NUCLEAR ENERGY; EFFECTS AND USES OF RADIATION

NUCLEAR ENERGY; EFFECTS AND USES OF RADIATION 31 Responses to Questions 1. (a) 137 56 n+ Ba? + γ Conserve nucleon number: 1+ 137 = A + 0. Thus, A = 138. Conserve charge: 0+ 56= Z + 0. Thus, Z = 56. This is 138 56Ba, or barium-138. 137 137 56 55 n+ Ba Cs +? Conserve nucleon number: 1+ 137 = 137 + A. Thus, A = 1. Conserve charge: 0+ 56= 55 + Z. Thus, Z = 1. This is 1 1H, or p (a proton). (c) 4 1 d+ H He +? Conserve nucleon number: + = 4 + A. Thus, A = 0. Conserve charge: 1+ 1= + Z. Thus, Z = 0. This is γ, or a gamma ray (a photon). (d) 197 α + 79 Au? + d Conserve nucleon number: 4+ 197 = A +. Thus, A = 199. Conserve charge: + 79= Z + 1. Thus, Z = 80. This is 199 80Hg, or mercury-199.. The reaction is Na+ H 3 He +?. The reactants have 1 protons and 4 nucleons, so they must 11 1 have 1 neutrons. Thus the products must have 1 protons and 1 neutrons as well. Since the alpha has protons and neutrons, the other product must have protons and neutrons. That nucleus is 0 Ne. Thus, the full equation is 4 0 Na + H He + Ne. 11 1 3. Neutrons are good projectiles for producing nuclear reactions because they are neutral and they are massive. If you want a particle to hit the nucleus with a lot of energy, then a more massive particle is the better choice. A light electron would not be as effective. Using a positively charged projectile like an alpha or a proton means that the projectile will have to overcome the large electrical repulsion from the positively charged nucleus. Neutrons can penetrate directly to the nucleus and cause nuclear reactions. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. 31-1

31- Chapter 31 4. For spontaneous radioactive decay reactions, Q > 0, answer. Otherwise, more energy must be added to make the reaction proceed. 5. The thermal energy from nuclear fission appears in the kinetic energy of the fission products (daughter nuclei and neutrons). In other words, the fission products are moving very fast (especially the neutrons, due to the conservation of momentum). 6. (a) Yes, since the multiplication factor is greater than 1 (f = 1.5), a chain reaction can be sustained. The difference would be that the chain reaction would proceed more slowly, and to make sure the chain reaction continued, you would need to be very careful about leakage of neutrons to the surroundings. 7. Uranium can t be enriched by chemical means because chemical reactions occur similarly with all of the isotopes of a given element. The number of neutrons does not influence the chemistry, which is primarily due to the valence electrons. Thus, trying to enrich uranium by chemical means, which means trying to increase the percentage of 35 38 9U in the sample compared with 9U, is impossible. 8. First of all, the neutron can get close to the nucleus at such slow speeds due to the fact that it is neutral and will not be electrically repelled by either the electron cloud or the protons in the nucleus. Then, once it hits a nucleus, it is held due to the strong nuclear force. It adds more energy than just kinetic energy to the nucleus due to E = mc. This extra amount of energy leaves the nucleus in an excited state. To decay back to a lower energy state, the nucleus may undergo fission. 9. For a nuclear chain reaction to occur in a block of porous uranium, the neutrons being emitted by the decays must be slowed down. If the neutrons are too fast, then they will pass through the block of uranium without interacting, effectively prohibiting a chain reaction. Water contains a much higher density of protons and neutrons than does air, and those protons and neutrons will slow down (moderate) the neutrons, enabling them to take part in nuclear reactions. Thus, if water is filling all of the porous cavities, then the water will slow the neutrons, allowing them to be captured by other uranium nuclei, and allow the chain reaction to continue, which might lead to an explosion.. Ordinary water does not moderate, or slow down, neutrons as well as heavy water; more neutrons will also be lost to absorption in ordinary water. However, if the uranium in a reactor is highly enriched, then there will be many fissionable nuclei available in the fuel rods. It will be likely that the few moderated neutrons will be absorbed by a fissionable nucleus, and it will be possible for a chain reaction to occur. 11. A useful fission reaction is one that is self-sustaining. The neutrons released from an initial fission process can go on to initiate further fission reactions, creating a self-sustaining reaction. If no neutrons were released, then the process would end after a single reaction and not be very useful. 1. Heavy nuclei decay because they are neutron-rich, especially after neutron capture. After fission, the smaller daughter nuclei will still be neutron-rich and relatively unstable and will emit neutrons in order to move to a more stable configuration. Lighter nuclei are generally more stable with approximately equal numbers of protons and neutrons; heavier nuclei need additional neutrons in order to decrease Coulomb repulsion between the protons. 13. The water in the primary system flows through the core of the reactor and therefore could contain radioactive materials, including deuterium, tritium, and radioactive oxygen isotopes. The use of a secondary system provides for isolation of these potentially hazardous materials from the external environment. Also, it is desirable to keep the radioactive materials as confined as possible to avoid Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-3 accidental leakage and contamination. In the electric generating portion of the system, waste heat must be given off to the surroundings, and keeping all of the radioactive portions of the energy generation process as far away from this release point as possible is a major safety concern. 14. Fission is the process in which a larger nucleus splits into two or more fragments, roughly equal in size, releasing energy. Fusion is the process in which smaller nuclei combine to form larger nuclei, also releasing energy. 