Solutions for Chapter 3 End-of-Chapter...

December 2004 ACS Chemistry Chapter 3 suggested solutions 1

Solutions for Chapter 3 End-of-Chapter Problems

Problem 3.1. The diagram in Figure 3.3 shows that the red light from the discharge source is diffracted less (bent less far from the incoming light beam) than the blue and violet light. This is what the legend for Figure 3.1 says is the case for the diffraction of light by a prism, so the diagram and figure are consistent. Note that the prism orientations are different in the two figures, so you have to consider this when interpreting them.

Problem 3.2. The diagram of a simple spectrograph (looking at it from above) that uses a transmission diffraction grating is:

The diagram shows the blue light diffracted less (bent less far from the incoming light beam) than the red light. This is what the legend for Figure 3.2 says is the case for the diffraction of light by a grating, so the diagram and figure are consistent.

Problem 3.3. Since the spectrum consisted of lines (discrete wavelengths rather than a continuum), the light source for this experiment must have been an atomic emission source, such as a discharge lamp like you used in Investigate This 3.4.

Problem 3.4. In flame tests, the metal ions, which are just atoms that have lost one or more electrons, in the ionic compounds tested emit light of only certain wavelengths, so only those colors of light reach our eyes. We see different colors from different ions because each emits a characteristic set of wavelengths. This is just a less sophisticated detection system than using a spectrometer to look at the wavelengths individually. Bunsen actually invented the Bunsen burner to provide these emissions as a light source for a spectrometer, which he also invented.

Problem 3.5. When we add up the elemental abundances in the universe from Figure 3.7, any element whose abundance is more than four orders of magnitude ( 0.0001) lower than the abundance of He adds negligibly to the atom count, so we neglect any element with a log(abundance) < 5. Converting the log(abundance) to abundance [= 10log(abundance)] for the values in Figure 3.7 gives the data in this table, where the abundances have been divided by 107, in order to make them easier to compare.

element H He O C N Ne Mg Si Fe S Ar

abund 3200 250 2.5 1.3 0.40 0.32 0.13 0.10 0.079 0.040 0.013

Origin of Atoms Chapter 3

2 ACS Chemistry Chapter 3 suggested solutions

The sum of the H and He is 3450 107 and the sum of all the others is 5 107. Thus, as a percentage of the total, H plus He are 99.9% [= (3450/3455) 100%]. The vast majority of atoms in the universe are still those produced shortly after the Big Bang.

Problem 3.6. From the solution to Problem 3.5, we see that there are 250 He atoms (or nuclei) for every 3200 H atoms. The 3200 H atoms have a mass of about 3200 u and the 250 He atoms (being four times more massive) have a mass of about 1000 u. He is 24% [= (1000/4200) 100%] of the total mass of the H plus He and is, therefore, about 24% of the total mass of the universe.

Problem 3.7. The reasoning used to find the mass of iron in the universe compared to the mass of carbon is the same as the reasoning in Problem 3.6. We convert the relative numbers of atoms to mass and compare the masses. From the table in the solution to Problem 3.5, we see that there are about 79 iron atoms for every 1300 carbon atoms. Multiplying by the relative atomic masses gives 4400 u Fe and 15600 u of C, so the mass of iron is about 28% the mass of carbon in the universe.

Problem 3.8. The reasoning required to find out how the mass of iron in the universe compares to the mass of all the other fourth period transition metals combined is the same as in Problem 3.7, except we have to sum the masses of the other nine transition metals and compare them to the mass of iron. This table gives the data you need to show that iron makes up 92% {= [(44 106)/(48 106)] 100%} of the mass of all the fourth period transition elements in the universe.

element abundance rel at mass, u mass, u Sc 32 45 1,400 Ti 2,500 48 120,000 V 320 51 16,000 Cr 20,000 52 1,040,000 Mn 10,000 55 550,000 Fe 790,000 56 44,000,000 Co 2,000 59 120,000 Ni 40,000 59 2,400,000 Cu 500 64 32,000 Zn 2,000 65 130,000

Problem 3.9. The ions, S2–, Cl–, K+, and Ca2+, all have the same number of electrons, 18.

Chapter 3 Origin of Atoms

ACS Chemistry FROG 3

Problem 3.10. The data in Table 3.1 show that the mass of the proton is 1.837 103 greater than the mass of the electron:

mass proton

mass electron

1.673 10 27 kg9.109 10-31kg

1.837 103

Problem 3.11. Isotopes are atoms or ions of the same element with identical atomic number, but different mass numbers. Isotopic nuclei all have the same number of protons, but different numbers of neutrons.

Problem 3.12. The mass number of an isotope is not a mass, but a count of the number of protons and neutrons in the nucleus of the isotopic atoms. Since the mass of each of these particles is almost the same and equal to about one atomic mass unit, u, the mass number is close to the mass of the nucleus in atomic mass units. This is also close to the mass of the isotopic atom, because, as you found in Problem 3.10, the mass of electrons is so small that it can be ignored when taking the mass number for the nucleus as an approximate value for the mass of the isotopic atom (in atomic mass units).

Problem 3.13. (a) The element that has an isotopic species with 30 protons, 35 neutrons, 28 electrons is atomic number 30 (= number of protons in the nucleus): Zn. To be more precise, we might write Zn-65 to specify the mass number (= number of protons + number of neutrons) of this isotope of zinc. (b) Since the species described in part (a) has two fewer electrons than protons, it has an overall positive charge and is a cation that we can write as Zn2+.

Problem 3.14. The mass, in grams of 25 protons is:

25 protons = (25 protons)1.673 10–24 kg

1 proton

= 4.183 10–23 kg

Problem 3.15. The atomic symbols for each of these isotopes is given in two forms: (a) Z = 19; A = 40 19

40 K K-40 (b) Z = 79; A = 197 79

197 Au Au-197 (c) Z = 54; A = 118 54

118 Xe Xe-118 (d) Z = 13; A = 28 15

28 Al Al-28 (e) Z = 53; A = 118 53

118 I I-118 (f) Z = 83; A = 189 83

189 Bi Bi-189



Problem 3.16. The number of protons (p), neutrons (n), and electrons (e) is given for each of these ions: (a) 58Ni+ 28 p; 30 n; 27 e (b) 32S2– 16 p; 16 n; 18 e (c) 65Zn2+ 30 p; 35 n; 28 e (d) 37Cl– 17 p; 20 n; 18 e (e) 55Mn7+ 25 p; 30 n; 18 e (f) 56Fe2+ 26 p; 30 n; 24 e

Problem 3.17. Completing the table in this problem (answers in red) requires you to use (1) the periodic table to correlate atomic numbers with elemental symbols, (2) the definitions of atomic number and mass number in terms of numbers of protons and neutrons in elemental nuclei, and (3) the relationship between numbers of protons and electrons in atoms and ions. There is actually enough information in the symbols in the seventh column to fill in the entire remainder of the table without any of the values being filled in.

