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# f5ch43 - CHAPTER 43 NUCLEAR PHYSICS 1147 CHAPTER 43 Answer...

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Unformatted text preview: CHAPTER 43 NUCLEAR PHYSICS 1147 CHAPTER 43 Answer to Checkpoint Questions 1. 2. 3. 90 As and 158 Nd a little more than 75 Bq (elapsed time is a little less than three half-lives) 206 Pb Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. less more neutrons than protons 240 U above less (a) 196 Pt; (b) no (a) on the N = Z line; (b) positrons; (c) about 120 (a) below; (b) below; (c) radioactive no yes yes A and C tie, then B (a) increases; (b) same no eect 7h 209 Po (d) 1148 CHAPTER 43 NUCLEAR PHYSICS 18. (a) all except 198 Au; (b) 132 Sn and 208 Pb Solutions to Exercises & Problems 1E Let K = 5:30 MeV = U = (1=40 )(q qCu =rmin ) and solve for the closest separation, rmin : 19 C)(8 99 9 2 2 q q rmin = 4 Cu = (2e)(29)(1:60 10:30 106:eV 10 N m =C ) 5 0K 14 m = 15:8 fm : = 1:58 10 In order for the particle to penetrate the gold nucleus the closest separation between the centers of mass of the two particles must be no more than r = rCu + r = 6:23 fm+1:8 fm = 8:03 fm. Thus the minimum energy K is given by 1 K = U = 4 q qAu 0 r 9 2 2 )(2e 19 = (8:99 10 N m8:=C 10)(79)(1:60 10 C) = 28:3 MeV : 15 m 03 Apply the conservation of energy and linear momentum to the collision. The results are given in Chapter 10, Eqs. 10-18 and 10-19. The nal speed of the particle is and that of the recoiloing gold nucleus is (a) 3P 2E m vf = m + mAu vi m Au m vAu;f = m 2+ m vi : Au 2 4 1 m v2 = 1 m 2m 2 vi = Ei (m mAu m )2 EAu;f = 2 Au Au;f 2 Au m + m Au + mAu 4(197 u) = (5:00 MeV) (4:00 uu)(4:00u)2 = 0:390 MeV : + 197 CHAPTER 43 NUCLEAR PHYSICS 1149 (b) 1 m v2 = 1 m m mAu 2 v2 = E m mAu Ef = 2 f 2 m + m i m + m i Au Au 2 4 00 u = (5:00 MeV) 4::00 u + 197 u = 4:61 MeV : 197 u 2 Note that Ef + EAu;f = Ei is indeed satis ed. 4E Since M = V / R3 , we get R / (M=)1=3 . Thus the new radius would be s 1=3 = (6:96 108 m) 1410 kg/m3 1=3 = 13 km : R = Rs 2 1017 kg/m3 (a) 6 protons (since Z = 6 for carbon). (b) 8 neutrons (since A Z = 14 6 = 8). 6E 5E Solve for A from Eq. 43-3: R A= R 0 3 3 6 fm = 1::2 fm 3 = 27 : 7E Locate a nuclide by nding the coordinate (N; Z ) of the corresponding point in Fig. 43-4. You will easily nd that all the nuclides listed in Table 43-1 are stable except the last two, 227 Ac and 239 Pu. (a) 142 Nd, 143 Nd, 146 Nd, 148 Nd, 150 Nd. (b) 97 Rb, 98 Sr, 99 Y, 100 Sr, 101 Nd, 102 Mo, 103 Tc, 105 Rh, 109 In, 110 Sn, 111 Sb, 112 Te. (c) 60 Zn, 60 Cu, 60 Ni, 60 Co, 60 Fe. 8E 1150 CHAPTER 43 NUCLEAR PHYSICS 9E (a) For 239 Pu, Q = 94e and R = 6:64 fm so Q2 U = 203" R = 3[94(1::60 10 5(6 64 10 0 = 1:15 109 eV = 1:15 GeV : 19 C)]2 (8:99 109 N m2 =C2 ) 15 m)(1:60 10 19 J/eV) (b) Since Z = 94 and A = 239, the electrostatic potential per nucleon is 1:15 GeV=239 = 4:81 MeV/nucleon and that per proton is 1:15 GeV=94 = 12:2 MeV/proton. These are of the same order of magnitude as the binding energy per nucleon. (c) The bending energy is signi cantly reduced by the electrostatic repulsion among the protons. From Fig. 43-4 we nd the neutron numbers of the two daughter nuclei to be 40 and 74, respectively. The two nuclei are 89 Y and 127 I. The number of neutrons left over is 235 127 89 = 19. 