15. Fossil fuel power plants are less expensive to construct, and the technology is well known. However, the mining of coal is dangerous and can be environmentally destructive, the transportation of oil can be damaging to the environment through spills, the production of power from both coal and oil contributes to air pollution and the release of greenhouse gases into the environment, and there is a limited supply of both coal and oil. Fission power plants produce no greenhouse gases and virtually no air pollution, and the technology is well known. However, they are expensive to build, they produce thermal pollution and radioactive waste, and when accidents occur they tend to be very destructive. Uranium is also dangerous to mine. Fusion power plants produce very little radioactive waste and virtually no air pollution or greenhouse gases. Unfortunately, the technology for large-scale sustainable power production is not yet known, and the pilot plants are very expensive to build. 16. Gamma particles penetrate better than beta particles because they are neutral and have no mass. Thus, gamma particles do not interact with matter as easily or as often as beta particles, allowing them to better penetrate matter. 17. The large amount of mass of a star creates an enormous gravitational attraction, which causes the gas to be compressed to a very high density. This high density creates very high pressure and high temperature. The high temperatures give the gas particles a large amount of kinetic energy, which allows them to overcome the Coulomb repulsion and then fuse when they collide. These conditions at the center of the Sun and other stars make the fusion process possible. 18. Stars, which include our Sun, maintain confinement of the fusion plasma with gravity. The huge amount of mass in a star creates an enormous gravitational attraction on the gas molecules, and this attractive force overcomes the outward repulsive forces from electrostatics and radiation pressure. 19. Alpha particles are relatively large and are generally emitted with relatively low kinetic energies. They are not able to penetrate the skin, so they are not very destructive or dangerous as long as they stay outside the body. If alpha emitters are ingested or inhaled, however, then the protective layer of skin is bypassed, and the alpha particles, which are charged, can do tremendous amounts of damage to the lungs and other delicate internal tissues due to ionizing effects. Thus, there are strong rules against eating and drinking around alpha emitters, and the machining of such materials, which produces fine dust particles that could be inhaled, can be done only in sealed conditions. 0. The absorbed dose measures the amount of energy deposited per unit mass of absorbing material and is measured in grays (Gy) or rads. The gray is the SI unit and is 1 J/kg. A rad is 0.01 Gy. The effective dose takes into account the type of radiation depositing the energy and is used to determine the biological damage done by the radiation. The effective dose is the absorbed dose multiplied by a relative biological effectiveness (RBE) factor. The effective dose is measured in rem, which are rad RBE, or sieverts (Sv), which are gray RBE. Sieverts is the SI unit, and 1 Sv = 0 rem. 1 Sv of any type of radiation does approximately the same amount of biological damage. 1. Radiation can kill or deactivate bacteria and viruses on medical supplies and even in food. Thus, the radiation will sterilize these items, making them safer for humans to use. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-4 Chapter 31. Allow a radioactive tracer to be introduced into the liquid that flows through the pipe. Choose a tracer that emits particles that cannot penetrate the walls of the pipe. Then check the pipe with a Geiger counter. When tracer radiation is found on the outside of the pipe (radiation levels will be higher at that point), the leak will have been located. Responses to MisConceptual Questions 1. (e) In nuclear reactions, energy is converted between mass, kinetic energy, and radiant energy, but the total energy is conserved. The momentum is conserved as particles are absorbed and released in the reactions. The net electric charge is conserved. That is, the total net charge before the reaction will always equal the net charge after the reaction. The new conserved quantity in this chapter is the conservation of nucleon number. In any reaction, the number of nucleons will remain unchanged.. Large nuclei have more neutrons per proton than smaller nuclei. In a fission reaction, the daughter nuclei are neutron-rich and emit β particles to convert excess neutrons into protons. 3. (c) The melting point, boiling point, and valence shells do not affect the critical mass for a chain reaction. The nuclear density is relatively constant for all substances. The critical mass is determined by the mass necessary for each reaction to produce, on average, one neutron that creates an additional reaction. If each fission can produce additional neutrons, then it is more likely that one of those neutrons will create an additional reaction and the critical mass can be smaller. 4. Fusion occurs in small nuclei because the binding energy per nucleon of larger nuclei is greater than the binding energies of the small nuclei. Fission occurs in very large nuclei because the binding energy per nucleon of the smaller nuclei is greater than the binding energy per nucleon of the very large nuclei. If the binding energy per nucleon always increased, then fusion could occur through all elements, but fission could not occur. 5. (a) Nuclear reactions occur when uranium-35 absorbs a neutron. Thermal neutrons are more likely to be absorbed by uranium-35, so the moderator is used to slow down the neutrons. 6. (e) Fission and fusion are sometimes thought of as different names for the same physical process, but this is incorrect. Fission occurs when energy is released as large nuclei are broken apart. Fusion occurs when energy is released as small nuclei are put together. 