f5ch44 - CHAPTER 44 ENERGY FROM THE NUCLEUS 1173 CHAPTER 44...

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Unformatted text preview: CHAPTER 44 ENERGY FROM THE NUCLEUS 1173 CHAPTER 44 Answer to Checkpoint Questions 1. 2. 3. (c) and (d) (a) no; (b) yes; (c) no (e) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. (a) (a) (b) (b), (e), (a), (c), (d) (a) 93 Sr; (b) 140 I; (c) 155 Nd (c) (c) (c), (a), (d), (b) (a) (d) (c) (c) Solutions to Exercises & Problems (a) The mass of a single atom of 235 U is (235 u)(1:661 10 27 kg/u) = 3:90 10 the number of atoms in 1:0 kg is (1:0 kg)=(3:90 10 25 kg) = 2:6 1024 . 1E 25 kg, so 1174 CHAPTER 44 ENERGY FROM THE NUCLEUS (b) The energy released by N ssion events is given by E = NQ, where Q is the energy released in each event. For 1:0 kg of 235 U, E = (2:561024 )(200106 eV)(1:6010 19 J/eV) = 8:2 1013 J. (c) If P is the power requirement of the lamp, then t = E=P = (8:19 1013 J)=(100 W) = 8:19 1011 s = 26; 000 y. The conversion factor 3:156 107 s/y was used to obtain the last result. 2E The energy released is (1:00 E = (239 u)(1:kg)(180 MeV) = 4:54 1026 MeV : 66 10 27 kg/u) If R is the ssion rate, then the power output is P = RQ, where Q is the energy released in each ssion event. Hence R = P=Q = (1:0 W)=(200 106 eV)(1:60 10 19 J/eV) = 3:1 1010 ssions/s. Use the conservation of A, N and Z . The missing entries are (by rows): 95 Sr, 95 Y, 134 Te, and 3. 5E 4E 3E At T = 300 K the average kinetic energy of the neutrons is 3 3 K = 2 kT = 2 (8:62 10 5 eV/K)(300 K) 0:04 eV : If mCr is the mass of a 52 Cr nucleus and mMg is the mass of a 26 Mg nucleus, then the disintegration energy is Q = (mCr 2mMg ) c2 = [51:94051 u 2(25:98259 u)] (931:5 MeV/u) = 23:0 MeV. 7E 6E Consider the process 98 Mo ! 49 Sc + 49 Sc. We have Q = (mMo 2mSc )c2 = [97:90541 u 2(48:95002 u)](932 MeV/u) = +5:00 MeV: CHAPTER 44 ENERGY FROM THE NUCLEUS 1175 8E The energy released is Q = (mU + mn mCs mRb 2mn )c2 = (235:04392 u 1:00867 u 140:91963 u 92:92157 u)(932 MeV/u) = 181 MeV : 9E (a) M R ss = ss N = nln 2 = m ln 2 ss U ss 23 g/mol) ln 2 (1 0 g)(6:02 10 = (235 :g/mol)(3: 1017 y)(365 d/y) = 16 ssions/day : 0 (b) The ratio is R = ss = 3:0 1017 y = 4:3 108 : R ss 7:0 108 y 10P The mass of 235 U in M = 1:0 kg of UO2 is :0%M 238 + M235 = 397%m (97%m0%m 3:0%m235 ) 238 + 3: 235 + 2m16 0 030(235)] = (3:0%)(1:0 kg)[0::97(238) ++ :2(16:0) = 0:02644 kg : 0:97(238) + 0 030(235) The total energy released is then 23 E = (0:02644 kg)(6:02 : 10 ==1mol)(200 MeV) = 2:17 1012 J : (0:235 kg/mol)(1 00 eV :60 10 19 J) The time this much energy can keep can keep a 100-W lamp burning is then 12 J 2:17 t = E = (100 W)(3: 10 107 s/y) = 690 y : P 15 11P (a) Consider the process 239 U + n !140 Ce + 99 Ru + Ne. We have N = Zf ZRu ZU = 58 + 44 92 = 10. So the number of beta-decay events is 10. Zi = ZCe + 1176 CHAPTER 44 ENERGY FROM THE NUCLEUS (b) Q = (mU + mn mCe mRu 10me )c2 = [238:05079 u + 1:00867 u 139:90543 u 98:90594 u 10(0:511 MeV)](932 MeV/u) = 226 MeV : 12P (a) If X represents the unknown fragment, then the reaction can be written 235 U + 1 n ! 