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- 1. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition Chapter 39. Nuclear Physics and the Nucleus The Elements 208 39-1. How many neutrons are in the nucleus of 82 Pb ? How many protons? What is the ratio N/Z? (N is the number of neutrons and Z is the number of protons.) A A = N + Z; N = A – Z = 126 neutrons; Z = 82 protons; = 1.54 Z 39-2. The nucleus of a certain isotope contains 143 neutrons and 92 protons. Write the symbol for this nucleus. 235 A = N + Z = 143 + 92 = 235; Z = 92: 92 U 39-3. From a stability curve it is determined that the ratio of neutrons to protons for a cesium nucleus is 1.49. What is the mass number for this isotope of cesium? N Z = 55; = 1.49; N = 1.49(55) = 81.95; A = N + Z ; A = 137 Z 39-4. Most nuclei are nearly spherical in shape and have a radius that may be approximated by 1 r = r0 A 3 r0 = 1.2 x 10−15 m 197 What is the approximate radius of the nucleus of a gold atom 79 Au ? r = (1.2 x10−15 m) 3 197 = 6.98 x 10-15 m ; r = 6.98 x 10-15 m 39-5. Study Table 39-4 for information on the several nuclides. Determine the ratio of N/Z for the following nuclides: Beryllium-9, Copper-64, and Radium 224. N Beryllium: A = 9; Z = 4; N = 9 – 4 = 5; = 1.25 Z 26
- 2. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition N 39-5. (Cont.) Copper: A = 64; Z = 29; N = 64 – 29 = 35; = 1.21 Z N Radium: A = 224; Z = 88; N = 224 – 88 = 136; = 1.55 Z The Atomic Mass Unit 39-6. Find the mass in grams of a gold particle containing two million atomic mass units? 1.66 x 10-27 kg m = 2 x 106 u ; m = 3.32 x 10-21 kg 1.00 u 39-7. Find the mass of a 2-kg copper cylinder in atomic mass units? In Mev? In joules? 1u m = 2 kg -27 ; m = 1.20 x 1027 u 1.6606 x 10 kg 931 MeV m = 1.204 x 1027 u ; m = 1.12 x 1030 MeV 1u 30 1.6 x 10-13 J m = 1.12 x 10 MeV ; m = 1.79 x 1017 J 1 MeV 39-8. A certain nuclear reaction releases an energy of 5.5 MeV. How much mass (in atomic mass units) is required to produce this energy? 1u m = 5.5 MeV ; m = 0.00591 u 931 MeV 39-9. The periodic table gives the average mass of a silver atom as 107.842 u. What is the average mass of the silver nucleus? (Recall that me = 0.00055 u ) For silver, Z = 47. Thus, the nuclear mass is reduced by the mass of 47 electrons. m = 107.842 u – 47(0.00055 u); m = 107.816 u 27
- 3. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition *39-10. Consider the mass spectrometer as illustrated by Fig. 39-3. A uniform magnetic field of 0.6 T is placed across both upper and lower sections of the spectrometer., and the electric field in the velocity selector is 120 V/m. A singly charged neon atom (+1.6 x 10-19 C) of mass 19.992 u passes through the velocity selector and into the spectrometer. What is the velocity of the neon atom as it emerges from the velocity selector? E 120, 000 V/m v= = ; v = 2 x 105 m/s B 0.6 T R *39-11. What is the radius of the circular path followed by the neon atom of Problem 19-10? B = 0.6 T into paper 1.66 x 10-27 kg −26 m = 19.992 u = 3.32 x10 kg 1u mv 2 mv (3.32 x 10-27 kg)(2 x 105 m/s) = qvB; R= ; R= ; R = 6.92 cm R qB (1.6 x 10-19 C)(0.600 T) Mass Defects and Binding Energy (Refer to Table 39-4 for Nuclidic Masses.) 