Substance # of protons

# of neutrons

# of electrons

Atomic number

Mass number

Nuclear symbol

rhodium 45 55 45 45 100 45100 Rh

iron 26 34 23 26 60 2660 Fe3+

radium 88 137 88 88 225 88225 Ra

sulfur 16 22 18 16 38 1638 S2–

fermium 100 148 100 100 248 100248 Fm

Problem 3.18. (a) The present background temperature of the universe, 3 K, is very cold. This corresponds to a Celsius temperature of –270 C [t C = (T – 273) K]. (b) If the Earth formed at a temperature just below 1000 K, there would have been no liquid water on its surface. The boiling point of water, 100 C, is 373 K [= 100 +273) K]. 1000 K is far above the normal boiling point of water. (c) We use the information given in this problem to find the kelvin temperature corresponding to a Fahrenheit temperature of 85 F. The Fahrenheit temperature is 53 F (= 85 F – 32 F) above the freezing point of water, 32 F = 273 K. We are told that a degree on the kelvin scale is 1.8 times the size of a degree on the Fahrenheit ( F) scale, so the 53 F difference is:

� F = 53 F = (53 F)1 K

1.8 oF

= 29 K = �� K

In kelvin, the difference in temperature between the freezing point of water and the temperature of interest is 29 K, so we have 85 F = 302 K (= 273 K + 29 K).



Problem 3.19. Familiar examples of condensation that are analogous to the formation of stars and planets when matter condensed upon cooling as the universe expanded are cloud formation as tiny droplets of water condense from water vapor and the deposition of dew or frost (liquid or solid water condensation on cool or cold surfaces).

Problem 3.20. We are to put these reactions in order from the one that takes place at the lowest temperature to the one that requires the highest temperature and explain our ordering:

(a) 22Ne10+ + 22Ne10+ 43K19+ + 1H+ (b) 13C3+ + e– 13C2+ (c) 12C6+ + 18O8+ 26Mg12+ + 4He2+ (d) 12C6+ + 3He2+ 13N7+ + 2H+

Reaction (b) is the combination of a cation and an electron; the reactants attract one another, so the reaction takes place at a relatively low temperature, <104 K. At somewhat higher temperatures atoms lose all their electrons. The other three reactions are nuclear fusions (requiring temperatures above about 107 K) with the loss of a lighter particle to form the product. To bring the nuclei together in the first place they have to have enough energy to overcome the repulsion between their positive charges. The larger the charges that have to be brought together, the higher the temperature required to get the nuclei moving fast enough. Thus we can see that increasing temperature is required going from (d), 6+ and 2+, to (c), 6+ and 8+, to (a), 10+ and 10+. The order of increasing temperature required for reaction is: (b) << (d) < (c) < (a).

Problem 3.21. A temperature of approximately 107 K is necessary for nuclei to have sufficient kinetic energy to sustain nuclear fusion.

Problem 3.22. All elements with atomic numbers up to iron (26), except H and He, are formed in stars by fusion reactions, like those shown in equations (3.6) through (3.9). He can also be formed in stars but a large part was formed just after the Big Bang before stars were formed. Heavier elements are formed by fusions involving large quantities of neutrons in supernovae explosions of dying stars, as illustrated in equations (3.10) and (3.11).

Problem 3.23. A good way to proceed to write a complete balanced nuclear reaction, when a partial reaction with a reactant or product omitted has been given, is to rewrite the reaction with all the atomic numbers and mass numbers shown explicitly. Then you can add them up on each side to find their values for the missing species and hence its identity. Showing the nuclear and other charges on the species is unnecessary, but provides a check on your answer.

(a) 14N + ____ 17O + p 7

14 N 7+ + ____ 817 O8+ + 1

1H+



To balance this reaction, the missing species has to have mass number = 4 and an atomic number = 2. The required species is He-4, that is, an alpha particle. The complete equation is:

714 N 7+ + 2

4 He2+ 817 O8+ + 1

1H+ As a check on our work, add the charges on each side and see that they are the same, 9+.

(b) 13C + neutron ____ + 6

13 C6+ + 01n0 ____ +

To balance this reaction, the missing species has to have mass number = 14 and an atomic number = 6. The required species is C-14. The complete equation is:

613 C6+ + 0

1n0 614 C6+ +

As a check on our work, add the charges on each side and see that they are the same, 6+.

(c) 1H + 1H 2H + ____ 1

1H+ + 11H+ 1

2 H + + ____ To balance this reaction, the missing species has to have mass number = 0 and an atomic number = 1. The required species is 1

0 e , that is, a positron. The complete equation is: 1

1H+ + 11H+ 1

2 H + + 10 e


(d) 20Ne + ____ 24Mg + 10

20 Ne10+ + ____ 1224 Mg12+ +

To balance this reaction, the missing species has to have mass number = 4 and an atomic number = 2. The required species is He-4, that is, an alpha particle. The complete equation is:

1020 Ne10+ + 2

4 He2+ 1224 Mg12+ +


(e) 20Ne + 4He ____ + 16O 10

20 Ne10+ + 24 He2+ ____ + 8

16 O8+ To balance this reaction, the missing species has to have mass number = 8 and an atomic number = 4. The required species is Be-8. The complete equation is:

1020 Ne10+ + 2

4 He2+ 48 Be4+ + 8

16 O8+ As a check on our work, add the charges on each side and see that they are the same, 12+.

(f) 27Al + 2H ____ + 28Al 13

27 Al13+ + 12 H + ____ +

1328 Al13+

To balance this reaction, the missing species has to have mass number = 1 and an atomic number = 1. The required species is H-1. The complete equation is:

1327 Al13+ + 1

2 H + 11H+ +

1328 Al13+


Problem 3.24. These incomplete nuclear reactions are completed and balanced as in the solutions to Problem 3.23.