83 Ge + A X ; 92 0 32 Z where A is the mass number and Z is the atomic number of the fragment. Conservation of charge yields 92 + 0 = 32 + Z , so Z = 60. Conservation of mass number yields 235 + 1 = 83 + A, so A = 153. Look in Appendix F or G for nuclides with Z = 60. You should nd that the unknown fragment is 153 Nd. 60 (b) Ignore the small kinetic energy and momentum carried by the neutron that triggers the ssion event. Then Q = KGe + KNd , where KGe is the kinetic energy of the germanium nucleus and KNd is the kinetic energy of the neodymium nucleus. Conservation of momentum yields pGe + pNd = 0, where pGe is the momentum of the germanium nucleus and pNd is the momentum of the neodymium nucleus. Since pNd = pGe , the kinetic energy of the neodymium nucleus is 2 2 m KNd = 2pNd = 2pGe = mGe KGe : m m Thus the energy equation becomes and Similarly, Nd Nd Nd m Q = KGe + mGe KGe = mNd + mGe KGe m Nd Nd Nd KGe = m m+ m Q = 153153 u83 u (170 MeV) = 110 MeV : u+ Nd Ge (c) The initial speed of the germanium nucleus is 83 u Ge KNd = m m+ m Q = 153 u + 83 u (170 MeV) = 60 MeV : Nd Ge 2KGe = m Ge s vGe = r 2(110 106 eV)(1:60 10 19 J/eV) = 1:60 107 m/s : (83 u)(1:661 10 27 kg=u) 2(60 106 eV)(1:60 10 19 J/eV) = 8:7 106 m/s : (153 u)(1:661 10 27 kg=u) The initial speed of the neodymium nucleus is vNd = r 2KNd = m Nd s CHAPTER 44 ENERGY FROM THE NUCLEUS 1177 13P (a) The electrostatic potential energy is given by 2 1 Z U = 4 R Xe ZSr e ; 0 Xe + RSr where ZXe is the atomic number of xenon, ZSr is the atomic number of strontium, RXe is the radius of a xenon nucleus, and RSr is the radius of a strontium nucleus. Atomic numbers can be found in Appendix F. The radii are given by R = (1:2 fm)A1=3 , where A is the mass number, also found in Appendix F. Thus RXe = (1:2 fm)(140)1=3 = 6:23 fm = 6:2310 15 m and RSr = (1:2 fm)(96)1=3 = 5:49 fm = 5:46 10 15 m. Hence the potential energy is C) U = (8:99 109 m/F) 6:23(54)(38)(1:60 :10 10 15 m = 4:08 10 15 m + 5 49 10 19 2 11 J : This is 252 MeV. (b) The energy released in a typical ssion event is about 200 MeV, roughly the same as the electrostatic potential energy when the fragments are touching. The energy appears as kinetic energy of the fragments and neutrons produced by ssion. 14P (a) The surface area a of a nucleus is given by a ' 4R2 ' 4[R0 A1=3 ]2 / A2=3 . Thus the fractional change in surface area is a = af ai = (140)3=2 + (96)2=3 ai ai (236)2=3 (b) Since V / R3 / (A1=3 )3 = A, we have V = Vf V Vi (c) U = Uf 1 = Q2 =RXe + Q2 =RSr 1 Xe Sr U Ui Q2 =RU U 2 (140) 1=3 + (38)2 (96) 1=3 = (54) 1 = 36% : (92)2 (236) 1=3 1 = 140 + 96 1 = 0 : 236 1 = +25% : If P is the power output, then the energy E produced in the time interval t (= 3 y) is E = P t = (200 106 W)(3 y)(3:156 107 s=y) = 1:89 1016 J, or (1:89 1016 J)=(1:60 10 19 J/eV) = 1:18 1035 eV = 1:18 1029 MeV. At 200 MeV per event, this means 15E 1178 CHAPTER 44 ENERGY FROM THE NUCLEUS (1:181029 )=(200 MeV) = 5:921026 ssion events occurred. This must be half the number of ssionable nuclei originally available. Thus there were 2(5:92 1026 ) = 1:18 1027 nuclei. The mass of a 235 U nucleus is (235 u)(1:661 10 27 kg/u) = 3:90 10 25 kg, so the total mass of 235 U originally present was (1:18 1027 )(3:90 10 25 kg) = 463 kg. 16E From Fig. 44-4 we see that about 330 per 1,330 neutrons produced by 235 U do not result in ssion. So the mass of 235 U should be (see 15E) (1330=1000)(463 kg) = 617 kg: 17E When a neutron is captured by 237 Np it gains 5:0 MeV, more than enough to oset the 4:2 MeV required for 238 Np to ssion. So 237 Np is ssionable by thermal neutrons. If R is the decay rate then the power output is P = RQ, where Q is the energy produced by each alpha decay. Now R = N = N ln 2= , where is the disintegration constant and is the half-life. The relationship = (ln 2)= was used. If M is the total mass of material and m is the mass of a single 238 Pu nucleus, then 1:00 N = M = (238 u)(1:661 kg m 10 Thus 27 kg/u) 18P = 2:53 1024 : NQ ln 2 = (2:53 1024 )(5:50 106 eV)(1:60 10 19 J/eV)(ln 2) = 558 W : P= (87:7 y)(3:156 107 s/y) 19P (a) Solve Qe from P = RQe : P mP Qe = P = N = M ln 2 R y)(3:15 7 66 10 27 kg/u)(0 = (90:0 u)(1::00 10 3 g)(ln 2)(1::93 W)(29 13 J/MeV)10 s/y) (1 60 10 = 1:2 MeV : (b) The amount of 90 Sr needed is 150 W M = (5:0%)(0:93 W/g) = 3:2 kg : CHAPTER 44 ENERGY FROM THE NUCLEUS 1179 20P The reaction series is as follows: 21P 238 U + n !239 U + 239 Np + e, 239 Np !239 Pu + e. (a) The TNT equivalent is (2:50 kg)(4:54 1026 MeV/kg) = 4:4 104 tons = 44 kton : 2:6 1028 MeV=106 tons (b) The total mass of plutonium in the bomb must exceed the critical mass to start the chain reaction. (a) The energy yield of the bomb is E = (66 10 3 Mton)(2:6 1028 MeV/Mton) = 1:72 1027 MeV. At 200 MeV per ssion event, (1:72 1027 MeV)=(200 MeV) = 8:58 1024 ssion events take place. Since only 4:0% of the 235 U nuclei originally present undergo ssion, there must have been (8:58 1024 )=(0:040) = 2:14 1026 nuclei originally present. The mass of 235 U originally present was (2:14 1026 )(235 u)(1:661 10 27 kg/u) = 83:7 kg. (b) Two fragments are produced in each ssion event so the total number of fragments is 2(8:58 1024 ) = 1:72 1025 . (c) One neutron produced in a ssion event is used to trigger the next ssion event, so the average number of neutrons released to the environment in each event is 1:5. The total number released is (8:58 1024 )(1:5) = 1:3 1025 . After each time interval tgen the number of nuclides in the chain reaction gets multiplied by k. The number of such time intervals that has gone by at time t is t=tgen . Thus the number of nuclides engaged in the chain reaction at time t is N (t) = N0 kt=tgen . Since P / N we have 23P 22P P (t) = P0 kt=tgen : Let P0 be the initial power output, P be the nal power output, k be the multiplication factor, t be the time for the power reduction, and tgen be the neutron generation time. Then according to the result of 23P P = P0 kt=tgen . Divide by P0 , then take the natural logarithm of both sides of the equation and solve for ln k. You should obtain 24P P ln k = tgen ln P : t 0 1180 CHAPTER 44 ENERGY FROM THE NUCLEUS Hence where k = e ; 10 3 P 350 = tgen ln P = 1:3 :6 s s ln 1200MW = 6:161 10 4 : t 2 MW 0 This yields k = 0:99938. 