 26.9764 u + 12me, 26.9764 + 12(5.49 × 10 −4 u) = 26.9830 u

(includes mass of 12 e−)

+ 11 H → 12 H + Δm = (2.01410 u − me + me) − 2(1.00782 u − me)

1 0
55. 1H +1 e;

Δm = 2.01410 − 2(1.00782) + 2(0.000549) = −4.4 × 10 −4 u for two protons reacting

When 2 mol of protons undergoes fusion, Δm = −4.4 × 10 −4 g.

ΔE = Δmc2 = −4.4 × 10 −7 kg × (3.00 × 108 m/s)2 = −4.0 × 1010 J

− 4.0 × 1010 J 1 mol

× = −2.0 × 1010 J/g of hydrogen nuclei
2 mol protons 1.01 g

+ 13 H → 42 He + 0 n; using atomic masses, the masses of the electrons cancel when

2 1
56. 1H
determining Δm for this nuclear reaction.

Δm = [4.00260 + 1.00866 − (2.01410 + 3.01605)] u = −1.889 × 10 −2 u

For the production of 1 mol of 42 He : Δm = −1.889 × 10 −2 g = −1.889 × 10 −5 kg

ΔE = Δmc2 = −1.889 × 10-5 kg × (2.9979 × 108 m/s)2 = −1.698 × 1012 J/mol

− 1.698 × 1012 J 1 mol
For 1 nucleus of 42 He : × = −2.820 × 10 −12 J/nucleus
mol 6.0221 × 10 nuclei
23

Detection, Uses, and Health Effects of Radiation

57. 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.

58. Not all of the emitted radiation enters the Geiger-Müller tube. The fraction of radiation
entering the tube must be constant.

59. Water is produced in this reaction by removing an OH group from one substance and an H
from the other substance. There are two ways to do this:
802 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

O O
18 18
i. CH3C OH + H OCH3 CH3C O CH3 + HO H

O O
18
ii. H3CCO H + H18O CH3 CH3CO CH3 + H OH

Because the water produced is not radioactive, methyl acetate forms by the first reaction
where all of the oxygen-18 ends up in methyl acetate.

60. The only product in the fast-equilibrium step is assumed to be N16O18O2, where N is the
central atom. However, this is a reversible reaction where N16O18O2 will decompose to NO
and O2. Because any two oxygen atoms can leave N16O18O2 to form O2, we would expect (at
equilibrium) one-third of the NO present in this fast equilibrium step to be N16O and two-
thirds to be N18O. In the second step (the slow step), the intermediate N16O18O2 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 N16O18O2 to NO when the NO2 product is
formed. The distribution of 18O in the product can best be determined by forming a
probability table.

N16O (1/3) N18O (2/3)

16
O (1/3) from N16O18O2 N16O2 (1/9) N18O16O (2/9)
18
O (2/3) from N16O18O2 N16O18O (2/9) N18O2 (4/9)

From the probability table, 1/9 of the NO2 is N16O2, 4/9 of the NO2 is N18O2, and 4/9 of the
NO2 is N16O18O (2/9 + 2/9 = 4/9). Note: N16O18O is the same as N18O16O. In addition,
N16O18O2 is not the only NO3 intermediate formed; N16O218O and N18O3 can also form in the
fast-equilibrium first step. However, the distribution of 18O in the NO2 product is the same as
calculated above, even when these other NO3 intermediates are considered.

61. 235
92 U + 01 n → 144 90 1 0
58 Ce + 38 Sr + ? 0 n + ? −1 e; to balance the atomic number, we need 4
beta-particles, and to balance the mass number, we need 2 neutrons.

62. 238
92 U + 01 n → 23992 U → −1 e +
0
93 Np →
239 0
−1 e + 239
94 Pu; plutonium-239 is the
fissionable material in breeder reactors.

