SM Chapter 20

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


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


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


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


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.

29. t1/2 = 5730 y; k = (ln 2)/t1/2; ln (N/N0) = kt; lnyr5730

t)2(ln

3.15

1.15 , t = 109 years

No; from 14

C dating, the painting was produced during the early 1900s.

30. yr5730

)yr000,15(6931.0

3.15

Nln,

t

t)6931.0(kt

N

Nln

t

2lnk

2/102/1

= 1.8

3.15

N = e

1.8 = 0.17, N = 15.3 × 0.17 = 2.6 counts per minute per g of C

If we had 10. mg C, we would see:

10. mg min

counts026.0

gmin

counts6.2

mg1000

g1

It would take roughly 40 minutes 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.

Energy Changes in Nuclear Reactions

31. ΔE = Δmc2, Δm

28

2223

2 m/s)10(3.00

/smkg103.9

c

EΔ 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.

32. 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.


33. From the text, the mass of a proton = 1.00728 amu, the mass of a neutron = 1.00866 amu, and

the mass of an electron = 5.486 × 104 amu.

Mass of Fe5626 nucleus = mass of atom mass of electrons = 55.9349 26(0.0005486)

= 55.9206 amu

;Fen30H26 5626

11

11 Δm = 55.9206 amu [26(1.00728) + 30(1.00866)] amu

= 0.5285 amu

ΔE = Δmc2 = 0.5285 amu

amu

kg106605.1 27 (2.9979 × 10

8 m/s)

2 = 7.887 × 1011 J

nucleons56

J10887.7

Nucleon

energyBinding 11

1.408 × 1012 J/nucleon

34. 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

amu, mp = 1.00728 amu,

and mn = 1.00866 amu.

Δm = 2.01410 amu 0.000549 amu (1.00728 amu + 1.00866 amu) = 2.39 × 103 amu

ΔE = Δmc2 = 2.39 × 103

amu × amu

kg106605.1 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

amu

ΔE = 9.10 × 103 amu ×

amu

kg106605.1 27 × (2.998 × 10

8 m/s)

2 = 1.36 × 1012

J

nucleons3

J1036.1

Nucleon

energyBinding 12

4.53 × 1013 J/nucleon

35. Let me = mass of electron; for 12

C (6e, 6p, 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 amu 6me [6(1.00782 me) + 6(1.00866)]; mass of electrons cancel.

Δm = 12.00000 [6(1.00782) + 6(1.00866)] = 0.09888 amu


ΔE = Δmc2 = 0.09888 amu

amu

kg106605.1 27 (2.9979 × 10

8 m/s)

2

= 1.476 × 1011 J

nucleons12

J10476.1

Nucleon

energyBinding 11

1.230 × 1012 J/nucleon

For 235

U (92e, 92p, 143n):

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

ΔE = Δmc2 = 1.9139 ×

amu

kg1066054.1 27 (2.99792 × 10

8 m/s)

2 = 2.8563 × 1010

J

nucleons235

J108563.2

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 20.9 of the text).

36. Let mLi = mass of 6Li nucleus; an

6Li nucleus has 3p and 3n.

0.03434 amu = mLi (3mp + 3mn) = mLi [3(1.00728 amu) + 3(1.00866 amu)]

mLi = 6.01348 amu

Mass of 6Li atom = 6.01348 amu + 3me = 6.01348 + 3(5.49 × 410 amu) = 6.01513 amu

(includes mass of 3 e)

37. 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 × kg106605.1

amu127

= 0.2399 amu = mass defect

Let mMg = mass of 27

Mg nucleus; an 27

Mg nucleus has 12 p and 15 n.

0.2399 amu = mMg (12mp + 15mn) = mMg [12(1.00728 amu) + 15(1.00866 amu)]

mMg = 26.9764 amu

Mass of 27

Mg atom = 26.9764 amu + 12me, 26.9764 + 12(5.49 × 410 amu) = 26.9830 amu

(includes mass of 12 e)


38. ;eHHH 01

21

11

11 Δm = (2.01410 amu - me + me) 2(1.00782 amu me)

Δm = 2.01410 2(1.00782) + 2(0.000549) = 4.4 × 104 amu for two protons reacting

When 2 mol protons undergo fusion, Δm = 4.4 × 104 g.

ΔE = Δmc2 = 4.4 × 107

kg × (3.00 × 108 m/s)

2 = 4.0 × 10

10 J

g01.1

mol1

protonsmol2

J100.4 10

= 2.0 × 1010

J/g of hydrogen nuclei

39. ;nHeHH 10

42

31

21 mass of electrons cancel when determining Δm for this nuclear

reaction.