7. (c) The primary fuel source for fusion is hydrogen, which is very plentiful. The by-product of nuclear fusion is helium, which is not radioactive. The problem with fusion is that the hydrogen nuclei repel each other; therefore, very high temperatures are necessary for them to get close enough to fuse. 8. (a) When the two hydrogen nuclei bind together to form helium, energy is released. This energy comes from the mass of the hydrogen nuclei, so the resulting helium nucleus has less mass than the mass of the two hydrogen nuclei. 9. (a) Gamma and beta radiation induce about the same amount of biological damage. Alpha radiation induces much more damage in a concentrated area than either gamma or beta radiation due to its relatively large mass and the slow speed of alpha radiation. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-5. (a) A single fission reaction releases about 00 MeV of energy. A single fusion reaction produces less than 5 MeV. Therefore, the fission reaction releases more energy. 11. (a) The fuel for fusion is hydrogen, which is a component of water. 1. (a, b) Radiation is composed of high-energy particles that can disrupt cells. If a small amount of radiation hits the correct molecules of a cell (such as DNA), then it can alter the cell, causing cancer. Therefore, any amount of radiation can be harmful. Radiation is produced in natural nuclear decays and is therefore part of the normal environment. Gamma radiation can be very penetrating, but alpha and beta radiation do not penetrate very far into living tissue. 13. (e) Cell damage due to radiation occurs from extended exposure to high levels of radiation. The damage can be minimized by having the technician work farther away from the radiation, work for a shorter time, and be shielded from the radiation. A radiation badge can warn the worker when the allowed radiation exposure has been reached. Therefore, all of the answers can help reduce radiation damage. 14. (d) X-rays, gamma rays, and beta rays all produce about the same amount of radiation damage. Neutrons and alpha particles produce more cell damage per unit energy than X-rays, gamma rays, or beta rays due to their larger masses and slower speeds. 15. The critical mass is the mass that must be present so that on average, a neutron from each fission is absorbed by another nucleus to instigate another fission. When more neutrons are released during each fission, another reaction is more likely with less mass present. Therefore, the plutonium has a smaller critical mass. Solutions to Problems 1. When aluminum absorbs a neutron, the mass number increases by one and the atomic number is 8 unchanged. The product nucleus is 13 Al. Since the nucleus now has an extra neutron, it will decay by β, according to this reaction: 8 8 13 14 β ve Al Si + +. Thus the product is. If the Q-value is positive, then no threshold energy is needed. 8 14 Si. Q = m c m3 c m c = (. 014 u) 3. 01609 u 1. 008665 u 931. 5 c 1H = 3.70 MeV MeV/ c He n [ ] Thus, no threshold energy is required, since the Q-value is positive. The final mass is less than the initial mass, so energy is released in this process. 3. A slow neutron has negligible kinetic energy. If the Q-value is positive, then the reaction is possible. Q = m38 c + m c m39 c = +... c 9 U 9 = 4.806 MeV MeV/ c n U [ 38.050788 u 1 008665 u 39 05494 u] 931 5 Thus, the reaction is possible. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-6 Chapter 31 4. (a) The product has 16 protons and 16 neutrons. Thus the reactants must have 16 protons and 16 neutrons. Thus, the missing nucleus has 15 protons and 16 neutrons, so it is 15 31 The Q-value tells us whether the reaction requires or releases energy. In order to balance the electrons, we use the mass of 1 1H for the proton. p S S Q = m c + m c m c 31 3 15 16 = [ 1.00785 u + 30.97376 u 31.97071 u] 931.5 c = 8.864 MeV 5. We assume that all of the particles are essentially at rest, so we ignore conservation of momentum. To make just the fluorine nucleus, the Q-value plus the incoming kinetic energy should add to 0. In order to balance the electrons, we use the mass of 1 1H for the proton. KE Q KE mpc m c m c m F nc + = + + = 0 18 18 8O 9 18 18 9F 8 18 9 KE p O n m c = + m c + m c m c = m =.438 MeV + [ 1.00785 u + 17.999160 u 1.008665 u] 931.5 c 4 = 1.6767873 MeV 1u = (1.6767873 MeV/ ) = 18.000937 u 931.5 MeV/ c 4 c F 6. (a) If the Q-value is positive, then no threshold energy is needed. n Na d Q = m c + m c m c m c 4 3 1 Mg 11 = [ 1.008665 u + 3.98504 u.989769 u.014 u] 931.5 c = 9.468 MeV Thus more energy is required if this reaction is to occur. The 18.00 MeV of kinetic energy is more than sufficient, so the reaction can occur. 18.00 MeV 9.468 MeV = 8.53 MeV of energy is released 7. (a) If the Q-value is positive, then no threshold energy is needed. In order to balance the electrons, we use the mass of 1 1 H for the proton and the mass of 4 He for the alpha particle. p Li He α Q = m c + m c m c m c 7 4 3 = [ 1.00785 u + 7.016003 u (4.00603 u) ] 931.5 c = 17.346 MeV Since the Q-value is positive, the reaction can occur. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-7 The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy. KEtotal = KEreactants + Q = 3.1 MeV + 17.346 MeV = 0.4 MeV 8. (a) If the Q-value is positive, then no threshold energy is needed. In order to balance the electrons, we use the mass of 1 1 H for the proton and the mass of 4 He for the alpha particle. α O p Q = m c + m c m c m c 14 17 7N 8 = [ 4.00603 u + 14.003074 u 16.99913 u 1.00785 u] 931.5 c = 1.19 MeV Thus, more energy is required if this reaction is to occur. The 9.85 MeV of kinetic energy is more than sufficient, so the reaction can occur. The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy. KEtotal = KEreactants + Q = 9.85 MeV 1.19 MeV = 8.66 MeV 9. In order to balance the electrons, we use the mass of 4 He for the alpha particle. α Q = m c + m c m 16 0 8O Ne = [ 4.00603 u + 15.994915 u 19.99440 u] 931.5 c = 4.730 MeV. The Q-value tells us whether the reaction requires or releases energy. d N n Q = m c + m c m c m c 13 14 6C 7 = [.014 u + 13.003355 u 14.003074 u 1.008665 u] 931.5 c = 5.36 MeV The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy. KEtotal = KEreactants + Q = 41.4 MeV + 5.36 MeV = 46.7 MeV 11. The 14 7 N absorbs a neutron, and 14 6 C is a product. Thus the reaction is 14 14 n+ 7N 6C +?. The reactants have 7 protons and 15 nucleons, which means 8 neutrons. Thus the products also have 7 protons and 8 neutrons, so the unknown product must be a proton. The reaction is 14 14 7 6 n+ N C+ p. In order to balance the electrons, we use the mass of 1 1H for the proton. n C p Q = m c + m c m c m c 14 14 7N 6 = [ 1.008665 u + 14.003074 u 14.0034 u 1.00785 u] 931.5 c = 0.66 MeV Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-8 Chapter 31 1. (a) The deuteron is 1 H, so the reactants have 4 protons and 8 nucleons. Therefore, the reactants have 4 neutrons. Thus the products must have 4 protons and 4 neutrons. That means that X must have 3 protons and 4 neutrons, so X is 3 7 Li. This is called a stripping reaction because the lithium nucleus has stripped a neutron from the deuteron. (c) The Q-value tells us whether the reaction requires or releases energy. In order to balance the electrons, we use the mass of 1 1H for the proton d Li p Q = m c + m c m c m c 6 7 3Li 3 = [.014 u + 6. 01513 u 7.016003 u 1.00785 u] 931.5 c = 5.07 MeV Since the Q-value is positive, the reaction is exothermic. 13. (a) This is called a pickup reaction because the helium has picked up a neutron from the carbon nucleus. The alpha is 4 He. The reactants have 8 protons and 15 nucleons, so they have 7 neutrons. Thus, the products must also have 8 protons and 7 neutrons. The alpha has protons and neutrons, so X must have 6 protons and 5 neutrons. Thus, X is 11 6 C. (c) The Q-value tells us whether the reaction requires or releases energy. In order to balance the electrons, we use the mass of 1 1 H for the proton and the mass of 4 He for the alpha particle. He C C α Q = m c + m c m c m c 3 1 11 6 6 = [ 3.01609 u + 1.000000 u 11.011434 u 4.00603 u] 931.5 c = 1.856 MeV Since the Q-value is positive, the reaction is exothermic. 14. The Q-value tells us whether the reaction requires or releases energy. In order to balance the electron count, we use the mass of 1 1 H for the proton and the mass of 4 He for the alpha particle. Q= m c + m7 c m4 c m c = +.. c 3 = 17.35 MeV MeV/ c p Li He α [ 1.00785 u 7 016003 u (4 00603) u] 931.5 The reaction releases 17.35 MeV. 15. The Q-value tells us whether the reaction requires or releases energy. In order to balance the electron count, we use the mass of 4 He for the alpha particle. α C n Q = m c + m c m c m c 9 1 4Be 6 = [ 4.00603 u + 9.01183 u 1.000000 u 1.008665 u] 931.5 c = 5.70 MeV The reaction releases 5. 70 MeV. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-9 16. The Q-value gives the energy released in the reaction, assuming that the initial kinetic energy of the neutron is very small. n 3 U Ba Kr n Q = m c + m c m c m c m c 35 141 9 9 56 36 = [ 1.008665 u + 35.043930 u 140.914411 u 91.96156 u 3(1.008665 u) ] 931.5 c = 173.3 MeV 17. The Q-value gives the energy released in the reaction, assuming the initial kinetic energy of the neutron is very small. n 1 U Sr Xe n Q = m c + m c m c m c m c 35 88 136 9 38 54 = [ 1.008665 u + 35.043930 u 87.90561 u 135.90714 u 1(1.008665 u) ] 931.5 c = 16.5 MeV 18. The power released is the energy released per reaction times the number of reactions per second. energy # reactions P = reaction s # reactions P 40 W 18 = = = 7.5 reactions/s s energy 6 19 (00 ev/reaction)(1.60 J/eV) reaction 6 18 8 reactions/s 19. Compare the energy per fission with the mass energy. energy per fission 00 MeV 4 1 = = 9.1 (9 )% mass energy mc (35 u)(931.5 MeV/ c ) c 10 0. Convert the 960 watts over a year s time to a mass of uranium. 7 35 960 J 3.156 s 1 MeV 1fission 0.35 kg 9U -4 35 3.696 kg 13 3 9U 1 s 1 yr 1.60 J 00 MeV 6.0 atoms = 0.4 g U 1. (a) The total number of nucleons for the reactants is 36, so the total number of nucleons for the products must also be 36. The two daughter nuclei have a total of 31 nucleons, so 5 neutrons must be produced in the reaction: 35 133 98 9 51 41 35 133 98 9 51 41 n 5 U Sb Nb n Q= m c + m c m c m c m c U+ n Sb+ Nb+ 5n. = [ 35.043930 u + 1.008665 u 13.91550 u 97.938 u 5(1.008665 u) ] 931.5 c = 171.1 MeV 35 9 Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31- Chapter 31. We assume as stated in Problems 18, 19, and 0 that an average of 00 MeV is released per fission of a uranium nucleus. 7 1 ev 1 MeV 1 nucleus 0.35 kg 7 (3 J) = 3.7 kg 19 6 00 MeV 3 1.60 J 1 ev 6.0 nuclei 35 9 U 3. Since the reaction is 34% efficient, the fission needs to generate (950/0.34) MW of power. Convert the power rating to a mass of uranium using the factor-label method. We assume that 00 MeV is released per fission, as in other Problems. 6 7 950 J 1atom 1eV 0.35 kg U 3.156 s 6 19 3 35 = 75 kg 9U 0.34 s 00 ev 1.60 J 6.0 atoms 1 yr 10 kg U 35 9 4. We find the number of collisions from the relationship E 1 n E ( ) collisions. n ln 6 1 n E0 0 ( ) ln 1 ln 1 =, where n is the number of 0 E 0.040 ev ln 1.0 ev E = E n = = = 4.58 5 collisions n 5. If the uranium splits into two roughly equal fragments, then each will have an atomic mass number of half of 36, or 118. Each will have a nuclear charge of half of 9, or 46. Calculate the electrical potential energy using a relationship derived in Example 17 7. The distance between the nuclei will be twice the radius of a nucleus, and the radius is given in Eq. 30 1. PE 19 1 QQ 1 9 (46) (1.60 C) 1 MeV = = (8.99 N m /C ) = 60 MeV 4πε 15 1/3 13 0 r (1. m)(118) 1.60 J This is about 30% larger than the nuclear fission energy released. 