63. 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 64 for a specific example.)
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 803

64. (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
65. The most abundant isotope is generally the most stable isotope. The periodic table predicts
that the most stable isotopes for exercises a-d are 39K, 56Fe, 23Na, and 204Tl. (Reference Table
19.2 of the text for potential decay processes.)
45
a. Unstable; K has too many neutrons and will undergo beta-particle production.

b. Stable

c. Unstable; 20Na 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 20Na nuclei. Alpha-particle production
usually occurs for heavy nuclei.

d. Unstable; 194Tl has too few neutrons and will undergo electron capture, positron
production, and/or alpha-particle production.

66. 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.

67. 214
83 Bi → −01 e + 214
84 Po → 42 He + 210
82 Pb → 0
−1e + 210
83 Bi → 0
−1e + 210
84 Po

The products of the various steps are 214

84 Po,
210 210 210
82 Pb, 83 Bi, and 84 Po.

3.7 × 1010 decays / s ln 2

68. a. 0.0100 Ci × = 3.7 × 108 decays/s; k =
Ci t1/ 2
3.7 × 108 decays  0.6931 1h 
Rate = kN, =  ×  × N, N = 5.5 × 1012 atoms of 38S
s  2.87 h 3600 s 
38
1 mol 38S 1 mol Na 2 SO 4
5.5 × 1012 atoms 38S × × = 9.1 × 10 −12 mol Na238SO4
6.02 × 10 atoms
23
mol 38S
38
−12 148.0 g Na 2 SO 4
9.1 × 10 38
mol Na2 SO4 × 38
= 1.3 × 10 −9 g = 1.3 ng Na238SO4
mol Na 2 SO 4
804 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

 0.01  − (0.6931) t
b. 99.99% decays, 0.01% left; ln  = −kt = , t = 38.1 hours ≈ 40 hours
 100  2.87 h

453.6 g 18 g C 1.6 × 10 −10 g 14

C 1 mol 14
C
69. N = 180 lb × × × × 14
lb 100 g body 100 g C 14 g C
6.022 × 10 23
nuclei 14
C
× 14
= 1.0 × 1015 carbon-14 nuclei
mol C
ln 2 0.693 1 yr 1d 1h
Rate = kN; k = = × × × = 3.8 × 10 −12 s −1
t1/2 5730 yr 365 d 24 h 3600 s

Rate = kN; k = 3.8 × 10 −12 s −1 (1.0 × 1015 14

C nuclei) = 3800 decays/s

A typical 180 lb person produces 3800 beta particles each second.

15.1 − (ln 2) t
70. t1/2 = 5730 yr; k = (ln 2)/t1/2; ln(N/N0) = −kt; ln = , t = 109 yr
15.3 5730 yr
No; from 14C dating, the painting was produced during the early 1900s.

71. The third-life will be the time required for the number of nuclides to reach one-third of the
original value (N0/3).
 N  − (0.6931)t  1  − (0.6931)t
ln  = −kt = , ln  = , t = 49.8 yr
 N0  t 1/2  3 31.4 yr

The third-life of this nuclide is 49.8 years.

72. ln(N/N0) = −kt; k = (ln 2)/t1/2 ; N = 0.001 × N0

 0.001 × N 0  − (ln 2) t
ln  = , ln(0.001) = −(2.88 × 10−5)t, t = 2 × 105 yr = 200,000 yr
 N 0  24,100 yr

 N  − (ln 2) t  0.17 × N 0 
73. ln  = −kt = , ln  = −(5.64 × 10 −2 )t, t = 31.4 yr
 0
N 12 . 3 yr  N 0 
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).

74. Δm = −2(5.486 × 10-4 u) = −1.097 × 10 −3 u

1.6605 × 10 −27 kg
ΔE = Δmc2 = −1.097 × 10 −3 u × × (2.9979 × 108 m/s)2
u
= −1.637 × 10 −13 J
Ephoton = 1/2(1.637 × 10 −13 J) = 8.185 × 10 −14 J = hc/λ
hc 6.6261 × 10 −34 J s × 2.9979 × 108 m/s
λ= = = 2.427 × 10 −12 m = 2.427 × 10 −3 nm
E 8.185 × 10 −14 J
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 805

4 × 109 J 1 mol 235 U 235 g 235 U

75. 20,000 ton TNT × × × = 940 g 235U ≈ 900 g 235U
ton TNT 2 × 1013 J mol 235 U
This assumes that all of the 235U undergoes fission.

