45 Applications of Nuclear Physics CHAPTER OUTLINE 45.1 Interactions Involving Neutrons 45.2 Nuclear Fission 45.3 Nuclear Reactors 45.4 Nuclear Fusion 45.5 Radiation Damage 45.6 Uses of Radiation * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ45.1 Answer (c) We compute the change in mass number A: 235 + – 137 – 96 = All the protons that start out in the uranium nucleus end up in the fission product nuclei OQ45.2 Answer (d) The best particles to trigger a fission reaction of the uranium nuclei are slow moving neutrons Fast moving neutrons may not stay in close proximity with a uranium nucleus long enough to have a good probability of being captured by the nucleus so that a reaction can occur Positively charged particles, such as protons and alpha particles, have difficulty approaching the target nuclei because of Coulomb repulsion OQ45.3 Answer (c) The total energy released was E = (17 × 103 ton) ( 4.2 × 109 J ton ) = 7.1 × 1013 J and according to the mass-energy equivalence, the mass converted was m= E 7.1 × 1013 J = = 7.9 × 10−4 kg = 0.79 g  g c ( 3.00 × 108 m s )2 1146 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1147 OQ45.4 The ranking is (b) > (c) > (a) > (d) See Table 45.1 for the RBE factors Dose (a) is rem Dose (b) is (1 rad × 10) = 10 rem Doses (c) and (d) are (1 rad × or 5) = to rem, but dose (d) is to the hands only (less mass has absorbed the radiation) If we assume that (a) and (b) as well as (c) were whole-body doses to many kilograms of tissue (more mass has absorbed the radiation), we find the ranking stated OQ45.5 Answer (c) The function of the moderator is to slow down the neutrons released by one fission so that they can efficiently cause more fissions OQ45.6 The ranking is Q1 > Q2 > Q3 > Because all of the reactions involve 108 nucleons, we can look just at the change in binding-energy-pernucleon as shown on the curve of binding energy The jump from lithium to carbon is the biggest jump (~ 5.4 → 7.7 MeV), and next the jump from A = 27 to A = 54 (~ 8.3 → 8.8 MeV), which is near the peak of the curve The step up for fission from A = 108 to A = 54 (~ 8.7 → 8.8 MeV) is smallest All the reactions result in an increase in binding-energy-per-nucleon, so both of the fusion reactions described and the fission reaction put out energy, so Q is positive for all Imagine turning the curve of binding energy upside down so that it bends down like a cross-section of a bathtub On such a curve of total energy per nucleon versus mass number it is easy to identify the fusion of small nuclei, the fission of large nuclei, and even the alpha decay of uranium, as exoenergetic processes The most stable nucleus is at the drain of the bathtub, with minimum energy OQ45.7 Answer (d) The particles lose energy by collisions with nuclei in the bubble chamber to make their speed and their cyclotron radii r = mv/qB decrease OQ45.8 Answer (b) The cyclotron radius is given by r = mv qB = ( 21 m2 v ) qB = 2mK qB K and B are the same for both particles, but the ratio m q is smaller for the electron; therefore, the path of the electron has a smaller radius, meaning the electron is deflected more OQ45.9 Answer (b) The nuclei must be energetic enough to overcome the Coulomb repulsion between them so that they can get close enough to fuse, and numerous enough for many collisions to occur in a short period of time so that the reaction produces more energy than it requires to operate © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1148 Applications of Nuclear Physics ANSWERS TO CONCEPTUAL QUESTIONS CQ45.1 The two factors presenting the most technical difficulties are the requirements of a high plasma density and a high plasma temperature These two conditions must occur simultaneously CQ45.2 For the deuterium nuclei to fuse, they must be close enough to each other for the nuclear forces to overcome the Coulomb repulsion of the protons—this is why the ion density is a factor The more time that the nuclei in a sample spend in close proximity, the more nuclei will fuse—hence the confinement time is a factor CQ45.3 The products of fusion reactors are generally not themselves unstable, while fission reactions result in a chain of reactions which almost all have some unstable products, because they