Δm = [4.00260 + 1.00866 - (2.01410 + 3.01605)] amu = 1.889 × 102 amu

For the production of 1 mol of :He42 Δm = 1.889 × 102

g = 1.889 × 105 kg

ΔE = Δmc2 = 1.889 × 105

kg × (2.9979 × 108 m/s)

2 = 1.698 × 10

12 J/mol

For one nucleus of :He42

nuclei100221.6

mol1

mol

J10698.123

12

= 2.820 × 1012

J/nucleus

40. Δm = 2(5.486 × 104 amu) = 1.097 × 103

amu

ΔE = Δmc2 = 1.097 × 10

-3 amu ×

amu

kg106605.1 27 × (2.9979 × 10

8 m/s)

2

= 1.637 × 1013 J

Ephoton = 1/2 (1.637 × 1013 J) = 8.185 × 1014

J = hc/λ

λ

J10185.8

m/s109979.2sJ106261.6

E

hc14

834

2.427 × 1012 m = 2.427 × 103

nm

Detection, Uses, and Health Effects of Radiation

41. 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 out because at high activities, radioactive particles are

entering the tube faster than the tube can respond to them.


42. Not all the emitted radiation enters the Geiger-Müller tube. The fraction of radiation entering

the tube must be constant for a meaningful measurement.

43. In order to sustain a nuclear chain reaction, the neutrons produced by the fission process 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 buildup causes a

violent explosion.

44. No, coal fired power plants also pose risks. A partial list of risks are:

Coal Nuclear

Air pollution Radiation exposure to workers

Coal mine accidents Disposal of wastes

Health risks to miners Meltdown

(black lung disease) Terrorists

45. Fission: Splitting of a heavy nucleus into two (or more) lighter nuclei.

Fusion: Combining two light nuclei to form a heavier nucleus.

The maximum binding energy per nucleon occurs at Fe. Nuclei smaller than Fe become

more stable by fusing to form heavier nuclei closer in mass to Fe. Nuclei larger than Fe form

more stable nuclei by splitting to form lighter nuclei closer in mass to Fe.

46. 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.

47. Moderator: Slows the neutrons to increase the efficiency of the fission reaction.

Control rods: Absorbs neutrons to slow or halt the fission reaction.

48. For fusion reactions, a collision of sufficient energy must occur between two positively

charged particles to initiate the reaction. This requires high temperatures. In fission, an

electrically neutral neutron collides with the positively charged nucleus. This has a much

lower activation energy.

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

50. Radiotracer: a radioactive nuclide introduced into an organism for diagnostic purposes whose

pathway can be traced by monitoring its radioactivity. 14

C and 32

P work well as radiotracers

because the molecules in the body contain carbon and/or phosphorus; they will be

incorporated into the worker molecules of the body easily, which allows monitoring of the

pathways of these worker molecules.


51. (i) and (ii) mean that Pu is not a significant threat outside the body. Our skin is sufficient to

keep out the α 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,

including the bone marrow, where red blood cells are produced. Once inside the body, α

particles can cause considerable damage.

52. Even though gamma rays penetrate human tissue very deeply, they are very small and cause

only occasional ionization of biomolecules. Alpha particles, because they are much more

massive, are very effective at causing ionization of biomolecules; these produce a dense trail

of damage once they get inside an organism.

53. A nonradioactive substance can be put in equilibrium with a radioactive substance. The two

materials can then be checked to see whether all the radioactivity remains in the original

material or if it has been scrambled by the equilibrium.

54. 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:

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.

55. Some factors for the biological effects of radiation exposure are:

a. The energy of the radiation. The higher the energy, the more damage it can cause.

b. The penetrating ability of radiation. The ability of specific radiation to penetrate human

tissue where it can do damage must be considered.

c. The ionizing ability of the radiation. When biomolecules are ionized, their function is

usually disturbed.

d. The chemical properties of the radiation source. Specifically, can the radioactive

substance be readily incorporated into the body, or is the radiation source inert

chemically so that it passes through the body relatively quickly.

90

Sr will be incorporated into the body by replacing calcium in the bones. Once incorporated, 90

Sr can cause leukemia and bone cancer. Krypton is chemically inert, so it will not be

incorporated into the body.

18+ HO HCH3C OCH3

O

18+ H OCH3CH3C OH

O

H OHCH3CO CH3

O

H O CH3 ++CH3CO H

O

ii.

i.

18 18


Additional Exercises

56. 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 years

The third-life of this nuclide is 49.8 years.

57. 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.

58. 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.

