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Chapter 3 Origin of Atoms
SOLUTIONS to Chapter 3 Problems GAMMA Version
Section 3.1. ; Spectroscopy and the Composition of Stars and the Cosmos
3.1. Figure 3.3 shows the parts of a simple spectrograph and the result when the light from a hydrogen discharge (hydrogen atom) lamp enters and is diffracted. Is the diagram and diffraction consistent with the information from Figure 3.1? Explain why or why not.Answer to 3.1: The diagram 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.
3.2. Draw a diagram of a simple spectrograph, modeled after Figure 3.3, that uses a transmission diffraction grating, such as shown in Figure 3.2, instead of a prism to diffract the incoming light beam. Be sure the diffraction results you show on the photographic film are consistent with the information from Figure 2.2.Answer to 3.2: A simple diagram, looking at the spectrometer from above, 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.
3.3. A spectrometer was used to analyze a source of light and the spectrum showed a series of emission lines. What can you concludeabout the light source? Explain.Answer to 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.
3.4. Flame tests can often be used to identify the metal ions in acompound. Robert W. Bunsen (German chemist andspectroscopist, 18111899) invented the bunsen burner in order tocreate a flame hot enough to cause light emission from metal ions in compounds that were placed in the flame. Two examplesareshown here: the greenish flame from barium compounds and thescarlet flame from strontium compounds. How do the emissionsfrom atoms discussed in Section 3.1 explain why flame testswork? Explain clearly.Answer to 3.4: The metal ions, which are just atoms that have lostone or more electrons, 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
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barium strontium
Chapter 3 Origin of Atoms
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.
3.5. More than 99% of all the atomic nuclei in the universe are hydrogen and helium. Are the
data in Figure 3.7 consistentwith this statement? Clearly explain why or why not.
Answer to 3.5: Any element whose abundance is more than four orders of magnitude
(0.0001) lower than He will add negligibly to the atom count, so we neglect any element with
a log(abundance) below 5. Convert the log(abundance) to abundance [= 10log(abundance)] for the
values in Figure 3.7. The results are given 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
The sum of the H and He is 3250 107 and the sum of all the others is 4.83 107 . Thus, as a percentage of the total , H plus He are 99.8% [= (3250/3255) 100%]. The vast majority of atoms in the universe are still those produced shortly after the Big Bang.
3.6. As the previous problem states, almost all the nuclei in the universe are hydrogen and helium. Assume that almost all the mass of the universe is also hydrogen and helium. About what percentage of the mass of the universe is helium? Clearly explain the reasoning you use to get your answer.Answer to 3.6: From the solution to the preceding problem, 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.
3.7. How does the mass of iron in the universe compare to the mass of carbon? Explain your reasoning.Answer to 3.7: The reasoning here is the same as in the preceding problem. 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.
3.8. How does the mass of iron in the universe compare to the mass of all the other fourth period transition metals, scandium through zinc, combined?Answer to 3.8: Same reasoning as in the previous problem, 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.
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Chapter 3 Origin of Atoms
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
Section 3.2. ; The Nuclear Atom
3.9. What do these ions, S2–, Cl–, K+, and Ca2+, have in common?
3.10. Answer to 3.9: All have the same number of electrons.
3.11. How does the mass of a proton compare to the mass of an electron?
Answer 3.1. Answer to 3.10: The mass of the proton is 1.837 X 103 greater than the mass of the electron.
massproton
masselectron
3.12. What is an isotope?Answer to 3.11: Forms of the same element with identical atomic number, but different massnumbers.
3.13. Why are electrons not included when calculating the mass number of an isotope?Answer to 3.12: As seen in problem 3.10, the mass of electrons is so small that it can be ignored.
3.14. If an isotope of an element has 30 protons, 35 neutrons, 28 electrons:(a) What is the element?(b) Is this a anion or a cation? Explain.Answer to 3.13: (a) Zn (b) Cation has two more protons than electrons.