have an excess of neutrons CQ45.4 The advantage of a fission reaction is that it can generate much more electrical energy per gram of fuel compared to fossil fuels Also, fission reactors not emit greenhouse gases as combustion byproducts like fossil fuels—the only necessary environmental discharge is heat The cost involved in producing fissile material is comparable to the cost of pumping, transporting, and refining fossil fuel The disadvantage is that some of the products of a fission reaction are radioactive—and some of those have long half-lives The other problem is that there will be a point at which enough fuel is spent that the fuel rods not supply power economically and need to be replaced The fuel rods are still radioactive after removal Both the waste and the “spent” fuel rods present serious health and environmental hazards that can last for tens of thousands of years Accidents and sabotage involving nuclear reactors can be very serious, as can accidents and sabotage involving fossil fuels CQ45.5 Fusion of light nuclei to a heavier nucleus releases energy Fission of a heavy nucleus to lighter nuclei releases energy Both processes are steps towards greater stability on the curve of binding energy, Figure 44.5 The energy release per nucleon is typically greater for fusion, and this process is harder to control CQ45.6 The excitation energy comes from the binding energy of the extra nucleon CQ45.7 Advantages of fusion: high energy yield, no emission of greenhouse gases, fuel very easy to obtain, reactor cannot go supercritical like a fission reactor and low amounts of radioactive waste © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1149 Disadvantages: requires high energy input to sustain reaction, lithium and helium are scarce, and neutrons released by the reaction cause structural damage to reactor housing CQ45.8 For each additional dynode, a larger applied voltage is needed, and hence a larger output from a power supply—“infinite” amplification would not be practical Nor would it be desirable: the goal is to connect the tube output to a simple counter, so a massive pulse amplitude is not needed If you made the detector sensitive to weaker and weaker signals, you would make it more and more sensitive to background noise CQ45.9 The hydrogen nuclei in water molecules have mass similar to that of a neutron, so that they can efficiently rob a fast-moving neutron of kinetic energy as they scatter it A neutron bouncing off a more massive nucleus would lose less energy, so it would continue to travel through the shield Once the neutron is slowed down, a hydrogen nucleus can absorb it in the reaction n + 11 H → 21 H SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 45.1 Interactions Involving Neutrons Section 45.2 Nuclear Fission *P45.1 The energy consumed by a 100-W lightbulb in a 1.0-h time period is ⎛ 3600 s ⎞ E = PΔt = ( 100 J/s ) ( 1.0 h ) ⎜ = 3.6 × 105 J ⎝ h ⎟⎠ The number of fission events, yielding an average of 208 MeV each, required to produce this quantity of energy is n= P45.2 E 3.6 × 105 J ⎛ MeV ⎞ = = 1.1 × 1016 −13 ⎟ ⎜ 208 MeV 208 MeV ⎝ 1.60 × 10 J ⎠ The mass of U-235 producing the same amount of energy as 000 kg of coal is ⎛ MeV ⎞ m = ( 3.30 × 1010 J ) ⎜ ⎝ 1.60 × 10−13 J ⎟⎠ 235 g ⎛ U-235 nucleus ⎞ ⎛ ⎞ ×⎜ ⎜ ⎟ 23 ⎝ ⎠ ⎝ 6.02 × 10 nucleus ⎟⎠ 200 MeV = 0.403 g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1150 P45.3 P45.4 Applications of Nuclear Physics Three different fission reactions are possible: n+ 235 92 U→ 90 38 Sr + 144 54 Xe + 01 n 144 54 Xe n+ 235 92 U→ 90 38 Sr + 143 54 Xe + 01 n 143 54 Xe n+ 235 92 U→ 90 38 Sr + 142 54 Xe + 01 n 142 54 Xe If the electrical power output of 1.00 GW is 40.0% of the power derived from fission reactions, the power output of the fission process is 1.00 GW = ( 2.50 × 109 J s ) ( 8.64 × 10 s d ) = 2.16 × 1014 J d 0.400 The number of fissions per day is ( 2.16 × 10 14 ⎞ eV ⎛ fission ⎞ ⎛ J d)⎜ ⎟ −19 ⎟ ⎜ ⎝ 200 × 10 eV ⎠ ⎝ 1.60 × 10 J ⎠ = 6.75 × 1024 d −1 This also is the number of 235U nuclei used, so the mass of 235U used per day is (6.75 × 10 24 ⎛ ⎞ 235 g mol nuclei d ) ⎜ 23 ⎝ 6.02 × 10 nuclei mol ⎟⎠ = 2.63 × 103 g d = 2.63 kg d In contrast, a coal-burning steam plant producing the same electrical power uses more than × 106 kg/d