20 *39-12. Calculate the mass defect and binding energy for the neon-20 atom 10 Ne . mD = [( ZmH + Nmn )] − M = [ 10(1.007825 u) + 10(1.008665 u)] − 19.99244 u mD = 20.016490 u –19.99244 u ; mD = 0.17246 u 931 MeV E = mD c 2 = (0.17246 u) = 160.6 MeV ; E = 161 MeV 1u 3 *39-13. Calculate the binding energy and the binding energy per nucleon for tritium 1 H . How much energy in joules is required to tear the nucleus apart into its constituent nucleons? mD = [( ZmH + Nmn )] − M = [ 1(1.007825 u) + 2(1.008665 u) ] − 3.016049 u 28
- 4. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 931 MeV 39-13. (Cont.) mD = 0.009106 u; E = mD c 2 = (0.009106 u) = 8.48 MeV 1u EB 8.48 MeV EB E = 8.48 MeV; = ; = 2.83 MeV/nucleon A 3 A Energy to tear apart: EB = 8.48 MeV(1.6 x 10-13 J/MeV) = 1.36 x 10-12 J *39-14. Calculate the mass defect of 7 Li . What is the binding energy per nucleon? 3 mD = [( ZmH + Nmn )] − M = [ 3(1.007825 u) + 4(1.008665 u) ] − 7.016930 u mD = 7.058135 u –7.016930 u ; mD = 0.041205 u 931 MeV E = mD c 2 = (0.041205 u) = 38.4 MeV ; E = 38.4 MeV 1u EB 38.4 MeV EB = ; = 5.48 MeV/nucleon A 7 A 12 *39-15. Determine the binding energy per nucleon for carbon-12 ( 6 C ). mD = [ 6(1.007825 u) + 6(1.008665 u) ] − 12.0000 u = 0.09894 u 931 MeV EB 92.1 MeV E = mD c 2 = (0.09894 u) ; = = 7.68 MeV/nucleon 1u A 12 197 *39-16. What is the mass defect and the binding energy for a gold atom 79 Au ? mD = [ 79(1.007825 u) + 118(1.008665 u) ] − 196.966541 u = 1.674104 u 931 MeV E = mD c 2 = (1.674104 u) = 1.56 GeV; 1u EB 1.56 GeV = = 7.91 MeV/nucleon A 197 29
- 5. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 120 *39-17. Determine the binding energy per nucleon for tin-120 ( 50 Sn ). mD = [ 50(1.007825 u) + 70(1.008665 u) ] − 119.902108 u = 1.09569 u EB (1.09569 u)(931 MeV/u EB = ; = 8.50 MeV/nucleon A 120 A Radioactivity and Nuclear Decay 39-18. The activity of a certain sample is rated as 2.8 Ci. How many nuclei will have disintegrated in a time of one minute? 3.7 x 1010s-1 Nuclei = 2.8 Ci (60 s); nuclei = 6.22 x 1012 nuclei 1 Ci 60 39-19. The cobalt nucleus 27 Co emits gamma rays of approximately 1.2 MeV. How much mass is lost by the nucleus when it emits a gamma ray of this energy? 1u E = 1.2 MeV ; m = 0.00129 u 931 MeV 39-20. The half-life of the radioactive isotope indium-109 is 4.30 h. If the activity of a sample is 1 mCi at the start, how much activity remains after 4.30, 8.60, and 12.9 h? t / T½ 1 1 t 4.3 h 1 R = R0 ; = = 1; R = (1 mCi) ; R = 0.5 mCi 2 T½ 4.3 h 2 t / T½ 2 1 t 8.6 h 1 R = R0 ; = = 2; R = (1 mCi) ; R = 0.25 mCi 2 T½ 4.3 h 2 t / T½ 3 1 t 12.9 h 1 R = R0 ; = = 3; R = (1 mCi) ; R = 0.125 mCi 2 T½ 4.3 h 2 30