3.15. What would be the mass, in grams, of 25 protons?Answer to 3.14: 4.18 X 1023 g
3.16. Write the atomic symbols for the following isotopes:(a) Z = 19; A = 40 (d) Z = 13; A = 28(b) Z = 79; A = 197 (e) Z = 53; A = 118(c) Z = 54; A = 118 (f) Z = 83; A = 189
Answer to 3.15: (a) 1940 K (b) 79
197 Au (c) 54118 Xe (d) 15
28 Al (e) 53118 I (f) 83
189 Bi
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3.17. How many protons, neutrons, and electrons do the following ions contain?(a) 58Ni+ (d) 37Cl–
(b) 32S2– (e) 55Mn7+
(c) 65Zn2+ (f) 56Fe2+
Answer 3.2. Answer to 3.16: Let p = protons; n = neutrons; e = electrons(a) 28 p; 30 n; 27 e
(b) 16 p; 16 n; 18 e
(c) 30 p; 35 n; 28 e
(d) 17 p; 20 n; 18 e
(e) 25 p; 30 n; 18 e
(f) 26 p; 30 n; 24 e
3.18. Complete the following table. Some of the substances are ions and some are atoms.
# of protons
# of neutrons
# of electrons
Atomicnumber
Massnumber
NuclearSymbol
45 55 45 45 100 45100 Rh
26 34 23 26 60 2660Fe3+
88 137 88 88 225 88225 Ra
16 22 18 16 38 1638S2
100 148 100 100 248 100248Fm
Answer to 3.17: See above table
Section 3.3. ; Evolution of the Universe: Stars
3.19. The following questions deal with the kelvin temperature scale.(a) Is 3 K, the present background temperature of the universe, hot or cold? Explain.(b) The Earth is thought to have formed at a temperature just below 1000 K. Would the Earth, at that temperature, have had liquid water on its surface? Explain.(c) If the average daytime high temperature for a city is 85 F, what is the temperature in kelvin? Hint: A temperature of 32 F is 273 K. A degree on the kelvin scale is 1.8 times the size of a degree on the fahrenheit (F) scale.
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Answer to 3.18: (a) 3 K is very cold (270 oC).(b) 1000 K = 727 oC which is far above the normal boiling point of water.(c) 85 oF is approximately 29 oC which would be 302 K.
3.20. According to the Big Bang theory stars, and eventually planets, formed when matter condensed upon cooling as the universe expanded. Give some examples of analogous phenomena showing condensation.Answer to 3.19: Examples of possible student answers might be:Cloud formation, dew deposition, frost, fat solidifying after cooking, any freezing process.
3.21. Put these reactions in order from the one that takes place at the lowest temperature to theone that requires the highest temperature. Explain the reasoning for each of your choices.(a) 22Ne10+ + 22Ne10+ 43K19+ + 1H+
(b) 13C3+ + e– 13C2+
(c) 12C6+ + 18O8+ 26Mg12+ + 4He2+
(d) 12C6+ + 3He2+ 13N7+ + 2H+
Answer to 3.20: 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 fastenough. 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 is: (b) << (d) < (c)< (a).
3.22. What kelvin temperature is necessary for nuclei to have sufficient kinetic energy to sustain nuclear fusion?Answer to 3.21: A temperature of approximately 107 K is what evidence suggest to be necessary to sustain nuclear fusion.
3.23. Where are the elements formed? Give examples of the processes by which elements can be formed.Answer to 3.22: All elements except H and He are formed in the center of stars by fusion reactions. He can also be formed in stars but a large part was formed just after the Big Bang before stars were formed.
Section 3.4. ; Nuclear Reactions
3.24. Complete the following nuclear reactions (nuclear charges have been omitted):(a) 14N + ____ 17O + p (d) 20Ne + ____ 24Mg + (b) 13C + neutron ____ + (e) 20Ne + 4He ____ + 16O(c) 1H + 1H 2H + ____ (f) 27Al + 2H ____ + 28AlAnswer to 3.23:
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Answer 3.3. (a) 714N 7+
Answer 3.4. (b) 613C6+
Answer 3.5. (c) 11H+
Answer 3.6. (d) 1020 Ne10 +
Answer 3.7. (e) 1020 Ne10 +
(f) 1327 Al13+
3.25. Complete the following nuclear reactions (nuclear charges have been omitted):(a) 97Tc 97Ru + ____ (d) 1H + 14N ____ + 4He(b) ____ + 4He 243Bk + n (e) n + 235U ____ + 94Sr + 2n(c) 249Cf + ____ 263Sg + 4n (f) 228Ra ____ + 228AcAnswer to 3.24:
