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 halflives) 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. 1018 and 1019. 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 satised.
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. 433: 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. 434. You will easily nd that all the nuclides listed in Table 431 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 signicantly reduced by the electrostatic repulsion among the protons. From Fig. 434 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 471) 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 200MeV electron. To resolve detail the wavelength should be smaller than the target, ideally a tenth of the diameter or less. 200MeV 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 dened 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) Dening the mass of 1 H to be exactly 1 does not result in any overall simplication. 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 dened 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 halflife 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 halflife 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 halflife the number of undecayed nuclei is reduced by half. At the end of one halflife N = N0 =2, at the end of two halflives N = N0 =4, and at the end of three halflives 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 halflife 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 halflife, 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 halflife 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 halflife 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 halflives 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 halflife 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 halflife, 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 halflife. 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 + n, 237 U !236 Pa + p, 235 Pa !234 Th + p. The net energy released is then 236 Pa !235 Pa + n, E2 = (m238U m237U mn )c2 + (m237U m236Pa mp )c2 + (m236Pa m235Pa mn )c2 + (m235Pa m234Th mp )c2 = (m238U 2mn 2mp m234Th ) = [238:05079 u 2(1:00867 u) 2(1:00783 u) 234:04363 u](932 MeV=c2 ) = 24:1 MeV : (c) The binding energy of the particle is Ea = (2mn + 2mp mHe )c2 = j 24:1 MeV 4:25 MeVj = 28:3 MeV:
50P (a) For the rst reaction Q1 = (mRa mPb mC )c2 = (223:01850 u 208:98107 u 14:00324 u)(931:5 MeV=c2 ) = 31:8 MeV ;
and for the second one Q2 = (mRa mRn mHe )c2 = (223:01850 u 219:00948 u 4:00260 u)(931:5 MeV=c2 ) = 5:98 MeV : (b) From U / q1 q2 =r we get q (82e)(6 0e) U1 U2 qqPb q C = (30:0 MeV) (86e)(2::0e) = 86 MeV : Rn He
The disintegration energies are
51E Q3 = (m235 U m232 Th m3 He ) c2 = (235:0439 u 232:0381 u 3:0160 u)(931:5 MeV=c2 ) = 9:50 MeV ; Q4 = (m235 U m231 Th m4 He ) c2 = (235:0439 u 231:0363 u 4:0026 u)(931:5 MeV=c2 ) = 4:66 MeV ; CHAPTER 43 NUCLEAR PHYSICS 1163 and Q5 = (m235 U m230 Th m5 He ) c2 = (235:0439 u 230:0331 u 5:0122 u)(931:5 MeV=c2 ) = 1:30 MeV :
Only the second decay process (the decay) is spontaneous, as it releases energy.
52E Let A X represent the unknown nuclide. The reaction equation is Z
A X + 1 n ! 0 e + 2 4 He : Z 0 2 1 Conservation of charge yields Z + 0 = 1 + 4 or Z = 3. Conservation of mass number yields A + 1 = 0 + 8 or A = 7. According to the periodic table in Appendix G, lithium has atomic number 3, so the nuclide must be 7 Li. 3
53E Let MCs be the mass of one atom of 137 Cs and MBa be the mass of one atom of 137 Ba. To 55 56 obtain the nuclear masses we must subtract the mass of 55 electrons from MCs and the mass of 56 electrons from MBa . The energy released is thus Q = [(MCs 55me ) (MBa 56me ) me ]c2 , where me is the mass of an electron. Once cancellations have been made, Q = (MCs MBa )c2 is obtained. Thus Q = (136:9071 u 136:9058 u) c2 = (0:0013 u)c2 = (0:0013 u)(931:5 MeV/u) = 1:21 MeV :
54E (a) The mass number A of a radionuclide changes by 4 in an decay and is unchanged in a decay. If the mass numbers of two radionuclides are given by 4n + k and 4n0 + k (where k = 0; 1; 2; 3) then the heavier one can decay into the lighter one by a series of (and ) decays, as their mass numbers dier by only an integer times 4. If A = 4n + k, then after decaying for m times, its mass number becomes A = 4n + k 4m = 4(n m) + k, still in the same chain. (b) 235 = 58 4 + 3 = 4n1 + 3; 236 = 59 4 = 4n2 ; 238 = 59 4 + 2 = 4n2 + 2; 239 = 59 4 + 3 = 4n2 + 3; 240 = 60 4 = 4n3 ; 245 = 61 4 + 1 = 4n4 + 1; 246 = 61 4 + 2 = 4n4 + 2; 249 = 62 4 + 1 = 4n5 + 1; 253 = 63 4 + 1 = 4n6 + 1.