of coal P45.5 First, the thorium is bombarded: 233 n  + 232    90 Th   →       90 Th Then, the thorium decays by beta emission: 233    90 Th   →    233 Pa +  −10 e + ν    91 Protactinium-233 has more neutrons than the more stable protactinium-231, so it too decays by beta emission: 233    91 P45.6 (a) Pa   →    233 U+    92 −1 e+ν The energy released is equal to the Q value, given by Q = ( Δm) c = [ mn + MU-235 − MBa-141 − MKr-92 − 3mn ] c with Δm = [1.008 665 u + 235.043 923 u − 140.914 4 u −91.926 2 u − ( 1.008 665 u )] = 0.185 993 u © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1151 Then, Q = ( 0.185 993 u )( 931.5 MeV u ) = 173 MeV (b) The fraction of rest energy transformed is f = P45.7 Δm 0.185 993 u = = 7.88 × 10−4 = 0.078 8% mi 236.05 u The energy released in the reaction is n+ 235 92 U → 88 38 Sr + 136 54 Xe + 12 10 n Q = ( Δm) c = ⎡ m 235 U − 11mn − m 88 Sr − m136 Xe ⎤ c 38 54 ⎣ 92 ⎦ = ⎡⎣ 235.043 923 u − 11( 1.008 665 u ) −87.905 614 u − 135.907 220 u ]( 931.5 MeV u ) = 126 MeV P45.8 In N collisions, the energy is reduced from 2.00 MeV to 0.039 eV: N ( 2.00 × 10 eV ) ⎛⎜⎝ 21 ⎞⎟⎠ ≤ 0.039 eV N 0.039 ⎛ 1⎞ ⎜⎝ ⎟⎠ ≤ 2.00 × 106 ⎛ 1⎞ ⎛ 0.039 ⎞ N ln ⎜ ⎟ ≤ ln ⎜ ⎝ 2⎠ ⎝ 2.00 × 106 ⎟⎠ ⎛ 2.00 × 106 ⎞ N ln ( ) ≥ ln ⎜ ⎝ 0.039 ⎟⎠ which gives N ≥ 25.6 → N = 26 P45.9 The mass defect is Δm = ( mn + MU ) − ( MZr + MTe + 3mn ) Δm = [ 1.008 665 u + 235.043 923 u − 97.912 u − 134.916 u − ( 1.008 665 u ) ⎤⎦ = 0.197 393 u The energy equivalent is ⎛ 931.5 MeV c ⎞ Δmc = ( 0.197 393 u ) c ⎜ ⎟⎠ = 184 MeV ⎝ u © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1152 P45.10 Applications of Nuclear Physics (a) At a concentration of c = mg/m3 = × 10–3 g/m3, the mass of uranium dissolved in the oceans covering two-thirds of Earth’s surface to an average depth of havg = km is mU = cV = c ( 23 A ) ⋅ hav = c ⎡⎣ 23 ( 4π RE2 ) ⎤⎦ ⋅ hav or g ⎞ ⎛ 2⎞ ⎛ mU = ⎜ × 10−3 4π ( 6.38 × 106 m ) ( × 103 m ) ⎟ ⎜ ⎟ ⎝ m ⎠ ⎝ 3⎠ = × 1015 g (b) Fissionable 235U makes up 0.700% of the mass of uranium computed above If we assume all of the 235U is collected and caused to undergo fission, with the release of about 200 MeV per event, the potential energy supply is E = ( number of = 235 U atoms )( 200 MeV ) 0.700 ⎛ mU ⎜ 100 ⎝ m 235U ⎞ ⎟ ( 200 MeV ) atom ⎠ and at a consumption rate of P = 1.5 × 1013 J/s, the time interval this could supply the world’s energy needs is Δt = E/P, or Δt = = 0.700 ⎛ mU ⎜ 100 ⎝ m 235U ⎞ ( 200 MeV ) ⎟ P atom ⎠ ⎛ kg ⎞ ⎤ 0.700 ⎡ × 1015 g ⎢ ⎥ 100 ⎢⎣ ( 235 u )( 1.66 × 10−27 kg u ) ⎜⎝ 103 g ⎟⎠ ⎥⎦ ⎡⎛ 200 MeV ⎞ ⎛ 1.60 × 10−13 J ⎞ ⎛ yr × ⎢⎜ ⎜ 13 ⎜ ⎟ ⎟ ⎝ ⎣⎝ 1.50 × 10 J s ⎠ ⎝ MeV ⎠ 3.16 × 10 ⎞⎤ ⎟ s ⎠ ⎥⎦ = × 103 yr (Compare this value to that in part (b) of Problem 17, which is a more realistic estimate of the time interval for the uranium that can be extracted reasonably from the Earth.) (c) The uranium comes from rocks and minerals dissolved in water and carried into the ocean by rivers (d) No Uranium cannot be replenished by the radioactive decay of other elements on Earth © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 P45.11 One kg of enriched uranium contains 3.40% uranium-235 is 235 92 1153 U, so the mass of m235 = 0.034 0(1 000 g) = 34.0 g In terms of number of nuclei, this is equivalent to ⎛ ⎞ N 235 = (34.0 g) ⎜ (6.02 × 1023  atoms/mol ) ⎝ 235 g/mol ⎟⎠ = 8.71 × 1022  nuclei If all these nuclei fission, the energy released is equal to ( 8.71 × 10 22  nuclei ) ( 200 × 106  eV/nucleus ) × ( 1.602 × 10 –19  J/eV ) = 2.79 × 1012  J Now, for the engine, efficiency = work output heat input or e = PΔr cosθ Qh So the distance the ship can travel per kilogram of uranium fuel is Δr = Section 45.3 *P45.12 (a) 0.200 ( 2.79 × 1012  J ) eQh = = 5.58 × 106  m P cos ( 0° ) 1.00 × 10  N Nuclear Reactors With a specific gravity of 4.00, the density of soil is ρ = 4.00 × 103 kg/m Thus, the mass of the top 1.00 m of soil is ⎡ 1m ⎞ ⎤ ⎛ m = ρV = ( 4.00 × 10 kg/m ) ⎢( 1.00 m )( 43 560 ft ) ⎜ ⎝ 3.281 ft ⎟⎠ ⎥⎦ ⎣ = 1.62 × 107 kg 3 At a rate of part per million, the mass of uranium in this soil is mU = (b) m 1.62 × 107 kg = = 16.2 kg 106 106 Since 0.720% of naturally occurring uranium is 235 92 235 92 U, the mass of U in the soil of part (a) is m 235 U = ( 7.20 × 10−3 ) mU = ( 7.20 × 10−3 ) ( 16.2 kg ) 92 = 0.117 kg = 117 