- 6. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 39-21. The initial activity of a sample containing 7.7 x 1011 bismuth-212 nuclei is 4.0 mCi. The half-life of this isotope is 60 min. How many bismuth-212 nuclei remain after 30 min? What is the activity at the end of that time? t / T½ ½ 1 t 30 min 1 N = N0 ; = = 0.5; N = (7.7 x 1011 ) = 5.44 x 1011 nucl. 2 T½ 60 min 2 t / T½ ½ 1 1 R = R0 ; R = (4 mCi) ; R = 2.83 mCi 2 2 *39-22. Strontium-90 is produced in appreciable quantities in the atmosphere during a nuclear explosion. If this isotope has a half-life of 28 years, how long will it take for the initial activity to drop to one-fourth of its original activity? n n 1 R 1 1 R = R0 ; = = ; n=2 2 R0 2 4 t t n= = = 2; t = 2(28 yr); t = 56 yr T½ 28 yr *39-23. Consider a pure, 4.0-g sample of radioactive Gallium-67. If the half-life is 78 h, how much time is required for 2.8 g of this sample to decay? When 2.8 g decay, that leaves 4 g – 2.8 g or 1.20 g remaining. n n 1 m 1.2 g 1 m = m0 ; = = 0.300; = 0.300 2 m0 4g 2 The solution is accomplished by taking the common log of both sides: −0.523 n log(0.5) = log(0.300); − 0.301n = −0.523; n= = 1.74 −0.301 t t n= = = 1.74; t = 1.74 (78 h) t = 135 h T½ 78 h 31
- 7. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition *39-24. If one-fifth of a pure radioactive sample remains after 10 h, what is the half-life? n 1 1 = ; 0.200 = (0.5) n ; n log(0.5) = log(0.2); − 0.301 n = −0.699 5 2 t 10 h 10 h n= = = 2.32; T½ = ; T½ = 4.31 h T½ T½ 2.32 Nuclear Reactions *39-25. Determine the minimum energy released in the nuclear reaction 19 9 F +1 H →4 He + 16 O + energy 1 2 8 The atomic mass of 19 9 F is 18.998403 u, 4 2 He = 4.002603 u, 1 1H = 1.007824 u. E= 19 9 F + 1H - 2 He - 16 O ; (Energy comes from mass defect) 1 4 8 E = 18.998403 u + 1.007825 u – 4.002603 – 15.994915 u = 0.0087 u 931 MeV E = (0.00871 u) ; E = 8.11 MeV 1u *39-26. Determine the approximate kinetic energy imparted to the alpha particle when Radium-226 decays to form Radon-222. Neglect the energy imparted to the radon nucleus. 226 88 Ra → 222 86 Rn + 4 He + Energy; 2 226 88 Ra = 226.02536 931 MeV E = 226.02536 u - 222.017531 u - 4.002603 u = 0.00523 u = 4.87 MeV 1u *39-27. Find the energy involved in the production of two alpha particles in the reaction 7 3 Li + 1H → 2 He + 2 He + energy 1 4 4 32
- 8. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 931 MeV E = 7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u = 17.3 MeV 1u *39-28. Compute the kinetic energy released in the beta minus decay of thorium-233. 233 90 Th → 233 91 Pa + 0 β + energy; +1 233 90 Th = 233.041469 u; 233 91 Pa = 233.040130 E = 233.041469 u - 233.040130 u - 0.00055 u = 0.000789 u 931 MeV E = 0.000789 u ; E = 0.735 MeV 1u *39-29. What must be the energy of an alpha particle as it bombards a Nitrogen-14 nucleus 17 producing 8 O and 1H? (17 O = 16.999130 u) . 1 8 4 2 He + 14 N + Energy → 7 17 8 1 O + 1H 931 MeV E = 16.999130 u + 1.007825 u - 14.003074 - 4.002603 u = 0.001278 u 1u E = 1.19 MeV This is the Threshold Energy for the reaction Challenge Problems *39-30. What is the average mass in kilograms of the nucleus of a boron-11 atom? 1.66 x 10-27 kg m = 11.009305 u ; m = 1.83 x 10-26 kg 1.00 u *39-31. What are the mass defect and the binding energy per nucleon for boron-11? mD = [( ZmH + Nmn )] − M = [ 5(1.007825 u) + 6(1.008665 u) ] − 11.009305 u mD = 11.09112 u –11.009305 u ; mD = 0.08181 u 931 MeV E = mD c 2 = (0.08181 u) = 76.2 MeV ; E = 76.2 MeV 1u 33