(a) 97Tc 97Ru + ____ 43

97 Tc43+ 4497 Ru44+ + ____

To balance this reaction, the missing species has to have mass number = 0 and an atomic number = –1. The required species is –1

0e– , that is, an electron (beta particle). The complete equation is:

4397 Tc43+ 44

97 Ru44+ + –10e–


(b) ____ + 4He 243Bk + n ____ + 2

4 He2+ 97243 Bk97+ + 0

1n0 To balance this reaction, the missing species has to have mass number = 240 and an atomic number = 95. The required species is 95

240 Am95 . The complete equation is: 95

240 Am95 + 24 He2+ 97

243 Bk97+ + 01n0


(c) 249Cf + ____ 263Sg + 4n 98

249 Cf98 + ____ 106263 Sg106 + 40

1n 0 To balance this reaction, the missing species has to have mass number = 18 and an atomic number = 8. The required species is 8

18 O8 . The complete equation is: 98

249 Cf98 + 818 O8 106

263 Sg106 + 401n0


(d) 1H + 14N ____ + 4He 1

1H+ + 714 N 7+ ____ + 2

4 He2+ To balance this reaction, the missing species has to have mass number = 11 and an atomic number = 6. The required species is 6

11C6 . The complete equation is: 1

1H+ + 714 N 7+ 6

11C6 + 24 He2+


(e) n + 235U ____ + 94Sr + 2n 0

1n0 + 92235 U 92 ____ + 38

94 Sr38 + 201n0

To balance this reaction, the missing species has to have mass number = 140 and an atomic number = 54. The required species is 54

140 Xe54 . The complete equation is: 0

1n0 + 92235 U 92 54

140 Xe54 + 3894 Sr38 + 20

1n0 As a check on our work, add the charges on each side and see that they are the same, 92+.

(f) 228Ra ____ + 228Ac 88

228 Ra88 ____ + 89228 Ac89

To balance this reaction, the missing species has to have mass number = 0 and an atomic number = –1. The required species is –1

0e– , that is, an electron (beta particle). The complete equation is:

88228 Ra88 –1

0e– + 89228 Ac89




Problem 3.25. The balanced nuclear equation for the beta decay of 24Na, including both mass and atomic numbers, is: 11

24 Na11+ 1224 Mg12+ + -1

0 e–

Problem 3.26. [The reactions in this problem were chosen because they make connections back to the chapter, rather than being “blue-sky.”] (a) Americium–241, which is used in smoke detectors, is extracted from spent nuclear reactor fuel rods where it has been produced by a p–n reaction of plutonium. In a p–n reaction, a proton reacts with a nucleus to produce a new nucleus and a neutron as products. Since p and n have the same mass number, the isotope of plutonium that produces Am-241 must be Pu-241. The balanced nuclear reaction is: 94

241 Pu94 + 11H+ 95

241 Am95 + 01n 0

(b) By the same reasoning as in part (a), the carbon isotope produced by an n–p reaction of nitrogen–14 in the atmosphere must be carbon-14. The balanced nuclear reaction is: 7

14 N 7 + 01n0 + 6

14 C6 + 11H+

Problem 3.27. The unbalanced nuclear equation for the reaction of 14N capturing a neutron and decaying to 3H (tritium) and another product is: + 7

14 N 7+ + 01n0 1

3H+ + ____ To balance this reaction, the missing species has to have mass number = 12 and an atomic number =6. The required species is 6

12 C6 . The complete equation is:

+ 714 N 7+ + 0

1n0 13H+ + 6

12 C6

Problem 3.28. These equations describe how the fusion of two 12C6+ can lead to the formation of the target nuclei: (a) 23Na11+ 6

12 C6+ + 612C6+ 11

23Na11+ + 11 H+

(b) 23Mg12+ 612 C6+ + 6

12C6+ 1223Mg12+ + 0

1n

(c) 20Ne10+ 612 C6+ + 6

12C6+ 1020Ne10+ + 2

4 He2+ or

612 C6+ + 6

12C6+ 1020Ne10+ + 1

3 H+ + 11 H+ or

612 C6+ + 6

12C6+ 1020Ne10+ + 2 1

2 H+

(d) 16O8+ 612 C6+ + 6

12C6+ 816 O8+ + 2 2

4 He2+ or

612 C6+ + 6

12C6+ 816 O8+ + 4

8 Be4+ or several other possibilities you can probably think of.



Problem 3.29. These equations describe how the fusion of two 16O8+ can lead to the formation of the target nuclei. Other reactions can also be written, as we have seen in the solution for Problem 3.28. For example, an alpha particle, 2

4 He2+ , could be replaced by two deuterium nuclei, 12 H + , or a

tritium nucleus, 13H+ , and proton, 1

1H+ , and a proton could be replaced by a neutron, 01n 0 , and a

positron, 10 e .

(a) 32S16+ 816 O8+ + 8

16 O8+ 1632 S16+ +

(b) 31P15+ 816 O8+ + 8

16 O8+ 1531 P15+ + 1

1H

(c) 28Si14+ 816 O8+ + 8

16 O8+ 1428 Si14+ + 2

4 He2+

(d) 24Mg12+ 816 O8+ + 8

16 O8+ 1224 Mg12+ + 2 2

4 He2+

Problem 3.30. These equations show how the fusion of 12C6+ and 16O8+ could produce the three isotopes of Mg (24Mg12+, 26Mg12+, and 27Mg12+). As in the solutions to Problems 3.28 and 3.29, you can write reactions with other products (in addition to the target nuclei). (a) 6

12 C6+ + 816O8+ 12

24 Mg12+ + 24 He2+

(b) 612 C6+ + 8

16O8+ 1226 Mg12+ + 21

1H

(c) 612 C6+ + 8

16O8+ 1227 Mg12+ + 0

1n + 2 10e+

Problem 3.31. The series of six successive beta particle emissions by which xenon-143 decays to a stable isotope, neodymium-143, are: 143Xe54+ 143Cs55+ + 0e– 143Cs55+ 143Ba56+ + 0e– 143Ba56+ 143La57+ + 0e– 143La57+ 143Hf58+ + 0e– 143Hf58+ 143Pr59+ + 0e– 143Pr59+ 143Nd60+ + 0e–

Problem 3.32. The skeleton of the “Kennewick Man” contains only 0.362 (36.2%) as much carbon-14 as is present in living organisms, so he must have died long enough ago that 0.638 (63.8%) of his carbon-14 has decayed. This datum, the decimal fraction of C-14 remaining, fn, can be used with equation (3.18) to find the number of C-14 half lives, n, since he died:

n = ln(fn )

–0.693 =

ln(0.632)–0.693

= –1.016–0.693

= 1.47 half lives

We find the age of the sample using the half life of C-14, 5730 yr:



age of sample = (1.47 half lives)5730 yr

1 half life

= 8420 yr

Kennewick Man died about 8420 years BP (before present) or about 6420 BCE.