g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1154 P45.13 Applications of Nuclear Physics In one minute there are N = 60.0 s = 5.00 × 10 fissions −3 1.20 × 10 s So the rate increases by a factor of ( 1.000 25 ) P45.14 (a) ⎛ 3V ⎞ For a sphere: V = π r → r = ⎜ ⎝ 4π ⎟⎠ 4π r A ⎛ 36π ⎞ = = =⎜ ⎟ V ( 3)π r r ⎝ V ⎠ (b) 50 000 = 2.68 × 105 13 , so 13 = 4.84V −1 For a cube: V = 3 →  = V , so A 62 = = = 6V −1 V   (c) ⎛V⎞ For a parallelepiped: V = 2a → a = ⎜ ⎟ ⎝ 2⎠ 13 , so 2 13 13 A ( 2a + 8a ) ⎛ 2⎞ ⎛ 250 ⎞ −1 = = = = ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = 6.30V V 2a a V V (d) The answers show that the sphere has the smallest surface area for a given volume and the brick has the greatest surface area of the three Therefore, The sphere has minimum leakage and the parallelepiped has maximum leakage P45.15 Recall the radius of a nucleus of mass number A is r = aA1/3, where a = 1.2 fm The center to center distance of the nuclei of helium (A = 4) and gold (A = 197) is the sum of their combined radii: r = ( 1.2 fm ) ( ) 13 + ( 1.2 fm ) ( 197 ) 13 = 8.9 fm = 8.9 × 10−15 m The electric potential energy is ke q1q2 r ( 8.99 × 109 N ⋅ m2 / C2 ) ( )(79)(1.60 × 10−19 C) e U = qV = = 8.9 × 10−15 m = 2.6 × 107 eV = 26 MeV P45.16 The power after three months is P = 10.0 MW = 1.00 × 107 J/s If each decay delivers 1.00 MeV = 1.60 × 10–13 J, then the number of decays/s = 1.00 × 107 J/s = 6.25 × 1019 Bq 13 1.60 ì 10 J â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 P45.17 (a) 1155 Do not think of the “reserve” as being held in reserve We are depleting it as fast as we choose The remaining current balance of irreplaceable 235U is 0.7% of the whole mass of uranium: ⎛ 103 kg ⎞ ⎛ 103 g ⎞ = 3.08 × 1010 g ( 0.007 00 )( 4.40 × 10 tons ) ⎜ ⎟ ⎜ ⎟ ⎝ ton ⎠ ⎝ kg ⎠ (b) The number of moles of 235U in the reserve is n= (c) m 3.08 × 1010 g = = 1.31 × 108 mole M 235 g/mole The number of moles found in part (b) corresponds to ⎛ 6.02 × 1023 atom ⎞ ⎛ nucleus ⎞ N = nN A = ( 1.31 × 108 mole ) ⎜ ⎟⎠ ⎜⎝ atom ⎟⎠ ⎝ mole = 7.89 × 1031 nuclei (d) We imagine each nucleus as fissioning, to release (7.89 × 10 (e) 31 −13 ⎛ 200 MeV ⎞ ⎛ 1.60 × 10 J ⎞ fissions ) ⎜ = 2.52 × 1021 J ⎝ fission ⎟⎠ ⎜⎝ MeV ⎟⎠ The definition of power is represented by P = (energy converted)/ Δt , so we have Δt = energy 2.52 × 1021 J yr ⎛ = = ( 1.68 × 108 s ) ⎜ 13 ⎝ 3.156 × 107 P 1.5 × 10 J/s ⎞ ⎟ s⎠ = 5.33 yr (f) P45.18 Fission is not sufficient to supply the entire world with energy at a price of $130 or less per kilogram of uranium Assuming that the impossibility is not that he can have this control over the process (which, as far as we know presently, is impossible), let’s see what else might be wrong The reaction can be written 141 94 n  +   235 92 U   →    57 La  +   35 Br  +  n ( n ) where n is the number of neutrons released in the fission reaction By balancing the equation for electric charge and number of nucleons, we find that n = If one incoming neutron results in just one outgoing neutron, the possibility of a chain reaction is not there, so this nuclear reactor will not work © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1178 Applications of Nuclear Physics (c) The number of decays per second is the decay rate R, and the energy released in each decay is Q Then the energy released per unit time interval is P = QR (d) The decay rate for all steps in the radioactive series in steady state is set by the parent uranium: ⎛ 7.00 × 10 g ⎞ N=⎜ 6.02 × 1023 nuclei mol ) ( ⎟ ⎝ 238 g mol ⎠ = 1.77 × 1026 nuclei The decay constant is λ= ln ln = = 1.55 × 10−10 T1 4.47 × 10 yr yr and the rate of decays is then ⎛ 1⎞ R = λ N = ⎜ 1.55 × 10−10 1.77 × 1026 nuclei ) ( ⎟ yr ⎠ ⎝ = 2.75 × 1016 decays yr so, P = QR = ( 51.7 MeV ) ( 2.75 × 1016 yr −1 ) ( 1.60 × 10−13 J MeV ) = 2.27 × 105 J yr (e) We know that dose in rem = dose in rad × RBE or 5.00 rem/yr = (dose in rad/yr)(1.10) giving (dose in rad/yr) = 4.55 rad/yr The allowed whole-body dose is then ⎛ 10−2 J kg ⎞ (70.0 kg )( 4.55 rad yr ) ⎜⎝ rad ⎟⎠ = 3.18 J yr P45.61 (a) The mass of the pellet is ⎡ 4π ⎛ 1.50 × 10− cm ⎞ ⎤ 4π 3 m = ρV = ρ r = ( 0.200 g cm ) ⎢ ⎜ ⎟⎠ ⎥ ⎢⎣ ⎝ ⎥⎦ = 3.53 × 10−7 g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1179 The pellet consists of equal numbers of 2H and 3H atoms, so the average