Answer 3.8. (a) 4397 Tc43+
Answer 3.9. (b) 95240 Am95+
Answer 3.10. (c) 98249 Cf 98+ + 8
18O8+
Answer 3.11. (d) 11H+
Answer 3.12. (e) 01 n + 92
235U92 +
(f) 88228 Ra88+
3.26. Write the balanced nuclear equation for the beta decay of 24Na. Include both mass and atomic numbers.
Answer to 3.25: 1124 Na11+
3.27. The p–n reaction is a common nuclear reaction. In a p–n reaction, a proton reacts with a nucleus to produce a new nucleus and a neutron as products.(a) The americium–241 used in smoke detectors, Investigate This 3.30, and Consider This 3.31, is extracted from spent nuclear reactor fuel rods. It is produced in the rods by a p–n reaction of plutonium. What isotope of plutonium is required? Write the balanced nuclear reaction that produces americium–241.(b) A carbon isotope is produced by a p–n reaction of nitrogen–14 in the atmosphere. Cosmic rays (radiation and particles from the Sun) are the source of the protons. What isotope of carbon is produced? Write the balanced nuclear reaction for its production.Answer to 3.26:(a) Pu241 is required and the easy way to get this answer is to write the balanced equation with everything known except the mass number of the Pu. Since p and n have the same mass number, the reactant and product nuclei must have the same mass number.(b) This is the same as part (a), except it is the product of unknown mass number that is desired. These reactions were chosen because they make connections back to the chapter, rather than being “bluesky.”
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3.28. When 14N captures a neutron, it decays to 3H and another product. Write the balanced nuclear equation for this reaction.
Answer to 3.27: 714N 7+ + 0
1n
3.29. Write equations to describe how the fusion of two 12C6+ can lead to the formation of:(a) 23Na11+ (c) 20Ne10+
(b) 23Mg12+ (d) 16O8+
Answer to 3.28:
(a) 612C6+ + 6
12C6+
(b) 612C6+ + 6
12C6+
(c) 612C6+ + 6
12C6+
(d)
612C6+ + 6
12C6+
3.30. Write equations to describe how the fusion of two 16O8+ can lead to the formation of:(a) 32S16+ (c) 28Si14+
(b) 31P15+ (d) 24Mg12+
Answer to 3.29:
Answer 3.13. (a) 816O8+ + 8
16O8+
(b) 816O8+ + 8
16O8+
(c) 816O8+ + 8
16O8+
(d) 816O8+ + 8
16O8+
3.31. Write equations to show how the three isotopes of Mg (24Mg12+, 26Mg12+, and 27Mg12+) areproduced from the fusion of 12C6+ and 16O8+.Answer to 3.30:
Answer 3.14. (a) 612C6+ + 8
16O8+
(b) 612C6+ + 8
16O8+
(c) 612C6+ + 8
16O8+
3.32. Xenon143 decays by a series of six successive beta particle emissions to a stable isotope. Write the series of decays and identify the stable isotope.Answer to 3.31:
143Xe54+ 143Cs55+ + 0e–
143Cs55+ 143Ba56+ + 0e–
143Ba56+ 143La57+ + 0e–
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Chapter 3 Origin of Atoms
143La57+ 143Hf58+ + 0e–
143Hf58+ 143Pr59+ + 0e–
143Pr59+ 143Nd60+ + 0e–
3.33. In 1996, the skeleton of an ancient hunter was found in the mud under the water near theshore of the Columbia river in Kennewick, Washington. A tiny sample from the skeleton (collagen in the bone) was analyzed for carbon14 and found to contain 0.362 as much of thisisotope as is present in living organisms. When did the “Kennewick Man” die?Answer to 3.32: This is a straightforward C14 half life problem.
number half lives = ln(0.361)/(–0.693) = 1.47 half livesage of sample = (1.47 half lives)(5730 yr∙halflife–1) = 8420 yrs
Kennewick Man died about 6420 BC (or 8420 years B.P. – before present)
3.34. At 8:15 a.m., a PET scan patient was injected with a compound containing fluorine 18. Assuming that none of the compound is excreted, what fraction of the F18 remains in the patient’s body at noon? Explain how you solve the problem. Hint: See Table 3.3.Answer to 3.33: We want to know what fraction of the F18 isotope remains undecayed after225 minutes (3 hours and 45 minutes from 8:15 a.m. to noon). Table 3.3 shows us that the half life of F18 is 110 minutes, so the elapsed time is equivalent to 2.05 half lives = n [= (225 min)/(110 min∙half life–1)]. We know that ln(fn) = –0.693n = (–0.693)∙(2.05) = –1.42fn = fraction F18 remaining = 0.242
3.35. Tiny quantities of iodine are essential for the proper functioning of our thyroid gland. A common treatment for patients with enlarged thyroid glands (hyperthyroidism) is ingestion ofa compound, such as NaI, containing iodine131. The iodine concentrates in the thyroid and beta particles produced by its decay kill the thyroid cells where it has accumulated.(a) Why are only thyroid gland cells killed by the beta emission? Hint: See Consider This 3.31.(b) Assuming that none of the I131 is excreted, how long will it take for the radioactivity todecay to 10% of its initial level? Hint: See Table 3.3.Answer to 3.34:(a) Beta emission, electrons, does not penetrate materials very far. Any electrons produced in Investigate This 3.30 were essentially completely blocked by a sheet of aluminum foil, as you learned in Consider This 3.31. Thus, the electrons emitted by the decay of I131 are absorbed by the cells very close to the source of the emission and this is within the thyroid gland, so it is thyroid gland cells that are killed.(b) We know that the fraction of radioactive isotope remaining is a function of the number of half lives that have elapsed. We can write the number of half lives for decay to 0.10 of the initial amount as: n = ln(fn)/(–0.693) = ln(0.10)/(–0.693) = 3.32 half lives. Table 3.3 shows that the half life for I131 is 8 days. Thus, the I131 will decay to 10% of its initial level in about 27 days.