11E 10E N - Z = 15 N - Z = 16 N - Z = 17 N - Z = 18 118 119 120 121 122 52 Te (14) 117 Te (15) 118 Te (16) 119 Te (17) 120 Te (18) 121 N - Z = 19 51 Sb (15) 116 Sb (16) 117 Sb (17) 118 Sb (18) 119 Sb (19) 120 N - Z = 20 A = 121 50 Sn (16) 115 Sn (17) 116 Sn (18) 117 Sn (19) 118 Sn (20) 119 N - Z = 21 A = 120 49 114 In (17) 115 In (18) 116 In (19) 117 In (20) 118 In (21) A = 119 48 Z N Cd (18) Cd (19) Cd (20) A = 115 Cd (21) A = 116 Cd (22) A = 117 A = 118 66 67 68 69 70 CHAPTER 43 NUCLEAR PHYSICS 1151 12E (a) For 55 Mn the mass density is 0:055 m = M = (4=3)[1:2 fm(55)1=kg/mol 1023 =mol) = 2:3 1017 kg/m3 ; 3 ]3 (6:02 V and for 209 Bi 0:209 kg/mol m = M = (4=3)[1:2 fm(209)1=3 ]3 (6:02 1023 =mol) = 2:3 1017 kg/m3 : V (b) For 55 Mn the charge density is 19 (25)(1: q = Ze = (4=3)[162 10 1C)]3 = 1:0 1025 C/m3 ; V : fm(55) =3 and for 209 Bi 19 C) q = Ze = (4(83)(1::6 10 1=3 ]3 = 8:8 1024 C/m3 : V =3)[1 2 fm(209) (c) Since V / R3 = (R0 A1=3 )3 / A, we expect m / A=V / A=A const. for all nuclides, while q / Z=V / Z=A should gradually decrease as A 2Z for large nuclides. The binding energy is given by E = [ZmH + (A Z )mn mPu ] c2 , where Z is the atomic number (number of protons), A is the mass number (number of nucleons), mH is the mass of a hydrogen atom, mn is the mass of a neutron, and mPu is the mass of a 239 Pu atom. In 94 principal, nuclear masses should have been used, but the mass of the Z electrons included in ZmH is canceled by the mass of the Z electrons included in mPu , so the result is the same. First calculate the mass dierence in atomic mass units: m = (94)(1:00783 u) + (239 94)(1:00867 u) (239:05216 u) = 1:94101 u. Since 1 u is equivalent to 931:5 MeV, E = (1:94101 u)(931:5 MeV/u) = 1808 MeV. Since there are 239 nucleons, the binding energy per nucleon is En = (1808 MeV)=(239) = 7:56 MeV. (a) The mass number A is the number of nucleons in an atomic nucleus. Since mp mn the mass of the nucleus is approximately Amp . Also, the mass of the electrons is neglegible since it is much less than that of the nucleus. So M Amp . (b) For 1 H the approximate formula gives M Amp = (1)(1:007276 u) = 1:007276 u. The actual mass is (see Table 47-1) 1:007825 u. The percent error committed is then = (1:007825 u 1:007276 u)=1:007825 u = 0:054%. Similarly = 5:0% for 7 Li, 0:81% for 31 P, 0:83% for 81 Br, 0:81% for 120 Sn, 0:78% for 157 Gd, 0:74% for 197 Au, 0:72% for 272 Ac, and 0:71% for 239 Pu. 14E 13E 1152 CHAPTER 43 NUCLEAR PHYSICS (c) No. In a typical nucleus the binding energy per nucleon is several MeV, which is a bit less than 1% of the nucleon mass times c2 . This is comparable with the percent error calculated in (b) so we need to use a more accurate method to calculate the nuclear mass. 15E d t= v = p d = R 2mn E 2E=mn r ' (1:2 10 = 4 10 16E 15 m)(100)1=3 s 22 s : 2(938 MeV) (5 MeV)(3:0 103 m/s)2 (a) The de Broglie wavelength is given by = h=p, where p is the magnitude of the momentum. According to the relativistic relationship between kinetic energy K and momentum p (see chapter 38), pc = K 2 + 2Kmc2 = (200 MeV)2 + 2(200 MeV)(0:511 MeV) = 200:5 MeV : Thus 1240 eV = hc = 200:5 10nm = 6:18 10 6 nm = 6:18 fm : 6 eV pc p p (b) The diameter of a copper nucleus, for example, is about 8:6 fm, just a little larger than the de Broglie wavelength of a 200-MeV electron. To resolve detail the wavelength should be smaller than the target, ideally a tenth of the diameter or less. 