molar mass is 2.50 and the total number of atoms is ⎛ 3.53 × 10−7 g ⎞ N=⎜ (6.02 × 1023 atoms mol ) ⎝ 2.50 g mol ⎟⎠ = 8.51 × 1016 atoms When the pellet is vaporized, the plasma will consist of 2N particles (N nuclei and N electrons) The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ The temperature of the plasma is found from E = ( 2N ) 23 kBT as ( ) 2.00 × 103 J E T= = = 5.68 × 108 K − 23 16 3NkB ( 8.51 × 10 ) 1.38 × 10 J K ( (b) ) Each fusion event uses nuclei, so N/2 events will occur From Equation 45.4, the energy released by one fusion event is 17.59 MeV, so the total energy released will be ⎛ 8.51 × 1016 ⎞ ⎛ N⎞ E=⎜ ⎟Q=⎜ ( 17.59 MeV )( 1.60 × 10−13 J MeV ) ⎟ ⎝ 2⎠ ⎝ ⎠ = 1.20 × 105 J = 120 kJ P45.62 (a) From the given equation, the ratio of the two intensities is I I e − µ2 x = = e − ( µ − µ1 ) x − µ1 x I1 I e (b) Substituting numerical values into the equation in part (a) gives I 50 = exp ⎡⎣ − ( 5.40 cm −1 − 41.0 cm −1 ) ( 0.100 cm ) ⎤⎦ = e 3.56 = 35.2 I100 (c) Here, x = 10.0 mm = 1.00 cm, and I 50 = exp ⎡⎣ − ( 5.40 cm −1 − 41.0 cm −1 )( 1.00 cm )⎤⎦ = e 35.6 I100 = 2.89 × 1015 Thus, a 1.00-cm-thick aluminum plate has essentially removed the long-wavelength x-rays from the beam P45.63 The momentum of the alpha particle and that of the neutron must add to zero, so their velocities must be in opposite directions with magnitudes related by   mnv n + mα v α = or (1.008 u)vn = (4.002 u) vα © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1180 Applications of Nuclear Physics At the same time, their kinetic energies must add to 17.6 MeV: 1 1 mn vn2 + mα vα2 = (1.008 7 u)vn2 + (4.002 6 u)vα2 2 2 = 17.6 MeV E= Substitute vα = 0.252 0vn to obtain E = ( 0.504 35 u ) vn2 + ( 0.127 10 u ) vn2 ⎛ ⎞ 1 u = 17.6 MeV ⎜ 2⎟ ⎝ 931.494 MeV/c ⎠ Solving for then gives = 0.018 9c = 0.173c = 5.19 × 107 m/s 0.631 45 Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic energy of the neutron from the classical equation, K= 2 ⎛ 931.494 MeV/c ⎞ mv = ( 1.008 7 u )( 0.173c ) ⎜ ⎟⎠ ⎝ 2 u = 14.1 MeV For a more accurate calculation of the kinetic energy, we should use relativistic expressions Conservation of energy for this reaction requires that En + Eα = ( mn c + K n ) + ( mα c + Kα ) = mn c + mα c + K [1] where K = 17.6 MeV is the total kinetic energy, and conservation of momentum for this reaction requires that   pn + pα = → pn = pα [2] From the relation between total energy, mass, and momentum of a particle, we have E = p c + ( mc ) p c = E − ( mc ) → [3] From equations [2] and [3], we may write pn2 c = pα2 c En2 − ( mn c ) = Eα2 − ( mα c ) En2 − Eα2 = ( mn c ) − ( mα c ) 2 (En − Eα )(En + Eα ) = ( mn c ) − ( mα c ) 2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1181 Substituting the above expression into equation [1] gives (En − Eα )( mn c + mα c + K ) = ( mn c ) − ( mα c ) En − Eα (m c ) − (m c ) = (m c + m c + K ) (m c ) − (m c ) − (m c + m c + K ) n n Eα = En n n 2 α 2 α 2 α 2 α Substituting this result back into equation [1] gives 2 ⎡ mn c ) − ( mα c ) ⎤ ( ⎥ = mn c + mα c + K En + ⎢En − 2 ⎢ ( mn c + mα c + K ) ⎥⎦ ⎣ (m c ) − (m c ) 2E = ( m c + m c + K ) + (m c + m c + K ) (m c + m c + K ) + (m c ) − (m c ) E = (m c + m c + K ) n n n α n n α 2 n n n α 2 α α 2 2 α 2 To find the kinetic energy of the neutron, we note that En = mn c + K n : En (m c = Kn (m c = n n + mα c + K ) + ( mn c ) − ( mα c ) 2 = mn c + K n ( mn c + mα c + K ) 2 + mα c + K ) + ( mn c ) − ( mα c ) 2 ( mn c + mα c + K ) 2 − mn c For K = 17.6 MeV, mn c = ( 1.008 u ) c ( 931.494 MeV c ⋅ u ) = 939.60 MeV and mα c = ( 4.002 u ) c ( 931.494 MeV c ⋅ u ) = 728.4 MeV we find that K n = 14.0 MeV P45.64 (a) The number of Pu nuclei in 1.00 kg is 6.02 × 1023 nuclei mol 000 g ) = 2.52 × 1024 nuclei ( 239.05 g mol The total energy is ( 25.2 × 10 23 ⎛ fission ⎞ ⎛ 200 MeV ⎞ nuclei ) ⎜ = 5.04 × 1026 MeV ⎝ nucleus ⎟⎠ ⎜⎝ fission ⎟⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1182 Applications of Nuclear Physics E = ( 5.04 × 1026 MeV ) ( 4.44 × 10−20 kWh MeV ) = 2.24 × 107 kWh or 22 million kWh (b) E = Δmc = ( 3.016 049 u + 2.014 102 u − 4.002 603 u − 1.008 665 u ) × ( 931.5 MeV u ) E = 17.6 MeV for each D-T fusion (c) En = ( total number of D nuclei ) ( 17.6 MeV ) ( 4.44 × 10−20 kWh/MeV ) ⎛ 6.02 × 1023 ⎞ ⎛ 000 g ⎞ En = ⎜ (17.6 MeV ) mol ⎟⎠ ⎜⎝ 2.014 g/mol ⎟⎠ ⎝ × ( 4.44 × 10−20 kWh/MeV )       = 2.34 × 108 kWh (d) ⎛ 4.20 eV ⎞ En = ( the number of C atoms in 1.00 kg ) × ⎜ ⎝ kg ⎟⎠ ⎛ 6.02 × 1026 ⎞ En = ⎜ 4.20 × 10−6 MeV ) ( 4.44 × 10−20 kWh/MeV ) ( ⎟ ⎝ 12 g ⎠ = 9.36 kWh (e) P45.65 (a) Coal is cheap at this moment in human history We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels Burning coal in the open puts carbon dioxide into the atmosphere, worsening global warming Plutonium is a very dangerous material to have sitting around We have 1.00 kg – (1.00 kg)(0.007 20) – (1.00 kg)(0.000 0500) = 0.993 kg of 238U, comprising ⎛ 6.02 × 1023 nuclei ⎞ ⎛ mol ⎞ N = ( 0.993 kg ) ⎜ ⎟⎠ ⎜⎝ 0.238 kg ⎟⎠ ⎝ mol = 2.51 ì 1024 nuclei â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1183 with activity R = λN = ln ( 2.51 × 1024 nuclei ) 4.47 × 109 yr yr Ci ⎛ ⎞⎛ ⎞ ×⎜ ⎟ ⎜ 10 −1 ⎝ 3.16 × 10 s ⎠ ⎝ 3.70 × 10 s ⎟⎠ = 3.3 × 10−4 Ci = 330 µCi We have (1.00 kg)(0.007 20) = 0.007 kg of 235U, comprising ⎛ 6.02 × 1023 nuclei ⎞ ⎛ mol ⎞ N = ( 0.007 kg ) ⎜ ⎟⎠ ⎜⎝ 0.235 kg ⎟⎠ ⎝ mol = 1.84 × 1022 nuclei with activity R = λN = ln 1.84 × 1022 nuclei ) ( 7.04 × 10 yr yr Ci ⎛ ⎞⎛ ⎞ ×⎜ ⎟ ⎜ ⎝ 3.16 × 10 s ⎠ ⎝ 3.70 × 1010 s −1 ⎟⎠ = 1.6 ì 105 Ci = 16 àCi We have (1.00 kg)(0.000 0500) = 5.00 × 10–5 kg of 234U, comprising ⎛ 6.02 × 1023 nuclei ⎞ ⎛ mol ⎞ N = ( 5.00 × 10−5 kg ) ⎜ ⎟⎠ ⎜⎝ 0.234 kg ⎟⎠ ⎝ mol = 1.29 × 1020 nuclei with activity R = λN = ln (1.29 × 1020 nuclei ) 2.44 × 105 yr yr Ci ⎛ ⎞⎛ ⎞ ×⎜ ⎟ ⎜ ⎝ 3.16 × 10 s ⎠ ⎝ 3.70 × 1010 s −1 ⎟⎠ = 3.1 × 10−4 Ci = 310 µCi (b) (c) The total activity is (330 + 16 + 310) µCi = 656 µCi, so the fractional contributions are, respectively, 330/656 = 50%, 16/656 = 2.4%, and 310/656 = 47% It is dangerous, notably if the material is inhaled as a powder With precautions to minimize human contact, however, microcurie sources are routinely used in laboratories © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1184 P45.66 Applications of Nuclear Physics (a) The number of molecules in 1.00 liter of water (mass = 000 g) is ⎛ 1.00 × 103 g ⎞ N=⎜ 6.02 × 1023 molecules mol ) ( ⎟ 18.0 g mol ⎝ ⎠ = 3.34 × 1025 molecules The number of deuterium nuclei contained in these molecules is ⎛ deuteron ⎞ N ′ = ( 3.34 × 1025 molecules ) ⎜ ⎝ 3 300 molecules ⎟⎠ = 1.01 × 1022 deuterons Since deuterons are consumed per fusion event, the number of N′ events possible is = 5.07 × 1021 reactions, and the energy released is Efusion = ( 5.07 × 1021 reactions ) ( 3.27 MeV reaction ) = 1.66 × 1022 MeV Efusion = ( 1.66 × 1022 MeV ) ( 1.60 × 10−13 J MeV ) = 2.65 × 10 J (b) In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is Efusion 2.65 × 109 J = = 78.0 times larger Egasoline 3.40 × 107 J P45.67 (a) At × 108 K, the average kinetic energy of a carbon atom is kBT = ( 1.5 ) ( 8.62 × 10−5 eV K ) ( × 108 K ) = × 10 eV Note that × 108 K is about 62 = 36 times larger than 1.5 × 107 K, the core temperature of the Sun This factor corresponds to the higher potential-energy barrier to carbon fusion compared to hydrogen fusion It could be misleading to compare it to the temperature ~ 108 K required for fusion in a low-density plasma in a fusion reactor (b) The energy released is Q = ⎡⎣ 2MC12 − MNe20 − MHe4 ⎤⎦ c Q = [ ( 12.000 000 u ) − 19.992 440 u − 4.002 603 u ] × ( 931.5 MeV u ) = 4.62 MeV © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1185 In the second reaction, Q = ⎡ 2MC12 − MMg24 ⎤ c ⎣ ⎦ Q = [ ( 12.000 000 u ) − 23.985 042 u ]( 931.5 MeV u ) = 13.9 MeV (c) The energy released is the energy of reaction of the number of carbon nuclei in a 2.00-kg sample, which corresponds to ⎛ 6.02 × 1023 atoms mol ⎞ ΔE = ( 2.00 × 10 g ) ⎜ ⎟⎠ 12.0 g mol ⎝ ⎛ 4.62 MeV fusion event ⎞ ⎛ kWh ⎞ ×⎜ ⎜⎝ ⎟ 19 ⎟ ⎝ nuclei fusion event ⎠ 2.25 × 10 MeV ⎠ (1.00 × 10 )( 4.62 ) kWh = ΔE = ( 2.25 × 10 ) 26 19 P45.68 1.03 × 107 kWh From Table 44.2 of isotopic masses, the half-life of 32P is 14.26 d Thus, the decay constant is λ= ln ln = = 0.048 d −1 = 5.63 × 10−7 s −1 T1 14.26 d and the initial number of nuclei is N0 = R0 5.22 × 106 decay s = = 9.28 × 1012 nuclei −7 −1 λ 5.63 × 10 s At t = 10.0 days, the number remaining is N = N e − λ t = ( 9.28 × 1012 nuclei ) exp ⎡⎣ − ( 0.048 d −1 ) ( 10.0 d ) ⎤⎦ = 5.71 × 1012 nuclei so the number of decays has been N0 – N = 3.57 × 1012 and the energy released