55E The decay scheme is n ! p + e + . The electron kinetic energy is a maximum if no neutrino is emitted. Then Kmax = (mn mp me )c2 , where mn is the mass of a neutron, 1164 CHAPTER 43 NUCLEAR PHYSICS mp is the mass of a proton, and me is the mass of an electron. Since mp + me = mH , where mH is the mass of a hydrogen atom, this can be written Kmax = (mn mH )c2 . Hence Kmax = (840 10 6 u)c2 = (840 10 6 u)(931:5 MeV/u) = 0:782 MeV.
(a)
56E = h = p 2 hc 2 2 = p 2 hc 2 p E (me c ) (me c + K ) me c2 1240 fm MeV =p = 9:0 102 fm : (0:511 MeV + 1:0 MeV)2 (0:511 MeV)2
(b) R = R0 A1=3 = (1:2 fm)(150)1=3 = 6:4 fm: (c) Since R the electron cannot be conned in the nuclide. (d) Yes. See part (c) above. The energy of the particles before and after the decay is Ei = mV c2 + me c2 EK and Ef = mTi c2 . Thus Q = Ei Ef = (mV mTi + me )c2 EK . Note that mV = mV 23me and mTi = mTi 22me so mV mTi + me = (mV 23me ) (mTi 22me )+ me = mV mTi . Thus Q = (mV mTi )c2 EK :
58P 57P Q = (mV mTi )c2 EK = (48:94852 u 48:94787 u)(932 MeV=c2 ) 5:47 keV = 600 keV : 59P (a) Use the result obtained in 57P, with the modication that no electron is at the initial stage of the process and one positron is present at the nal stage. This means that there is an additional term me c2 me c2 = 2me c2 in the expression for Q = Ei Ef . So Q = (mC mTi )c2 2me c2 = (mC mTi 2me )c2 :
(b) Q = [11:011434 u 11:009305 u 2(0:0005486 u)](931:5 MeV/u) = 0:961 MeV: This compares favorably with the maximum energy of the positron of 0:960 MeV. CHAPTER 43 NUCLEAR PHYSICS 1165 60P (a) The rate of heat production is
3 3 3 dE = X R Q = X N Q = X ln 2 (1:00 kg)fi Q i dt i=1 i i i=1 1 i i i=1 i mi (1:00 kg)(ln 2)(1:60 10 13 J/MeV) (4 10 6 )(51:7 MeV) = (3:15 107 s/y)(1:661 10 27 kg/u) (238 u)(4:47 109 y) (13 10 6 )(42:7 MeV) + (4 10 6 )(1:31 MeV) + (208 u)(1:41 1010 y) (40 u)(1:28 109 y) = 1:0 10 9 W : (b) P = (2:7 1022 kg)(1:0 10 9 W/kg) = 2:7 1013 W:
61P Since the electron has the maximum possible kinetic energy, no neutrino is emitted. Since momentum is conserved, the momentum of the electron and the momentum of the residual sulfur nucleus are equal in magnitude and opposite in direction. If pe is the momentum of the electron and pS is the momentum of the sulfur nucleus, then pS = pe . The kinetic energy KS of the sulfur nucleus is KS = p2 =2MS = p2 =2MS , where MS is the mass of the e S sulfur nucleus. Now the electron's kinetic energy Ke is related to its momentum by the 2 relativistic equation (pe c)2 = Ke + 2Ke me c2 , where me is the mass of an electron. Thus
