Physics Term 3 STPM Chapter 25 Nuclear Physics 295 25 9. If mn = mass of neutron mp = mass of proton mN = mass of nucleus Binding energy, E = (Δm) c 2 = [Σmn + Σmp – mN] c 2 10. Definition: The binding energy of a nucleus is the energy required to separate completely all the nucleons in the nucleus. 11. The binding energy per nucleon of a nucleus is a measurement of the stability of the nucleus. Binding energy per nucleon = Binding energy of nucleus Nucleon number of nucleus 12. Nuclei which are more stable have higher values of binding energy per nucleon. 10 50 100 150 200 250 8 6 4 2 0 Binding energy per nucleon (MeV) Nucleon number, A H-2 He-4 Fe-56 U-238 Release of energy Release of energy Figure 25.1 13. Figure 25.1 shows the variation of binding energy per nucleon with the nucleon number, A. The maximum value is about 9 MeV for nucleus of nucleon number around 60. 14. If the binding energy per nucleon is high, the nucleons are at a lower energy state. In a nuclear reaction, if the binding energy per nucleon of the product nuclei is higher, then energy is released. 15. Examples of such reaction are: (a) Radioactive decay of U-238. When U-238 decays by α-emission, the products of the decay have smaller nucleon numbers. From the graph in Figure 18.3, when the nucleon number decreases from 238, the binding energy per nucleon increases.
Physics Term 3 STPM Chapter 25 Nuclear Physics 296 25 Hence, the nucleons of the resulting nucleus are in lower energy states. This results in the release of energy. (b) Fusion of deuterium (H-2) During the fusion of two deuterium nuclei to from a larger nucleus, the larger nucleus has greater binding energy per nucleons. The nucleons are in a lower energy state. Hence, energy is released. Example 1 Calculate the binding energy per nucleon of a helium-4 ( 4 2 He) nucleus. Given mn, mass of neutron = 1.0087u mp , mass of proton = 1.0078u mHe mass of 4 2 He nucleus = 4.0026u Solution: Mass defect of He-4 nucleus Δm = (2 mp + 2 mn) – mHe = 2(1.0078 + 1.0087) – 4.0026u = 0.0304u Binding energy, E = (Δm)c2 = (0.0304)(1.66 × 10–27)(3.0 × 108 )2 J = 0.0304 × (1.66 × 10–27) × (3.0 × 108 )2 (1.6 × 10–19) × 106 MeV = 28.3 MeV Binding energy per nucleon = 28.3 4 MeV = 7.10 MeV Example 2 Find in MeV, the energy equivalent for a mass of 1 u. Solution: 1 u = 1.66 × 10–27 kg Using E = mc 2 = (1.66 × 10–27) × (3.0 × 108 )2 J = (1.66 × 10–27) (3.0 × 108 )2 (1.6 × 10–19) × 106 MeV = 934 MeV Exam Tips When using the equation E = (Δm) c 2 Δm must be in kg. To convert Δm in u to kg, multiply by (1.66 × 10–27) Since 1u = (1.66 × 10–27) kg Exam Tips Unless otherwise stated, you can use 1 u ; 934 MeV in calculation of binding energy. Find the mass defect (Δm) in u. Then binding energy = (Δm)(934 MeV)
Physics Term 3 STPM Chapter 25 Nuclear Physics 297 25 Example 3 By definition, the mass of an atom of 12 6C is exactly 12u. Find the total mass of the constituents particles of a 12 6 C atom, expressing your answer in u. Explain why your answer is not exactly 12u? Rest mass of: electron, me = 9 × 10–31 kg proton, mp = 1.6726 × 10–27 kg neutron, mn = 1.6750 × 10–27 kg Atomic mass unit, u = 1.6606 × 10–27 kg Solution: Total mass of constituents = 6mp + 6mn + 6me = 6 [1.6726 + 1.6750 + 9 × 10–4] 10–27 kg = 20.091 × 10–27 kg = 20.091 × 10–27 1.6606 × 10–27 u = 12.10u The total mass of the constituents (12.10u) is more than the mass of the C-12 atom because the nucleons are bounded in the nucleus. The nucleons are in a lower energy state compared to when they are outside the nucleus. Example 4 A nuclide is represented by the symbol 81 35Br. (a) Write down (i) the number of protons, (ii) the number of neutrons in the nucleus. (b) The mass of the nucleus is 80.8971u. If the masses of a proton, and a neutron are 1.0073u and 1.0087u respectively, deduce the binding energy per nucleon. (c) Explain the implication of low binding energy per nucleon on the stability of a nucleus. Solution: (a) (i) Number of proton = 35 (ii) Number of neutron = (81 – 35) = 46 (b) Total mass of necleons = 35 (1.0073)u + 46 (1.0087)u = 81.6557u mass defect, Δm = (81.6557 – 80.8971)u = 0.7586 u = 0.7586 (1.66 × 10–27) kg Binding energy, E = (Δm) c2 = (0.7586 × 1.66 × 10–27) (3.0 × 108 )2 = 1.13 × 10–10 J Binding energy per nucleon = 1.13 × 10–10 81 = 1.40 × 10–12 J (c) A nucleus with low binding energy per nucleon is less stable because the nucleons are in a higher energy state compared to nucleons in nucleus with high binding energy per nucleon.
Physics Term 3 STPM Chapter 25 Nuclear Physics 298 25 Quick Check 1 1. The mass of an atom of 12 6C is 12.0000 u, mass of an atom 1 1H is 1.0078 u, and mass of the neutron 1 0n is 1.0087 u. What is the mass defect of the 12 6C nucleus? A 0.0100u C 0.0990u B 0.0054u D 0.1040u 2. The mass of an atom 202 80Hg of is 201.970 617 u, mass of an atom 1 1H is 1.007 825 u, and mass of the neutron 1 0n is 1.008 665 u. What is the binding energy per nucleon of the 202 80Hg nucleus in meV? (1 u ≡ 934 MeV). A 7.83 MeV C 12.8 MeV B 7.92 MeV D 14.4 MeV 3. The binding energy per nucleon may be used as a measurement of the stability of a nucleus. This quantity A is directly proportional to the neutron/ proton ratio of the nucleus B is a maximum for nuclides in the middle of the Periodic table C increases uniformly as the proton number increases D has maximum values for nuclei of the inert gases 4. A nucleus which has nucleon number A and proton number Z is of mass M. If the masses of the proton, and neutron are p and n respectively, the binding energy of the nucleus is directly proportional to A pA + nZ – M B pZ + nA – M C pA + n (Z – A) – M D pZ + n (A – Z) – M 5. The fi gure shows a graph of the binding energy per nucleon for a number of naturally-occuring nuclides plotted against their nucleon number. 40 80 120 160 200 240 10 8 6 4 2 0 Binding energy per nucleon /MeV Nucleon number H 2 1 Na 23 11 AI 27 13 Br 81 35 U 238 92 Which of the following deduction from the graph is correct? A 1 2 H is the most stable nuclide B 27 13Al will not decay to 23 11Na C 238 92U is the stable end-nuclide for a radioactive series D Energy will be released when 82 35Br undergoes fusion with another nuclues 6. The figure below is a graph of binding energy per nucleon against nucleon number. Binding energy per nucleon (Me V) Nucleon number 8.7 7.4 0 56 238 What is the binding energy for the nucleus 238 92U? A 309 MeV C 681 MeV B 487 MeV D 1 761 MeV 7. The masses of 16 8 O nucleus, 1 1 H nucleus, and neutron are 15.995u, 1.0078u and 1.0087u respectively. What is the binding energy per nucleon of the nucleus 16 8 O? A 6.40 × 10–13 J C 2.56 × 10–12 J B 1.28 × 10–12 J D 1.51 × 10–10 J 8. The mass equivalent of one joule of energy is A 1.1 × 10–17 kg C 3.0 × 10–8 kg B 3.3 × 10–9 kg D 9.0 × 10–16 kg 9. A nuclide is more stable if A the number of proton equals to the number of neutron. B the binding energy is high. C there are more neutron than proton. D the binding energy per nucleon is high.
Physics Term 3 STPM Chapter 25 Nuclear Physics 299 25 10. The masses of the nucleus 12 6 C and 13 6 C are m1 and m2 respectively. If mp = mass of proton and mn = mass of neutron, which of the following is correct? A mp = m2 – m1 B mn = m2 – m1 C mn > m2 – m1 D mp < m2 – m1 11. The binding energy per nucleon is maximum for nucleon numbers between A 1 and 20 B 60 and 80 C 100 and 120 D 220 and 240 12. If the mass defect of a nucleus is Δm, and E is the binding energy of the nucleus, which of the following is true? A E is the kinetic energy of Δm B E is the potential energy of Δm C E is the internal energy of Δm D E is the energy equivalent of Δm 13. What is the energy equivalent of a mass of 0.025u? A 1.41 MeV C 23.34 MeV B 2.25 MeV D 43.20 MeV 14. Bombardment of a certain material with α-particles produces an emission which penetrates lead, ejects protons from paraffin wax, and travels at speed up to 5 × 107 m s–1. What does this emission consist of? A X-rays C Protons B Neutrons D Electrons 15. The graph shows how the binding energy per nucleon varies with the nucleon number for naturally occuring nuclides. Binding energy per nucleon/ p J Nucleon number 1.4 1.3 64 156 What is the binding energy of the nuclide 156 64Gd? A 15.6 pJ B 83 pJ C 90 pJ D 203 pJ 16. Find the mass equivalent for (a) one joule of energy, (b) 934 MeV of energy. 17. Deuterium is represented by the symbol 2 1 H. Name the nucleons which constitute the nucleus. Use the data below to calculate the binding energy of the deuterium nucleus. Mass of deuterium atom = 2.01410 u Mass of hydrogen atom = 1.00783 u Rest mass of neutron, mn = 1.00867 u 18. (a) Explain what is meant by (i) nucleon number A, (ii) proton number Z, (iii) binding energy per nucleon of a nucleus. (b) (i) Explain why the sum of the rest mass of the nucleon is always greater than the mass of the nucleus. (ii) The mass of the nucleus of 52 24Cr is 51.940509u. The rest mass of a neutron is 1.008665u and that of a proton is 1.007276u. Calculate the binding energy per nucleon of 52 24Cr . 19. (a) Write an equation representing Einstein mass-energy relationship. Explain the symbols you use. (b) Write an expression for the binding energy of the nucleus Z A X in terms of mp, mass of proton, mn mass of neutron, mX mass of the nucleus Z AX, and c the speed of light. 20. (a) Explain what is meant by (i) nucleon. (ii) nucleon number. (iii) proton number. (b) Explain why the mass of a nucleus is less than the sum of the masses of its constituents.