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3.36. Rubidium87 decays by beta emission with a half life of 4.9 1010 years.(a) Write the balanced nuclear reaction equation for the decay of Rb87.(b) A rock sample from Greenland was found to have a Sr87/Rb87 mass ratio of 0.056. How old is the rock sample? Explain how you find the age and clearly state the assumptions you make.Answer to 3.35:(a) 87Rb37+ 87Sr38+ + e–
(b) Assume that the sample contained only Rb87 when it was formed, so that all the Sr87 is a result of the subsequent Rb87 decay. Assume that all the Rb87 nuclei initially present are still present as either Rb87 or Sr87 nuclei. Let the initial mass of Rb87 be m. If a fraction x of the Rb87 nuclei remains unreacted, then the mass of Rb87 remaining is x∙m and the mass of Sr87 formed is (1 – x)∙m. The Sr87/Rb87 mass ratio is [(1 – x)∙m]/[x∙m] = (1 – x)/x. Thus, for our sample of Greenland rock:
(1 – x)/x = 0.056 x = 1/1.056 = 0.95number of half lives = ln(0.95)/(–0.693) = 0.074 half lives
age of sample = (0.074 half lives)(4.9 1010 yr∙halflife–1) = 3.6 109 yrs
Section 3.5. ; Nuclear Reaction Energies
3.37. What is binding energy?Answer to 3.36: The energy required to break apart a nuclei into individual protons and neutrons.
3.38. What is the difference between fusion and fission? How might you predict whether a nucleus would undergo fission or fusion?Answer to 3.37: Fusion is when two or more nuclei “stick” together to form something larger, fission is when a heavy element breaks up into smaller nuclei. Light elements tend to undergo fusion, heavy elements tend to undergo fission. Each will become more stable and acquire a higher nuclear binding energy.
3.39. Calculate the binding energy per nucleon for:(a) 20Ne10+ (nuclear mass = 33.18913 10–24 g)(b) 28Si14+ (nuclear mass = 46.45681 10–24 g)(c) Which nucleus is more stable? Explain your answer.Answer to 3.38:
Answer 3.15. Mass loss per nucleus = (mass of protons + mass of neutrons) – mass of nucleus.
For:
1020 Ne10 + = [10
1.5552
To calculate the binding energy per nucleon, use the formula, E=mc2.
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E = 1.552
For 1428Si14 +
, the binding energy is calculated in a similar manner. The answer is
7.4276
3.40. For the elemental nuclei, 6Li3+ (nuclear mass = 9.98561 x 1024 g) and 56Fe26+ (nuclear mass = 92.8585 x 1024 g), calculate the binding energy in kilojoules(a) per nucleus(b) per mole(c) per nucleonAnswer to 3.39:Mass loss per nucleus = (mass p = mass n) – mass nucleus For Li6 = (3(1.672623 x 1024) + 3(1.674929 x 1024) ) – (9.98850 x 1024)
= 0.0541 x 1024g
(a) Binding energy per nucleus = E=mc2= 0.0541 x 1024g x 1000g/kg x(3.00x108 m/s)2
= 4.88 x 1012J (kg.m2s2)
(b) Binding energy per mole = 4.88 x 1012J/nucleus x 6.022 x 1023 nuclei/mol
2.94 x 1012 J/mol
(c) Binding energy per nucleon = (2.94 x 1012 J/mol )/64.9 x1011J/mol.nuc
The same calculation for Fe56 using 26 protons and 30 neutrons
Mass loss per nucleus = 0.8526 1024g
(a) Binding energy per nucleus = 7.67 x1011J(b) Binding energy per mole = 4.62 x 1013 J/mol(c) Binding energy per nucleon = 8.25 x1011J/mol.nuc
The binding energy for Fe56 is larger than that of Li just as you would expect since Fe is themost stable element.