2 (pe c)2 = Ke + 2Ke me c2 = (1:71 MeV)2 + 2(1:71 MeV)(0:511 MeV) KS = 2M c2 2MS c2 2(32 u)(931:5 MeV/u) S 5 MeV = 78:4 eV ; = 7:84 10 where me c2 = 0:511 MeV was used. (a) The mass of a 238 U atom is (238 u)(1:661 10 24 g/u) = 3:95 10 22 g, so the number of uranium atoms in the rock is NU = (4:20 10 3 g)=(3:95 10 22 g) = 1:06 1019 . The mass of a 206 Pb atom is (206 u)(1:661 10 24 g) = 3:42 10 22 g, so the number of lead atoms in the rock is NPb = (2:135 10 3 g)=(3:42 10 22 g) = 6:24 1018 . (b) If no lead was lost there was originally one uranium atom for each lead atom formed by decay, in addition to the uranium atoms that did not yet decay. Thus the original number of uranium atoms was NU0 = NU + NPb = 1:06 1019 + 6:24 1018 = 1:68 1019 . (c) Use NU = NU0 e t ;
62E 1166 CHAPTER 43 NUCLEAR PHYSICS where is the disintegration constant for the decay. It is related to the halflife by = (ln 2)= . Thus N N 1 t = ln N U = ln 2 ln N U U0 U0 9 y 1:06 1019 10 ln 9 = 4:47 ln 2 1:68 1019 = 2:98 10 y :
63E Solve for t from R = R0 e t : 1 ln R0 = 5730 y ln 15:3 5:00 = 1:61 103 y : t= R ln 2 63:0 1:00 64P The original amount of 238 U the rock contains is given by m0 = met = (3:70 mg) e(ln 2)(260106 y)(4:47109 y) = 3:85 mg :
Thus the amount of lead produced is m0 = (m0 m 206 m) m206 = (3:85 mg 3:70 mg) 238 = 0:132 mg : 238 65P We can nd the age t of the rock from the masses of 238 U and 206 Pb. The initial mass of 238 U is 238 mU0 = mU + 206 mPb so mU = mU0 e U t = (mU + m238 Pb =206)e (t ln 2)=U . Solve for t: t = lnU2 ln mU + (238=206)mPb mU 9y 4:47 10 ln 1 + 238 0:15 mg = ln 2 206 0:86 mg 9 y: = 1:18 10
Now, for the decay of 40 K, the initial mass of 40 K is mK0 = mK + (40=40)mAr = mK + mAr CHAPTER 43 NUCLEAR PHYSICS 1167 so Solve for mK : mK = mK0 e K t = (mK + mAr )e K t : e K t mK = mAre K t = emtAr 1 K 1 6 mg = 1:7 mg : = (ln 2)(1:18101:y)=(1:25109 y) 9 e 1 66P (8700 R = (3:7 10disintegrations/min)=C = 3:92 10 9 Ci = 145 Bq : 10 disintegrations/s)
i The decay rate R is related to the number of nuclei N by R = N , where is the disintegration constant. The disintegration constant is related to the halflife by = (ln 2)= , so N = R= = R= ln 2. Since 1 Ci = 3:7 1010 disintegrations/s,
10 1 Ci)(2 4 N = (250 Ci)(3:7 10 s ln 2 :7 d)(8:64 10 s/d) = 3:11 1018 : The mass of a 198 Au atom is M = (198 u)(1:661 10 24 g/u) = 3:29 10 22 g so the mass required is NM = (3:11 1018 )(3:29 10 22 g) = 1:02 10 3 g = 1:02 mg.
68E 67E The annual does is (20 h)(52 week/y)(7:0 Sv/h) = 7:3 mSv:
69E (a) The energy absorbed is E = DM = (24 10 3 rad)(10 mJ/kg)(75 kg) = 18 mJ :
(b) The equivalent does is D RBE = (2:4 10 4 Gy)(12) = 2:9 mSv = 0:29 rem. The dose equivalent is the product of the absorbed dose and the RBE factor, so the absorbed dose is (dose equivalent)=( RBE) = (250 10 6 Sv)=(0:85) = 2:94 10 2 rad. But 1 rad = 10 mJ/kg, so the absorbed dose is (2:94 10 2 )(10 10 3 J/kg) = 2:94 10 4 J/kg. To
70P 1168 CHAPTER 43 NUCLEAR PHYSICS obtain the total energy received, multiply this by the mass receiving the energy: E = (2:94 10 4 J/kg)(44 kg) = 1:3 10 2 J = 13 mJ.