25P From E = Ptgen = NQ we get the number of free neutrons: (500 106 W)(1:0 10 3 s) N = Ptgen = (181 MeV)(1:60 10 13 J/MeV) = 1:7 1016 : Q Use the formula from 23P: P (t) = P0 kt=tgen = (400 MW)(1:0003)(5:00 min)(60 s/min)=(30:0 ms) = 8:03 103 MW : 26P 27P (a) mnvni neutron before collision atom m nvnf neutron atom after collision mu The solution to u is u = 2mn vni =(mn + m). Thus the fractional loss in neutron kinetic energy is Kn = K n 1 mu2 2 1 mn v 2 ni 2 Apply the conservation of energy and momentum: 8 < mn vni = mn vnf + mu 1 2 1 2 1 : mn v 2 = ni 2 mn vnf + 2 mu : 2 m = m n u vni 2 m = m n 2mn mn + m 2 4 n = (m mmm)2 : + n CHAPTER 44 ENERGY FROM THE NUCLEUS 1181 (b) For hydrogen for deuterium for carbon and for lead Kn = 4(1:0 u)(1:0 u) = 1:0 ; Kn (1:0 u + 1:0 u)2 Kn = 4(2:0 u)(1:0 u) = 0:89 ; Kn (1:0 u + 2:0 u)2 Kn = 4(12 u)(1:0 u) = 0:28 ; Kn (1:0 u + 12 u)2 Kn = 4(207 u)(1:0 u) = 0:019 : Kn (1:0 u + 207 u)2 (c) The number of collisions, n, must satisfy K = K0 (1 Kn =Kn )n . Solve for n: 1 00 ) n = ln(1ln(K=K0=K ) = ln(0:025 eV=0::89)MeV) 8 : Kn n ln(1 28E See Sample Problem 44-3. We have N5 (t) = N5 (0) e (5 8 )t ; N8 (t) N8 (0) or 1 ln N5 (t) N8 (0) t= N8 (0) N5 (0) 8 5 1 = (1:55 9:85)10 10 y 1 ln[(0:0072)(0:15) 1 ] = 3:6 109 y : 29E (a) P av = (15 109 W y)=(200; 000 y) = 7:5 104 W = 75 kW: (b) 27 kg/u)(15 9 W y)(3 7 M = mU Etotal = (235 u)(1:66 10 MeV)(1:6 1010 J/MeV):15 10 s/y) 13 Q (200 = 5:8 103 kg : Let t be the present time and t = 0 be the time when the ratio of 235 U to 238 U was 3:0%. Let N235 be the number of 235 U nuclei present in a sample now and N235; 0 be the number 30P 1182 CHAPTER 44 ENERGY FROM THE NUCLEUS present at t = 0. Let N238 be the number of 238 U nuclei present in the sample now and N238; 0 be the number present at t = 0. The law of radioactive decay holds for each specie, so N235 = N235; 0 e 235 t and N238 = N238; 0 e 238 t : Divide the rst equation by the second to obtain r = r0 e (235 238 )t ; where r = N235 =N238 (= 0:0072) and r0 = N235; 0 =N238; 0 (= 0:030). Solve for t: to obtain t = 1 ln rr : 235 238 0 Now use 235 = (ln 2)=235 and 238 = (ln 2)=238 , where 235 and 238 are the half-lives, t = ( 235238) ln 2 ln rr 238 235 0 8 y)(4:5 109 y) (7:0 10 0:0072 = 1:7 109 y : = (4:5 109 y 7:0 108 y) ln 2 ln 0:030 31P The nuclei of 238 U can capture neutons and beta-decay. With large amount of neutrons available due to the ssion of 235 U the probability for this process is substantially increased, resulting in a much higher decay rate for 238 U, causing the depletion of 238 U and relative enrichment of 235 U. 32E The height of the Coulomb barrier is taken to be the value of the kinetic energy K each nucleus