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3.41. WEB Chap 3, Sect 3.5.14.(a) Draw a picture of two nuclei that would undergo nuclear fusion, for example C12 and He4. Be sure to indicate the components of the nucleus as protons, neutrons, or electrons. Draw the resulting nucleus, after the nuclei have undergone fusion.(b) Why would this type of fusion occur? Explain clearly.(c) Why would these nuclei not undergo fission? Explain clearly.Answer to 3.40:(a) A picture of two nuclei, C12 and He4, undergoing fusion to give O16 is shown here with protons denoted 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, O16, has a greater binding energy per nucleon (–77 107 kJ∙mol–1) than either of the starting nuclei, C12 (–74 107 kJ∙mol–1) and He4 (–68 107 kJ∙mol–1). The overall energy change for the reaction is:
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 C12 were to split into two Li6 nuclei, the energy change would be approximately:
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 Figure3.18 are for the most abundant isotope of each element and Li7 is the most abundant isotope of lithium. Different isotopes of the same element do not usually differ a greatdeal in binding energy per nucleon, so a small difference will not affect the conclusion that this is a very unfavorable reaction energetically.
3.42. (a) The mass of a Xe142 nucleus is 235.63075 10–24 g. What is its binding energy in kJ∙mol–1?(b) Use your result from part (a) and the data in Figure 3.20 to calculate the energy released in this fission reaction:
235U92+ + 1n0 90Sr38+ + 142Xe54+ + 41n0
(c) How does your result in part (b) compare to the energy for the fission reaction in Figure 3.20? What might you conclude about the energies released in the fission reactions of U235?Explain.Answer to 3.41:(a) The Xe142 binding energy is the energy change for this imagined reaction:
54p + 88n 142Xem = 235.63075 10–27 kg – [54∙(1.67262 10–27 kg) + 88∙(1.67493 10–27 kg)]
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= –2.0846 10–27 kgE (mole) = (–2.0846 10–27 kg)∙(3.00 108 cm∙s)2∙(6.022 1023 nuclei∙mol–1)
= 1.130 1011 kJ∙mol–1 = 11300 107 kJ∙mol–1
(b) The nuclei in the fission reaction in this problem are the same as in Figure 3.20, except for Xe142 here in place of Xe143 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) + (11300 107 kJ∙mol–1)] – (17250 107 kJ∙mol–1)= 1610 107 kJ∙mol–1
(c) Within the roundoff 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 U235 nucleus always produces about the same amount of energy, regardless of the way the nucleussplits. The next problem provides some further data.
3.43. The energy change (in kJ∙mol–1) for one of the collisioninduced nuclear fission reactions in Figure 3.21 is shown in Figure 3.20. The binding energies per nucleon for Kr 92,Rb 89, Cs 144, and Ba 141 are, respectively, –82.3, –83.7, –79.4, and –80.5 107 kJ∙mol–1.(a) What are the energy changes for the other nuclear fission reactions in Figure 3.21?(b) What might you conclude about the energies released in the fission reactions of U235? Explain.Answer to 3.42:(a) We solve this problem exactly as is illustrated in Figure 3.20 for fission to yield Sr90 and Xe143. (For ease in writing, we will omit the 107 factor as well as the units and include them only in the final result.)
For Kr92 + Ba141: E = [92∙(–82.3) + 141∙(––80.5)] – 235∙(–73.4)= –1670 107 kJ∙mol–1
For Rb89 + Cs144: E = [89∙(–83.7) + 144∙(––79.4)] – 235∙(–73.4)= –1630 107 kJ∙mol–1
(b) Again, just as in the previous problem, we find that the energy released in the fission of one mole of U235 is in the range 1600 to 1700 107 kJ∙mol–1, and the conclusion we reached,that the fission of U235 produces about the same amount of energy, no matter what the product nuclei, is strengthened.