6. The reaction rate is proportional to the number of neutrons causing the reactions. For each fission, the number of neutrons will increase by a factor of 1.0004, so in 00 milliseconds, the number of 00 neutrons will increase by a factor of (1.0004) = 1.5. n 7. KE 3 3 3 7 16 16 1eV = kt = (1.38 J/K)( K) = 4 J = 4 J 1.60 19 J 3000 ev 8. The Q-value gives the energy released in the reaction. H H He n Q = m c + m c m c m c 3 4 1 1 = [.014 u + 3.016049 u 4.00603 u 1.008665 u] 931.5 c = 17.59 MeV Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-11 9. Calculate the Q-value for the reaction H+ H 3 He+ n. He n Q = m c m c m c 3 1H 1 1 = [ (. 014 u) 3.01609 u 1.008665 u] 931.5 c = 3.7 MeV 30. For the reaction in Eq. 31 6a, if atomic masses are to be used, then one more electron needs to be added to the products side of the equation. Notice that charge is not balanced in the equation as written. The balanced reaction is 1 H 1 H + + H+ e + v + e. 1 1 1 1 1 1 + H H e e Q = m c m c m c m c = [ (1. 00785 u).014 u (0. 000549 u) ] 931.5 c = 0.419 MeV 0.4 MeV For the reaction in Eq. 31 6b, use atomic masses, since there would be two electrons on each side. H H He Q = m c + m c m c 1 3 1 1 = [ 1.00785 u +.014 u 3.01609 u] 931.5 c = 5.4940 MeV 5. 49 MeV For the reaction in Eq. 31 6c, use atomic masses, since there would be two electrons on each side. He He H Q = m c m c m c 3 4 1 1 = [ (3.01609 u) 4.00603 (1.00785 u) ] 931.5 c = 1.8594 MeV 1.86 MeV 31. (a) Eq. 31 8a: Eq. 31 8b: Eq. 31 8c: 4.03 MeV 1 u 1 kg 3 = 6.03 MeV/g (.014 u) 7 1.66 kg 00 g 3.7 MeV 1 u 1 kg 3 = 4.89 MeV/g (.014 u) 7 1.66 kg 00 g 17.59MeV 1u 1kg 4 =.11 MeV/g (.014 u + 3.016049 u) 7 1.66 kg 00 g Uranium fission (00 MeV per nucleus): 00 MeV 1 u 1 kg 3 = 5.13 MeV/g (35 u) 7 1.66 kg 00 g Eq. 31 8a: Eq. 31 8b: Eq. 31 8c: 5.13 6.03 513. 489. 513. 11. 3 3 3 3 3 4 = 0851. = 5. = 043. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-1 Chapter 31 3. Calculate the Q-value for the reaction 38 39 9 + 9 U n U. n U Q = m c + m c m c 38 39 9 U 9 = [ 38.050788 u + 1.008665 u 39. 05494 u] 931.5 c = 4.806 MeV 33. The reaction of Eq. 31 8b consumes deuterons and releases 3.3 MeV of energy. The amount of energy needed is the power times the elapsed time, and the energy can be related to the mass of deuterium by the reaction. 3 13 3 J 7 s 1 MeV d.014 kg 960 (1 yr) 3.156 s yr 1.60 J 3.3MeV 6.0 d 4 = 3.93 kg = 0.39 g 34. (a) The reactants have a total of 3 protons and 7 neutrons, so the products should have the same. After accounting for the helium, there are 3 neutrons and 1 proton in the other product, so it must be tritium, 3 6 1 4 3 1H. The reaction is 3Li + 0n He + 1H. The Q-value gives the energy released. Li n He H Q = m c + m c m c m c 6 1 4 3 3 0 1 = [ 6.01513 u + 1.008665 u 4.00603 u 3.016049 u] 931.5 c = 4.784 MeV 35. Assume that the two reactions take place at equal rates, so they are both equally likely. Then from the reaction of 4 deuterons, there would be a total of 7.30 MeV of energy released, or an average of 1.85 MeV per deuteron. A total power of 1150 MW = 3485 MW must be obtained from the fusion 0.33 reactions to provide the required 1150-MW output, because of the 33% efficiency. We convert the power to a number of deuterons based on the energy released per reacting deuteron and then convert that to an amount of water using the natural abundance of deuterium. 3485 MW 3485 6 J 3600 s 1 MeV 1 d 1 H atom s 1 h 13 1.60 J 1.85 MeV 0.000115 d's 1 HO molecule 0.018 kg HO Hatoms 3 6.0 molecules = 5586 kg/h 5600 kg/h 36. We assume that the reactants are at rest when they react, so the total momentum of the system is 0. As a result, the momenta of the two products are equal in magnitude. The available energy of 17.57 MeV is much smaller than the masses involved, so we use the nonrelativistic relationship between momentum and kinetic energy, KE p = p = m m KE. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-13 KE KEn KE He total He n KE He He nken 4 4 4 4 ( ) KE nke He He n KE He He n KEtotal KE He 4 4 4 4 4 KE He 4 + = = 17.57 MeV p = p m = m m = m m = m mn 1.008665 = = 17.57 MeV = 3.536 MeV 3.5 MeV m + m 4.00603 + 1.008665 4 He KEn KEtotal KE He KEtotal n = = 17.57 MeV 3.54 MeV = 14.03 MeV 14 MeV 4 If the plasma temperature were significantly higher, then the approximation of 0 kinetic energy being brought into the reaction would not be reasonable. Thus the results would depend on plasma temperature. A higher plasma temperature would result in higher values for the energies. 37. In Eq. 31 8a, 4.00 MeV of energy is released for every deuterium atoms. The mass of water can be converted to a number of deuterium atoms. 1 (1.00 kg H O) = 7.69 d nuclei 3 4 6.0 HO H 1.15 d 0018kgHO 1HO 1H. 1 6 19 4.00 ev 1.60 J 9 (7.69 d nuclei) =.46 J d atoms 1 ev As compared to gasoline: 9.46 J 50 more than gasoline 7 5 J 1 13 38. (a) We follow the method of Example 31 9. The reaction is 6C+ 1H 7N+ γ. We calculate the potential energy of the particles when they are separated by the sum of their radii. The radii are calculated from Eq. 30 1. = = 4 (1. m)(1 1 ) 1.60 J =.19 MeV 19 1 QQ C H 9 (6)(1)(160. C) 1MeV KE total (8.99 N m /C ) πε 15 1/3 1/3 13 0 rc + r H + For the d t reaction, Example 31 9 shows that KE total = 0.45 MeV. Find the ratio of the two energies. KE C.19 MeV = = 4.9 KE d t 0.45 MeV The carbon reaction requires about 5 times more energy than the d t reaction. The kinetic energy is proportional to the temperature by KE = 3 kt. Since the kinetic energy has to increase by a factor of 5, so does the temperature. Thus we estimate 9 T 1.5 K. 39. Because the RBE of alpha particles is up to 0 and the RBE of X-rays is 1, it takes 0 times as many rads of X-rays to cause the same biological damage as alpha particles. Thus the 350 rads of alpha particles is equivalent to 350 rads 0 7000 rads = of X-rays. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-14 Chapter 31 40. Use Eq. 31 b to relate Sv to Gy. From Table 31 1, the RBE of gamma rays is 1, so the number of Sv is equal to the number of Gy. Thus 4.0 Sv = 4.0 Gy. 41. The biological damage is measured by the effective dose, Eq. 31 a, using Table 31 1. (7 rads fast neutrons) = ( x rads slow neutrons) 5 7 rads x = = 144 rads slow neutrons 5 4. A gray is 1 joule per kilogram, according to Eq. 31 9. (.5 J/kg) 65 kg = 16.5 J 160 J 43. (a) Since the RBE is 1, the effective dose (in rem) is the same as the absorbed dose (in rad). Thus the absorbed dose is 1.0 rad or 0.0 Gy. A Gy is 1 J/ kg. 1J/kg 1eV 1p (0.01 Gy) (0.0 kg) = 1.0 p 1Gy 19 6 1.60 J 1. ev 44. The counting rate will be 85% of 35% of the activity. 6 3.7 decays/s 1 β (0.035 Ci) (0.35)(0.85) = 385.3 counts/s 390 counts/s 1 Ci 1 decay 1 45. The two definitions of roentgen are 1.6 ion pairs/g produced by the radiation and the newer definition of 0.878 J/kg deposited by the radiation. Start with the current definition and relate them by the value of 35 ev per ion pair. 19 1 R = (0.878 J/kg)(1 kg/00 g)(1 ev/1.60 J)(1 ion pair/35 ev) 1 = 1.567 ion pairs/g The two values of ion pairs per gram are within about % of each other. 46. We approximate the decay rate as constant and find the time to administer 3 Gy. If that calculated time is significantly shorter than the half-life of the isotope, then the approximation is reasonable. If 1.0 mci delivers about mgy/min, then 1.6 mci would deliver 16 mgy/min. dose 3 Gy 1 day dose = rate time time = = 1.39 days 1.4 days rate 3 = 16 Gy/min 1440 min This is only about % of a half-life, so our approximation is reasonable. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-15 47. Since the half-life is long (5730 yr), we can consider the activity as constant over a short period of time. Use the definition of the curie from Section 31 5. 6 3.70 decays/s 4 ΔN 0.693 (.50 Ci) = 9.5 decays/s = = N 1Ci 1 Δt T 1 ΔN T 4 5730 yr 7 16 N = = (9.5 decays/s) (3.156 s/yr) =.414 nuclei Δt 0.693 0.693 16 0.0140 kg.414 nuclei = 5. 3 6.0 nuclei 61 kg = 0.561 μg 48. Each decay releases one gamma ray of energy 1 kev. Half of that energy is deposited in the body. The activity tells at what rate the gamma rays are released into the body. We assume that the activity is constant. 6 γ /s 16 1 (1.55 Ci) 3.70 (86, 400 s/day)(0.50)(1 kev/ γ )(1.60 J/keV) Ci 65 kg 7 J/kg 7 Gy = 7.44 7.4 day day 49. Use the dose, the mass of the beef, and the energy per electron to find the number of electrons. 3 1J/kg 1MeV 1e 16 16 (4. 5 Gy) (5kg) = 8.78 e 9 e 1Gy 13 1.60 J 1.6MeV 50. (a) According to Appendix B, 131 53I decays by beta decay. 131 131 53 54 I Xe+ + v β The number of nuclei present is given by Eq. 30 4, with N = 0.05 N0. λt ln NN / 0 T1/ln NN / 0 (8.0 d) ln 0.05 N = N0e t = = = = 34.58 d 35 d λ 0. 693 0.693 (c) The activity is given by ΔN/ Δ t = λn. This can be used to find the number of nuclei, and then the mass can be found. ΔN/ Δ t = λn 3 ΔN/ Δ t ( T1/ )( ΔN/ Δt) (8.0 d)(86, 400 s/d)(1 Ci)(3.70 decays/s) N = = = λ 0693. 0693. 13 13 0.131 kg 1 = 3. 69 nuclei; m = 3.69 nuclei = 8 kg 3 6.0 nuclei 51. The activity is converted to decays per day, then to energy per year, and finally to a dose per year. The potassium decays by gammas and betas, according to Appendix F. Gammas and betas have an RBE of 1, so the number of Sv is the same as the number of Gy, and the number of rem is the same as the number of rad. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-16 Chapter 31 1 Ci decays/s 3600 s L 6 decays 000 3.70 (1 h) 0.5 = 1.598 L 1Ci h day day 13 6 decays days MeV 1. 60 J 5 J 1.598 365 (0.) 1.5 = 1.40 day yr decay MeV yr (a) For the adult, use a mass of 60 kg. 5 J 1 1Gy 1Sv Effective dose = 1.40 yr 60 kg 1 J/kg 1 Gy 5 7 mrem =.33 Sv/yr =.33 mrem/yr Sv 7 Sv/yr or mrem/yr mrem.33 year 4 Fraction of allowed dose = times the allowed dose mrem 0 year For the baby, the only difference is that the mass is times smaller, so the effective dose is times bigger. The results are as follows: 6 Sv/yr, 0.mrem/yr, and 3 times the allowed dose 5. (a) The reaction has Z = 86 and A = for the parent nucleus. The alpha has Z = and A = 4, so the 18 daughter nucleus must have Z = 84 and A = 18. That makes the daughter nucleus 84 Po. From Fig. 30 11, polonium-18 is radioactive. It decays via both alpha and beta decay. The half-life for the alpha decay is 3.1 minutes. (c) The daughter nucleus is not a noble gas, so it is chemically reacting. It is in the same group as oxygen, so it might react with many other elements chemically. (d) The activity is given by Eq. 30 3b, 1 ln R = λ N = N. T Δ N 0. 693 0.693 9 6.0 nuclei = N = (1.4 g) Δ t T (3. 835 d)(86, 400 s/d) g 6 6 1Ci 4 = 7.964 decays/s 8.0 Bq =. Ci 3.70 Bq 1/ 3 To find the activity after 1 month, use Eq. 30 5. 0693. 0693. t (30 d) 6 (3. 835 d) 4 ΔN ΔN T1/ = e = (7.964 decays/s) e = 3.465 decays/s Δt Δt 0 4 1Ci 7 3. 5 Bq = 9.4 Ci 370. Bq Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-17 53. The frequency is given in Section 31 9 as 4.58 MHz. Use that to find the wavelength. c.998 m/s c = fλ λ = = = f 6 4.58 Hz 8 7.041 m This lies in the radio wave portion of the spectrum. 54. (a) The mass of fuel can be found by converting the power to energy to number of nuclei to mass. 6 7 3.156 s 1 MeV 1 fission atom 0.35 kg 13 3 (0 J/s)(1 yr) 1yr 1.60 J 00MeV 6.0 atom = 808.5 kg 8 kg The product of the first five factors above gives the number of U atoms that fission. 7 90 6 3.156 s 1 MeV 1 fission atom 38Sr nuclei 0 06(0 J/s)(1 yr) 13 # =. 1yr 1.60 J 00MeV 6 90 38 = (1.4 ) Sr nuclei The activity is given by the absolute value of Eq. 30 3b. Δ N 0. 693 0.693 6 16 decays = λn = N = (1.4 ) = 9.389 Δ t T 7 (9 yr)(3.156 s/yr) s 1/ 16 1Ci 6 = (9.389 decays/s) =.5 Ci 3.70 decays/s 55. (a) The reaction is 9 Be + 4 He n +?. There are 6 protons and 13 nucleons in the reactants, so 4 there must be 6 protons and 13 nucleons in the products. The neutron is 1 nucleon, so the other product must have 6 protons and 1 nucleons. Thus, it is 1 6 C. Be He n C Q = m c + m c m c m c 9 4 1 4 6 = [ 9.01183 u + 4.00603 u 1.008665 u 1.000000 u] 931.5 c = 5.70 MeV 56. If KE = kt, then to get from kelvins to kev, use the Boltzmann constant. It must be put in the proper units. 