Problem 3.33. Figuring out how much fluorine-18 remains in a patient’s body three hours and forty-five minutes (225 minutes from 8:15 a.m. to noon) after a compound containing F-18 is injected, is a half-life problem. From Table 3.3, we find that the half life of F-18 is 110 min. Thus, about 2.05 half lives = n [= (225 min)/(110 min·half life–1)] have elapsed. We can use equation (3.18) to find the fraction, fn, of the original F-18 that is left: ln(fn) = –0.693n = (–0.693)·(2.05) = –1.42 fn = eln(fn) = e–1.42 = 0.24 The fraction of F-18 remaining undecayed is 0.24 (24%) of the amount that was injected.

Problem 3.34. (a) Beta particles (electrons) do not penetrate materials very far. Any electrons produced by the Am-241 in Investigate This 3.30 would have been completely blocked by a sheet of aluminum foil, as you learned in Consider This 3.31. The electrons emitted by the decay of I-131 in the thyroid gland are absorbed by the cells very close to the source of the emission and this is within the gland, so it is thyroid gland cells that are killed. (b) We know that the fraction of a radioactive isotope remaining in a sample is a function of the number of half lives that have elapsed from when the sample was prepared. We can use equation (3.18) to find the number of half lives, n, for decay to 0.10 of the initial amount as:

n = ln(fn )

–0.693 =

ln(0.10)–0.693

= –2.30–0.693

= 3.32 half lives

Table 3.3 shows that the half life for I-131 is 8 days. The time required for a sample of I-131 to decay to 10% of its initial level is:

decay time = (3.32 half lives)8 day

1 half life

27 days

Problem 3.35. (a) The balanced nuclear reaction equation for the decay of Rb-87by beta emission is: 87Rb37+ 87Sr38+ + e–

(b) We can use the half life for this reaction, 4.9 1010 years, to date the rock sample from Greenland that has a Sr-87/Rb-87 mass ratio of 0.056. We begin by assuming that the sample contained only Rb-87 when it was formed, so that all the Sr-87 present now is a result of the subsequent Rb-87 decay. We further assume that all the Rb-87 nuclei initially present are still present as either Rb-87 or Sr-87 nuclei. Let the initial mass of Rb-87 be m. If a fraction f of the Rb-87 nuclei remains unreacted, then the mass of Rb-87 remaining is f·m and the mass of Sr-87 formed is (1 – f)·m. The Sr-87/Rb-87 mass ratio is [(1 – f)·m]/[f·m] = (1 – f)/f. For our sample of Greenland rock, we have (1 – f)/f = 0.056



Solving this equation for f, gives f = 0.95, the fraction of Rb-87 nuclei left undecayed. We can use equation (3.18) to find the number of half lives corresponding to this fraction undecayed:

n = ln(f )

–0.693 =

ln(0.95)–0.693

= –0.051–0.693

= 0.074 half lives

The age of the rock sample is:

age = (0.074 half lives)4.9 1010 yr1 half life

= 3.6 109 yrs = 3.6 billion years

Problem 3.36. Nuclear binding energy is the energy required to break apart a nucleus into individual protons and neutrons. It is found by converting the mass difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus to energy via the Einstein mass-energy equivalence relationship.

Problem 3.37. Fusion is the reaction between two or more nuclei to form a product that is more massive than any of the reacting nuclei. Fission is the reaction of a heavy elemental nucleus breaking up into smaller nuclei of approximately equal mass plus two or more neutrons. (Nuclei that emit alpha or beta particles are not undergoing fission. These emissions are characterized by well-defined half lives that are essentially independent of the environment of the nucleus. Fissions are usually initiated by collision of the fissionable nucleus with a neutron.) Elements lighter than iron tend to undergo fusion, heavier elements tend to undergo fission. In both cases the products of the reaction have a higher nuclear binding energy than the reactants, so the products are more stable.

Problem 3.38. To calculate the binding energy per nucleon for an elemental nucleus, subtract the sum of the masses of the protons and neutrons that compose the nucleus from the experimental mass of the nucleus: mass loss per nucleus = m = mass of nucleus – (mass of protons + mass of neutrons) Convert the mass difference to energy via the Einstein mass-energy equivalence relationship ( E = mc2) and divide the result by the number of nucleons to get the binding energy per nucleon. Binding energies can be given per nucleon or per mole of nucleons. (a) For 20Ne10+ (nuclear mass = 33.18913 10–27 kg) we find: m = (33.18913 10–27 kg) – [10(1.672623 10–27 kg) + 10(1.674929 10–27 kg)] = –0.28639 10–27 kg binding energy = E = (–0.28639 10–27 kg·nucleus–1)(2.9979 108 m·s–1)2 = –2.5739 10–11 J·nucleus–1

binding energy·nucleon–1 = –2.5739 10–11 J·nucleus–1

20 nucleon·nucleus–1 = –1.2870 10–12 J·nucleon–1

For a mole of nucleons, we have:



molar binding energy·nucleon–1 = (–1.2870 10–12 J·nucleon–1) 6.0221 1023 nucleon1 mol nucleon

= –7.7504 1011 J·(mol nucleon)–1

(b) For 28Si14+ (nuclear mass = 46.45681 10–27 kg) we find: m = (46.45681 10–27 kg) – [14(1.672623 10–27 kg) + 14(1.674929 10–27 kg) = –0.40892 10–27 kg binding energy = E = (–0. 40892 10–27 kg·nucleus–1)(2.9979 108 m·s–1)2 = –3.6751 10–11 J·nucleus–1

binding energy·nucleon–1 = –3.6751 10–11 J·nucleus–1

28 nucleon·nucleus–1 = –1.3125 10–12 J·nucleon–1

For a mole of nucleons, we have:

molar binding energy·nucleon–1 = (–1.3125 10–12 J·nucleon–1) 6.0221 1023 nucleon1 mol nucleon

= –7.9040 1011 J·(mol nucleon)–1

(c) The silicon-28 nucleus has a higher binding energy per nucleon than neon-20, so the silicon nucleus is more stable. Both elements have atomic numbers less than iron (26) and we know, Figure 3.18, that the elements preceding iron in the periodic table increase in nuclear stability as the atomic number increases. Thus, the greater stability of the silicon nucleus (compared to neon) is consistent with this trend.