59. 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


60. Mass of nucleus = atomic mass – mass of electron = 2.01410 amu – 0.000549 amu

= 2.01355 amu

urms =

2/1

M

RT3

=

1/2711

)g1000/kg1(g01355.2

)K104)(molKJ3145.8(3

= 7 × 105 m/s

KEavg =

amu

kg1066.1amu01355.2

2

1mu

2

1 272 (7 × 10

5 m/s)

2 = 8 × 1610 J/nucleus

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

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/s3800C)nuclei101.0(s103.8 1415112

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

62. 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. ;e2HeH4 01

42

11 ; Δm = 4.00260 amu 2me + 2me [4(1.00782 amu me)]

Δm = 4.00260 4(1.00782) + 4(0.000549) = 0.02648 amu 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 × 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

63. 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.

64. Characteristic frequencies of energies emitted in a nuclear reaction suggest that discrete

energy levels exist in the nucleus. Extra stability of certain numbers of nucleons and the

predominance of nuclei with even numbers of nucleons suggests that the nuclear structure

might be described by using quantum numbers.

Challenge Problems

65. mol I = counts100.5

minImol1

min

counts3311

= 6.6 × 10

11 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 × 1029

66. a. From Table 11.1: 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 ZrO2H2O + H2O + 4 e

3 H2O(l) + Zr(s) 2 H2(g) + ZrO2H2O(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

atm0.1

)K1273)(molKatmL08206.0)(mol1019.2(

P

nRTV

114

2.3 × 106 L H2

e. Probably yes; less radioactivity overall was released by venting the H2 than what would

have been released if the H2 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.

67. 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:

100nucleitotal)100.2100.6(

nucleiU100.243

2384

= 77%

238U and 23%

235U

68. 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 because 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

sdecays/107.3

pCi

Ci101

L

pCi0.4 1012

= 0.15 decays s1

L1

k = s3600

h1

h24

d1

d82.3

6391.0

t

2ln

2/1

= 2.10 × 16 s10

N = 16

11

s1010.2

Lsdecays15.0

K

rate

= 7.1 × 10

4 atoms

222Rn/L

atoms106.02

Rnmol1

L

Rn atoms107.123

2222224

= 1.2 × 1910 mol

222Rn/L

69. 21 H +

21 H → 4

2 He; Q for 21 H = 1.6 ×

1910 C; mass of deuterium = 2 amu.

E = r

)QQ(C/mJ100.9 2129

= m102

)C106.1(C/mJ100.915

21929

= 1 × 1310 J per alpha particle

KE = 1/2 mv2; 1 × 1310 J = 1/2 (2 amu × 1.66 × 2710 kg/amu)v

2, v = 8 × 10

6 m/s


From the kinetic molecular theory discussed in Chapter 5:

urms =

2/1

M

RT3

where M = molar mass in kilograms = 2 × 310 kg/mol for deuterium

8 × 106 m/s =

2/1

3

11

kg102

)T)(molKJ3145.8(3

, T = 5 × 109 K

70. Total activity injected = 86.5 × 103

Ci

Activity withdrawn OHmL

Ci108.1

OHmL2.0

Ci106.3

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

g6.453

lb1

mL

OHg0.1OHmL108.4 2

24

= 71%

71. a. For a gas, uavg = 8RT/πR where M is the molar mass in kg. From the equation, the

lighter the gas molecule, the faster is the average velocity. Therefore, 235

UF6 will have

the greater average velocity at a certain temperature because it is the lighter molecule.

b. From Graham’s law (see Section 5.7 of the text):

g/mol03.349

g/mol05.352

)UF(M

)UF(M

UFforratediffusion

UFforratediffusion

6235

6238

6238

6235

= 1.0043

Each diffusion step increases the 235

UF6 concentration by a factor of 1.0043. To

determine the number of steps to get to the desired 3.00% 235

U, we use the following

formula:

6

238

6235

N

6238

6235

UF0.97

UF00.3)0043.1(

UF3.99

UF700.0

original ratio final ratio

where N represents the number of steps required.


Solving (and carrying extra sig. figs.):

(1.0043)N =

9.67

9.297 = 4.387, N log(1.0043) = log(4.387)

N = 310863.1

6422.0

= 345 steps

Thus 345 steps are required to obtain the desired enrichment.

c. 98500

15265358.1

UF

UF,

152610000.1

1526)0043.1(

UF

UF

6238

6235

5

100

6238

6235

original ratio final ratio

6

238

6235

UF

UF = 1.01 × 10

2 = initial 235

U to 238

U atom ratio

72. 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 amu

E = mc2 = 0.5631 amu ×

amu

kg106605.1 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 =

)s/m10998.290.0(kg10109.9

sJ10626.6831

34

= 2.7 × 1210 m

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SM Chapter 20

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