71P (a) 3 g)(6:02 23 N0 = (2:5 10 239 g/mol 10 =mol) = 6:3 1018 : (b) N = N0 (1 e t ln 2= ) h 18 ) 1 e = (6:3 10 = 2:5 1011 :
(12 h) ln 2=(24;100 y)(8760 h/y)
i (c) E = (95%)E N = (95%)(5:2 MeV)(2:5 1011 )(1:6 10 (d) D = E=m = 197:6 mJ=85 kg = 2:3 10 mJ/kg = 2:3 mGy: (e) Dbio = D RBE = (2:3 mGy)(13) = 30 mSv:
72E 13 J/MeV) = 0:20 J : (a) (b) No, since t 10
73E 10 h E ' t = 4:14 (10 2
14 s 15 eV s 22 s) ' 6:6 MeV : (see Sample Problem 4310). 2 Solve for T from K = 3 kT : 2(5: 10 T = 2K = 3(8:62 00 MeV) 3k 10 5 eV/K) = 3:87 10 K : 74E Compare both the proton and neutron numbers of the nuclides given with the magic nucleon numbers. (a) 18 O, 60 Ni, 92 Mo, 144 Sm, 207 Pb. (b) 40 K, 91 Zr, 121 Sb, 143 Nd. (c) 13 C, 40 K, 49 Ti, 205 Tl, 207 Pb. CHAPTER 43 NUCLEAR PHYSICS 1169 A generalized formation reaction can be written X+ x ! Y, where X is the target nucleus, x is the incident light particle, and Y is the excited compound nucleus (20 Ne). Assume X is initially at rest. Then conservation of energy yields 75P mX c2 + mx c2 + Kx = mY c2 + KY + EY ;
where mX , mx , and mY are masses, Kx and KY are kinetic energies, and EY is the excitation energy of Y. Conservation of momentum yields px = pY :
Now KY = p2 =2mY = p2 =2mY = (mx =mY )Kx , so x Y mX c2 + mx c2 + Kx = mY c2 +
and mx K + E mY x Y (a) Let x represent the alpha particle and X represent the 16 O nucleus. Then (mY mX mx )c2 = (19:99244 u 15:99491 u 4:00260 u)(931:5 MeV/u) = 4:722 MeV and 19 u K = 19:99244:99244:00260 u ( 4:722 MeV + 25:0 MeV) = 25:4 MeV : u 4 (b) Let x represent the proton and X represent the 19 F nucleus. Then (mY mX mX )c2 = (19:99244 u 18:99841 u 1:00783 u)(931:5 MeV/u) = 12:85 MeV and 19 u K = 19:99244:99244:00783 u ( 12:85 MeV + 25:0 MeV) = 12:8 MeV : u 1 (c) Let x represent the photon and X represent the 20 Ne nucleus. Since the mass of the photon is zero we must rewrite the conservation of energy equation: if E is the energy of the photon, then E + mX c2 = mY c2 + KY + EY . Since mX = mY , this equation becomes E = KY + EY . Since the momentum and energy of a photon are related by p = E =c, the conservation of momentum equation becomes E =c = pY . The kinetic 2 energy of the compound nucleus is KY = p2 =2mY = E =2mY c2 . Substitute this result Y into the conservation of energy equation to obtain Kx = m mYm (mY mX mx )c2 + EY :
Y x E2 E = 2m c2 + EY :
Y This quadratic equation has the solutions E = mY c2 (mY c2 )2 2mY c2 EY : p 1170 CHAPTER 43 NUCLEAR PHYSICS If the problem is solved using the relativistic relationship between the energy and momentum of the compound nucleus only one solution would be obtained, the one corresponding to the negative sign above. Since mY c2 = (19:99244 u)(931:5 MeV/u) = 1:862 104 MeV, E = (1:862 104 MeV) = 25:0 MeV : p (1:862 104 MeV)2 2(1:862 104 MeV)(25:0 MeV) The kinetic energy of the compound nucleus is very small; essentially all of the photon energy goes to excite the nucleus. For each of the three decay processes, label the 20 Ne with subscript 0, and the larger and smaller nuclides resulting from the decay process with sbuscripts 1 and 2. Then the conseration of energy gives Q = K1 + K2 = E0 E1 E2 = (m0 m1 m2 )c2 and conservation of linear momentum gives jp1 j = jp2 j, which, upon noting that p2;2 = 1 2 2 2 (K1;2 + E1;2 )2 E1;2 , can be written as K1 + 2K1 E1 = K2 + 2K2 E2 . Solve for K1 : 2 2 K1 = Q +E E2 Q : 2 0 (a) Now Q = (m0 m1 m2 )c2 = (19:99244 u 18:00094 u 2:01410 u)(932 MeV/u) + 25:0 MeV = 3:9368 MeV so 2 2 Kd = K1 = Q +E E2 Q 2 0 2 + 2(18:00094 u)(932 MeV/u)(3 = (3:9368 MeV) 2(19:99244 u)(931:5 MeV/u) :9368 MeV) = 3:55 MeV : (b) Now Q = (m0 m1 m2 )c2 = (19:99244 u 19:00188 u 1:00867 u)(931:5 MeV/u) + 25:0 MeV = 8:1215 MeV so 2 2 Kn = K1 = Q +E E2 Q 2 0 2 5 MeV/u)(8 = (8:1215 MeV) + 2(19:00188 u)(931:MeV/u) :1215 MeV) 2(19:99244 u)(931:5 = 7:72 MeV :