Physics Term 3 STPM Chapter 25 Nuclear Physics 300 25 25.(a) (i) Define the atomic mass unit, u. (ii) Use the definition of u to obtain the conversion factor from u to kg. (b) An atom of magnesium has 12 electrons each of mass 0.00055u, 12 protons each of mass 1.00728u, 13 neutrons each of mass 1.00866u. (i) What are the constituents of the magnesium nucleus, and what is its nucleon number? (ii) Which of the numbers mentioned in (b) determines that the atom is magnesium and not any other atom? (iii) The mass of the magnesium atom is 24.98584u. Calculate the sum of the mass constituents of an atom of magnesium. Explain for any difference with the mass of the atom. (iv) Calculate in joule, the binding energy of the atom. 26. (a) Explain what is meant when two nuclides are said to be isotopes. (b) (i) State the number of neutrons and the number of protons contained in a single nucleus of the nuclide 14 6C. (ii) Write the symbol for another possible isotope of carbon. (iii) Give a similarity, and a difference between the two isotopes of carbon. (c) An approximate relationship between the radius R of a nucleus and its nucleon number A is R/m = 1.2 × 10–15 A 1 — 3 Estimate (i) the number of nucleons per unit volume of the nucleus, (ii) the density of a nucleus. (Take the proton mass mp and neutron mass mn as 1.7 × 10–27 kg). 21. (a) What is meant by binding energy per nucleon of a nucleus? (b) Sketch a labelled graph to show the variation with nucleon number of the binding energy per nucleon. (c) Explain in terms of binding energy why energy is released during a decay of the radioactive isotope uranium-238. 22. In a mass spectrometer, the magnetic flux density is 0.95 T and the electric field intensity is 3.5 × 106 V m–1. If the specific charge q m of the hydrogen ion H+ is 9.65 × 107 C kg–1, calculate the radii of the tracks in the magnetic field of (a) the H+ ion. (b) the He2+ ion. 23. What is the velocity of positive ions which are undeflected in a region where an electric field of intensity 8.1 × 104 V m–1 and a magnetic field of flux density 0.370 T are acted? How must be two fields be orientated so that the ions are not deflected? 24. (a) What is meant by (i) the atomic mass unit, u. (ii) mass defect of a nucleus. (iii) binding energy of a nucleus. Write an expression relating (ii) and (iii). (b) The mass of 20 10 Na nucleus is 19.99244u. The rest masses of a neutron and a proton are 1.008665u and 1.007825u respectively. (i) Calculate the binding energy of 20 10Na nucleus. (ii) Hence, find the binding energy per nucleon of 20 10Na. (c) (i) Sketch a graph to show how the binding energy per nucleon depends on the nucleon number. (ii) Explain the shape of the graph for heavy nucleus. (iii) Fusion of two deuteron produces an isotope of helium 3 2 He. Write an equation to represent the fusion reaction. Show how the reaction can be explained using the graph you sketched.
Physics Term 3 STPM Chapter 25 Nuclear Physics 301 25 25.2 Radioactivity Learning Outcomes Students should be able to: • explain radioactive decay as a spontaneous and random process • defi ne radioactive activity • state and use the exponential law dN dt = −λN for radioactive decay • defi ne decay constant • derive and use the formula N = N0 e−λt • defi ne half-life, and derive the relation λ = In 2 t 1/2 • solve problems involving the applications of radioisotopes as tracers in medical physics Radioactive Decay 2008/P1/Q48, 2009/P1/Q49, 2010/P1/Q49, 2008/P2/Q8, 2011/P2/Q8 1. Radioactivity is the spontaneous and random emission of radiations from uranium and other radioactive materials. 2. Radioactive radiations are emitted when an unstable nucleus decays. The radiations are α-particles, β-particles and γ-rays. 3. A nucleus is unstable when there are either (a) too many protons in the nucleus for stability, or (b) too many neutrons in the nucleus for stability. 4. The decaying process of an unstable nucleus is spontaneous – that is unplanned, cannot be predicted and independent of physical conditions such as pressure and temperature. 5. The process is random because the probability of a nucleus decaying at a given instant is the same for all nuclei in the sample. 6. The rate of decay or activity of a radioactive sample, – dN dt , is directly proportional to the number of radioactive nuclei N in the sample. Rate of decay, – dN dt ∝ N dN dt = –λN where λ is a constant, known as the decay constant of the radioactive nuclide. 7. Decay constant, λ = N – dN dt = Rate of decay Number of radioactive nuclei in the sample Hence, the decay constant of a radioactive nuclide is the probability that a radioactive nucleus in the sample would decay in one second. Info Physics The word ‘radioactive’ was introduced by Marie Curie and her husband, Pierre in 1898 after they successfully extract polonium and radium from ore of uranium. 2013/P3/Q14,Q17, 2014/P3/Q15, Q17, 2015/P3/Q13, 2017/P3/Q15
Physics Term 3 STPM Chapter 25 Nuclear Physics 302 25 8. From the equation, dN dt = –λN dN dt = –λ dt Integrating, and when t = 0, N = N0 , and when t = t, N = N, ∫N0 N dN N = –λ ∫0 t dt ln [N] N N0 = –λ [t]t 0 ln N N0 = –λt N N0 = e–λt N = N0 e–λ t This equation shows that the number of radioactive nuclei in a sample decreases exponentially with time. 9. Figure 25.2 shows the graph of N, number of radioactive nuclei against time, t. 0 N0 N = N0e–λt 2 N0 N0 16 Number of radioactive nuclei, N Time, t N0 4 N0 8 t 1 2 2t 1 2 3t 1 2 4t 1 2 5t 1 2 t 1 2 = Half-life Figure 25.2 10. The half-life (t1) 2 of a radioactive nuclide is the time taken for half of the number of nuclei in a sample of the radioactive nuclide to decay. The half-life (t1) 2 of a radioactive nuclide is also the time taken for a sample of the nuclide to decay to half of its initial number of nuclei.
Physics Term 3 STPM Chapter 25 Nuclear Physics 303 25 11. The time taken for (a) N0 radioactive nuclei to decay to 1 2 N0 is t1 2 , (b) 1 2 N0 radioactive nuclei to decay to 1 2 1 2 N0 or 1 4 N0 is t1 2 , (c) 1 4 N0 radioactive nuclei to decay to 1 2 1 4 N0 or 1 8 N0 is t1 2 , and so on. 12. Using the equation N = N0 e –λt when t = t1 2 , N = N0 2 then N0 2 = N0 e –λt — 1 2 2 = eλt — 1 2 ln 2 = λ t1 2 Half-life, ln 2 λ t1 2 = 13. Table 25.1 shows the half-life of some radioactive nuclides. Table 25.1 Half-life Nuclide Half life, t1 2 238 92 U 4.5 × 109 years 222 86 Rn 3.8 days 212 83 Bi 61 minutes 234 90 Th 24 days 14. The S.I. unit for activity or rate of decay is the becquerel (Bq). 1 Bq = Rate of decay of 1 nucleus per second Example 5 Thorium-234 has a half-life of 24 days. The initial activity of a sample of thorium-234 is 8.0 × 104 Bq. (a) What is the activity of the sample after 72 days? (b) How long does the sample take for its activity to become 3.0 × 104 Bq? Solution: (a) Half-life, t1 2 = 24 days t = 72 days = 3t1 2
Physics Term 3 STPM Chapter 25 Nuclear Physics 304 25 Hence activity after t = 3t1 2 is dN dt = 1 2 3 dN dt = 1 8 × (8.0 × 104 ) = 1.0 × 104 Bq (b) dN dt 0 = 8.0 × 104 Bq, dN dt = 3.0 × 104 Bq Suppose time taken, t = xt1 2 then dN dt = 1 2 x dN dt 0 3.0 × 104 = 1 2 x (8.0 × 104 ) (2x ) = 8.0 3.0 = 2.67 x = lg 2.67 lg 2.0 = 1.42 Time taken, t = 1.42 × 24 days = 34.1 days Alternative method: The activity A = dN dt at any time t can be written as A = A0 e–λt (identical to N = N0 e–λt ) A = 3.0 × 104 Bq, A0 = 8.0 × 104 Bq, λ = ln 2 t 1 2 3.0 × 104 = (8.0 × 104 ) e–λt eλt = 8.0 3.0 = 2.67 λt = ln 2.67 ln 2 t 1 2 t = ln 2.67 Time taken, t = ln 2.67 ln 2.0 24 days = 34.1 days Example 6 The half-life of a certain radioactive nuclide is such that 7 8 of a given quantity decays in 12 days. What is the fraction remains undecayed after 24 days?
