chapter 19 the nucleus: a chemist's view · chapter 19 the nucleus: a chemist's view...

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




a. eNiCo 01

6028

6027 b. MoeTc 97

4201

9743

c. eRuTc 01

9944

9943 d. HeUPu 4

223592

23994




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.


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.


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.


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


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:


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


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


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.


−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:


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


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.


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.


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


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


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


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


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

chapter 19 the nucleus: a chemist's view · chapter 19 the nucleus: a chemist's view...

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