3.44. WEB Chap 3, Sect 3.4.3.(a) Draw a nucleus of a carbon12 atom. Clearly label the components of the nucleus as protons, neutrons, or electrons.(b) Draw a picture of what this nucleus would look like if it separately underwent each of the main types of radioactive decay: alpha, beta, gamma and positron emission. Identify whatthe resulting elemental nucleus would be in each case.(c) Which type of radiation do you think C12 would be like likely to emit (if any). Give an example of an elemental nucleus that would be more likely than C12 to undergo each type of radioactive decay.Answer to 3.43:
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(a) and (b) The C12 nucleus is shown in each of the four reactions asked for in part (b). In these diagrams, the proton is represented by a dark circle and the neutron by a lighter gray circle.
alpha emission:
beta emission:
positron emission:
gamma emission:
(c) Carbon12 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 carbon12, 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 — N13, for example. Beta emission tends to occur from isotopes that have more than the normal number of protons — C14, for example. Alpha particle emission is generally a property of the heavier elements (beyond Fe) where the resulting nuclei (including the He4) 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 metastable nuclei decays.
3.45. The mass of a He3 nucleus is 5.00642 10–24 g. Is this reaction endothermicor exothermic?
23He 2+ + 2
3He 2+ 24 He 2+ + 21
1H +
How much energy is required or released per gram of He3 that reacts?Answer to 3.44: 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 H1 nucleus, 1.67262 10–24 g. Thus, the mass loss and energy change in this reaction 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)∙(3.00 108 cm∙s)2 = –2.0646 10–12 JThis energy change is for 10.0 10–24 g of He3 reacting. One gram is 1023 times this amount of reactant, so the energy change for one gram of reactant is –2.06 1011 J = –20.6 107 kJ.
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3.46. In March 2002, scientists from Oak Ridge National Laboratory reported that they had subjected a sample of acetone, CH3C(O)CH3, in which all the hydrogens had been replaced with deuterium, 2H, to intense cavitation (formation and collapse of bubbles), and had detected the formation of neutrons and tritium, 3H, in the sample. A tentative explanation for this observation was that fusion of deuterium nuclei, which can occur in two equally probable ways, was taking place at the high temperatures and pressures in the collapsing bubbles:
2H + 2H 3H + p (i)2H + 2H 3He + n (ii)
(a) What are the energy changesfor reactions (i) and (ii)? The masses of H2, H3, and He3nuclei are, respectively, 3.34357 10–24 g, 5.00736 10–24 g, and 5.00642 10–24 g.(b) Which set of products is more stable? Explain the reasoning for your answer.Answer to 3.45:(a) (i) m = (5.00736 10–27 kg + 1.67262 10–27 kg) – 2(3.34359 10–27 kg)
= –0.00720 10–27 kgE = (–0.00720 10–27 kg)∙(3.00 108 cm∙s)2 = –6.48 10–13 J
= 4.04 MeV {1.01 MeV H3 + 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 kgE = (–0.00583 10–27 kg)∙(3.00 108 cm∙s)2 = –5.25 10–13 J
= 3.28 MeV { 0.82 MeV He3 + 2.45 MeV n = 3.27 MeV}
(b) The products of reaction (i), tritium (H3) and a proton, are more stable, since more energy is released in this reaction. The reactants are the same in both reactions, so the releaseof 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 H3 and a He3 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).
Section 3.6. ; Cosmic Elemental Abundance and Nuclear Stability
3.47. (a) The nuclear masses of helium4 and beryllium8are 6.644655 10–24 g and 13.28949 10–24 g. What are the m and E (in kJ∙mol–1) for this reaction:
4He2+ + 4He2+ 8Be4+
(b) Beryllium8 decays by splitting into two alpha particle with a half life of about 7 10–17 s.Does your result in part (a) help you understand this very short lifetime? Explain why or whynot.(c) What is E (in kJ∙mol–1) for this reaction:
4He2+ + 8Be4+ 12C6+
(d) The sum of the two reactions in this problem is reaction equation (3.7) in the text. What is E for reaction (3.7)? Explain how you get your result.(e) The text that accompanies reaction equation (3.7) says the fusion involves “almostsimultaneous collision of three helium4 nuclei.” How is the information in this problem
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related to this statement? Do your results in this problem help to explain the relative abundance of beryllium in the universe? Explain your responses.Answer to 3.46:(a) 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 cm∙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 He4 to form a Be8 is endothermic, which is unfavorable. The reverse reaction is favored and helps to explain why the Be8 nucleus falls apart so rapidly.(c) The mass of the C12 nucleus, 19.92101 10–24 g, is given in Check This 3.42, so we have:m = (19.92101 10–27 kg) – [(6.644655 10–27 kg) + (13.28949 10–27 kg)]
= –0.01314 10–27 kgE = (–0.01314 10–27 kg)∙(3.00 108 cm∙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 to get the overall reaction of three He4 to yield C12, 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 roundoff uncertainty in Check This 3.50(e) The half life of the Be8 nucleus is so short that the collision of the third He4 has to occur within about 10–16 seconds of the Be8 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.