- 9. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition EB 76.2 MeV EB = ; = 6.92 MeV/nucleon A 11 A *39-32. Find the binding energy per nucleon for Thallium-206. mD = [ 81(1.007825 u) + 125(1.008665 u) ] − 205.976104 u = 1.740846 u 931 MeV EB 1621 MeV E = mD c 2 = (1.740846 u) ; = = 7.87 MeV/nucleon 1u A 206 *39-33. Calculate the energy required to separate the nucleons in mercury-204. mD = [ 80(1.007825 u) + 124(1.008665 u) ] − 203.973865 u = 1.7266 u 931 MeV E = mD c 2 = (1.7266 u) ; EB = 1610 MeV 1u *39-34. The half-life of a radioactive sample is 6.8 h. How much time passes before the activity drops to one-fifth of its initial value? n R 1 1 = = ; 0.200 = (0.5) n ; n log(0.5) = log(0.2); − 0.301 n = −0.699 R0 5 2 0.699 t 6.8 h 6.8 h n= = 2.32; n = = = 2.32; T½ = ; T½ = 2.93 h 0.301 T½ T½ 2.32 *39-35. How much energy is required to tear apart a deuterium atom? (1 H = 2.014102 u) 2 mD = [ 1(1.007825 u) + 1(1.008665 u) ] − 2.014102 u = 0.002388 u 931 MeV E = mD c 2 = (0.002388 u) ; EB = 2.22 MeV 1u 34
- 10. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition *39-36. Plutonium-232 decays by alpha emission with a half-life of 30 min. How much of this substance remains after 4 h if the original sample had a mass of 4.0 g? Write the equation for the decay? t / T½ 8 1 t 4h 1 m = m0 ; = = 8.00; m = (4 g) ; m = 15.6 mg 2 T½ 0.5 h 2 232 94 Pu → 228 92 U + 4 He + energy 2 *39-37. If 32 x 109 atoms of a radioactive isotope are reduced to only 2 x 109 atoms in a time of 48 h, what is the half-life of this material? n N 2 x 109 1 = 9 = ; 0.0625 = (0.5) n ; n log(0.5) = log(0.0625); − 0.301 n = −1.204 N 0 32 x 10 2 1.204 t 48 h 48 h n= = 4.00; n = = = 4.00; T½ = ; T½ = 12.0 h 0.301 T½ T½ 4.00 *39-38. A certain radioactive isotope retains only 10 percent of its original activity after a time of 4 h. What is the half-life? n N 1 1 = = ; 0.100 = (0.5) n ; n log(0.5) = log(0.100); − 0.301 n = −1.00 N 0 10 2 1.00 t 4h 4h n= = 3.32; n = = = 3.32; T½ = ; T½ = 72.3 min 0.301 T½ T½ 3.32 *39-39. When a 6 Li nucleus is struck by a proton, an alpha particle and a produce nucleus are 3 released. Write the equation for this reaction. What is the net energy transfer in this case? 6 3 Li + 1H → 2 He + 3 He + energy 1 4 2 35
- 11. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 931 MeV E = 6.015126 u + 1.007825 u - 4.002603 u - 3.016030 = 0.00432 u = 4.02 MeV 1u *39-40. Uranium-238 undergoes alpha decay. Write the equation for the reaction and calculate the disintegration energy. 