Chapter 30 - CHAPTER 30 Quick Quizzes 1. on 4. 109 J t = 1....

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C HAPTER 30 Quick Quizzes 1. (c). The total energy released was ( ) ( ) 39 30 10 ton 4. 01 0 Jton 1. 21 1 4 J E × . The mass equivalent of this quantity of energy is ( ) 14 3 2 2 8 0 J 31 0 kg ~ 1 g 3. ms E m c × == = × × 2. (a). This reaction fails to conserve charge and cannot occur. 3. (b). This reaction fails to conserve charge and cannot occur. 4. (c), (e). The proton and the electron each have spin 1 2 s = . The two possible resultant spins after decay are 1 (spins aligned) or 0 (spins anti-aligned). Neither equal the spin of a neutron, 1 2 s = , so spin angular momentum is not conserved. The are no leptons present before the proposed decay and one lepton (the electron) present after decay. Thus, the decay also fails to conserve lepton number. 441
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CHAPTER 30 Problem Solutions 30.1 1 235 141 92 1 0 56 36 0 n U Ba + Kr + 3n +→ To conserve both charge and the total number of nucleons, it is seen that this reaction must yield 3 neutrons in addition to the other fission products. 30.2 The energy released in the reaction is 12 3 5 9 8 1 3 5 1 40 52 0 Zr + Te + () 98 135 22 UZ r 2 n Qm cm mm m ⎡⎤ =∆ = ⎣⎦ T e c ( ) ( ) 235.043 924 u 2 1.008 665u 97.912 0 u 134.908 7 u 931.5 M eV u =−− = 192 MeV 30.3 The energy released in the reaction is 1 88 136 1 38 54 0 Sr + Xe + 12 n US r 11 n = X e c ( ) ( ) 11 1.008 665 u 87.905 618 u 135.907 215u =− − = 126 30.4 Three different fission reactions are possible: 1 90 144 1 09 23 8 5 4 0 nUS rX e 2 +→+ + n Xe 1 143 1 4 0 e 3 n 3 5 9 0 1 4 2 1 4 0 e 4 n 142 30.5 (a) With a specific gravity of 4.00, the density of soil is 3 4000 kg m ρ = . Thus, the mass of the top 1.00 m of soil is ( ) 2 27 3 kg 1 m 4000 1.00 m 43 560 f t1 . 62 10 kg m3 . 281 ft mV ⎛⎞ == = × ⎢⎥ ⎜⎟ ⎝⎠ 442
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CHAPTER 30 At a rate of 1 part per million, the mass of uranium in this soil is then 7 U 66 1.62 10 kg 10 m m × == = 16.2 kg (b) Since 0.720% of naturally occurring uranium is 235 , the mass of in the soil of part (a) is U U ( ) ( )( ) 33 U U 7.20 0.117 kg mm −− = = 117 g 30.6 At 40.0% efficiency, the useful energy obtained per fission event is ( ) ( ) 13 11 0.400 200 M eV event 1.60 J M eV 1.28 J event event E = × The number of fission events required each day is then ( ) ( ) 94 24 1.00 10 J s 8.64 10 s d 6.75 eventsd t N E ×× ℘⋅ × Each fission event consumes one atom. The mass of this number of atoms is U U atom mN m = ( ) ( ) ( ) 27 events d 235.044 u 1.66 kg u × = 2.63 kg d In contrast, a coal-burning steam plant producing the same electrical power uses more than 6 61 0 k gd × of coal. 30.7 The mass of in 1.0-kg of fuel is 0.017 kg, and the number of nuclei is U U () ( ) 22 0.017 kg 4.35 235.044 u1 .66 m N m × At 208 MeV per fission event and 20% efficiency, the useful energy available from this number of fission events is ( ) ( ) ( ) events 208 M eV event 1.60 J M ev 0.20 2. 91 0 J E ⎡⎤ × = × ⎣⎦ 443
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CHAPTER 30 From , the distance the ship can travel on this 1.0-kg of fuel is drag Work F s E =⋅ = 11 6 5 2. 91 0 J 0 m 1. 01 0 N E s F × == = × = 3 0 km × × (or about 1800 miles) 30.8 The total energy released was ( ) ( ) 39 20 10 ton 4. 0 Jton 8.
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This homework help was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.

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Chapter 30 - CHAPTER 30 Quick Quizzes 1. on 4. 109 J t = 1....

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