200-MeV electrons are perhaps at the lower limit in energy for useful probes. Let p p ' h=x ' h=R, we get 2 (1240 MeV fm)2 h2 E = 2pm ' 2mR2 = 2(938 MeV)[1:2 pm(100)1=3 ]2 ' 30 MeV : 17E 18E (a) In terms of the original value of u, the newly de ned u is greater by a factor of 1.007825. So the mass of 1 H would be 1:000000 u, the mass of 12 C would be (12:000000=1:007825) u = 11:90683 u, and the mass of 238 U would be (238:050785=1:007825) u = 236:2025 u. (b) De ning the mass of 1 H to be exactly 1 does not result in any overall simpli cation. CHAPTER 43 NUCLEAR PHYSICS 1153 (a) Since the nuclear force has a short range, any nucleon interacts only with its nearest neighbors, not with more distant nucleons in the nucleus. Let N be the number of neighbors that interact with any nucleon. It is independent of the number A of nucleons in the nucleus. The number of interactions in a nucleus is approximately NA, so the energy associated with the strong nuclear force is proportional to NA and therefore proportional to A itself. (b) Each proton in a nucleus interacts electrically with every other proton. The number of pairs of protons is Z (Z 1)=2, where Z is the number of protons. The Coulomb energy is therefore proportional to Z (Z 1). (c) As A increases, Z increases at a slightly slower rate but Z 2 increases at a faster rate than A and the energy associated with Coulomb interactions increases faster than the energy associated with strong nuclear interactions. Let f24 be the abundance of 24 Mg, f25 be the abundance of 25 Mg, and f26 be the abundance of 26 Mg. Then the entry in the periodic table for Mg is 24:312 = 23:98504f24 + 24:98584f25 + 25:98259f26 . Since there are only three isotopes, f24 + f25 + f26 = 1. Solve for f25 and f26 . The second equation gives f26 = 1 f24 f25 . Substitute this expression and f24 = 0:7899 into the rst equation to obtain 24:312 = (23:98504)(0:7899) + 24:98584f25 + 25:98259 (25:98259)(0:7899) 25:98259f25 . The solution is f25 = 0:09303. Then f26 = 1 0:7899 0:09303 = 0:1171. 78:99% of naturally occurring magnesium is 24 Mg, 9:30% is 25 Mg, and 11:71% is 26 Mg. (a) The rst step: 4 H p !3 H. The work needed is E1 = (m3H + mp m4He )c2 = (3:01605 u + 1:00783 u 4:00260 u)(931:5 MeV/u) = 19:8 MeV: The second step: 3 He n !2 H. The work needed is E2 = (m2H + mp m3He )c2 = (2:01410 u + 1:00867 u 3:01605 u)(931:5 MeV/u) = 6:26 MeV: The third step: 2 H p !1 H. The work needed is E3 = (mp + mp m2H )c2 = (1:00783 u+ 1:00783 u 2:01410 u)(931:5 MeV/u) = 2:22 MeV: (b) The total binding energy is E = E1 +E2 +E3 = 19:8 MeV+6:26 MeV+2:22 MeV = 28:3 MeV. (c) E=A = 28:3 MeV=4 = 7:07 MeV. 22P 21P 20P 19P The nuclear reaction in question is written as n + p !2 H+ . Conservation of energy gives mn c2 + mp c2 = m2 H c2 + E , or mn = m2H mp + E2 c = 1:0087 u : = 2:01410 u 1:00783 u + 2:2233 MeV 938 MeV/u 1154 CHAPTER 43 NUCLEAR PHYSICS The nuclear energy released per copper nucleus is E = (29mp + 34mn mCu )c2 . The number of 63 Cu nuclei in a penny is N = M=mCu . So the total energy required is 23P M Etotal = N E = m (29mp + 34mn mCu )c2 Cu + 34(1:00867 u) 9260 u](931:5 MeV/u) = 3:0 g[(29)(1:00783 u)9260 u)(1:661 10 62:kg/u) 27 (62: 25 MeV : = 1:6 10 (a) For 1 H, = (1:00783 u 1:00000 u)(931:5 MeV=c2 ) = 7:29 MeV: (b) For the neutron, = (1:00867 u 1:00000 u)(931:5 MeV=c2 ) = +8:07 MeV: (c) For 120 Sn, = (119:9022 