76. 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.

77. Mass of nucleus = atomic mass – mass of electron = 2.01410 u – 0.000549 u = 2.01355 u
1/2
 3(8.3145 J/K • mol)(4 × 10 7 K) 
1/2
 3 RT  
urms =   =   = 7 × 105 m/s
 M   2.01355 g(1 kg/1000 g) 

1 1  1.66 × 10 −27 kg 
KEavg = mu 2 =  2.01355 u ×  (7 × 105 m/s)2 = 8 × 10 −16 J/nuclei

2 2  u 

We could have used KEave = (3/2)RT to determine the same average kinetic energy.

78. 1
1H + 01n → 2 11H + 01n + 1
−1H; mass 1
−1H = mass 11H = 1.00728 u = mass of proton = mp

Δm = 3mp + mn − (mp + mn) = 2mp = 2(1.00728) = 2.01456 u

1.66056 × 10 −27 kg
ΔE = Δmc2 = 2.01456 amu × × (2.997925 × 108 m/s)2
amu
ΔE = 3.00660 × 10−10 J of energy is absorbed per nuclei, or 1.81062 × 1014 J/mol nuclei.

The source of energy is the kinetic energy of the proton and the neutron in the particle
accelerator.

79. All evolved oxygen in O2 comes from water and not from carbon dioxide.

80. 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.
806 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

ChemWork Problems
81. The equations for the nuclear reactions are:

239
94 Pu → 235
92 U + 42 He ; 214
82 Pb → 214
83 Bi + 0
−1 e ; 27 Co → 28 Ni
60 60
+ 0
−1e

99
43Tc → 99
44 Ru + 0
−1 e ; 239
93 Np → 239
94 Pu + 0
−1 e

ln 2 0.69315 1h
82. a. t1/2 = (ln 2)/k, k = = × = 6.42 × 10 −5 s −1
t 1/2 3.00 h 3600 s
6.022 × 10 23 nuclei
b. Rate = kN = 6.42 × 10 −5 s −1 × 1.000 mol ×
mol
= 3.87 × 1019 decays/s
83. 12.5% of 60Co remains. This decay represents 3 half-lives:

100% 50% 25% 12.5%

t1/2 t1/2 t1/2

Time = 3 × 5.26 yr = 15.8 yr

84. Because 1 mol 87Sr is produced per mol 87Rb decayed:

1 mol Sr 1 mol Rb 86.90919 g Rb
87
Rb decayed = 3.1 μg Sr × × × = 3.1 μg 87Rb
86.90888 g Sr mol Sr mol Rb
Original mass 87Rb present = 109.7 μg + 3.1 μg = 112.8 μg 87Rb
 N  − (ln 2)t  109.7 μg  − 0.693(t)
ln  = −kt = , ln  = , t = 1.9 × 109 yr
 N0  t 1/2  112.8 μg  4.7 × 1010 yr
The rock is 1.9 × 109 yr old.

12 Mg : mass defect = Δm = mass of 12 Mg nucleus − mass of 12 protons − mass of 12

For 24 24
85.
neutrons. The mass of the 24Mg nucleus will equal the atomic mass of 24Mg minus the mass
of the 12 electrons in an 24Mg atom.

Δm = 23.9850 u − 12(0.0005486 u) − [12(1.00728 u) + 12(1.00866 u)] = −0.2129 u

1.6605 × 10 −27 kg
ΔE = Δmc2 = −0.2129 u × × (2.9979 × 108 m/s)2 = −3.177 × 10−11 J
u
The binding energy of 24Mg is 3.177 × 10−11 J.

86. a. True; 16 minutes represent two half-lives, so 8.0 min is one half-life. In two half-lives, a
first order substance decreases from 100% to 25% of its original amount. That is what
happened here.

b. True; alpha-particle production does predominantly occur for heavy nuclides. But heavy
nuclides also undergo other types of decay processes.
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 807

c. False; as Z increases, stable nuclei have more neutrons than protons. So the ratio of
protons/neutrons decreases as Z increases.

d. False; for stable light nuclides, there are about equal numbers of protons and neutrons.