Problem 3.39. For the elemental nuclei, 6Li3+ (nuclear mass = 9.98561 10-24 g) and 56Fe26+ (nuclear mass = 92.8585 10-24 g), we are asked to calculate the binding energy in kilojoules (a) per nucleus, (b) per mole [of nuclei], and (c) per nucleon [also per mole of nucleons]. The procedure for these calculations is outlined in the solution to Problem 3.38.

For 6Li3+: 9.98561 10-27 kg m = (9.98561 10–27 kg) – [3(1.672623 10–27 kg) + 3(1.674929 10–27 kg) = –0.05705 10–27 kg (a) The binding energy per nucleus is: binding energy per nucleus = E = (–0.05705 10–27 kg·nucleus–1)(2.9979 108 m·s–1)2 = –5.1273 10–12 J·nucleus–1 = –5.1273 10–15 kJ·nucleus–1 (b) The binding energy per mole of nuclei is:

molar binding energy·nucleus–1 = (–5.1273 10–15 kJ·nucleus–1) 6.0221 1023 nuclei1 mol nuclei

= –3.0877 109 kJ·(mol nuclei)–1

(c) The binding energy per nucleon is:

binding energy·nucleon–1 = –5.1273 10–15 kJ·nucleus–1

6 nucleon·nucleus–1 = –8.5455 10–16 kJ·nucleon–1



molar binding energy·nucleon–1 = (–8.5455 10–16 kJ·nucleon–1) 6.0221 1023 nucleon1 mol nucleon

= –5.1462 108 kJ·(mol nucleon)–1 = –51.462 107 kJ·(mol nucleon)–1

For 56Fe26+: 92.8585 10-27 kg m = (92.8585 10–27 kg) – [26(1.672623 10–27 kg) + 30(1.674929 10–27 kg) = –0.87757 10–27 kg (a) The binding energy per nucleus is: binding energy per nucleus = E = (–0.87757 10–27 kg·nucleus–1)(2.9979 108 m·s–1)2 = –7.8871 10–11 J·nucleus–1 = –7.8871 10–14 kJ·nucleus–1 (b) The binding energy per mole of nuclei is:

molar binding energy·nucleus–1 = (–7.8871 10–14 kJ·nucleus–1) 6.0221 1023 nuclei1 mol nuclei

= –4.7497 1010 kJ·(mol nuclei)–1

(c) The binding energy per nucleon is:

binding energy·nucleon–1 = –7.8871 10–14 kJ·nucleus–1

56 nucleon·nucleus–1 = –1.4084 10–15 kJ·nucleon–1

molar binding energy·nucleon–1 = (–1.4084 10–15 kJ·nucleon–1) 6.0221 1023 nucleon1 mol nucleon

= –8.4816 108 kJ·(mol nucleon)–1 = –84.816 107 kJ·(mol nucleon)–1

Note that the binding energy per nucleon for Fe-56 is larger than that for Li-6, just as you would expect since Fe has the most stable elemental nucleus.

Problem 3.40. (a) A picture of two nuclei, C-12 and He-4, undergoing fusion to give O-16 is shown here, as in the Web Companion, Chapter 3, Section 3.5, pages 1-4, with protons represented by dark circles and neutrons by lighter gray circles (no electrons are involved):

(b) This fusion can occur because, as Figure 3.18 shows, the resultant nucleus, O-16, has a greater molar binding energy per nucleon (–77 107 kJ·mol–1) than either of the starting nuclei, C-12 (-74 107 kJ·mol–1) and He-4 (–68 107 kJ·mol–1). The overall energy change, �E, for the reaction is approximately: E = 16·(–77 107 kJ·mol–1) – [12·(–74 107 kJ·mol–1) + 4·(–68 107 kJ·mol–1)] = –72 107 kJ·mol–1 This is the result represented in Figure 3.19 and differs slightly from the values obtained in Check This 3.50 due to rounding off of the binding energy values.



(c) Fission of these nuclei would lead to products with less binding energy per nucleon and hence an overall positive energy change, that is, a highly endothermic reaction. For example, if C-12 were to split into two Li-6 nuclei, the energy change would be approximately: E = 2·6·(–54 107 kJ·mol–1) – 12·(–74 107 kJ·mol–1) = 240 107 kJ·mol–1 This enormous positive energy is approximate, because the binding energies in Figure 3.18 are for the most abundant isotope of each element and Li-7 is the most abundant isotope of lithium. Different isotopes of the same element do not usually differ a great deal in binding energy per nucleon, so a small difference will not affect the conclusion that this is a very unfavorable reaction energetically. You found the molar binding energy per nucleon for Li-6 (–51 107 kJ·mol–1) in Problem 3.39. Using this value in this problem, we get 276 107 kJ·mol–1 as the energy change for the C-12 fission reaction—an even more unfavorable result.

Problem 3.41. (a) The Xe-142 (nuclear mass = 235.63075 10–27 kg) binding energy is the energy change for this imagined reaction: 54p + 88n 142Xe m = 235.63075 10–27 kg – [54·(1.672623 10–27 kg) + 88·(1.674929 10–27 kg)] = –2.0846 10–27 kg E (mole) = (–2.0846 10–27 kg)·( 2.9979 108 m·s)2·(6.0221 1023 nuclei·mol–1) = –1.128 1011 kJ·mol–1 = –11280 107 kJ·mol–1

(b) The nuclei in the fission reaction in this problem are the same as in Figure 3.20, except for Xe-142 here in place of Xe-143 in the figure. Thus, the binding energies are the same for the other nuclei and the energy change for the reaction here is: E = [(–7560 107 kJ·mol–1) + (–11280 107 kJ·mol–1)] – (–17250 107 kJ·mol–1) = –1590 107 kJ·mol–1

(c) Within the round-off uncertainties of the two values, the one here and the one in Figure 3.20 are the same. With these limited data, we might conclude that the fission of a U-235 nucleus always produces about the same amount of energy, regardless of the way the nucleus splits. The next problem provides some further data.