Physics Term 3 STPM Chapter 25 Nuclear Physics 305 25 Solution: If 7 8 of a given quantity decays in 12 days, 1 8 remains. Hence N = 1 8 N0 = 1 2 3 N0 time taken is 3t1 2 = 12 days t1 2 = 4 days When t = 24 days = 6t1 2 N = 1 2 6 N0 Fraction remains = 1 2 6 = 1 64 Example 7 The isotope 234 90 Th has a half-life of 24 days and decays to 234 91 Pa. What is the time taken for 90% of a sample of 234 90 Th to become 234 91 Pa? Solution: When 90% has decay, 10% of Th remains Hence, N N0 = 1 10 Using N = 1 2 x N0 2x = N0 N = 10 x = lg 10 lg 2 Time taken, t = xt1 2 = lg 10 lg 2 24 days = 79.7 days Example 8 A radioactive source contains 1.0 × 10–6 g of plutonium-239. The source emits 2 300 α-particles per second. Calculate the half-life of plutonium-239.
Physics Term 3 STPM Chapter 25 Nuclear Physics 306 25 Solution: 1 mole (or 239 g) of plutonium contains 6.02 × 1023 atoms Number of plutonium atoms in 1.0 × 10–6 g N = (1.0 × 10–6) 239 × (6.02 × 1023) atoms Rate of decay, – dN dt = 2 300 s–1 Using – dN dt = λN, λ = ln 2 t 1 2 – dN dt = ln 2 t 1 2 N t1 2 = (ln 2) N dN dt – = (ln 2) × (1.0 × 10–6)(6.02 × 1023) 2 300 × 239 = 7.59 × 1011 s Example 9 A radioactive source contains 1.0 × 10–4 g of uranium-238 and its average rate of decay is 1.2 s–1. The safety limit of radioactive activity is 10 µCi. Calculate the maximum mass of uranium-238 which may be used without shielding. (1 Ci = 3.7 × 1010 Bq) Solution: Activity = 1.2 s–1 = 1.2 3.7 × 1010 Ci = 3.24 × 10–5 µCi Activity of 3.24 × 10–5 µCi is produced by 1.0 × 10–4 g of uranium-238. Hence activity of 10 µCi is due to 10 3.24 × 10–5 × (1.0 × 10–4) g of U-238 = 31 g Example 10 A particular medical application requires a radioactive source with an activity of 3.90 × 103 Bq at the beginning of the treatment. The nuclide used has a half-life of 1.80 × 105 s and is prepared 1 week (6.05 × 105 s) before the treatment commence. What should be the activity of the source at the time of preparation?
Physics Term 3 STPM Chapter 25 Nuclear Physics 307 25 Solution: Half-life t1 2 = 1.80 × 105 s Time t = 6.05 × 105 s = 6.05 × 105 1.80 × 105 × t1 2 = 3.361 t1 2 Using dN dt = 1 2 3.361 dN dt 0 Initial activity, dN dt 0 = 23.361 × (3.90 × 103 ) = 4.01 × 104 Bq Example 11 2 0 1.0 2.0 3.0 4.0 5.0 6.0 4 6 8 10 12 14 Activity / × 103 Bq Time / Days The graph shows the variation of the activity of a sample of radon with time. The sample was in a tube and initially there were 5.0 × 109 radioactive radon atoms when the tube was inserted into the body of a patient. To provide enough dosage for the therapy, the tube was removed after 6 days. From the graph, (a) fi nd the half-life of radon, (b) determine the number of α-particles emitted by the sample in 6 days, (c) if the sample were inserted into the patient’s body one day after being prepared, how long should it be left in the body in order to provide for the dosage required as in (b) above? (d) state any changes to the average intensity of the radiation, (e) give two advantages of using a α-emitter compared to a β-emitter or γ-emitter for the therapy.
Physics Term 3 STPM Chapter 25 Nuclear Physics 308 25 Solution: (a) From the graph, initial activity, A0 = 6 × 103 Bq When activity = A0 2 = 3 × 103 Bq, time = 4.0 days Hence half-life, t1 2 = 4.0 days (b) At time t = 0, activity dN dt 0 = 6.0 × 103 Bq Number of radioactive atoms, N0 = 5.0 × 109 Using dN dt ∝ N (6.0 × 103 ) ∝ 5.0 × 109 ............................. ① At time t = 6.0 days, activity dN dt = 2.1 × 103 Bq (from graph) Hence, (2.1 × 103 ) ∝ N ..............................➁ ② ① : N 5.0 × 109 = 2.1 × 103 6.0 × 103 Number of radioactive atom remaining, N = 1.75 × 109 Number of α-particles emitted in 6.0 days = (5.0 × 109 – 1.75 × 109 ) = 3.25 × 109 (c) One day after the sample was prepared, t = 1.0 day from the graph, activity dN dt ′ 0 = 4.9 × 103 Bq Number of radioactive atoms in sample N′ 0 = 4.9 × 103 6.0 × 103 (5.0 × 109 ) N ∝ dN dt = 4.08 × 109 Dosage required = 3.25 × 109 (from (b) above) Hence, fi nal number of radioactive atoms = (4.08 – 3.25) × 109 = 8.3 × 108 Using dN dt ∝ N Activity when N = 8.3 × 108 is dN dt = (8.3 × 108 ) (5.00 × 109 ) × (6.0 × 103 ) = 996 Bq From the graph, when activity = 996 Bq time = 10.1 days
Physics Term 3 STPM Chapter 25 Nuclear Physics 309 25 Hence the sample must be left in the body for (10.1 – 1.0) days = 9.1 days (d) The intensity of radiation decreased as the same number of α-particles were emitted over a longer period, 9.1 days compared to 6.0 days. (e) Advantages of α-emitter. • higher ionising power of α-particles compared to β-particles or γ-rays. Destruction of cancerous cells more effective. • short range of α-particles are useful for destroying cancerous cells locally, leaving the non-cancerous cells which are further away unharmed. Example 12 The nuclide 90 38 Sr is a β-emitter of half-life 28 years. In the year of 2003 a sample of Sr–90 emits a certain number of β-particles per second. (a) In the year 2059, how long will the sample take to emit the same number of β-particles it emitted in one second in 2003? (b) In which year will the sample take one full year to emit the same number of β-particles it emitted in one second in 2003? Solution: (a) Time t = (2059 – 2003) years = 56 years = 2t1 2 (t1 2 = 28 years) Rate of decay = 1 2 2 dN dt 0 = 1 4 of rate of decay in 2003 Hence the time taken to emit, the same number of β-particles = 4 s. (b) 1 year = (365 × 24 × 60 × 60)s = 2n s n = 24.9 If in 2003, the rate of decay, dN dt 0 = A s–1 The fi nal rate of decay, dN dt = A 1 year = A (365 × 24 × 60 × 60) s–1 = A 224.9 = 1 224.9 dN dt 0 Hence, time taken t = 24.9 t1 2 = 24.9 × 28 = 697 years later or in the year (2003 + 697) = 2700
Physics Term 3 STPM Chapter 25 Nuclear Physics 310 25 Example 13 The nuclide 24 11Na is radioactive. It is produced at a constant rate k from the stable isotope 23 Na in a nuclear reactor. (a) Construct an equation relating δN, the small increases in the number of 24 Na atoms, in time δt to λ the decay constant of 24 Na, and k, the rate of production of 24 Na. (b) Then show that the number of 24 Na atoms present in the reactor will eventually tend to a constant value. Solution: (a) Using – dN dt = λN in time δt, the decrease in the number of 24 Na atoms – dN = λN δt In time δt, the increase in the number of 24 Na atom = k δt Hence the net increase in the number of 24 Na atoms in the sample δN = k δt – λN δt = (k – λN) δt (b) From δN = (k – λN) δt dN (k –λN) = dt Integrating dN (k –λN) ∫ 0 N = ∫ 0 t dt – 1 λ [ln (k – λN)]N 0 = [t] t 0 – 1 λ ln (k – λN) = t ln (k – λN) = –λt k – λN = e–λt N = k – e–λt λ When t → ∞, e–λt → 0, N = k λ (constant) Quick Check 2 1. The mass of a sample of radium-226 is 1.00 g. The half life of radium-226 is 1620 years. What is the activity of the sample of radium-226? A 3.61 × 1010 Bq C 8.66 × 1011 Bq B 5.21 × 1010 Bq D 1.25 × 1012 Bq 2. A sample of phosphorous-32 (P-32) has an activity of 55 MBq. The decay constant of P-32 is 5.6 × 10–7 s–1. What is the mass of the sample? A 5.8 × 10–7 g C 5.2 × 10–9 g B 1.8 × 10–8 g D 1.9 × 10–9 g 3. The equation dN dt = –λN gives the rate of decay of a radioactive sample which contains N radioactive atoms. Which of the following statements about λ is correct? A λ δt gives the fraction of atoms present which will decay in the next small time interval δt
Physics Term 3 STPM Chapter 25 Nuclear Physics 311 25 B λ is equal to the half-life of the sample C λ is the number of atoms left after a time equal to e seconds D λ is the probability that any one atom will not decay after one second 4. What is the relation between the decay constant λ and the half-life t 1 2 of a radioactive isotope? A λ = —– 1 t 1 2 C λ = —–– ln 2 t 1 2 B λ = t 1 2 D λ = t 1 —–– 2 ln 2 5. In 1.0 minute a sample of radioactive isotope decays to 1 16 of its initial number of atoms. The decay constant of the isotope is A 2.2 × 10–2 s–1 C 6.7 × 10–2 s–1 B 4.6 × 10–2 s–1 D 1.5 × 10 s–1 6. At time t = 0, some radioactive gas is injected into a sealed tube. At time = T, some more of the same gas is injected into the tube. Which one of the following graphs best represents the variation of the logarithm of the activity A of the gas with time t? A T t In A 0 B T t In A 0 C T t In A 0 D T t In A 0 7. The radioactive decay of a certain nuclide is governed by the equation dN dt = –λN where λ = 2.4 × 10–8 s–1. What is the half-life of the nuclide? A 2.9 × 107 s C 1.2 × 10–8 s B 1.3 × 107 s D 3.4 × 10–8 s 8. The half-life of a certain radioactive nuclide is such that 7 8 of a given quantity decays in 12 days. What is the total fraction remains undecayed after 24 days? A 1 16 C 1 64 B 1 32 D 1 128 9. The graph represents the decay of a sample of radioactive nuclide X to a stable nuclide Y. The half-life of X is t, the growth curve for Y intersects the decay curve for X after time T. T Time X Y 0 Number of atoms What is the time T ? A 1 2 t C ln t 2 B t D ln (2t) 10. A sample contains a radioactive nuclide X, and another sample contains a radioactive nuclide Y. After a time interval 3 4 of X and 7 8 of Y have decayed. What is the value of the ratio ? half-life of X half-life of Y A 3 7 C 7 6 B 2 3 D 3 2