76P (c) Now Q = (m0 m1 m2 )c2 = (19:99244 u 16:99913 u 3:01603 u)(931:5 MeV/u) + 25:0 MeV = 3:8250 MeV so 2 2 KHe = K1 = Q +E E2 Q 2 0 2 5 MeV/u)(3 = (3:8250 MeV) + 2(16:99913 u)(931:MeV/u) :8250 MeV) 2(16:99913 u)(931:5 = 3:26 MeV : CHAPTER 43 NUCLEAR PHYSICS 1171 (a) Consider the process 209 Bi !208 Pb + p. The energy required for the process is E1 = (mPb + mp mBi )c2 = (207:9767 u + 1:00783 u 208:9804 u)(931:5 MeV=c2 ) = 3:85 MeV: 0 To remove a proton from the lled shell of 208 Pb (208 Pb !207 Tl + p) we need E1 = (mTl + mp mPb )c2 = (206:9767 u + 1:00783 u 207:9767 u)(931:5 MeV/u) = 7:95 MeV > E1 = 3:85 MeV. (b) Now the process is 209 Pb !208 Pb + n. The energy required is E2 = (207:9774 u + 1:00807 u 208:9811 u)(931:5 MeV/u) = 3:98 MeV: To remove a neutron from the lled 0 shell of 208 Pb (208 Pb !207 Pb + n) we need E2 = (206:9759 u + 1:00867 u 207:9767 u) (931:5 MeV/u) = 7:33 MeV > E2 = 3:98 MeV. 0 0 Both results (E1 > E1 and E2 > E2 ) are expected. (a) The binding energy is E1 = (m90 + mn m91 )c2 = (89:90471 u + 1:00867 u 90:90564 u)(931:5 MeV/u) = 7:21 MeV. (b) The binding energy for the next neutron is E2 = (m89 + mn m90 )c2 = (88:90890 u+ 1:00867 u 89:90471 u)(931:5 MeV/u) = 12:0 MeV: (c) The binding energy per nucleon is E3 = (40mp +51mn m91 )c2 =A = [(40(1:00783 u)+ 51(1:00867 u 90:90564 u)(931:5 MeV/u)=91 = 8:69 MeV. From the calculation above we see that E1 < E3 < E2 . This is expected. The extra neutron outside the lled shell is relatively easy to remove, so E1 is less than the average binding energy per nucleon (E2 ). To remove a neutron from a lled shell, however, is more dicult, which is why E3 < E2 . (a) Consider the process 121 Sb !120 Sn + p. The energy needed is E1 = (mSn + mp mSb )c2 = (119:9022 u + 1:00783 u 120:9038 u)(931:5 MeV/u) = 5:80 MeV. (b) Consider the process 120 Sn !119 In + p. The energy needed is E2 = (mIn + mp mSn )c2 = (118:9058 u + 1:00783 u 119:9022 u)(931:5 MeV/u) = 10:6 MeV. (a) From the decay series we know that the N210 , the amount of 210 Pb nuclei, changes because of two decays: the decay from 226 Ra into 210 Pb at the rate R226 = 226 N226 , and the decay from 210 Pb into 206 Pb at the rate R210 = 210 N210 . The rst of these decays causes N210 to increase while the second one causes it to decrease. Thus
80 79P 78P 77P dN210 = R 226 R210 = 226 N226 210 N210 : dt (b) Set dN210 =dt = R226 R210 = 0 to obtain R226 = 1 : R210 1172 CHAPTER 43 NUCLEAR PHYSICS (c) From R226 = 226 N226 = R210 = 210 N210 , we obtain N226 = 210 = 226 = 1:60 103 y = 70:8 : N210 226 210 22:6 y
(d) Since only 1:00% of the 226 Ra remains the ratio R226 =R210 is 1:00% of that of the equilibrium state computed in part (b). Thus the ratio is (1:00%)(1) = 0:0100: (e) Similar to part (d) above, since only 1:00% of the 226 Ra remains the ratio N226 =N210 is 1:00% of that of the equilibrium state computed in part (c), or (1:00%)(70:8) = 0:708: (f) Since the actual value of N226 =N210 is 0.09, which much closer to 0.0100 than to 1, the sample of the lead pigment cannot be 300 years old. So Emmaus is not a Vermeer. ...
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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.
 Fall '07
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 Physics

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