238 92 U→ 224 90 4 Th + 2 He + energy E = 238.05079 u - 234.04363 u - 4.002603 u = 0.00456 u 931 MeV E = 0.00456 u ; E = 4.24 MeV 1u *39-41. A 9-g sample of radioactive material has an initial activity of 5.0 Ci. Forty minutes later, the activity is only 3.0 Ci. What is the half-life? How much of the pure sample remains? n R 3 Ci 1 = = ; 0.600 = (0.5) n ; n log(0.5) = log(0.6); − 0.301 n = −0.222 R0 5 Ci 2 0.222 t 40 min 40 min n= = 0.737; n = = = 0.737; T½ = ; T½ = 54.3 min 0.301 T½ T½ 0.737 n 0.737 1 1 m = m0 = (9 g) m = 5.40 g 2 2 Critical Thinking Problems *39-42. Nuclear fusion is a process that can produce enormous energy without the harmful byproducts of nuclear fission. Calculate the energy released in the following nuclear fusion reaction: 3 2 H + 3 He →2 He +1 H+1H + energy 2 4 1 1 931 MeV E = 2(3.016030 u) - 4.002603 u - 2(1.007825 u) = 0.013807 u 1u 36
- 12. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition E = 12.9 MeV *39-43. Carbon-14 decays very slowly with a half-life of 5740 years. Carbon dating can be accomplished by seeing what fraction of carbon-14 remains, assuming the decay process began with the death of a living organism. What would be the age of a chunk of charcoal if it was determined that the radioactive C-14 remaining was only 40 percent of what would be expected in a living organism? n R 1 = 0.40 = ; 0.400 = (0.5) n ; n log(0.5) = log(0.4); − 0.301 n = −0.399 R0 2 0.399 t t n= = 0.737; n = = = 1.32; t = 1.32(5740 yr); t = 7590 yr 0.301 T½ 5740 yr *39-44. The velocity selector in a mass spectrometer has a magnetic field of 0.2 T perpendicular to an electric field of 50 kV/m. The same magnetic field is across the lower region. What is the velocity of singly charged Lithium-7 atoms as they leave the selector? If the radius of the path in the spectrometer is 9.10 cm, find the atomic mass of the lithium atom? E 50, 000 V/m v= = ; v = 2.5 x 105 m/s B 0.2 T mv 2 qBR (1.6 x 10-19 C)(0.2 T)(0.091 m) R = qvB; m = ; m= ; R v (2.5 x 105 m/s) 1u m = 1.165 x 10-26 kg -27 ; m = 7.014 u B = 0.2 T into paper 1.66 x 10 kg *39-45. A nuclear reactor operates at a power level of 2.0 MW. Assuming that approximately 200 MeV of energy is released for a single fission of U-235, how many fission processes are occurring each second in the reactor? 37
- 13. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 2 x 106 J/s 1.25 x 1019 MeV/s P= = 1.25 x 1019 MeV/s; = 6.25 x 10216 fissions/s 1.6 x 10-19 J/MeV 200 MeV/fission 14 *39-46. Consider an experiment which bombards 7 N with an alpha particle. One of the two product nuclides is 1 H . The reaction is 1 4 2 He +14 N → Z X +1 H 7 A 1 What is the product nuclide indicated by the symbol X? How much kinetic energy must the alpha particle have in order to produce the reaction? Conservation of nucleons means: 4 2 He +14 N + enery → 7 17 8 O +1 H 1 E = 16.99913 u + 1.007825 u - 4.002603 u - 14.003074 u = 0.001282 u 931 MeV E = 0.001282 u ; E = 1.19 MeV 1u *39-47. When passing a stream of ionized lithium atoms through a mass spectrometer, the radius of the path followed by 7 Li (7.0169 u) is 14.00 cm. A lighter line is formed by the 6 Li 3 3 6 (6.0151 u). What is the radius of the path followed by the 3 Li isotopes? mv 2 mv = qvB; R = ; (See Problems 39-10 and 39-11) R qB m1 R1 m2 R1 (6.0151 u)(14 cm) = ; R2 = = ; R2 = 6.92 cm m2 R2 m1 7.0169 u 38

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