must initially have if they are to come to rest when their surfaces touch. If R is the radius of a nucleus, conservation of energy yields 1 e2 2K = 4 2R ; 0 1 e2 = (8:99 109 m/F) (1:60 10 19 C)2 = 2:74 10 14 J : K = 4 4R 4(0:80 10 15 m) 0 This is 171 keV. (Note: a useful formula, which you can verify yourself, is e2 =40 = 1:44 fm MeV, which leads to the answer to this excercise immediately.) so CHAPTER 44 ENERGY FROM THE NUCLEUS 1183 33E (a) 30 MeV. (b) 6 MeV. Both the answers above can be found in the text. The total energy released in the fusion of M = 1:0 kg of deuterium is (1:0 kg)(3: 26 E = MQ = 2(2:0 u)(1:66 27 MeV) 2md 10 27 kg/u) = 4:92 10 MeV : Thus the time t this much energy could keep a 100-W light bulb burning is 26 13 t = E = (4:92 10 MeV)(1:60 10s/y) MeV/J) 3 104 y : P (100 W)(3:15 107 35E 34E (a) The voltage V required satis es 2eV = U = 170 keV, or V = 170 kV. (b No, it is not dicult to achieve. (c) Only a very small amount of deuterons may be accelerated in a beam. The amount of energy generated would not be enough for a power plant. 36P From the expression for n(K ) given we may write n(K ) / K 1=2 e K=kT . Thus n(K ) = K 1=2 e (K K )=kT ) K n(K 5:00 keV 1=2 e (5:00 keV 1:94 keV)=[(8:6210 = 1:94 keV = 0:151 : 37P 5 eV/K)(1:50107 K)] The radius of each lithium nucleus is R = R0 A1=3 . Thus the minimum kinetic energy each nucleus must have to overcome the Coulomb barrier is 2 1 1 1 ( 2 K = 1 U = 2 4" (Ze) = 4" 4RZe)1=3 2 0 2R 0 0A 19 C)]2 (8:99 109 N m2 =C2 ) = [3(1:60 10 4(1:2 10 15 m)(7)1=3 (1:60 10 13 J/MeV) = 1:41 MeV : 1184 CHAPTER 44 ENERGY FROM THE NUCLEUS (a) The energy distibution is given in the expression for n(K ) in 36P. Set dn(K )=dK = 0 for Kp : 1:13n K 3=2 e K=kT dn(K ) 1 = 0; dK K =Kp = (kT )3=2 2K 1=2 kT K =Kp which gives Kp = 1 kT . Numerically for T = 1:5 107 K we nd 2 1 1 Kp = 2 kT = 2 (8:62 10 5 eV/K)(1:5 107 K) = 0:65 keV ; in good agreement with Fig. 44-10. (b) Recall from Chapter 20 that 38P vp = s r 2RT = M r 2RT = mN A r 2kT ; m where k = R=NA was used. At T = 1:5 107 K 7 10 23 :5 vp = 2(1:38 1:67 J/K)(1kg 10 K) = 5:0 105 m/s : 10 27 (c) 1 2 1 K jv=vp = 2 mvp = 2 m r 2kT 2 m = kT : In Fig. 44-11, let Q1 = 0:42 MeV, Q2 = 1:02 MeV, Q3 = 5:49 MeV, and Q4 = 12:86 MeV, then for the overall proton-proton cycle 39E Q = 2Q1 + 2Q2 + 2Q3 + Q4 = 2(0:42 MeV + 1:02 MeV + 5:49 MeV) + 12:86 MeV = 26:7 MeV : 40E If mHe is the mass of an atom of helium and mC is the mass of an atom of carbon, then the energy released in a single fusion event is Q = (3mHe mC ) c2 = [3(4:0026 u) (12:0000 u)] (931:5 MeV/u) = 7:27 MeV : Note that 3mHe contains the mass of 6 electrons and so does mC . The electron masses cancel and the mass dierence calculated is the same as the mass dierence of the nuclei. CHAPTER 44 ENERGY FROM THE NUCLEUS 1185 (a) From H = 35% = np mp we get the proton density np : 5 3 np = 35% = (35%)(1:5 1027kg/m ) = 3:1 1031 m 3 : mp 1:67 10 kg (b) From Chapter 20 we know that n = 2:44 1025 m 3 . Thus np = 3:14 1031 m 3 = 