u 120:0000 u)(931:5 MeV=c2 ) = 91:0 MeV: 25P 24P the mass of the atom containing the nucleus of interest. If the masses are given in atomic mass units then mass excesses are de ned by H = (mH 1)c2 , n = (mn 1)c2 , and = (m A)c2 . This means mH c2 = H + c2 , mn c2 = n + c2 , and mc2 = + Ac2 . Thus E = (Z H + N n ) + (Z + N A)c2 = Z H + N n , where A = Z + N was used. For 197 Au, Z = 79 and N = 197 79 = 118. Hence 79 If a nucleus contains Z protons and N neutrons, its binding energy is E = (ZmH + Nmn m)c2 , where mH is the mass of a hydrogen atom, mn is the mass of a neutron, and m is E = (79)(7:29 MeV) + (118)(8:07 MeV) ( 31:2 MeV) = 1560 MeV : This means the binding energy per nucleon is En = (1560 MeV)=(197) = 7:92 MeV. 26E The number of atoms remaining is N = N0 e t = N0 e (t ln 2)= = (48 1019 )e (26 h)(ln 2)=6:5 h = 3:0 1019 : 27E Let and solve for t : t = 2 = 2(140 d) = 280 d. N = N0 e t = N0 e (t ln 2)= = N0 4 CHAPTER 43 NUCLEAR PHYSICS 1155 (a) Since 60 y = 2(30 y) = 2 the fraction left is 2 2 = 1=4. (b) Since 90 y = 3(30 y) = 3 the fraction left is 2 3 = 1=8. (a) The decay rate is given by R = N , where is the disintegration constant and N is the number of undecayed nuclei. Initially R = R0 = N0 , where N0 is the number of undecayed nuclei at that time. You must nd values for both N0 and . The disintegration constant is related to the half-life by = (ln 2)= = (ln 2)=(78 h) = 8:89 10 3 h 1 . If M is the mass of the sample and m is the mass of a single atom of gallium, then N0 = M=m. Now m = (67 u)(1:661 10 24 g/u) = 1:113 10 22 g and N0 = (3:4 g)=(1:113 10 22 g) = 3:05 1022 . Thus R0 = (8:89 10 3 h 1 )(3:05 1022 ) = 2:71 1020 h 1 = 7:5 1016 s 1 . (b) The decay rate at any time t is given by 29E 28E R = R0 e t : where R0 is the decay rate at t = 0. At t = 48 h, t = (8:89 10 3 h 1 )(48 h) = 0:427 and R = (7:53 1016 s 1 ) e 0:427 = 4:9 1016 s 1 : 30E (a) The half-life and the disintegration constant are related by = (ln 2)=, so = (ln 2)=(0:0108 h 1 ) = 64:2 h. (b) At time t the number of undecayed nuclei remaining is given by N = N0 e t = N0 e Substitute t = 3 to obtain (ln 2)t= : In each half-life the number of undecayed nuclei is reduced by half. At the end of one halflife N = N0 =2, at the end of two half-lives N = N0 =4, and at the end of three half-lives N = N0 =8 = 0:125N0 . (c) Use N = N0 e t : 10:0 d is 240 h, so t = (0:0108 h 1 )(240 h) = 2:592 and N =e N0 3 ln 2 = 0:125 : N = e 2:592 = 0:0749 : N0 1156 CHAPTER 43 NUCLEAR PHYSICS (a) For 238 U we have R0 = 12=s = N0 so = R0 =N0 = (12 s 1 )=(2:5 1018 ) = 4:8 10 18 s 1 : (b) = (ln 2)= = (ln 2)=(4:8 10 18 s 1 ) = 1:4 1017 s ' 4:6 109 y: 32E 31E (a) The number of nuclei is 6 2 M N = m = (239 u)(1: 10 10kg27 kg/u) = 5:04 1018 : 66 Pu (b) 18 )(ln ln R = N = N 2 = (2:41(5:04 4 10 :15 2) 7 =y) = 4:60 106 s 1 : 10 y)(3 10 33E The rate of decay is given by R = N , where is the disintegration constant and N is the number of undecayed nuclei. In terms of the half-life the disintegration constant is = (ln 2)= , so 10 1 7 R N = R = ln 2 = (6000 Ci)(3:7 10 s /Ci)(5:27 y)(3:156 10 s/y) ln 2 22 nuclei : = 5:3 10 (a) Assume that the chlorine in the sample had the naturally occurring isotopic mixture, so the average mass number was 35:453, as given in Appendix F. Then the mass of 226 Ra was m = 226 + 226 :453) (0:10 g) = 76:1 10 3 g : 2(35 The mass of a 226 Ra