Problem 3.42. (a) We solve this problem exactly as is illustrated in Figure 3.20 for fission to yield Sr-90 and Xe-143. (For ease in writing, we will omit the 107 factor as well as the units and include them only in the final result.) For fission that produces Kr-92 (–82.3 107 kJ·mol–1) and Ba-141 (–80.5 107 kJ·mol–1), we get: E = [92(–82.3) + 141(–80.5)] – 235(–73.4) = –1670 107 kJ·mol–1

For fission that produces Rb-89 (–83.7 107 kJ·mol–1) and Cs-144 (–79.4 107 kJ·mol–1), we get: E = [89(–83.7) + 144(–79.4)] – 235(–73.4) = –1630 107 kJ·mol–1



(b) Again, just as in Problem 3.41, we find that the energy released in the fission of one mole of U-235 is in the range about 1600 to 1700 107 kJ·mol–1, and the conclusion we reached, that the fission of U-235 produces about the same amount of energy, no matter what the product nuclei, is strengthened.

Problem 3.43. (a) and (b) The C-12 nucleus is shown here in each of the four reactions represented in the Web Companion, Chapter 3, Section 3.4, page 3: alpha, beta, gamma, and positron emission. In these diagrams, protons are represented by dark circles and neutrons by lighter gray circles, as they are in the Companion.

alpha emission:

beta emission:

gamma emission:

positron emission:

(c) Carbon-12 is a stable isotope, that is, it does not emit any form of radioactivity. From Figure 3.18, you can see that emission of any of the three particles above leads to products that are of higher energy than carbon-12, so this explains its stability. Positron (or electron capture) and beta emission would be more likely from some isotope of B or N, where, in each case, a more stable elemental nucleus is formed. Positron emission (or electron capture) tend to take place from isotopes that have fewer than the normal (stable) number of neutrons: N-13, for example. Beta emission tends to occur from isotopes that have fewer than the normal number of protons: C-14, for example. Alpha particle emission is generally a property of the heavier elements (beyond Fe) where the resulting nuclei (including the He-4) are more stable – lower energy – than the parent. Gamma emission may accompany any of the other emissions or be the only emission, as in electron capture or decays from metastable nuclei, such as technetium-99m.

Problem 3.44. The mass of a He-3 nucleus is 5.00642 10–24 g. From Worked Example 3.41 we get the mass of a 4He2+ nucleus, 6.64466 10-24 g, and from Table 3.1, the mass of an H-1 nucleus, 1.67262 10–24 g. Thus, for the reaction, 2

3He2+ + 23He2+ 2

4 He 2+ + 211H +, the mass loss and

energy change are: m = [2·(1.67262 10–27 kg) + 6.64466 10–27 kg] – 2·(5.00642 10–27 kg) = –0.02294 10–27 kg E = (–0.02294 10–27 kg)·(2.9979 108 m·s)2 = –2.065 10–12 J



This exothermic energy change is for 10.01 10–24 g [= 2·(5.00642 10–27 kg)] of He-3 reacting. One gram is 1023 times this amount of reactant, so the energy change for one gram of reactant is: E (per gram) = (–2.065 10–12 J)(1023) = –2.07 1011 J = –20.7 107 kJ.

Problem 3.45. (a) Two possible fusion reactions of deuterium are: 2H + 2H 3H + p (i) 2H + 2H 3He + n (ii) The energy changesfor reactions (i) and (ii), given that the masses of H 2, H 3, and He-3 nuclei are, respectively, 3.34357 10–24 g, 5.00736 10–24 g, and 5.00642 10–24 g, are:

(i) m = (5.00736 10–27 kg + 1.67262 10–27 kg) – 2(3.34359 10–27 kg) = –0.00720 10–27 kg E = (–0.00720 10–27 kg)·(3.00 108 m·s)2 = –6.48 10–13 J [This problem is based on the “benchtop fusion” article by R.P. Taleyarkhan, C.D. West, J.S. Cho, R.T. Lahey, Jr., R.I. Nigmatulin, and R.C. Block, “Evidence for Nuclear Emissions During Acoustic Cavitation,” Science 2002, 295, 1868-1873, with a commentary by F.D. Becchetti on page 1850 and a News article by Charles Seife on pages 1808-9. In the News article, the experimentally known kinetic energies of the emitted particles from these fusions is given in MeV (1MeV = 1.6022 10–13 J) Here, E from the mass loss is converted to MeV and compared with the energies cited in the News article, in brackets, {}.]

E = (–6.48 10–13 J) 1 MeV1.6022 10–13 J

= 4.04 MeV

{particle kinetic energies: 1.01 MeV H-3 + 3.02 MeV p = 4.03 MeV}

(ii) m = (5.00642 10–27 kg + 1.67493 10–27 kg) – 2(3.34359 10–27 kg) = –0.00583 10–27 kg E = (–0.00583 10–27 kg)·(3.00 108 m·s)2 = –5.25 10–13 J

E = (–5.25 10–13 J) 1 MeV1.6022 10–13 J

= 3.28 MeV

{particle kinetic energies: 0.82 MeV He-3 + 2.45 MeV n = 3.27 MeV}

(b) The products of reaction (i), tritium (H-3) and a proton, are more stable, since more energy is released in this reaction. The reactants are the same in both reactions, so the release of more energy means the products of reaction (i) are lower energy (more stable) than the products of reaction (ii). Note that the mass of a H-3 and a He-3 nucleus are very similar (differ by 0.00092 10–27 kg), but the neutron is a good deal heavier than the proton (by 0.00231 10–27 kg), so more of the mass of the reactants is converted to energy in reaction (i).