Physics Term 3 STPM Chapter 25 Nuclear Physics 312 25 11. At time t = 0, the mass of a radioactive element of half-life 5.0 minutes in a container is 32 mg. At time =10 minutes, 4 mg of the same radioactive element is injected into the container. What is the mass of the radioactive element in the container at time t =15 minutes? A 2 mg C 6 mg B 4 mg D 8 mg 12. The half-life of I-131 is 8 days. What is the fraction of a sample of I-131 had decayed after 48 days? A 3 4 C 1 8 B 5 6 D 63 64 13. (a) The initial number of radioactive nuclei in a sample is N0 , and its decay constant is l. Starting from the equation for the rate of decay dN dt = –lN deduce the expression for the number of radioactive nuclei in the sample N after a time t. (b) Technetium-99 is used as a radioactive tracer in medicine. It has a half live of 2.16 × 104 s. The mass of a sample of technetium-99 is 40.5 µg. Calculate (i) the initial activity of the sample. (ii) the activity of the sample after 1.0 hour. 14. P-32 is a β-emitter with decay constant 5.6 × 10–7 s–1. For a particular application, the initial decay rate must yield 4.0 × 107 β-particles every second. What is the mass of P-32 which will give this decay rate? 15. (a) Write an expression relating the rate of decay – dN dt of a sample of N radioactive nuclei to the decay constant λ. What are the dimensions of λ? (b) Initially, a sample contains 1.0 × 106 radioactive nuclei of half-life t. Estimate how many radioactive nuclei will remain after a time 0.5t. 16. Time / minutes Activity / minute–1 10 28 0 20 000 12 000 The above graph shows the decay of a sample of radioactive isotope. Calculate the half-life of the isotope. 17. The table shows how C, the count rate measured in a sample of radioactive gas, varies with time t. Time t/s Count rate C/s–1 0 20 40 60 80 100 120 140 160 180 200 400 600 800 1 000 1 200 1 400 86.1 67.8 52.9 45.2 33.2 24.1 20.1 17.8 13.7 9.8 8.3 3.6 2.4 2.2 2.6 2.3 2.5 (a) Describe briefly how the count rate C could be measured. (b) Use the data to tabulate R, the count rate due to the radioactive gas alone. (c) (i) Plot a graph of ln (R/s–1) against t/s. (ii) Deduce the decay constant and half-life of the radioactive gas. (iii) Explain why a graphical method is appropriate to handle the data and why a logarithmic graph is used in this case. (d) The gas considered is present in many houses at a constant concentration. How is this possible when the gas has such a short half-life?
Physics Term 3 STPM Chapter 25 Nuclear Physics 313 25 Uses of Radioisotopes Carbon-14 Dating 1. Carbon-14 (14 6C) is an isotope of carbon which is radioactive and has a half-life of 5 700 years. 2. Carbon-14 is formed in the atmosphere by the reaction between nitrogen (14 7N) and cosmic rays from outer space. 14 7 N + 1 0 n → 14 6C + 1 1 H (from cosmic rays) 3. The ratio of C-14 to C-12 in the atmosphere is small (about 1 : 1012) but is constant. 4. Carbon-14 in carbon dioxide of the atmosphere is absorbed by living things through photosynthesis and respiration. 5. Hence, the ratio of C-14 to C-12 in a living thing is the same as that in the atmosphere. 6. When a living thing, a plant or an animal, dies, the process of exchanging carbon dioxide with the atmosphere stops, and carbon-14 in the dead plant or animal decreases. 7. By measuring the activity of carbon-14 in an archeology specimen, such as a piece of bone, the archeology age of the specimen can be estimated. 8. The activity of carbon-14 in the atmosphere and in living things is 0.26 Bq per gram of carbon. If the activity of carbon-14 from an archeology specimen is 0.13 Bq, which is half the activity of ‘life’ carbon, than the archeological age of the sample is 5 700 years (t1 2 of C-14). Example 14 One gram of carbon from a ancient piece of wood gives a count rate of 700 hour –1. Whereas 1.00 g of carbon from a living plant gives a count rate of 900 hour –1. Calculate the archeological age of the piece of wood, if the half-life of carbon-14 is 5 700 years. Solution: Initial activity of 1.00 g of carbon A0 = 900 hour –1 Activity of 1.00 g of carbon after time = t A = 700 hour –1 Using A = 1 2 x A0 700 = 1 2 x 900 2 x = 9 7 x = lg 1.286 lg 2 = 0.3628 Time t = x t 1 2 = 0.3628 (5 700) = 2 070 years
Physics Term 3 STPM Chapter 25 Nuclear Physics 314 25 Radioactive Dating 1. Radioactive elements which have long half-lifes are used to determine the geological age of rocks. 2. In the uranium-238 series, 238 92 U decays and produces the following nuclides: 238 α β β α α α 92 U → 234 90 Th → 234 91 Pa → 234 92 U → 230 90 Th → 226 88 Ra → 222 86 Rn → ... → 206 82 Pb 3. The half-life of uranium-238 is 4.5 × 109 years. After a long period, radioactive equilibrium is achieved. 4. When radioactive equilibrium is achieved, the rate at which a nuclide in the series decays is the same as the rate it is formed by the decay of the nuclide before it in the series. 5. Hence, the rate of decay of all members of the series is the same, and is also equal to the rate at which 206 82 Pb which is stable is formed. 6. Consequently, the quantity of uranium-238 in a piece of uranium ore decreases with time, and the quantity of Pb-206 in the ore increases. The quantity of the other nuclides in the uranium series remains constant. 7. By measuring the ratio of Pb-206 to U-238 in a sample of the uranium ore, the geological age of the sample can be estimated. 8. When radioactive equilibrium is achieved, rates of decay of all nuclides in the uranium series are equal. dN dt U = dN dt Th = dN dt Pa = ........ λUNU = λTh NTh = λPaNPa = ........ λ = —–– ln 2 t 1 2 NU t U = NTh t Th = NPa t Pa = ........ where λU, λTh, λPa, ... are the decay constants of U, Th, Pa, ... t u, t Th, t Pa, ... are the half-lifes of U, Th, Pa respectively. Hence, —– N t 1 2 = constant or N ∝ t 1 2 . The number of atoms of a nuclide in the ore is directly proportional to its half-life. There are much more atoms of U-238 (t1 2 = 4.5 × 109 years) compared to the number of atoms of Rn-222 (t1 2 = 4 days) in a piece of uranium ore.
Physics Term 3 STPM Chapter 25 Nuclear Physics 315 25 Example 15 (a) Plot an accurate graph to show how the number of atoms of a radioactive element varies with time. For the time-axis, extend it to fi ve half-lifes, and the other axis, state the number of atoms as a fraction of the initial number of atoms N0 . (b) K-40, an isotope of potassium, has a half-life of 1.37 × 109 years and decays to Ar-40. In a particular sample of Moon rock, the ratio of potassium atoms to argon atoms is 1 : 7. Estimate the age of the rock, assuming that originally there was no argon present. State one other assumption which you have made in your calculation. (c) In another sample of rock, the value of the ratio is 1 : 4. Using your graph to estimate the age of this sample. Solution: (a) 0 0.2N0 0.4N0 0.6N0 0.8N0 N0 Number of atoms Time t 1 2 2t 1 2 3t 1 2 4t 1 2 5t 1 2 6t 1 2 (b) Number of K atoms Number of Ar atoms = 1 7 Hence Number of K atoms Total number of Ar and K atoms = 1 7 + 1 = 1 8 Number of K atoms remaining = 1 8 N0 = 1 23 N0 Age of rock, t = 3 t 1 2 = 3 × (1.37 × 109 ) years = 4.1 × 109 years Assumption: K-40 is not formed by the decay of another radioactive isotope in the rock or Ar-40 is stable
Physics Term 3 STPM Chapter 25 Nuclear Physics 316 25 (c) Number of K atoms Number of Ar atoms = 1 4 Number of K atoms Total number of Ar and K atoms = 1 4 + 1 = 1 5 Number of K atoms remaining = 1 5 N0 From the graph, when number of atoms = 0.20N0 t = 2.3 t 1 2 Age of rock sample = 2.3 × (1.37 × 109 ) years = 3.2 × 109 years Example 16 When a meteor is in space, bombardment of cosmic rays produces an isotope of argon 39 18Ar which is radioactive and has a half-life of 264 years, and tritium 3 1H which has a half-life of 12 years. Both the isotopes remain in the meteor. The production of these isotopes stops when the meteorite falls on the surface of the Earth. In a meteorite which has just landed, the ratio of 3 1H atoms to 39 18Ar atoms is 40. Calculate the value of this ratio now for a meteorite which landed on the Earth’s surface. (a) 12 years ago, (b) 264 years ago. (222 ≈ 4 × 106 ) Solution: (a) At time, t = 0 Number of 3 1 H atoms Number of 39 18 Ar atoms = N0 N′ 0 = 40 At time, t = 12 years = t 1 2 of 3 1 H = 1 22 t 1 2 of 39 18 Ar 264 12 = 22 Number of 3 1 H atoms, N = 1 2 N0 Number of 39 18 Ar atoms, N′ = 1 2 N′ 0 ≈ N′ 0 Hence N N′ = 1 2 N0 N′ 0 = 1 2 N0 N′ 0 = 1 2 × 40 = 20
Physics Term 3 STPM Chapter 25 Nuclear Physics 317 25 (b) At time t = 264 years = 22 t 1 2 of 3 1 H = t 1 2 of 39 18 Ar Number of 3 1 H atoms, N = 1 2 22 N0 Number of 39 18 Ar atoms, N′ = 1 2 N′ N N′ = 1 2 22 N0 1 2 N′ 0 = 2 4 × 106 (40) = 2 × 10–5 Medical Uses 2010/P2/Q14 1. A radioactive tracer is a chemical compound having at least one radioactive element. Radioactive tracer is used to trace the path of the radioisotope through a system. 2. Radioactive tracers are used for diagnostic purposes in medicine. The radioactive tracer enters the body by injection or orally. 3. The tracer which is a short-lived isotope emits gamma-rays from within the body. The gammarays are detected using a gamma camera which can view organs from many different angles. The camera builds up an image from the points from which radiation is emitted; this image is enhanced by a computer and viewed by a physician on a monitor for indications of abnormal conditions. 4. The most common radioactive isotope used in radioactive tracers is technetium-99m. It has a half-life of six hours, and emits gamma rays that provide the necessary information. 5. Other radioactive tracers include iodine-131 for thyroid conditions, iron-59 study metabolism in the spleen and potassium-42 for potassium in the blood. 6. Rapidly dividing cells in cancerous growth are particularly sensitive to damage by radiation and can be eliminated by irradiating with gamma-rays from an external radioactive cobalt-60 source. 7. Internal radionuclide therapy is administered by planting a small radiation source, usually a alpha or beta emitter, in the target area. Example iodine-131 is used to treat thyroid cancer. 8. Advantages of this are • Short-range and localised to the target tumour, • gives less overall radiation to the body, and • is cost effective. 9. Radioisotopes that emit very energetic short range alpha-particles are administered at the centre of a cancerous growth. The energetic alpha-particles are more effective in destroying cancer cells and leave the healthy cells which are further away unharmed. Lead-212 is used for treating pancreatic and ovarian cancers.