nucleus is (226 u)(1:661 10 24 g/u) = 3:75 10 22 g, so the number of 226 Ra nuclei present was N = (76:1 10 3 g)=(3:75 10 22 g) = 2:0 1020 . (b) The decay rate is given by R = N = (N ln 2)= , where is the disintegration constant, is the half-life, and N is the number of nuclei. The relationship = (ln 2)= was used. Thus (2:03 1020 ln R = (1600 y)(3156 )1072s/y) = 2:8 109 s 1 : : 34P CHAPTER 43 NUCLEAR PHYSICS 1157 35P The amount decayed is m = mjtf =16:0 h mjti =14:0 h = m0 (1 e ti ln 2= ) m0 (1 e tf ln 2= ) = m0 (e tf ln 2= e ti ln 2= ) h i = (5:50 g) e (16:0 h=12:7 h) ln 2 e (14:0 h=12:7 h) ln 2 = 0:256 g : 36P (a) Use R = R0 e t and nd t: 1 :28 t = ln R0 = ln 2 ln R0 = 14ln 2 d ln 3050 = 59:5 d : R R 170 (b) The factor is R0 = et = et ln 2= = e(3:48 d=14:28 d) ln 2 = 1:18 : R Label the two isotopes with subscripts 1 (for 32 P) and 2 (for 33 P). Initially R01 = 1 N01 = (1=9:00)R02 = 2 N02 =9:00. At time t we have R1 = R01 e 1 t and R2 = R02 e 2 t . Set R1 =R2 = 9:00 to nd (R01 =R02 )e (1 2 )t = 9:00. Solve for t: 37P R R01 02 t = ln 9:0001 = ln(2= =9:00R= ) R02 ln 1 ln 2 2 1 2 2] 9 = ln 2[(14:ln[(1=1 :00) :3 d) 1 ] = 209 d : 3 d) (25 1 The number N of undecayed nuclei present at any time and the rate of decay R at that time are related by R = N , where is the disintegration constant. The disintegration constant is related to the half-life by = (ln 2)= , so R = (N ln 2)= and = (N ln 2)=R. Since 15:0% by mass of the sample is 147 Sm, the number of 147 Sm nuclei present in the sample is (0: N = (147 u)(1150)(1:00 g) g/u) = 6:143 1020 : :661 10 24 38P 1158 CHAPTER 43 NUCLEAR PHYSICS Thus = (6:143 10 1 ) ln 2 = 3:55 1018 s = 1:12 1011 y : 120 s 20 39P The amount of 239 Pu that has decayed is mPu = m(1 e t ln 2= ) = (12:0 g) 1 h e (20;000 y=24;100 y) ln 2 i = 5:25 g : Thus the amount of helium produced is u mHe = 4:00u 5:25 g = 0:0878 g = 87:8 mg : 239 40P Solve for m from R0 = N0 = (m=mK )(ln 2)= : m = mK R0 ln 2 = (40 u)(1:661 10 = 6:6 10 4 kg = 0:66 g = 660 mg : 41P 27 kg/u)(1:70 105 disintegrations/s)(1:28 109 y) (ln 2)(1 y=3:15 107 s) The number of 90 Sr atoms needed to produce the decay rate of R = 74; 000=s is given by R = N = (M=m)(a=A), where M = 400 g, m is the mass of the 90 Sr nucleus, A = 2000 km2 , and a is the area in question. Solve for a: m a=A M 6 2 )(90 g/mol)(29 y)(3 15 107 = (2000 10 m (400 g)(6:02 1023 =:mol)(ln 2)s/y)(74; 000=s) = 7:3 10 2 m2 = 730 cm2 : R = AmR M ln 2 42P The following are ln R vs t plots for both 108 Ag and 110 Ag. The half-life of each of the isotopes can be obtained from the slope of the respective plot, similar to Sample Problem 434. The combined decay rate is R = R108 + R110 = R0;108 e 108 t + R0;110 e 110 t . You can CHAPTER 43 NUCLEAR PHYSICS 1159 see that taking the natural log of both sides will not produce an equation for a straignt line in the ln R vs t plot. 13 108 Ag 10 11 20 110 Ag 12 lnR 10 lnR 0 9 8 0 200 400 600 800 10 0 200 400 600 800 t (s) t (s) 43P If N is the number of undecayed nuclei present at time t, then where R is the rate of production by the cyclotron and is the disintegration constant. The second term gives the rate of decay. Rearrange the equation slightly and integrate: Z dN = R N ; dt where N0 is the number of undecayed nuclei present at time t = 0. This yields 1 ln R N = t : Solve for N : dN = Z t dt ; N0 R N 0 R N0 N After many half-lives