Problem 3.46. (a) For the reaction 4He2+ + 4He2+ 8Be4+, with nuclear masses of helium-4 and beryllium-8 6.644655 10–24 g and 13.28949 10–24 g, respectively, we have:



m = (13.28949 10–27 kg) – 2·(6.644655 10–27 kg) = 0.00018 10–27 kg E = (0.00018 10–27 kg)·(3.00 108 m·s)2·(6.022 1023 nuclei·mol–1) = 9.8 109 J·mol–1 = 0.98 107 kJ·mol–1

(b) The reaction of two He-4 to form a Be-8 is endothermic, which is unfavorable. The reverse exothermic decay reaction is favored and helps to explain why the Be-8 nucleus falls apart so rapidly. (c) The mass of the C-12 nucleus, 19.92101 10–24 g, is given in Check This 3.42, so, for the reaction 4He2+ + 8Be4+ 12C6+, we have: m = (19.92101 10–27 kg) – [(6.644655 10–27 kg) + (13.28949 10–27 kg)] = –0.01314 10–27 kg E = (–0.01314 10–27 kg)·(3.00 108 m·s)2·(6.022 1023 nuclei·mol–1) = –7.119 1011 J·mol–1 = –71.19 107 kJ·mol–1

(d) When we sum the two reactions in parts (a) and (c) to get the overall reaction of three He-4 to yield C-12, we can also sum the energy changes for each reaction to get the overall energy change: overall E = (0.98 107 kJ·mol–1) + (–71.19 107 kJ·mol–1) = –70.21 107 kJ·mol–1 This is the result represented in Figure 3.19 and obtained as well, within round-off uncertainty in Check This 3.50 (e) The half life of the Be-8 nucleus is so short that the collision of the third He-4 has to occur within about 10–16 seconds of the Be-8 formation, or it will have fallen apart. This is what is meant by “almost simultaneous collision.” The formation of beryllium nuclei by this pathway is impossible, so stable isotopes have to be formed in other reactions of less abundant nuclei, which makes the abundance of beryllium in the universe quite low, as you see in Figure 3.7. The low beryllium abundance is, ultimately, a result of the low binding energy of its nucleus.

Problem 3.47. (a) The sequence of balanced nuclear reactions represented by 12C6+ 13N7+ 13C6+ 14N7+ 15O8+ 15N7+ (12C6+ + 4He2+) is: 12C6+ + 1H+ 13N7+ 13N7+ + 0e– 13C6+ 13C6+ + 1H+ 14N7+ 14N7+ + 1H+ 15O8+ 15O8+ + 0e– 15N7+ 15N7+ + 1H+ 12C6+ + 4He2+ The first three of these reactions are ones we wrote in Check This 356(b) to show how odd atomic number nuclei might be formed in stars. (b) The product of each reaction is the reactant in the next reaction, so they all get used up in the sequence. A C-12 is required to start the sequence, but another is formed in the last step, so there is no net loss or gain of C-12. The net change in the reaction is obtained by adding all the reactions and canceling species that appear on both sides of the sums. The result is:



41H+ + 20e– 4He2+ This equation is balanced in mass number, atomic number, and charge. This equation is formally “equivalent” to reaction equation (3.6), because, if you add two positrons to each side of the equation here, the electrons and positrons annihilate one another on the reactant side, giving reaction equation (3.6). (c) As we said in part (b), C-12 is required to start the sequence but is then regenerated at the end, so it does not appear in the net reaction equation. This is exactly how a catalyst works. It takes part in a reaction to facilitate it, as it here facilitates the formation of helium from hydrogen, but is not used up. Of course any of the intermediate products can take part in other reactions and be lost to the cycle, as we suggested in Check This 3.56(b) when we considered the formation of N-14 as a way to begin synthesis of odd atomic number nuclei. Still, in massive stars with nuclei at high density, this cycle is a large component of the fusion reactions producing more He-4 from H-1.

Problem 3.48. (a) We are asked to find out whether the ratio of hydrogen to oxygen atoms from Figure 3.27 is consistent with the composition of water. Converting the logarithmic data in Figure 3.27 to numbers of atoms of hydrogen and oxygen (per million total), we get 6.3 105 atoms of H and 2.5 105 atoms of O (which is 88% of all the atoms in the body). Since water has two atoms of H per atom of O, we require at least this 2:1 ratio for the data to be consistent with the composition of water. We see that the ratio is actually 2.5:1, reflecting the composition of other molecules that are also present, but confirming that the abundance is consistent with the presence of a lot of water in the composition of the human body. (b) Using the same logic as in part (a), we look at the number of atoms of O and Si from the data for the Earth’s crust in Figure 3.27. There are 6.3 105 atoms of O and 2.0 105 atoms of Si (which is 83% of all the atoms in the crust). The ratio of O atoms to Si atoms in the crust is a little over 3:1. The ratio in sand, silicon dioxide, is 2:1, so the composition is consistent with the idea that the crust is largely silicon dioxide (with other oxides and silicate minerals, which contain oxygen and silicon in various higher ratios).

Problem 3.49. (a) We have seen, equation (3.27), that U-238 decays by a series of alpha and beta emissions to give Pb-206 and we have seen that the ratio of lead to uranium is used as a method of dating rocks that contain uranium. It is not surprising, therefore, to find that ores containing uranium would also contain lead-206. Other stable isotopes of lead are also formed by nucleosynthesis in supernovae, so the Earth contains these as well and apparently ores that do not have uranium, but contain lead (such as galena, an important lead ore) contain a heavier lead isotope, Pb-208. (b) The relative atomic masses in our periodic tables are averages that account for the relative amounts of the various stable isotopes of an element . Lead has four stable isotopes, Pb-204 (very minor), Pb-206, Pb-207, and Pb-208. If the amounts of Pb-206 and Pb-208 were the same, we would expect the average atomic mass of a sample to be 207 u. Since the experimental value is larger than 207 u, we can conclude that, on average, there is more Pb-208 than Pb-206 on Earth. The relative fractions of the four isotopes are, respectively, 0.014, 0.241, 0.221, and 0.524.



Problem 3.50. By analogy with the light spectrometer that disperses light, the mass spectrometer disperses mass. Note that the diagram in Figure 3.29 shows that low and high mass ions are separated in space when they reach the detector. The dispersing element in this form of mass spectrometer is the magnet. The pathways of ions moving in a magnetic field are curved or bent by the field. The more massive ions are harder for the field to deflect, so their pathway is less curved, as shown on the diagram.

Problem 3.51. The reasoning in this solution is based on the plots of 18O in otoliths, “ear stones,” from contemporary fish and from 6000-6500 year-old fossil fish. [The data are from “Otolith 18O record of mid-Holocene sea surface temperatures in Peru,” C. Fred T. Andrus, Douglas E. Crowe, Daniel H. Sandweiss, Elizabeth J. Reitz, and Christopher S. Romanek, Science 2002, 295, 1508-1511. The plots were scanned from the article, rearranged to fit across the page, and somewhat modified by removing precision bars and model values and redoing the captions and legends for the sake of the problem.]