Physics Term 3 STPM Chapter 25 Nuclear Physics 318 25 Industrial Uses 1. Thickness control: A β-emitter is placed on one side of the material, such as paper, plastic or other thin materials which the β-particles can penetrate. A GM tube is placed on the other side to detect the β-particles. The GM-tube is connected to the machinery. When the count rate detected by the GM tube is higher or lower than normal the machinery is automatically shut off. 2. Radioactive tracers are used to measure the wear and tear of machine parts and the effectiveness of lubricants. 3. Industrial radiography: Gamma-rays from a sealed sources such as Co-60 is directed to the object to be checked. A photographic fi lm is placed on the other side. More radiation will pass through if there are cracks, breaks or other fl aws in the metal parts and are recorded on the fi lm. Flaws in welding of pipes used to carry natural petroleum gas are detected using this method. Example 17 A sealed capsule containing the radioisotope phosphorus-32 (P-32), a beta-particle emitter is implanted into a patient’s tumor. Half-life of P-32 is 14.3 days. The initial activity is 6.50 MBq. Calculate (a) the initial number of P-32 atoms in the capsule, (b) the activity of the sample after 10 days. (c) the number of beta-particles absorbed by the tumor in 10 days. State two reasons why P-32 is suitable for the treatment described above. Solution: (a) Initial activity, A0 = −λN0 λ = ln 2 t1 2 Initial number of P-32 atoms, N0 = A0 t1 2 ln 2 = (6.50 × 106 ) 14.3 × (24 × 3600) ln 2 = 1.15 × 1013 (b) t = 10 days = 10 14.3 t1 2 = 0.6993 t1 2 Activity after 10 days = A0 20.6993 = 6.50 20.6993 MBq = 4.00 MBq (c) Number of radioactive atoms, N ∝ A, activity N ∝ 4.00 MBq 1.15 × 1013 ∝ 6.50 MBq Hence, after 10 days, number of P-32 atoms, N = 4.00 6.50 (1.15 × 1013) = 7.08 × 1012 Each P-32 nucleus that decays emits one beta-particle. Number of beta-particles absorbed = (11.5 – 7.08) × 1012 = 4.42 × 1012
Physics Term 3 STPM Chapter 25 Nuclear Physics 319 25 Reasons P-32 is suitable 1. It has a half-life of 14.3 days which is of the same order as the treatment duration. 2. It emits beta-particles which is short range, hence healthy tissue further from the capsule are not damaged. Beta-particles have greater ionizing power than gamma-rays. Example 18 Techtenium-99m (Tc-99m) is radioactive and has a half-life of 6.00 hours. It is used in diagnostic imaging. A freshly prepared sample of Tc-99m is administered intravenously to the patient. After a short period the patient is placed under a detector connected to a TV monitor as shown in the fi gure below. TV monitor Detector Patient (a) What is the type of radioactive radiation produced when Tc-99m decays that enable it to be detected? Explain your answer. (b) What is the advantage of the short half-life of Tc-99m? Solution: (a) Gamma-rays has high penetrating power. Gamma-rays emitted from TC-99m in the body is able to exist the body and detected by the detector. (b) After a few hours, the activity of Tc-99m that remains will be negligible and would not cause any damage. Quick Check 3 1. The ratio of C-14 to C-12 in living things has a constant value during life, but the ratio decreases after death because the C-14 is not replaced. The half-life of C-14 is 5 600 years. The C-14 content of a 5 g sample of living wood has a activity of 100 per minute. A 10 g sample of ancient wood has an activity of 50 per minute. What is the age of the sample? A 1 400 years B 2 800 years C 11 200 years D 16 800 years 2. The activity of C-14 in a piece of ancient wood is about 25% the activity of C-14 in living wood. If the half-life of C-14 is 5 700 years, an estimation of the age of the ancient wood is A 11 400 years. B 17 100 years. C 22 800 years. D 28 500 years. 3. Radioactive carbon-14 dating was used to estimate the age of an archaelogical specimen. The following count rates were obtained:
Physics Term 3 STPM Chapter 25 Nuclear Physics 320 25 Specimen Count rate 5 g of living wood 160 counts per minute 5 g of archaelogical specimen 70 counts per minute No sample 40 counts per minute If the half-life of carbon-14 is 5 700 years, what is an estimation of the archaelogical age of the specimen? A 1 400 years B 11 000 years C 13 000 years D 23 000 years 4. Radium-226 occurs in appreciable quantities in a mineral of great geological age, despite the fact that its half-life is only a very small fraction of the age of the mineral. This is because A the radioactive decay of radium-226 is exponential and the activity can never become zero. B Radium-226 is constantly being formed by the decay of a longer-lived isotope. C the concentration of radium-226 is constantly regenerated by radioactive fall-out from the atmosphere. D the rock surrounding the radium-226 slows down the escape of α-particles and γ-rays. 5. A radioactive isotope is added to the water in pipes buried about half-metre below the surface of a road. Which sort of isotope should be chosen? Emitter Half-life A α a few hours B β several years C γ a few hours D γ several years 6. A patient who is undergoing cancer therapy is given 10 minutes of irradiation from a radioactive source that has a half-life of 2 years. If the same source is to be used for the same treatment 2 years later, how long the patient should be irradiated? A 5 minutes C 14 minutes B 7 minutes D 20 minutes 7. The graph below shows the number of α-particles N, emitted per second by a radioactive sample as a function of time t. 1 2 3 4 5 6 t / s In N 0 1 2 3 Which one of the following gives the relationship between N and t? A N = 0.5e–3t C N = 3e–2t B N = 2e–6t D N = 20e–0.5t 8. The half-life of a certain radioactive isotope is 8 hours. What is the total fraction of a sample would have decayed after 4 hours? A 0.25 C 0.50 B 0.29 D 0.71 9. A radioactive sample X consists of 3.2 × 1010 atoms of a nuclide of half-life 20 minutes. A second sample consists of 8.0 × 109 atoms of another radioactive nuclide of half-life 30 minutes. After how many minutes will the number of active atoms in the two samples be equal? A 60 minutes C 120 minutes B 90 minutes D 180 minutes 10. A radioactive isotope has a decay constant λ. If NA is Avogadro constant, what is the activity of a sample which consists of n moles of the isotope? A nλNA C λn NA B λNA n D λ nNA 11. A radioactive source has a half-life of 4 hours. In the absence of this source, a constant count rate of 20 s–1 is recorded. When the source is placed at a fixed distance from the counter, the average count rate is 200 s–1. What average count rate is expected with the source in the same position 8 hours later? A 45 s–1 C 55 s–1 B 50 s–1 D 65 s–1
Physics Term 3 STPM Chapter 25 Nuclear Physics 321 25 12. A piece of ancient bone has an average C-14 activity of 12 unit per gram. If the C-14 activity of a piece of living bone is 36 unit per gram, estimate the archeological age of the bone. (Half-life of C-14 is 5 700 years) 13. Living material contains a constant but small fraction of C-14. After death, the activity of C-14 in the material decreases. The half-life of C-14 is 5.7 × 103 years. In the year 2003, a small sample of carbon from a living material gives a count rate of 12 per minute above the background count. What would be the count rate if the same quantity of carbon is taken from an archaeological material which is (a) from the year 53 A.D.? (b) from the year 1120 A.D? 14. (a) Define (i) decay constant and (ii) half-life of a radioisotope. Write an expression relating to decay constant and half-life. (b) Thorium-232 is an alpha-emitter and has a half-life of 1.41 × 1010 years. A sample of the radioisotope has an activity of 5.2 ×106 Bq. Calculate (i) decay constant, in s–1, of Th-232. (ii) the mass of Th-232 in the sample. (iii) the time taken for the activity to reduce by 10%. (c) Discuss whether dust that has been contaminated with thorium-232 will bring any health hazard. 15. A GM tube is used to investigate the radioactive decay of a sample. The corrected count rate N at time t is given below. t/min N/min–1 7 150 4 450 1 790 446 112 0 50 150 300 450 Estimate the half-life of the radioactive sample. 16. The figure below is the graph of ln R against time t, where R is the count rate of a radioactive source. In (R / min–1) 250 0 t / minutes 6 Calculate the half-life of the radioactive source. 17. The half-life of 60 27Co is 5.26 years. Calculate (a) the decay constant of 60 27Co, (b) the activity of 1.00 gram of 60 27Co. 18. The half-life of a radioactive nuclide is 3 500 s. A sample of this nuclide is used in an experiment. At the end of the experiment, the activity of the sample is 8.0 × 105 Bq. Safety regulations permit a radioactive sample to be disposed with domestic waste provided that the activity is 3.7 × 104 Bq or less. Find the minimum time after the end of the experiment for which the source should be retained before disposal. 19. (a) In a certain meteorite rock, the ratio of the number of 38 92U atom to the number of lead 206 82 Pb atoms attributable to radioactive decay of 238 92 U is 1.17. If the half-life of uranium is 4.5 × 109 years, estimate the age of the rock. (b) Wood from a sunkened ship has an activity of 0.12 Bq per gram due to carbon-14. Whereas living wood gives an activity of 0.20 Bq per gram. If the half-life of carbon-14 is 5 700 years, estimate the age of the ship. 20. In a procedure to determine the volume of a patient’s blood, 4 ml of a solution containing a radioactive sodium isotope of half-life 15 hours is injected into his blood-stream. The activity of the solution injected is 3 000 counts per minute. After 3 hours, 12 ml of blood is drawn from the patient and the activity is found to be 6 counts per minute. (a) Estimate the volume of the patient’s blood. (b) Discuss any assumptions made in the determination and suggest how they might be checked.