the exponential is small and the second term can be neglected. Then N = R=, regardless of the initial value N0 . At times that are long compared to the half-life the rate of production equals the rate of decay and N is a constant. (a) The sample is in secular equilibrium with the source and the decay rate equals the production rate. Let R be the rate of production of 56 Mn and let be the disintegration constant. According the result of Problem 43, R = N after a long time has passed. Now N = 8:88 1010 s 1 , so R = 8:88 1010 s 1 . 44P N = R + N0 R e t : 1160 CHAPTER 43 NUCLEAR PHYSICS (b) They decay at the same rate as they are produced, 8:88 1010 s 1 . (c) Use N = R=. If is the half-life, then the disintegration constant is = (ln 2)= = (ln 2)=(2:58 h) = 0:269 h 1 = 7:46 10 5 s 1 , so N = (8:88 1010 s 1 )=(7:46 10 5 s 1 ) = 1:19 1015 . (d) The mass of a 56 Mn nucleus is (56 u)(1:661 10 24 g/u) = 9:30 10 23 g and the total mass of 56 Mn in the sample at the end of the bombardment is Nm = (1:19 1015 )(9:30 10 23 g) = 1:11 10 7 g = 0:111 g. 45P M = (ln 2)(1:00 mg)(6:02 1023 =mol) = 3:66 107 s 1 : m (1600 y)(3:15 107 s/y)(226 g/mol) (b) Since 1600 y 3:82 d the time required is t 3:82 d. (c) It is decaying at the same rate as it is produced, or R = 3:66 107 s 1 . (d) From RRa = RRn and R = N = (ln 2= )(M=m), we get Rn mRn M = (3:82 d)(1:00 10 3 g)(222 u) = 6:42 10 9 g : MRn = mRa Ra (1600 y)(365 d/y)(226 u) Ra R = N = ln 2 46E (a) The electrostatic potential is given by 9 2 2 1 U (r) = 4 q qTh = (8:99 10 rN m C )(2)(90)(1:60 10 (1:60 10 25 J/MeV) 0 r = 259 MeV fm : r The plot is as follows. 30 25 20 19 C)2 U (MeV) 15 10 5 0 0 20 40 60 80 100 r (fm) CHAPTER 43 NUCLEAR PHYSICS 1161 47E and N = e t = e (ln 2)t= ; N0 where is the disintegration constant and (= (ln 2)=) is the half-life. The time for half the original 238 U nuclei to decay is 4:5 109 y. For 244 Pu at that time (ln 2)t = (ln 2)(4:5 109 y) = 38:99 8:0 107 y N = e 38:99 = 1:2 10 N0 17 : The fraction of undecayed nuclei remaining after time t is given by For 248 Cm at that time and (ln 2)t = (ln 2)(4:5 109 y) = 9170 3:4 105 y For any reasonably sized sample this is less than 1 nucleus and may be taken to be zero. Your calculator probably cannot evaluate e 9170 directly. Treat it as (e 91:70 )100 . Energy and momentum are conserved. Assume the residual thorium nucleus is in its ground state. Let K be the kinetic energy of the alpha particle and KTh be the kinetic energy of the thorium nucleus. Then Q = K + KTh . Assume the uranium nucleus is initially at rest. Then conservation of momentum yields 0 = p + pTh , where p is the momentum of the alpha particle and pTh is the momentum of the thorium nucleus. Both particles travel slowly enough that the classical relationship between momentum and energy can be used. Thus KTh = p2 =2mTh , where mTh is the mass of the thorium nucleus. Th Substitute pTh = p and use K = p2 =2m to obtain KTh = (m =mTh )K . Thus m K = 1 + m K = 1 + 4:00 u (4:196 MeV) = 4:27 MeV : Q = K + m mTh 234 u Th (a) The nuclear reaction is written as 238 U ! 234 Th + 4 He. The energy released is E1 = (mU mHe mTh )c2 = (238:05079 u 4:00260 u 234:04363 u)(931:5 MeV/u) = 4:25 MeV : 49P 48P N =e N0 9170 = 3:31 10 3983 : 1162 CHAPTER 43 NUCLEAR PHYSICS (b) The reaction series is now 238 U !237 U...
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