1:3 106 : n 2:44 1025 m 3 42P 41E Q1 = (2mp m2 me )c2 = [2(m1 me ) (m2 me ) me ]c2 = [2(1:007825 u) 2:014102 u 2(0:0005486 u)](932 MeV/u) = 0:42 MeV ; Q2 = (m2 + mp m3 )c2 = (m2 + mp m3 )c2 = (2:014102 u) + 1:007825 u 3:016029 u)(932 MeV/u) = 5:49 MeV ; Q3 = (2m3 + m4 2mp )c2 = (2m3 + m4 2mp )c2 = [2(3:016029 u) + 4:002603 u 2(1:007825 u)](932 MeV/u) = 12:86 MeV : 43P (a) 26 E2 = M2 Q2 = (235(1:0 kg)(200 MeV) mU u)(1:66 10 27 kg/u) = 5:1 10 MeV : You see that E1 =E2 ' 8. (b) kg)(26: MeV) 1Q E1 = Mm 1 = (1:0:67 107 27 kg) = 4:0 1027 MeV : 4 p 4(1 (a) Let Ms be the mass of the Sun at time t and E be the energy radiated to that time. Then the power output is P = dE=dt = (dMs =dt)c2 , where E = Ms c2 was used. At the present time dMs = P = 3:9 1026 W = 4:33 109 kg/s : dt c2 (3:00 108 m/s)2 44P 1186 CHAPTER 44 ENERGY FROM THE NUCLEUS (b) Assume the rate of mass loss remained constant. Then the total mass loss is M = (dMs =dt) t = (4:3 109 kg/s)(4:5 109 y)(3:156 107 s/y) = 6:15 1026 kg. The fraction lost is Ms = 6:15 1026 kg 4 M + M 2:0 1030 kg + 6:15 1026 kg = 3:1 10 : s s 45P (a) Since two neutrinos are produced per proton-proton cycle, the rate of neutrino producion R satis es P 2(3:9 1026 R = 2Q = (26:7 MeV)(1: 10 W)J/MeV) = 1:8 1038 s 1 : 13 6 (b) 2 38 1 6 4 106 e R; Earth = R 4R2 = (1:8 10 s ) 1::5 1011m des 4 m 2 = 8:2 1028 s 1 : 46P (a) The mass of a carbon atom is (12:0 u)(1:661 10 27 kg/u) = 1:99 10 26 kg, so the number of carbon atoms in 1:00 kg of carbon is (1:00 kg)=(1:99 10 26 kg) = 5:02 1025 . The heat of combustion per atom is (3:3 107 J/kg)=(5:02 1025 atom/kg) = 6:58 10 19 J/atom. This is 4:1 eV/atom. (b) In each combustion event two oxygen atoms combine with one carbon atom, so the total mass involved is 2(16:0 u) + (12:0 u) = 44 u. This is (44 u)(1:661 10 27 kg/u) = 7:31 10 26 kg. Each combustion event produces 6:58 10 19 J so the energy produced per unit mass of reactants is (6:58 10 19 J)=(7:31 10 26 kg) = 9:0 106 J/kg. (c) If the Sun were composed of the appropriate mixture of carbon and oxygen, the number of combustion events that could occur before the Sun burns out would be (2:0 1030 kg)=(7:31 10 26 kg) = 2:74 1055 . The total energy released would be E = (2:74 1055 )(6:58 10 19 J) = 1:80 1037 J. If P is the power output of the Sun, the burn time would be t = E=P = (1:80 1037 J)=(3:9 1026 W) = 4:62 1010 s. This is about 1500 y. 47P (a) The products of the carbon cycle are 2e+ + 2 + 4 He, the same as that of the protonproton cycle. (b) Qcarbon = Q1 + Q2 + + Q6 = (1:92+1:19+7:55+7:30+1:73+4:97) MeV = 24:7 MeV, which is the same as that for the proton-proton cycle: Qp p = 26:7 MeV 2(1:02 MeV) = 24:7 MeV. CHAPTER 44 ENERGY FROM THE NUCLEUS 1187 The mass of the hydrogen in the Sun's core is MH = (35%)(Ms =8). Thus the time it takes for the hydrogen to be entirely consumed is (0:35)(2:0 1030 t = MH = (6:2 1011 kg/s)(3:15 kg=8) s/y) = 5 109 y : R 107 49P 48P (b) The rate is (a) From E = NQ = (M=4mp )Q we get the energy per kilogram of hydrogen