Physics Term 3 STPM Chapter 25 Nuclear Physics 322 25 25.3 Nuclear Reaction Learning Outcomes Students should be able to: • state and apply the conservation of nucleon number and charge in nuclear reactions • apply the principle of mass-energy conservation to calculate the energy released (Q – value) in a nuclear reaction • relate the occurrence of fi ssion and fusion to the graph of binding energy per nucleon against nucleon number • explain the conditions for a chain reaction to occur • describe a controlled fi ssion process in a reactor • describe a nuclear fusion process which occurs in the Sun. 20 40 60 80 100 Proton number, Z Number of neutrons, N = (A – Z ) 0 20 40 60 80 100 120 N = Z Stable nuclides β - emitter α - emitter Figure 25.3 1. Figure 25.3 shows the graph of neutron number, N against the proton number, Z for stable nuclide. 2. The plot of neutron number N against proton number Z for the stable nuclides lies within a small region as shown in Figure 25.3. 3. For proton number Z less than 20, the value of N Z for stable nuclides is 1. 4. The value N Z increases to 1.5 for large value of Z. 5. Nuclides with proton number greater than 82 are radioactive. The largest proton number for natural occuring isotopes is 92, which is 92U. 6. When the plot of (N, Z) for a nuclide is above the region of stable nuclides (Figure 25.3), the nuclide is radioactive and decays by β-emission. 7. If the plot of (N, Z) for a nuclide is below the region of stable nuclides, the nuclide is also radioactive and decays by α-emission. 8. Radioactive decay is a nuclear reaction. Changes in the structure of the nucleus occurs when an unstable nucleus decays. The electrons that orbit round the nucleus have no role in radioactive decay. 9. If a nucleus has more protons than necessary for stability, the nucleus becomes unstable. In its effort to attain stability, the nucleus decays by emitting an α-particle. 2012/P1/Q48/2012/P2/Q14
Physics Term 3 STPM Chapter 25 Nuclear Physics 323 25 10. When a nuclide A ZX decays by α-emission, 2 protons and 2 neutrons are ejected together as 4 2 He or α-particle. The nuclear reaction is represented by the equation A ZX → A – 4 Z – 2Y + 4 2 He (α-particle) 11. In all nuclear reactions, • charge or proton number is conserved • mass or nucleon number is conserved • energy is conserved 12. Examples of radioactive decay by α-emission is 238 92 U → 234 90 Th + 4 2 He + γ (α-particle) (γ-ray) 230 90 Th → 226 88Ra + 4 2 He + γ 226 88Ra → 222 86Rn + 4 2 He 13. Emission of gamma-ray (γ-ray) does not result in any changes to the proton number and nucleon number because γ-ray is an electromagnetic radiation without any charge. 14. When a nucleus has to many neutrons necessary for stability, it becomes unstable and the nucleus decays by emitting a β-particle. 15. A β-particle is an electron, and since there are no electron in the nucleus the emission of a β-particle is explained as follows: one of the neutrons in the nucleus undergoes the change as shown by the equation: 1 0 n → 1 1 H + 0 –1 e (neutron) (proton) (β-particle) 16. The proton 1 1H remains in the nucleus, and the electron 0 –1 e is emitted as a β-particle. 17. Hence, when a nucleus decays by β-emission • the number of neutron decreases by 1 • the number of proton increases by 1 (proton number) • the nucleon number remains unchanged A Z P → A Z + 1Q + 0 –1 e (β-particle) Example of β-decay: (a) 64 29Cu → 64 30 Zn + 0 –1 e (β-particle) (b) 40 19 K → 40 20Ca + 0 –1 e (β-particle) Example 19 The nuclide 210 81X decays in four successive processes to A 82 Y. Each of the decay process involved the emission of a α-particle or a β-particle. Determine the value of A.
Physics Term 3 STPM Chapter 25 Nuclear Physics 324 25 Solution: Let n = number of α-particles emitted Hence number of β-particles emitted = (4 – n) 210 81X → A 82 Y + n4 2 He + (4 –n) 0 –1 e .............................. ① Since proton number is conserved. 81 = 82 + 2n + (–1)(4 – n) 3n = 3 n = 1 Equation ① becomes 210 81X → A 82 Y + 4 2 He + 3 0 –1 e Since nucleon number is conserved, 210 = A + 4 A = 206 Example 20 The fi rst part of the decay series of the artifi cially isotope neptunium 237 93Np involves the following sequence of emission : α, β, α, α, β, α. Illustrate these changes on a plot of N, the neutron number, against Z, the atomic number. Use the table below to identify the last nuclide in this portion of the decay series. Element Bi Po At Rn Fr Ra Ac Th Pa U Atomic number 83 84 85 86 87 88 89 90 91 92 Solution: α β α α β α 237 Np → 233 Pa → 233 U → 229 Th → 225 Ra → 225 Ac → 221 Fr 093 091 092 090 088 089 087 237 93Np 233 91Pa 233 91U 233 92Th 229 88Ra 225 89Ac 221 87Fr Z 93 91 92 90 88 89 87 N 144 142 141 139 137 136 134 86 87 134 132 136 138 140 142 144 88 89 90 91 92 93 N Z The last nuclide is 221 87 Fr
Physics Term 3 STPM Chapter 25 Nuclear Physics 325 25 Quick Check 4 1. The decay of 226 88 Ra is represented by the equation, 226 88Ra → X + α + γ where X is the daughter nuclide, α is an α-particle and γ is a photon of γ-ray. The proton number of X is A 86 C 222 B 88 D 224 2. Which one of the following combinations of radioactive decay results in the formation of an isotope of the original nucleus? A α and four β B α and two β C α and β D two α and β 3. A radioactive isotope A Z X decays by emitting successively a β-particle, an α-particle, and a β-particle. The last nuclide produced is represented by A A – 2 Z – 2Y C A – 4 ZY B A – 2 ZY D A – 4 Z + 1Y 4. During the decay of the nuclide 224 82Y to 220 82Y, the emissions consist of A one α-particle and one β-particle B one α-particle and two β-particles C two α-particles and one β-particle D two α-particles and two β-particles 5. When a radioactive nuclide decays successively by emitting a β-particle, an α-particle and γ-ray, the changes in the nucleon number and proton number are: A B C D Change in nucleon number –2 –3 –3 –4 Change in proton number –3 –2 –3 –1 6. When a radioactive isotope 226 88 X decays to a stable isotope, fi ve α-particles and four β-particles are emitted. What are the proton number Z, and nucleon number N of the stable nuclide? A B C D Z 78 78 82 82 N 202 210 202 206 7. 238 92U decays through a series of transformations to a final stable nuclide. The particles emitted in the successive transformation are α, β, β, α, α Which nuclide is not produced during this series of transformations? A 228 88Ra C 234 91 Pa B 230 90 Th D 234 92U 8. The isotope which decays by β-emission to produce 111 48Cd is A 111 47Ag C 110 47Ag B 111 49 In D 112 50 Sn 9. A radioactive isotope of an element Z undergoes a series of decay until a stable isotope of the element Z is produced. What is the ratio of the number of β-particles emittted to the number of α-particle emitted? A 1 : 1 B 1 : 2 C 1 : 4 D 2 : 1 10. A stationary thorium 220 90 Th nucleus emits an α-particle of kinetic energy Eα. What is the kinetic energy of the daughter nucleus? A Eα 108 C Eα 54 B Eα 110 D Eα 55 11. A stationary U-238 nucleus decays by emitting an α-particle and the energy released is T. 238 92 U → 234 90 Th + 4 2 He What is the kinetic energy of the α-particle? A slightly less than 1 2 T B T 2 C T D slightly less than T
Physics Term 3 STPM Chapter 25 Nuclear Physics 326 25 12. (a) Sketch a graph to show how the number of neutron N varies with the number of proton Z for stable nuclei. On your graph, show the line N = Z and denote the approximate valeus of N and Z on the axis of the graph. (b) What can be deduced from your graph about the number of neutrons and protons in stable nuclides? Energy in Nuclear Reactions 1. The decay of radium-226 by the emission of an α-particle is represented by the equation: 226 88Ra → 222 86Rn + 4 2 He + γ (α-particle) (γ-ray) (a) The proton number is conserved. Since the number of proton before the reaction, 88, is equal to that after the reaction, 86 + 2 = 88, charge in the nucleus is conserved. (b) The nucleon or mass number is conserved. Since mass and energy are equivalent, then energy is also conserved. 2. Accurate measurement gives the masses below. Mass of Ra-226 nucleus, mRa = 226.0254u Mass of Rn-222 nucleus, mRn = 222.0175u Mass of α-particle, mα = 4.0026u Total mass after decay = mRn + mα = 226.0201u The mass difference of the reaction Δm = (226.0254 – 226.0201)u = 0.0053u The energy equivalent of this mass difference = (Δm)c2 = (0.0053 × 1.66 × 10–27)(3.0 × 108 )2 = 7.92 × 10–13 J 3. The energy is released in the form of the kinetic energy of the daughter nucleus (Rn), kinetic energy of α-particle and energy of a photon of γ-ray. 4. The energy released in a nuclear reaction is known as the reaction energy, Q. The Q value of the decay of radon-226 is 7.92 × 10-13 J or 4.95 MeV and is positive. 5. Reaction energy, Q = (Dm)c2 where Dm is the mass difference of the reaction. In the general nuclear reaction a + X → Y + b mass difference Dm = [(ma + mX ) – (mY + mb )] Reaction energy, Q = [(ma + mX ) – (mY + mb )] c2 6. If the Q value is positive, the reaction is known as exothermic. If the Q value is negative, the reaction is known as endothermic. Endothermic reaction will not occur unless the bombarding particle has kinetic energy greater than Q. The minimum energy required for such a reaction is known as the threshold energy. 7. A nuclear reaction can only occur if the sum of the masses of the products is less than the total masses before the reaction. 8. The greater the mass difference, the greater the possibility of the nuclear reaction occuring. Info Physics The world largest nuclear power plant by capacity (8212 MWh) is Kashiwazaki-Kariwa, About 220 km northwest of Tokyo.