3 J 1 ev 1 kev 8 k = 1.381 = 8.60 kev/k K 19 1.60 J 00 ev 57. From Eq. 13 9, the average speed of a gas molecule (root-mean-square speed) is inversely proportional to the square root of the mass of the molecule, if the temperature is constant. We assume that the two gases are in the same environment and therefore at the same temperature. We use UF 6 molecules for the calculations. υ 35 m38 9 UF6 9 UF6 38. 05 + 6(19. 00) = = = 1.0043:1 υ m 35. 04 + 6(19. 00) 38 35 9 UF6 9 UF6 Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-18 Chapter 31 58. (a) We assume that the energy produced by the fission was 00 MeV per fission, as in Problems 18, 19, and 0. 1 5 J 1 MeV 1 fission atom 0.35 kg 13 3 (0 kilotons TNT) 1 kiloton 1.60 J 00 MeV 6.0 atoms = 1.0 kg 1 kg Use E = mc. 1 5 J (0 kilotons TNT) 1 kiloton E 3 E = mc m = = = 1.11 kg 1 g c 8 (3.0 m/s) 59. The effective dose (in rem) is equal to the actual dose (in rad) times the RBE factor. dose(rem) = (3 mrad/yr X-ray, γ -ray)(1) + (3.4 mrad/yr)() = 66 mrem/yr 60. Because the RBE factor for gamma rays is 1, the dose in rem is equal in number to the dose in rad. Since the intensity falls off as r (the square of the distance), the exposure rate times r will be constant. 5.0 rem 1 rad 1 yr 1 week 3 rad Allowed dose = =.747 yr 1 rem 5 weeks 35 h h 3 rad rad.747 r = 4.8 (1 m) h h rad 4.8 ( 1m) h r = = 4.180 m 4. m 3 rad.747 h 4 61. (a) The reaction is of the form? He+ 86Rn. There are 88 protons and 6 nucleons as products, so there must be 88 protons and 6 nucleons as reactants. The parent nucleus is 6 88 Ra. 6 4 88 + 86 Ra He Rn If we ignore the KE of the daughter nucleus, then the KE of the alpha particle is the Q-value of the reaction. KE m c m c m c α = 6 4 88Ra He 86 Rn MeV/ c = [ 6. 054 u 4. 00603 u. 017578 u] 931.5 c = 4.871 MeV Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-19 (c) From momentum conservation, the momentum of the alpha particle will be equal in magnitude to the momentum of the daughter particle. At the energy above, the alpha particle is not relativistic, so KEα α p = p = m α a makea. 931.5 pα = makea = (4.00603 u) (4.871 MeV) = 190.6 MeV/ c 1 The momentum of the daughter nucleus is the same as that of the alpha, 190.6 MeV/ c. α = daughter pα KE daughter = = mdaughter mdaughter (d) Since p p daughter, p. pα (190.6 MeV/ c) KE daughter m daughter = = = 8.78 MeV 931.5 (.0 u) 1 Thus we see that our original assumption of ignoring the kinetic energy of the daughter nucleus is valid. The kinetic energy of the daughter is less than % of the Q-value. 6. This heat of combustion is 6. MeV/4 hydrogen atoms. 13 7 6. MeV 1.60 J 1 H atom 1 u 14 = 6.6 J/kg 4 H atoms 1 MeV 1.00785 u 1.66 kg This is about 7 times the heat of combustion of coal. 63. (a) The energy is radiated uniformly over a sphere with a radius equal to the orbit radius of the Earth. 11 6 6 (1300 W/m )4 (1.496 m) 3.656 W 3.7 W π = (c) After subtracting the energy of the neutrinos, the reaction of Eq. 31 7 releases 6. MeV for every 4 protons consumed. 6 J 4 protons 1 MeV 38 3.656 = 3.489 protons/s s 6. MeV 13 1.6 J 38 3.5 protons/s Convert the Sun s mass to a number of protons and then use the above result to estimate the Sun s lifetime. 30 1proton 1s 1yr 11.0 kg 1.1 yr 7 38 7 1.673 kg 3.489 protons 3.156 s 64. For the net proton cycle, Eq. 31 7, there are two neutrinos produced for every four protons consumed. Thus the net number of neutrinos generated per second from the Sun is just half the value of protons consumed per second. That proton consumption rate is calculated in Problem 63b. 38 38 (3.43 protons/s)( v/4p) = 1.71 v/s Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31-0 Chapter 31 Assume that the neutrino distribution is spherically symmetric, centered at the Sun. The fraction that would pass through the area of the ceiling can be found by a ratio of areas, assuming that the ceiling is perpendicular to the neutrino flux. But since the ceiling is not perpendicular, a cosine factor is included to account for the angle difference, as discussed in Eq. 1 1. Finally, we adjust for the one-hour duration, assuming that the relative angle is constant over that hour. 38 180 m 0 v 11 (1.71 /s) (cos 44 )(3600 s) =.8 4 π (1.496 m) v 65. Use the common value of 00 MeV of energy released per fission, as used in Problems 18, 19, and 0. Multiply that by the number of fissions, which is 5.0% of the number of U-38 atoms. 13 35 00 MeV 1.60 J 9 Unuclei 38 35 0.05.0 kg 38 9 U 1 nucleus of 1MeV 9 U ( ) 9 U nuclei Total energy = 3 38 6.0 nuclei of 9 U nuclei 38 0.38 kg 9 U 1 1 = 8.7 J 8 J 66. (a) The energy released is given by the Q-value. Q = m1 c m4 c = [ (1. 000000 u) 3.98504 u] 931.5 c 13.93 MeV 6C 1Mg = The total KE of the two nuclei must equal their PE when separated by 6.0 fm. KE KE 1 QQ 1 = 4πε r 0 19 1 1 QQ (6)(1. 60 C) 1 1 9 13 4πε 15 0 r 60. m = = (9.0 N m /C ) = 6.91 J 13 19 6 = 6. 91 J(1 ev/1.60 J)(1 MeV/ ev) = 4.3 MeV 4.3 MeV (c) The kinetic energy and temperature are related by Eq.13 8. KE 13 3 KE 6.91 J kt T 3 3 3 = = = = 3.3 K k 1.38 J/K 67. (a) A Curie is 3.7 decays/s. 6 (0. Ci)(3.7 decays/s) = 3700 decays/s The beta particles have an RBE factor of 1. We calculate the dose in Gy and then convert to Sv. The half-life is over a billion years, so we assume that the activity is constant. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-1 13 7 1 (3700 decays/s)(1.4 MeV/decay)(1.60 J/MeV)(3.156 s/yr) 65 kg 4 4 4 = 4.04 J/kg/yr = 4.04 Gy/yr = 4.0 Sv/yr This is about 4 4.04 Sv/yr = 011. or 11% of the background rate. 3 3.6 Sv/yr 68. The surface area of a sphere is 4 π r. 7 7. 3 = = = 1.4 6 πrearth π Activity.0 Ci ( 0 Ci)(3.7 decays/s) decays/s m 4 4 (6.38 m) m Q = 3m c m c = 3(4. 00603 u) 1. 000000 u 931.5 c 7.74 MeV He 6 C = 69. 4 1 [ ] 70. Since the half-life is 30 years, we assume that the activity does not change during the 1.4 hours of exposure. We calculate the total energy absorbed and then calculate the effective dose. The two energies can be added directly since the RBE factor for both gammas and betas is about 1. 6 decays s (1. Ci) 3. 7 (1. 4 h) 3600 s 1h 5 Energy = = 3.043 J 3 ev 19 J 850 1.60 decay ev 5 3. 