3.48. A nuclear reaction sequence that occurs in some stars is:12C6+ 13N7+ 13C6+ 14N7+ 15O8+ 15N7+ (12C6+ + 4He2+)
(a) Write the balanced nuclear reactions for each step of the sequence. Hint: Consider fusions involving protons, 1H, and electron capture decays, as in Check This 3.39. Compare your reactions to those you proposed in Check This 3.56(b).(b) Show that the net result of the series of reactions you wrote in part (a) is equivalent to nuclear reaction equation (3.6).(c) The series of reactions you wrote in part (a) is usually referred to as the CNO (carbonnitrogenoxygen) cycle and carbon12 is said to catalyze the formation of helium from hydrogen. Explain why carbon is assigned this role. Hint: A catalyst facilitates a reaction, butis not itself consumed in the net reaction.Answer to 3.47:(a) The sequence of reactions is:
12C6+ + 1H+ 13N7+ 13N7+ + 0e– 13C6+ 13C6+ + 1H+ 14N7+
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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 C12 is required to start the sequence, but another is formed in the last step, so there is no net loss or gain of C12. 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), C12 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 N14 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 He4 from H1.
Section 3.7. ; Formation of Planets: The Earth
3.49. (a) Since 70% of the mass of your body is water, we might say that you are mostly water, and the data in Figure 3.27 confirm that hydrogen and oxygen atoms are the most abundant in the human body. Is the ratio of these atoms consistent with the compositionof water? Explain why or why not.(b) Some might say that the crust of the Earth is mostly sand, silicon dioxide, SiO2. Do the data in Figure 3.27 confirm this suggestion? Explain why or why not.Answer to 3.48:(a) 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 theratio 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.(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
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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).
3.50. During the early part of the twentieth century, chemists spent a good deal of effort determining accurate values for relative atomic masses. When lead was studied, the metal from different ores had different atomic masses. From ores that also contained uranium, the relative atomic mass was close to 206 u, but from other ores, the value was close to 208 u.(a) What would you suggest as the reason for the difference in the leadin different ores? Is your explanation consistent with the information in this problem and this chapter? Explain your answers clearly.(b) The relative atomic mass of lead shown in the periodic table on the end papers of the book is 207.2 u. Clearly explain how you account for this value.Answer to 3.49:(a) We have seen, equation (3.27), that U238 decays by a series of alpha and beta emissions to give Pb206 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 lead206. Other stable isotopes of lead are also formed by nucleosynthsis 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, Pb208.(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, Pb204 (very minor), Pb206, Pb207, and Pb208. If the amounts of Pb206 and Pb208 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 Pb208 than Pb206 on Earth. The relative fractions of the four isotopes are, respectively, 0.014, 0.241, 0.221, and 0.524.
Section 3.9. ; EXTENSION — Isotopes: Age of the Universe and a Taste of Honey
3.51. Figure 3.29 shows a schematic diagram of an instrument called a mass spectrometer, which is named by analogy with the light spectrograph (or spectrometer) represented in Figure 3.3. A light spectrometer uses a prism or diffraction grating to disperse light into its component wavelengths. What does the mass spectrometer disperse? What is the part of the mass spectrometer that is responsible for the dispersion? Explain.Answer to 3.50: By analogy with the light spectrometer that disperses light, the mass spectrometer disperses mass. Note that the diagram shows that low and high mass ions are separated in space when they react 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.
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3.52. Since the temperature of the ocean surface has such a large impact on the Earth’s climate, this is one of the variables whose history scientists wish to determine as they develop climate models. Most of the approaches to discovering this history make use of differences in stable isotope ratios. Fossil catfish from the shores of Peruwere the focus of one such study. These catfish deposit calcium carbonate as otoliths, “ear stones,” which growthroughout the life of the fish in layers (like the rings of a tree), so annual variations can be determined. The study examined the 18O in otoliths from contemporary fish and from 60006500 yearold fossil fish from a site on the northern coast of Peru and one from further south.The data are shown in this figure.
(a) During an El Niño year, the sea surface temperature (SST) of the eastern Pacific Ocean off the coast of South America gets warmer than usual. How does 18O vary with temperature in these calcium carbonate otoliths? Clearly explain the reasoning for your answer.(b) The authors conclude that “the most plausible explanation for the archaeological 18Ootolith values is warmer summer SSTs at [site X] and nearly tropical conditions near [site Y] in the early midHolocene [the last 11000 years].” Which of the sites represented in the figure is site X and which site Y? Clearly explain the reasoning for your choices.(c) Do the conditions at the sites you identified in part (b) make sense in terms of their geographic locations? Explain.Answer 3.51: (a) Since the 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 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 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 nonEl Niño years. Thus, the summer SSTs at the more southern site 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.