Challenge Problems

ln 2  N  − (0.693)t
87. k= ; ln  = − kt =
t1 / 2  0
N t1/2

 N  − (0.693)(4.5 × 10 9 yr) N
For 238U: ln  = = − 0.693, = e −0.693 = 0.50
 N0  4.5 × 10 yr
9
N0

 N  − (0.693)(4.5 × 10 9 yr) N
For 235U: ln  = = − 4.39, = e − 4.39 = 0.012
 0
N 7.1 × 10 8
yr N 0

If we have a current sample of 10,000 uranium nuclei, 9928 nuclei of 238U 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.
N N 9928 nuclei
For 238U: = 0.50, N 0 = = = 2.0 × 10 4 238
U nuclei
N0 0.50 0.50
N 72 nuclei
For 235U: N 0 = = = 6.0 × 103 235
U nuclei
0.012 0.012
So 4.5 billion years ago, the 10,000-nuclei sample of uranium was composed of 2.0 × 104
238
U nuclei and 6.0 × 103 235U nuclei. The percent composition 4.5 billion years ago would
have been:
2.0 × 10 4 238
U nuclei
× 100 = 77% 238U and 23% 235U
(6.0 × 10 + 2.0 × 10 ) total nuclei
3 4

88. Total activity injected = 86.5 × 10−3 Ci

3.6 × 10 −6 Ci 1.8 × 10 −6 Ci
Activity withdrawn = =
2.0 mL H 2 O mL H 2 O
Assuming no significant decay occurs, then the total volume of water in the body multiplied
by 1.8 × 10−6 Ci/mL must equal the total activity injected.
1.8 × 10 −6 Ci
V× = 8.65 × 10−2 Ci, V = 4.8 × 104 mL H2O
mL H 2 O
Assuming a density of 1.0 g/mL for water, the mass percent of water in this 150-lb person is:
1.0 g H 2 O 1 lb
4.8 × 10 4 mL H 2 O × ×
mL 453.6 g
× 100 = 71%
150 lb
808 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

89. 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:
5.0 × 103 cpm
0.10 mL × = 5.0 × 102 cpm
mL
Assuming that the total activity is uniformly distributed only in the rat’s blood, the blood
volume is:
48 cpm
V× = 5.0 × 102 cpm, V = 10.4 mL = 10. mL
mL

90. a. From Table 18.1: 2 H2O + 2 e− → H2 + 2 OH− E° = − 0.83 V

E ocell = E oH 2O − E oZr = − 0.83 V + 2.36 V = 1.53 V

Yes, the reduction of H2O to H2 by Zr is spontaneous at standard conditions because

E ocell > 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 × 105 J = −590. kJ
0.0591
E = E° − log Q; at equilibrium, E = 0 and Q = K.
n
0.0591 4(1.53)
E° = log K, log K = = 104, K ≈ 10104
n 0.0591
1000 g 1 mol Zr 2 mol H 2
d. 1.00 × 103 kg Zr × × × = 2.19 × 104 mol H2
kg 91.22 g Zr mol Zr
2.016 g H 2
2.19 × 104 mol H2 × = 4.42 × 104 g H2
mol H 2
nRT (2.19 × 10 4 mol)(0.08206 L atm/mol • K)(1273 K)
V= = = 2.3 × 106 L H2
P 1.0 atm
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.
12 12
91. a. C; it takes part in the first step of the reaction but is regenerated in the last step. C is
not consumed, so it is not a reactant.
13
b. N, 13C, 14N, 15O, and 15N are the intermediates.

c. 4 11 H → 4
2 He + 2 +01 e ; Δm = 4.00260 u − 2 me + 2 me − [4(1.00782 u − me)]
CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW 809

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 × 10 −5 kg × (2.9979 × 108 m/s)2 = −2.380 × 1012 J

− 2.380 × 1012 J
For 1 mol of protons reacting: = −5.950 × 1011 J/mol 1H
4 mol 1 H

92. a. 238
92 U → 222 4 0
86 Rn + ? 2 He + ? −1 e ; 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 222Rn decays.

b. Alpha-particles cause significant ionization damage when inside a living organism.