043 J 0 rad 5 5 dose = = 4.909 rad 4.9 rem 6 kg 1 J/kg 71. The actual power created by the reaction of Eq. 31 8a must be 00 MW/0.33 = 3030 MW. We convert that to grams of deuterium using Eq. 31 8a. 3 7 13 3 Mass d 3 03 9 J d nuclei 1 MeV kg d 3.156 s =. s 4. 03MeV 1.60 J 6.0 nuclei 1yr = 985.4 kg d 990 kg d 7. In the first scenario, slow neutrons have an RBE of 5, from Table 31 1. The maximum dose for a radiation worker is given in the text as 50 msv/yr, so this worker has been exposed to 50 msv. Use Eq. 31 b to convert the Sv to Gy and then convert that to J by the definition of Gy. 3 1Gy 1J/kg Energy = Gy mass = (50 Sv) (65 kg) = 0.65 J 5Sv 1Gy In the second scenario, the RBE changes to a value of for the protons. 3 1Gy 1J/kg Energy = Gy mass = (50 Sv) (65 kg) = 1.6 J Sv 1Gy Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31- Chapter 31 73. (a) There are 9 protons in the reactants, so X must have 9 38 = 54 protons. There are 36 total nucleons in the reactants, so X must have 36 9 3 = 141 total nucleons. Thus, X is 141 54 Xe. If the reaction is barely critical, then every nucleus that fissions leads to another nucleus that fissions. Thus one of the produced neutrons causes another fission. The other two either escape the reactor vessel or are absorbed by some nucleus without causing a fission. (c) The masses needed are 141 9 54Xe: 140.9665 u and 38Sr: 91.9138 u. Use those to calculate the Q-value. Q = ( m35 1 9 141 1 9U + m m m 3 m ) c 0 n 38Sr 54 Xe 0 n = [ 35. 043930 u + 1. 008665 u 91. 9138 u 140. 9665 u 3(1. 008665 u) ] c = (0. 18891 u) 931. 5 = 176. 0 MeV 74. The half-life of the strontium isotope is 8.90 years. Use that with Eq. 30 5 to find the time for the activity to be reduced to 15% of its initial value. 0693. 0693. 0693. t t t T T T ΔN ΔN ΔN ΔN e e e Δt Δt 0 Δt 0 Δt 0 0.693 T1/ln(0.15) (8.90 yr)ln(0.15) ln(0.15) = t t = = 79 yr T 0. 693 0.693 1/ 1/ 1/ = 015. = 015. = 1/ 75. Source B is more dangerous than source A because of its higher energy. Since the sources have the same activity, they both emit the same number of gammas. Source B can deposit twice as much energy per gamma and therefore causes more biological damage. Source C is more dangerous than source B because the alphas have an RBE factor up to 0 times larger than the gammas. Thus a number of alphas may have an effective dose up to 0 times higher than the effective dose of the same number of like-energy gammas. So from most dangerous to least dangerous, the ranking of the sources is C > B > A. We might say that source B is twice as dangerous as source A, and source C is 0 times more dangerous than source B. 76. The whole-body dose can be converted into a number of decays, which would be the maximum number of nuclei that could be in the Tc sample. The RBE factor of gammas is 1. 3 1kg 1eV 17 50 mrem = 50 mrad (50 rad) (55 kg) = 1.719 ev 0 rad 19 1.60 J 17 1 effective γ γ decays 1 nucleus 1 (1.719 ev) =.455 nuclei 3 140 ev 1 effective γ 1γ decay This then is the total number of decays that will occur. The activity for this number of nuclei can be calculated from Eq. 30 3b. 1 Δ N 0. 693N 0.693(. 455 decays) 1 Ci Activity = = λn = = Δ t T 1/ (6 h)(3600 s/h) 3.70 decays/s 3 =.19 Ci mci Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

Nuclear Energy; Effects and Uses of Radiation 31-3 Solutions to Search and Learn Problems 1. (a) Three problems that must be overcome to make a functioning fission nuclear reactor: (i) the neutrons must be moderated (slowed) so they will react with the fissionable nuclei; (ii) the neutrons must be prevented from being absorbed by the wrong nuclei; (iii) the neutrons must be prevented from escaping the reaction vessel. Five environmental problems or dangers resulting from a nuclear fission reactor: (i) thermal pollution the warming of the environment around the nuclear reactor; (ii) the need for safe storage of fuel before using it in the reactor; (iii) the safe disposal of radioactive waste; (iv) radiation damage to the power plant itself; (v) accidental release of radiation into the environment. (c) Breeder reactors make fissionable material that can be used in fission bombs.. (a) For small nuclei, the binding energy per nucleon increases as the number of nucleons increases. When two small nuclei are fused together, the total energy of the resulting nucleus is smaller than the energy of the two initial nuclei, with the excess energy released in the fusion process. The initial proton proton reaction has a much smaller probability of occurring than the other reactions. Since the other reactions can take place relatively quickly compared with the first reaction, it determines the time scale for the full fusion process. (c) Iron (Fe). (d) The force of gravity between all of the atoms in the Sun (or star) holds it together. (e) (i) A tokamak uses magnetic fields to confine the fusion material. (ii) Lasers are used to heat the fusion material so quickly that the nuclei cannot move away before they fuse (inertial confinement). 3. The oceans cover about 70% of the Earth, to an average depth of approximately 4 km. The density of 3 the water is approximately 00 kg/m. The mass of water is found by multiplying the volume of water by its density. Mass of water = 0. 7(surface area)(depth)(density) 6 3 1 = (0. 7)4 π (6.38 m) (4000 m)(00 kg/m ) = 1.43 kg water Dividing the mass by the molar mass of water and multiplying by Avogadro s number gives the number of water molecules. There are two hydrogen atoms per water molecule. Multiplying the number of hydrogen atoms by the percentage of deuterium gives the number of deuterium atoms. If the reactions of Eqs. 31 8a and 31 8b are carried out at the same rate, then 4 deuterons would produce 7.30 MeV of energy. Use that relationship to convert the number of deuterons in the oceans to energy. 13 43 7.30 MeV 1.60 J 30 30 (1. d) = 3.1 J 3 J 4d 1MeV Based on Chapter 6, Problem 3, this is about 30 billion times the annual energy consumption of the United States. Copyright 014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.