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3.53. Potassium is relatively abundant, Figures 3.7 and 3.25, so geologists make extensive useof argonpotassium dating. Potassium40 is radioactive and decays by two pathways: beta emission and electron capture. 10.7% of the decay occurs by the electron capture pathway. Argon40 formed in the electron capture reaction is a gas and can escape from molten rock, but is trapped in the crystal lattice, if it is formed after the rock solidifies. Thus, Ar40/K40 datingindicates the age of the sample since it solidified.(a) Write balanced nuclear reactions for the two modes of K40 decay. Hint: See Check This 3.39.(b) Assume that exactly 100 micrograms (g) of K40 are present in a sample when it solidifies. After one K40 half life, 1.25 109 years, how many micrograms of Ar40 will havebeen formed? What is the Ar40/K40 mass ratio at this time? Hint: 10.7% of the K40 that reacted has become Ar40.(c) Calculate the Ar40/K40 mass ratio after two half lives have elapsed. Hint: The Ar40 formed during the second half life adds to that formed during the first.(d) The Ar40/K40 mass ratio as a function ofthe age of the rock (time since it solidified) isshown in this plot. Do your results from parts(b) and (c) fall on the curve? Explain why orwhy not?(e) A rock sample returned from the surface ofour moon had an Ar40/K40 mass ratio of1.05. What is the age of the sample? What doesthis result suggest about the age of the moon?Explain.Answer 3.52:(a) The nuclear equations are (reference toCheck 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 K40 are left and 5.35 g (= 0.107 50 g) of Ar40 have been formed by the 50 g of K40 that decayed. The Ar40/K40 mass ratio is 0.107 (= 5.35 g/50 g).(c) After a second half life has elapsed, 25g of K40 are left and 2.675 g (= 0.107 25 g) of Ar40 have been formed by the 25 g of K40 that decayed. The total amount of Ar40 that has accumulated is 8.025g (5.35 g + 2.675 g), so the Ar40/K40 mass ratio is 0.321 (= 8.025 g/25 g).(d) One half life for K40 is 125 109 years and we see from the plot 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 iswhat 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 Ar40/K40 mass ratio asa function of age of the sample.
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(e) A sample with an Ar40/K40 mass ratio of 1.05 is about 4.6 109 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 4.6 billion years old.
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3.54. WEB Chap 3, Sect 3.59.The plot of activity as a function of time for the radioactive decay of indium113m on pages 7, 8, and 9 of this section of the WebCompanion is low resolution and difficult to read. Another similarset of data for this decay are given here.(a) Plot these data, as in the Web Companion, and use your plot to determine the half life for the decay. How does your valuecompare with the one from the Companion? Hint: One approach,which requires some mathematical manipulation, is to use agraphing calculator or computer plotting program to plot the data,find the equation of the exponential curve through the data, anduse the equation of this curve to find the half life.(b) Apply equation (3.28) to these data (with uranium replaced byindium) and plot the data to give a linear plot. Use the equation ofthe line to find the half life and compare your value with the onefrom the Web Companion.Answer 3.53:In preparation for the various approaches to this problem, the data and some manipulations ofthe data are given in this table:
time, min activity, counts per 30 s fraction remaining ln(fraction)
0 68372 1.0000 0.0000
30 54852 0.8023 0.2203
60 45457 0.6648 0.4082
90 36901 0.5397 0.6167
120 29964 0.4382 0.8250
150 24570 0.3594 1.0234
185 18936 0.2770 1.2839
229 15020 0.2197 1.5156
240 12790 0.1871 1.6763
(a) When the data are plotted, as in the Web Companion, the plot here is obtained. At time zero the activity is 68000 counts. When the activity has dropped to 34000 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 50000 to 25000 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.
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WEB
time, min activity,counts
per 30 s
0 6837230 5485260 4545790 36901
120 29964150 24570185 18936229 15020240 12790
Chapter 3 Origin of Atoms
0
10000
20000
30000
40000
50000
60000
70000
0 50 100 150 200 250time, 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
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0 50 100 150 200 250time, 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 righthand term. First we take its logarithmand 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)
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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 equationfor 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 (which 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
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0 50 100 150 200 250time, 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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