Because the half-life of 222Rn is relatively short, a significant number of alpha-particles
will be produced when 222Rn is present (even for a short period of time) in the lungs.

c. 222
86 Rn → 42 He + 218 84 Po; 84 Po → 2 He + 82 Pb; polonium-218 is produced when
218 4 214

218
radon-222 decays. Po is a more potent alpha-particle producer since it has a much
shorter half-life than 222Rn. In addition, 218Po 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.

4.0 pCi 1 × 10 −12 Ci 3.7 × 1010 decays/sec

d. Rate = kN; rate = × × = 0.15 decays/s•L
L pCi Ci
ln 2 0.6391 1d 1h
k= = × × = 2.10 × 10 −6 s −1
t1 / 2 3.82 d 24 h 3600 s
rate 0.15 decays/s • L
N= = = 7.1 × 104 222Rn atoms/L
k 2.10 × 10 −6 s −1
7.1 × 10 4 222
Rn atoms 1 mol 222
Rn
× = 1.2 × 10 −19 mol 222Rn/L
L 6.02 × 10 23
atoms

33 counts 1 mol I • min

93. Moles of I− = × = 6.6 × 10 −11 mol I−
min 5.0 × 10 counts
11

6.6 × 10 −11 mol I −

[I−] = = 4.4 × 10 −10 mol/L
0.150 L

Hg2I2(s) ⇌ Hg22+(aq) + 2 I−(aq) Ksp = [Hg22+][I−]2

Initial s = solubility (mol/L) 0 0
Equil. s 2s

From the problem, 2s = 4.4 × 10 −10 mol/L, s = 2.2 × 10 −10 mol/L.

Ksp = (s)(2s)2 = (2.2 × 10 −10 )(4.4 × 10 −10 )2 = 4.3 × 10 −29

810 CHAPTER 19 THE NUCLEUS: A CHEMIST’S VIEW

94. 2
1H + 2
1H → 4
2 He; Q for 21 H = 1.6 × 10 −19 C; mass of deuterium = 2 u.
9.0 × 10 9 J • m/C 2 (Q1Q 2 ) 9.0 × 10 9 J • m/C 2 (1.6 × 10 −19 C) 2
E= =
r 2 × 10 −15 m
= 1 × 10 −13 J per alpha particle

KE = 1/2 mv2; 1 × 10 −13 J = 1/2 (2 u × 1.66 × 10 −27 kg/u)v2, v = 8 × 106 m/s

From the kinetic molecular theory discussed in Chapter 5:

1/ 2
 3RT 
urms =   , where M = molar mass in kilograms = 2 × 10 −3 kg/mol for deuterium
 M 
1/2
 3(8.3145 J/K • mol)(T) 
8 × 10 m/s = 
6
 , T = 5 × 109 K
 2 × 10 −3 kg 

Integrative Problems

10 Ne → 107 Bh + ?; this equation is charge balanced, but it is not mass balanced.

249
95. 97 Bk + 22 267

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:

10 Ne →
249
97 Bk + 22 267
107 Bh + 4 01 n

 N  − (0.6931) t  11  − (0.6931) t
ln  = −kt = , ln = , t = 62.7 s
 N0  t1 / 2  199  15.0 s

Bh: [Rn]7s25f146d5 is the expected electron configuration.

96. 58
26 Fe+ 2 01 n → 27
60
Co + ?; in order to balance the equation, the missing particle has no
mass and a charge of 1−; this is an electron.
60
An atom of 27 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

1.6605 × 10 −27 kg
∆E = ∆mc2 = − 0.5631 u × × (2.9979 × 108 m/s)2 = − 8.403 × 10 −11 J
u
Binding energy 8.403 × 10 −11 J
= = 1.401 × 10 −12 J/nucleon
Nucleon 60 nucleons

The emitted particle was an electron, which has a mass of 9.109 × 10 −31 kg. The deBroglie
wavelength is:
h 6.626 × 10 −34 J s
λ = = −31
= 2.7 × 10 −12 m
mv 9.109 × 10 kg × (0.90 × 2.998 × 10 m/s)
8