is E = ( 3.57 × 1012 ) ( 700 keV ) ( 1.60 × 10−16 J keV ) = 0.400 J If this energy is absorbed by 100 g of tissue, the absorbed dose is ⎛ 0.400 J ⎞ ⎛ rad ⎞ Dose = ⎜ = 400 rad −2 ⎝ 0.100 kg ⎟⎠ ⎜⎝ 10 J kg ⎟⎠ P45.69 (a) The thermal power transferred to the water is Pw = 0.970 (waste heat): Pw = 0.970 ( 065 MW − 000 MW ) = 2.00 ì 109 J s â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1186 Applications of Nuclear Physics rw is the mass of water heated per hour: 2.00 × 109 J s ) ( 3600 s h ) ( Pw rw = = = 4.92 × 108 kg h c ( ΔT ) ( 4186 J kg ⋅ °C ) ( 3.50 °C ) Then, the volume used per hour is 4.91 × 108 kg h = 4.92 × 105 m h 1.00 × 103 kg m (b) The 235U fuel is consumed at a rate ⎛ 065 × 106 J s ⎞ ⎛ kg ⎞ ⎛ 600 s ⎞ rf = ⎜ ⎜ ⎟ = 0.141 kg h 10 ⎝ 7.80 × 10 J g ⎟⎠ ⎜⎝ 000 g ⎟⎠ ⎝ h ⎠ P45.70 We add two electrons to both sides of the given reaction Then, 11 H atom → 24 He atom + 2ν , where Q = ( Δm) c = ⎡⎣ ( 1.007 825 u ) − 4.002 603 u ⎤⎦ ( 931.5 MeV u ) = 26.7 MeV or Q = ( 26.7 MeV ) ( 1.60 × 10−13 J MeV ) = 4.28 × 10−12 J The proton fusion rate is then rate = power output 3.85 × 1026 J s = energy per proton ( 4.28 × 10−12 J ) ( protons ) = 3.60 × 1038 protons s Challenge Problems P45.71 The initial specific activity of 59Fe in the steel is ⎛ 20.0 àCi 100 àCi 3.70 ì 10 Bq =⎜ ( R m)0 = 0.200 ⎟ kg kg àCi = 3.70 ì 106 Bq kg The decay constant of 59Fe is λ = ln ⎛ d ⎞ ⎜ ⎟ 45.1 d ⎝ 24 h ⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1187 After 000 h, the activity is R ⎛ R ⎞ −λ t =⎜ ⎟ e m ⎝ m⎠ ⎡ ⎛ ln ⎞ ⎛ d ⎞ ⎤ = ( 3.70 × 106 Bq kg ) exp ⎢ − ⎜ 1 000 h ) ⎥ ( ⎟ ⎜ ⎟ ⎣ ⎝ 45.1 d ⎠ ⎝ 24 h ⎠ ⎦ = 1.95 × 10 Bq kg The activity of the oil is ⎛ 800 ⎞ Roil = ⎜ Bq liter ⎟ (6.50 liters) = 86.7 Bq ⎝ 60.0 ⎠ Therefore, oil = 86.7 Bq Roil = = 4.44 × 10−5 kg ( R m) 1.95 × 10 Bq kg So that the wear rate is P45.72 (a) 4.45 × 10−5 kg = 4.44 × 10−8 kg h 000 h The number of fissions occurring in the zeroth, first, second, …, nth generation is N , N K, N K , …, N K n The total number of fissions that have occurred up to and including the nth generation is N = N + N K + N K + …+ N K n = N ( + K + K + …+ K n ) Note that the factoring of the difference of two squares, a2 – = (a + 1) (a – 1), can be generalized to a difference of two quantities to any power, a − = ( a + a + 1) ( a − 1) a n+1 − = ( a n + a n−1 + …+ a + a + 1) ( a − 1) Thus, K n + K n−1 + …+ K + K + = and (b) N = N0 K n+1 − K−1 K n+1 − K−1 The number of U-235 nuclei is ⎞ 1u ⎛ atom ⎞ ⎛ N = ( 5.50 kg ) ⎜ = 1.41 × 1025 nuclei ⎟ −27 ⎜ ⎟ ⎝ 235 u 1.66 ì 10 kg â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1188 Applications of Nuclear Physics We solve the equation from part (a) for n, the number of generations: N ( K − 1) = K n+1 − N0 N ( K − 1) + = K n ( K ) N0 ⎛ N ( K − 1) ⎞ ⎛ N ( K − 1) N + ⎞ nln K = ln ⎜ = ln ⎜ + 1⎟ − ln K ⎟ ⎝ ⎠ K ⎝ N0 ⎠ ln ( 1.41 × 1025 ( 0.1) 1020 + 1) ln ( N ( K − 1) N + 1) n= −1= −1 ln K ln 1.1 = 99.2 Therefore time must be allotted for 100 generations: Δtb = 100 ( 10 × 10−9 s ) = 1.00 × 10−6 s = 1.00 µs (c) The speed of sound in uranium is v= 150 × 109 N m B = = 2.83 × 103 m s = 2.83 km/s 3 ρ 18.7 × 10 kg m (d) From the definitions of volume and mass density, V = m πr = , ρ and ⎛ 3m ⎞ r=⎜ ⎝ 4πρ ⎟⎠ 13 ⎛ ⎞ ( 5.5 kg ) =⎜ 3 ⎟ ⎝ 4π ( 18.7 × 10 kg m ) ⎠ 13 = 4.13 × 10−2 m then, the time interval is given by Δtd = (e) r 4.13 × 10−2 m = = 1.46 × 105 s = 14.6 às v 2.83 ì 10 m s 14.6 µs is greater than µs, so the entire bomb can fission The destructive energy released is (1.41 × 10 25 ⎛ 200 × 106 eV ⎞ ⎛ 1.60 × 10−19 J ⎞ nuclei ) ⎜ ⎟⎠ eV ⎝ fissioning nucleus ⎟⎠ ⎜⎝ ⎛ ton TNT ⎞ = 4.51 × 1014 J ⎜ ⎝ 4.20 × 109 J ⎟⎠ = 1.08 × 105 ton TNT = 108 kilotons of TNT © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1189 What if? If the bomb did not have an “initiator” to inject 1020 neutrons at the moment when the critical mass is assembled, the number of generations would be n= ln ( 1.41 × 1025 ( 0.1) + 1) ln 1.1 ( − = 582.4 ) requiring 583 10 × 10−9 s = 5.83 µs This time is not very short compared with 14.6 µs, so this bomb would likely release much less energy P45.73 (a) EI = 10.0 eVis the energy required to liberate an electron from a dynode Let ni be the number of electrons incident upon a dynode, each having gained energy eΔV as it was accelerated