Physics Term 3 STPM Chapter 25 Nuclear Physics 327 25 Example 21 The relative atomic mass Ar , of a number of nuclide are listed below: Nuclide Ar 4 2 He 4.0026 23 11 Na 22.9898 27 13 Al 26.9815 Discuss whether it is possible for 27 13 Al to emit spontaneously an α-particle? Solution: If 27 13 Al emits an α-particle, the reaction would be represented by 27 13 Al → 23 11 Na + 4 2 He 26.9815u 22.9898u 4.0026u 26.9924u Since the total mass after the reaction is greater than that before the reaction, the reaction does not occur. Example 22 An atom of 28 13 Al decays to 28 14 Si by emitting a particle and a γ-ray photon. (a) Write an equation to represent the nuclear reaction. Identify the particle emitted. (b) Explain why the 28 14 Si particle is positively charged. (c) Show that the total mass of 28 14 Si particle and the particle emitted is the same as the mass of an atom of 28 14 Si. (d) Calculate the total energy released when an atom of 28 13 Al decays. (e) The γ-ray photon emitted has a wavelength of 6.99 × 10–11 m. Calculate the energy of the γ-photon. Hence, deduce the total kinetic energy of the products of the decay. [mass of an atom of = 27.98191u, mass of an atom of 28 14 Si = 27.97693u] Solution: (a) 28 13 Al → 28 14 Si + 0 –1 e 0 –1 e : β-particle (b) 13p 14p 13e 13e 15n 14n + – 28 13 Al → 28 14 Si + 0 –1 e
Physics Term 3 STPM Chapter 25 Nuclear Physics 328 25 As shown in the above fi gure, the 28 14 Si particle has 14p in the nucleus and 13e orbiting the nucleus. 13e around the 28 13 Al nucleus are not involved in the radioactive decay. Since 28 14 Si particle has 1 proton more than electrons, it is positively charged. (c) Total mass of 28 14 Si particle and β-particle = mass of nucleus (14p and 14n) + mass of 13 electrons + mass of 1 electron (β-particle) Mass of 28 14 Si atom = mass of nucleus (14p + 14n) + mass of 14 electrons which is equal to mass of 28 14 Si particle. (d) Mass difference of reaction, Δm = Mass of an atom of 28 13 Al – mass of 28 14 Si particle + (mass of β-particles) = mass of an atom of 28 13 Al – mass of 28 14 Si atom (from (c) above) = (27.98191 – 27.97693)u = 0.00498u Total energy released = (Δm) c2 = (0.00498)(1.66 ×10–27)(3.0 × 108 )2 = 7.44 × 10–13 J (e) Energy of γ-photon = hc λ = (6.63 × 10–34)(3 × 108 ) 6.99 × 10–11 = 2.85 × 10–15 J Total kinetic energy of 28 14 Si particle and β-particle = (7.44 × 10–13 – 2.85 × 10–15) J = 7.41 × 10–13 J Bombardment with α-particles 1. Among the nuclear reaction produced by the bombardment of α-particles are (a) 14 7 N + 4 2 He → 17 8 O + 1 1 H (α-particle) (proton) Proton was discovered by Rutherford using this reaction. (b) 9 4Be + 4 2 He → 12 6C + 1 0 n Neutron was discovered by Chadwick using this reaction. 2. Nuclear reactions produced by bombardment of α-particles are limited to small nucleus. This is because the approaching α-particle have to overcome the barrier potential as it approaches the nucleus which is positively charged. The α-particle which is also positively charged experiences an increasing repulsive force as in approaches the nucleus. Figure 25.4 He2+ Barrier potential Potential well Diameter of nucleus
Physics Term 3 STPM Chapter 25 Nuclear Physics 329 25 3. For the α-particle to enter the nucleus, it must have suffi cient kinetic energy for it to overcome the barrier potential. Example 23 A nitrogen 14 7N nucleus changes to the oxygen nucleus 17 8O and a proton when bombarded by an α-particle of suffi cient energy. (a) Write an equation representing the nuclear reaction, showing the proton number and nucleon number of each of the particle involved. (b) Find the minimum kinetic energy of the α-particle necessary for the reaction to occur. [mass of 14 7 N = 2.32530 × 10–26 kg, mass of 17 8 O = 2.82282 × 10–26 kg mass of proton, mp = 0.16735 × 10–26 kg mass of α-particle, mα = 0.66466 × 10–26 kg] Solution: (a) 14 7 N + 4 2 He → 17 8 O + 1 1 H (proton) (b) 14 7 N + 4 2 He → 17 8 O + 1 1 H (2.32530 + 0.66466) × 10–26 kg (2.82282 + 0.16736) × 10–26 kg = 2.98996 × 10–26 kg = 2.99018 × 10–26 kg mass difference, Δm = (2.99018 – 2.98996) × 10–26 kg = 2.2 × 10–30 kg For the reaction to occur, the minimum energy of the α-particle = energy equivalent of the mass difference = (Δm) c 2 = (2.2 × 10–30)(3.0 × 108 )2 = 1.98 × 10–13 J Bombardment Using Protons from a Particle Accelerator 1. The success of nuclear reaction using bambardment of α-particle is limited by the energy of the α-particles produced by radioactive decay. 2. Particle accelerators such as linear accelerator and cyclotrons are used to accelerate charged particles usually protons to a high velocity. 3. The fast protons are used to bombard nucleus. 4. Example of such a reaction is 7 3 Li + 1 1 H → 2(4 2 He) (proton) This reaction was fi rst done by Cockroft and Walton. It was the fi rst time that a nucleus was split into two.