(a) Since the sea surface temperature (SST) is higher (warmer seas) during an El Niño year and the 18O is lower in these years (the arrows on the contemporary data plot), the correlation for the data is lower 18O when the SST is higher. (b) “Nearly tropical conditions” means a higher temperature all year round and we see that, at the northern site, the 18O values are all quite low, indicating a relatively high SST all year round. Thus, the northern site is site Y, the one with the nearly tropical conditions about 6000 years ago. The more southern site shows much greater seasonal variation in 18O, and hence in SST, but most of the summer values are at least as low as for a modern El Niño year with the winter values more like the modern non-El Niño years. Thus, the summer SSTs at the more southern site, site X, 6000 years ago were warmer than the modern average, but the seasonal swings were much larger. (c) Since Peru is in the southern hemisphere, the more northern site that was studied is nearer the equator, about 4 south latitude, and it makes sense that this site should have more tropical conditions. The southern site is at about 9 south latitude.

Problem 3.52. (a) The nuclear equations for K-40 radioactive decay are (reference to Check This 3.39 is a reminder about how to write electron-capture reactions): 40K19+ + e– 40Ar18+ (electron capture) 40K19+ 40Ca20+ + e– (beta emission)



(b) After one half life, 50 g of K-40 are left in a sample that initially contained 100 �g of K-40 and 5.35 g (= 0.107 50 g) of Ar-40 have been formed by the 50 g of K-40 that

decayed. The Ar-40/K-40 mass ratio is 0.107 (= 5.35 g50 g

).

(c) After a second half life has elapsed, 25 g of K-40 are left and 2.675 g (= 0.107 25 g) of Ar-40 have been formed by the 25 g of K-40 that decayed. The total amount of Ar-40 that has accumulated is 8.025 g (5.35 g + 2.675 g), so the Ar-40/K-40 mass ratio is 0.321

(= 8.025 g

25 g

).

(d) One half life for K-40 is 1.25 109 years and we see from the plot given in the problem that the mass ratio is just over 0.10 at this age, which is what we calculated in part (b). Two half lives are 2.50 109 years and we see from the plot that the mass ratio is somewhat over 0.30 at this age, which is what we calculated in part (c). The results calculated for one and two half lives fall on the curve, so you can have confidence that the curve does represent the Ar-40/K-40 mass ratio as a function of age of the sample. (e) A sample with an Ar-40/K-40 mass ratio of 1.05 is about 4.3 109 years old. We know that the moon is heavily marked with craters formed by impacts from solar system debris. These impacts liquefied the rock at the impact area and the “splashing” of the liquid formed the craters. Liquefaction releases any argon (and other gases in the rock), so the age of the sample probably represents the age since some impact occurred. Thus, the moon is probably more than 4.3 billion years old.

Problem 3.53. In preparation for the various approaches to this problem, the data from the problem and some manipulations of the data are given in this table:

time, min activity, counts per 30 s fraction remaining

ln(fraction)

0 68,372 1.0000 0.0000 30 54,852 0.8023 -0.2203 60 45,457 0.6648 -0.4082 90 36,901 0.5397 -0.6167

120 29,964 0.4382 -0.8250 150 24,570 0.3594 -1.0234 185 18,936 0.2770 -1.2839 229 15,020 0.2197 -1.5156 240 12,790 0.1871 -1.6763

(a) When the data are plotted, as in the Web Companion, Chapter 3, Section 3.5, pages 5-9, the plot here is obtained. At time zero the activity is about 68,000 counts. When the activity has dropped to 34,000 counts, the time is just over 100 minutes, which is consistent with the 102 minutes in the Web Companion. If we look at the time required to drop from 50,000 to 25,000



counts, we go from about 45 minutes to just before 150 minutes. Again the half life is a little over 100 minutes. You may have tried other possibilities.

0

10000

20000

30000

40000

50000

60000

70000

0 50 100 150 200 250

acti

vity

, cou

nts

per

30 s

time, min

activity = 68,011 * exp( -0.006813*t )

The analyses we did in the text chapter were based on the fraction of the radioactive isotope remaining and we can plot this, if we take the ratio of each measured activity to the initial activity. These are the data in the third column of our table and plotted on this graph.

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 50 100 150 200 250

frac

tion

rem

aini

ng

time, min

fraction = 0.995 * exp( -0.006813*t)

How can we interpret the equation of the curve (given with the plot) through the points. Equation (3.17) gave the fraction, fn, as a function of number of half lives, n: fn = (0.5)n Let’s do a little mathematical manipulation on the right-hand term. First we take its logarithm and get: nln(0.5) = –0.695n. Then we take the antilogarithm of this result to undo the logarithm and get us back to where we started. The result is e–0.695n, so we end up with a new expression for fn: fn = e–0.695n



If we express the number of half lives as the ratio of the time that has passed, t, to the half life, , we have: n = t/ . Substitute this into the expression for fn to get: fn = e(–0.695/ )t = exp[(–0.693/ )t] (another way the exponential is often written) This equation says that the fraction of isotope remaining is an exponential function of the time. The coefficient on the exponential should be unity and the “decay constant” (units of time–1 to cancel the units of t) in the exponential should be –0.693/ . The exponential equation for the points on the graph has a coefficient of 0.995, which is almost unity, and the decay constant is –0.006813 min–1 (since t is in minutes) which gives = 102 min [= –0.693/(–0.006813 min–1)]. Note that the decay curve in the first graph (activity vs. time) has exactly the same exponential dependence on time, so the “raw” data can also be used to get the half life by this method. (b) The suggested equation (3.28) is one that gives the logarithm of the fraction of isotope remaining as a function of time: ln(fn) = (–0.693/ )t This is exactly what you get by taking the logarithm of both sides in the final equation in part (a)). The advantage of this approach is that the plot should be a straight line with an intercept of zero and a slope equal to the decay constant, –0.693/ . The fourth column in the table gives the data for this plot.

-1.8

-1.6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0 50 100 150 200 250

ln(f

ract

ion

of a

ctiv

ity)

time, min

ln(fraction) = -0.006813*t – 0.00528

The equation of the straight line does have an intercept that is nearly zero and the decay constant, not surprisingly, is identical to the one we got from the exponential plot. All our approaches give the same value for the half life, 102 min, just as you find in the Web Companion.

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