to this dynode The number of electrons that will be freed from this ΔV dynode is N i = ni e EI At the first dynode, ni = and N1 = (b) (1) e (100 V ) = 10.0 eV 101 electrons For the second dynode, ni = N1 = 101, so (101 )e ( 100 V ) N2 = = 102 10.0 eV At the third dynode, ni = N2 = 102 and N3 = (102 )e ( 100 V ) = 103 10.0 eV Observing the developing pattern, we see that the number of electrons incident on the nth dynode is nn = N n −1 = 10 n −1 , so for the seventh and last dynode is n7 = N = 106 (c) The number of electrons incident on the last dynode is n7 = 106 The total energy these electrons deliver to that dynode is given by E = ni e ( ΔV ) = 106 e ( 700 V − 600 V ) = 10 eV © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1190 Applications of Nuclear Physics ANSWERS TO EVEN-NUMBERED PROBLEMS P45.2 0.403 g P45.4 2.63 kg/d P45.6 (a) 173 MeV; (b) 0.078 8% P45.8 26 P45.10 (a) × 1015 g; (b) ×103 yr; (c) The uranium comes from rocks and minerals dissolved in water and carried into the ocean by rivers; (d) No P45.12 (a) 16.2 kg; (b) 117 g P45.14 (a) 4.84V−1/3; (b) 6V−1/3; (c) 6.30V−1/3; (d) The sphere has minimum leakage and the parallelepiped has minimum leakage P45.16 6.25 × 1019 Bq P45.18 By balancing the equation for electric charge and number of nucleons, we find that n = If one incoming neutron results in just one outgoing neutron, the possibility of a chain reaction is not there, so this nuclear reactor will not work P45.20 (a) 4.56 × 10−24 kg ⋅ m/s; (b) 0.145 nm; (c) This size has the same order of magnitude as an atom’s outer electron cloud, and is vastly larger than a nucleus P45.22 3.07 × 1022 events P45.24 (a) E = 144Z1Z2 where E is in keV; (b) The energy is proportional to each atomic number; (c) Take Z1 = and Z2 = 59 or vice versa This choice minimizes the product Z1 Z2; (d) 144 keV for both, according to this model P45.26 (a) 3.24 fm; (b) 444 keV; (c) P45.28 (a) 1.66 × 107 J; (b) 6.45 kg P45.30 (a) 137 N ; (b) 136 C ; (c) 147 N ; (d) 158 O ; (e) 157 N ; (f) 126 C ; (g) The original carbon-12 nucleus is returned so the overall reaction is 11 H   →    24 He vi ; (d) 740 keV; (e) possibly by tunneling ( ) P45.32 (a) 2.5 mrem/x-ray; (b) The technician’s occupational exposure is high compared to background radiation; it is 38 times 0.13 rem/yr © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 45 1191 ln f ln ( ) ; (b) − µ µ P45.34 (a) P45.36 18.8 J P45.38 It would take over 24 days to raise the temperature of the water to 100°C and even longer to boil it, so this technique will not work for a 20-minute coffee break! P45.40 1.14 rad P45.42 3.96 × 10–4 J/kg P45.44 (a) See P45.44(a) for full explanation; (b) P45.46 (a) 8.68 MeV; (b) The particles must have enough kinetic energy to overcome their mutual electrostatic repulsion so that they can get close enough to fuse P45.48 (a) 103 Pa; (b) × 109 m3; (c) × 1012 J; (d) ~1014 J; (e) ~ 104 ton TNT P45.50 (a) 27.6 min; (b) 30 ± 27% P45.52 (a) See P45.52(a) for full explanation; (b) 177 MeV; (c) KBr = 112.0 MeV, KLa = 65.4 MeV; (d) vBr = 15.8 Mm/s, vLa = 9.30 Mm/s P45.54 232 yr P45.56 482 Ci, less than the fission inventory by on the order of a hundred million times P45.58   P45.60 (a) 15.4 cm; (b) 51.7 MeV; (c) The number of decays per second is the decay rate R, and the energy released in each decay is Q Then the energy released per unit time interval is P = QR; (d) 2.27 × 105 J/yr; (e) 3.18 J/yr P45.62 (a) See P45.62(a) for full explanation; (b) 35.2; (c) 2.89 × 1015 P45.64 (a) 2.24 × 107 kWh; (b) 17.6 MeV for each D-T fusion; (c) 2.34 × 108 kWh; (d) 9.36 kWh; (e) Coal is cheap at this moment in human history We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels Burning coal in the open puts carbon dioxide into the atmosphere, worsening global warming Plutonium is a very dangerous material to have sitting around P45.66 (a) 2.65 × 109 J; (b) 78.0 times larger R λ mN A E MU-235 ⎡⎣ cw ( 100 − Tc ) + Lv  + c s (Th  – 100 ) ⎤⎦ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1192 Applications of Nuclear Physics P45.68 400 rad P45.70 3.60 × 1038 protons/s P45.72 (a) See P45.72(a) for full explanation; (b) 1.00 µ s; (c) 2.83 km/s; (d) 14.6 µ s; (e) 108 kilotons of TNT © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part