consumed to be 13 J/MeV) = 6:3 1014 J/kg : E Q " = M = 4m = (26:2 MeV)(1:60 10 10 kg) 4(1:67 27 p 26 10 W R = P = 6:33:91014 J/kg = 6:2 1011 kg/s : " (c) From Es = Ms c2 we get P = dEs =dt = c2 dMs =dt, or dMs = P = 3:9 1026 W = 4:3 109 kg/s : dt c2 (3:0 108 m/s)2 The reason why R > dMs =dt is because as hydrogens are consumed their mass is mostly turned into that of heliums, which are still in the Sun. (d) The time it takes is (0 10%)(2:0 1030 kg) :10%M t = 0dM =dts = (4:3 :109 kg/s)(3:15 107 y/s) = 1:5 1010 y : s 50P Since the mass of a helium atom is (4:00 u)(1:661 10 27 kg/u) = 6:64 10 27 kg, the number of helium nuclei originally in the star is (4:6 1032 kg)=(6:64 10 27 kg) = 6:92 1058 . Since each fusion event requires three helium nuclei, the number of fusion events that can take place is N = 6:92 1058 =3 = 2:31 1058 . If Q is the energy released in each event and t is the conversion time, then the power output is P = NQ=t and 58 6 eV)(1 19 t = NQ = (2:31 10 )(7:275 10 1030 W:60 10 J/eV) = 5:07 1015 s : P :3 This is 1:6 108 y. 51P (a) Q = (5m2H m3He m4He m1H 2mn )c2 = [5(2:014102 u) 3:016029 u 4:002603 u 1:007825 u 2(1:008665 u)](938 MeV/u) = 24:9 MeV. 1188 CHAPTER 44 ENERGY FROM THE NUCLEUS (b) The total energy released from the fusion is 0% E = NQ = 305:m2 M Q : H Thus the rating is E R = 2:6 1028 MeV/megaton TNT (30:0%)(500 = 5(2:0 u)(1:66 10 27 kg/u)(2kg)(24:9 MeV) :6 1028 MeV/megaton TNT) = 8:65 megaton TNT : 52E In Eq. 44-9 Q = (2m2H m3He mn )c2 = [2(2:014102 u) 3:016049 u 1:008665 u](932 MeV/u) = 3:27 MeV ; in Eq. 44-10 Q = (2m2H m3H m1H )c2 = [2(2:014102 u) 3:016049 u 1:007825 u](932 MeV/u) = 4:03 MeV ; and in Eq. 44-11 Q = (m2H + m3H m3He mn )c2 = (2:014102 u) + 3:016049 u 4:002603 u 1:008665 u](932 MeV/u) = 17:59 MeV : 53P Conservation of momentum gives pHe +pn = 0, and conservation of energy gives p2 =2mHe + He p2 =2mn = Q. Solve for KHe and Kn : n 2 008665 u)(17 59 MeV) KHe = 2pHe = m mn Q = (1::008665 u + 4::002603 u = 3:541 MeV ; mHe 1 n + mHe 2 002603 u)(17 59 MeV) He Q Kn = 2pn = m m+ m = (4::008665 u + 4::002603 u = 14:05 MeV : mn 1 n He CHAPTER 44 ENERGY FROM THE NUCLEUS 1189 Since 1:00 L of water has a mass of 1:00 kg, the mass of the heavy water in 1:00 L is 0:0150 10 2 kg = 1:50 10 4 kg. Since a heavy water molecule contains one oxygen atom, one hydrogen atom and one deuterium atom, its mass is (16:0 u + 1:00 u + 2:00 u) = 19:0 u or (19:0 u)(1:661 10 27 kg/u) = 3:16 10 26 kg. The number of heavy water molecules in a liter of water is (1:50 10 4 kg)=(3:16 10 26 kg) = 4:75 1021 . Since each fusion event requires two deuterium nuclei, the number of fusion events that can occur is N = 4:75 1021 =2 = 2:38 1021 . Each event releases energy Q = (3:27 106 eV)(1:60 10 19 J/eV) = 5:23 10 13 J. Since all events take place in a day, the power output is 21 )(5 23 P = NQ = (2:38 10:64 :104 10 t 8 s 13 J) 54P = 1:44 104 W = 14:4 kW : ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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