Physics Term 3 STPM Chapter 25 Nuclear Physics 330 25 Bombardment with Neutron 1. A neutron is able to enter a nucleus easily because being uncharged, it does not encounter any barrier potential as it approaches the nucleus. (Figure 20.3) 2. Hence a neutron is able to enter a large nucleus. 3. When a neutron approaches a nucleus, the neutron is attracted to the nucleus. 4. Neutron has low ionising power. The electrons in the energy shells are not attracted to the neutron and does not collide with the neutron as the neutron approaches the nucleus. 5. Slow neutrons are more effective than fast neutrons in initiating nuclear reactions. 6. To obtain slow neutrons, thermal neutrons are used. Thermal neutrons are neutrons which are slowed down by being in thermal equilibrium with its surrounding which is mantained at a defi nite temperature. The kinetic energy of the thermal neutron is E = 3 2 kT where k = Boltzmann’s contstant = 1.38 × 10–23 J K–1 T = thermodynamic temparature of neutron’s surrounding Quick Check 5 1. In a nuclear reaction, energy equivalent to 10–11 kg of matter is released. The energy released is A 9.0 × 10–6 J B 3.0 × 10–3 J C 4.5 × 105 J D 9.0 × 105 J 2. The isotope 226 88 Ra decays into 222 86 Rn with the emission of an α-particle and γ-ray photon of frequency v. Which one of the equations implies that energy is conserved? A 226 = 222 + 4 B [mRa – mRn ]c 2 = hv C 1 2 mRnu2 Rn = hv + 1 2 mαuα 2 D [mRa – (mRn + mα)]c2 = hv + 1 2 mαuα 2 + 1 2 mRnu2 Rn (Where appropriate, m represents the mass of a particle, u its speed, c the speed of light) 3. Which one of the following particles, can most easily enter a nucleus? A proton C neutron B α-particle D electron 4. When a nitrogen nucleus 14 7N is bombarded with a neutron, the product is a nucleus of 14 6C and A proton C electron B neutron D positron 5. During a radioactive decay by β-emission which of the following quantities is conserved? A Charge B Number of proton C Number of neutron D Kinetic energy 6. A 238 92 U nucleus decays to a nucleus W by successively emitting the following particles: α, β, β. What is the ratio of the number of neutrons to the number of protons in the nucleus W? A 1.52 C 1.60 B 1.54 D 1.62 07. 7 3 Li + 1 1 p → 2 ( 4 2 He) + energy The equation above represents the bombardment of lithium nuclei with protons. Figure 25.5 1 0n Potential well Diameter of nucleus
Physics Term 3 STPM Chapter 25 Nuclear Physics 331 25 If c is the speed of light, and the masses of the particles are : 7 3 Li : mL 4 2 He : mH 1 1 p : mP What is the net energy released during the reaction? A [(mL + mP ) – 2 mH] c 2 B [3mH – (mL + mP )] c 2 C [2 mH + mL + mP ] c 2 D [2 mH – mP ]c 2 8. A high energy α-particle collides with a 14 7 N nucleus to produce a 17 8 O nucleus. What could be the other products of this collision? A a γ-photon only B a γ-photon and a β-particle C a γ-photon and a neutron D a γ-photon and a proton 9. A stationary 214 84 Po nucleus emits an α-particle spontaneously to produce the nuclide lead (Pb). Write the atomic number and mass number of the nuclide lead. Explain in general terms, how energy is conserved in the process. 10. The nuclide 40 19 K decays to 40 20 Ca. (a) Write an equation to represent the decay, and identify the particle emitted. (b) The mass of an atom of 40 19 K in 39.964001u and the mass of an atom of 40 20 Ca is 39.962582u. Calculate the energy released in the decay. 11. Complete the following nuclear equations. (a) 232 90 Th → + Ra (b) 14 7 N + 4 2 He → 17 8 O + (c) 4 Be + → 12 6 C + n (d) 7 Li + 1 → 24 (e) 235 U + → 56Ba + 93 36 Kr + 3 12. 13 6 C bombarded by protons of kinetic energy 2.50 MeV in an effort to produce 13 7 N. Show quantitatively whether the effort will be successful. [Mass of 13 6 C = 13.003355u, 13 7 N = 13.005739u, proton = 1.007825u, neutron = 1.008665u.] 13. 14 7 N + 4 2 He → 17 8 O + 1 1 H In the reaction represented by the above equation, nitrogen is bombarded with α-particles of kinetic energy 7.59 MeV and the kinetic energy of the oxygen nucleus and proton are 0.52 MeV and 5.88 MeV respectively. Calculate the mass of the α-particle in u, the unified atomic mass. [mass of 1 1 H = 1.007825u [ mass of 1 0 n = 1.008665u [ mass of 14 7 N = 14.003074u [ mass of 17 8 O = 16.999134u] 14. (a) Describe the changes that occur within the nuclei when they decay by (i) α-emission (ii) β-emission (iii) γ-emission (b) (i) Calculate the energy released when an atom of 64 29 Cu decays to 64 30 Zn. Mass of 64 29 Cu atom = 63.92976u Mass of 64 30Zn atom = 63.92914u One atomic mass unit, u is equivalent to 931.5 MeV (ii) If the 64 29 Cu nucleus was stationary, is it possible that the nucleus of 64 30 Zn is also stationary? Explain your answer. 15. (a) Explain why protons in a nucleus can exist together despite the fact that protons are positively charged. Hence, explain why the mass of a nucleus is less than the sum of the masses of its constituents. (b) An α-particle travelling at 3.0 × 107 m s–1 collides with a stationary nitrogen nucleus. The reaction is represented by the equation 14 7 N + 4 2 He → 17 8 O + 1 1 H The masses of the nuclei are 14 7 N : 13.9993u 4 2 He : 4.0015u 17 8 O : 16.9947u 1 1 H : 1.0073u The particles move in a straight line as shown in the figure below. The speed of the proton after collision is 6.0 × 107 m s–1.
Physics Term 3 STPM Chapter 25 Nuclear Physics 332 25 Nuclear Fission 2012/P3/Q15, 2013/P3/Q15, 2016/P3/Q15, 2017/P3/Q20 1. Nuclear fission is the splitting of a nucleus into two small nuclei. 2. Energy is released during nuclear fission because • the bigger nucleus has lower binding energy per nucleon. The nucleons in the bigger nucleus are at a higher energy state. • the products of nuclear fission are two smaller nuclei which have higher binding energy per nucleon. The nucleon in the smaller nuclei are at a lower energy state. • The difference in the binding energies is released as energy during the reaction. 3. The equation below represents a fission reaction of uranium-235 nucleus caused by a slow neutron. The masses of the nuclei are as shown. 235 92 U + 1 0 n → 92 36 Kr + 141 56Ba + 3 1 0 n + energy 235.043925 u 1.008665 u 91.926153 u 140.914406 u 4. The products are nuclei of 92 36 Kr, 141 56Ba, three secondary neutrons and energy. 5. Total mass before the reaction = (235.043925 u + 1.008665 u) = 236.052590 u Total mass after the reaction = 91.926153 u + 140.914406 u + 3(1.008665 u) = 235.866554 u Mass difference of the reaction, ∆m = (236.052590 u - 235.866554 u) = 0.186036 u Therefore energy released, E = (∆m)c 2 = (0.186036 u) (———— 934MeV 1u ) = 174 MeV 6. If the three secondary neutrons are decelerated and the quantity of U-235 used is above the critical size, then the chain reaction as shown in Figure 25.6 is obtained. 3.0 x 107 m s-1 6.0 x 107 m s-1 α-particle N H v O (i) Calculate the change in mass which takes place in this nuclear reaction. (ii) Use your answer in (b(i) to calculate the minimum kinetic energy needed by the a-particle to cause the nucelar reaction. Hence, show that the above reaction is possible using a-particle of speed 3.0 × 107 m s–1. (iii) Use the principle of conservation of momentum to calculate v, the velocity of the oxygen nucleus after the collision.
Physics Term 3 STPM Chapter 25 Nuclear Physics 333 25 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n U-235 U-235 U-235 U-235 U-235 U-235 U-235 U-235 U-235 U-235 U-235 U-235 U-235 Figure 25.6 7. In a chain reaction the energy released in a short time is tremendous, and if the chain reaction is not controlled an explosion is produced. This happens when an atomic bomb is exploded. 8. In a nuclear reactor, the chain reaction is controlled so that energy is released at a constant rate. 9. Most nuclear reactors use water as moderator to slow down the secondary neutrons. 10. Cadmium are used in control rods to absorb some of the secondary neutrons. 11. Chain reaction do not occur in natural uranium ore because (a) natural uranium contains only 0.7% U-235, the remaining 99.3% is U-238 which is nonfi ssionable. Enriched uranium used in nuclear reaction contains more than 3% U-235. (b) Most of the secondary neutrons leak out from the uranium ore. (c) Some of the secondary neutrons are captured by U-238 nuclei. (d) Number density of U-235 nuclei is not suffi cient to sustain chain reaction. Example 24 Assuming that fi ssion of an atom of U-235 releases 3 × 1011 J of energy and the end product is an atom of Pu-239, calculate to the nearest month, a nuclear reactor of output power 100 MW would take to produce 10 kg Pu-239. (Take Avogadro number = 6 × 1023 mol–1, 1 month = 2.6 × 106 s)
Physics Term 3 STPM Chapter 25 Nuclear Physics 334 25 Solution: The fi ssion of an atom of U-235 released 3 × 1011 J of energy and one atom of Pu-239. In 1 mol or (239 × 10–3) kg of Pu-239, there are 6 × 1023 atoms. In 10 kg of Pu-239, number of atoms = (6 × 1023) × 10 239 × 10–3 Energy released = 6 × 1023 × 10 239 × 10–3 (3 × 1011) J Time taken to produce 10 kg of Pu = energy produced power output of reactor = (6 × 1023) × 10 × (3 × 1011) (239 × 10–3) × (100 × 106 ) × (2.6 × 106 ) months = 3 months Nuclear Fusion 2008/P2/Q14, 2009/P2/Q14, 2011/P2/Q14, 2012/P1/Q49, 2012/P2/Q8 1. When two small nuclei combine to form a larger nucleus, the process is known as nuclear fusion. 2. The binding energy per nucleon curve shows that a small nucleus has low binding energy per nucleon. The nucleons are in a higher energy state. 3. The nucleus produced by nuclear fusion has higher binding energy per nucleons. Hence the nucleons are in a lower energy state. 4. The difference in the energies of the nucleons is released during fusion. 5. 2 1 H + 2 1 H → 3 2 He + 1 0 n + energy (deuterium) The above equation represents the fusion of two deuterium nuclei, 2 mD. The total mass after fusion (mHe + mn) is less than the total mass of two deuterium nucleus 2 mD. The mass difference of the reaction (Δm) = (2mD – (mHe + mn) The energy equivalent of (Δm) E = (Δm) c 2 is in the form of the total kinetic energy of 3 2 He and 1 0 n and energy released of about 27 MeV. 6. Two nuclei would only fuse together if they have suffi cient kinetic energy to overcome the electrostatic repulsion between the two positively charged nuclei. 7. A method of achieving the high kinetic energy is to raise the temperature of the deuterium to about 108 K. This process is known as thermonuclear reaction. 8. Thermonuclear fusion occurs on Earth when a hydrogen bomb is exploded. The high temperature necessary to initiate the fusion reaction is provided by the explosion of a fi ssion bomb. 9. The Sun is able to radiate energy at a rate of about 4 × 1026 W due to thermonuclear fusion which continuously occurs in the sun. 10. The thermonuclear fusion in the Sun is known as the proton-proton cycle as illustrated in Figure 25.7.
Physics Term 3 STPM Chapter 25 Nuclear Physics 335 25 Step 1: 1 Two protons (1 1 H) fuse to form deuteron 1 1 H + 1 1 H → 2 1 H + e+ (position) + v (neutrino) Step 2: 2 Deuteron (2 1 H) fuses with another proton to form Helium-3 1 1 H + 2 1 H → 3 2 He + γ (gamma-ray) Step 3: 3 Two helium-3 nuclei fuse to form helium-4 and two protons 3 2 He+ 3 2 He → 4 2 He (alpha particle) + 1 1 H + 1 1 H 1 H 1 H 1 H 1 H e+ v γ v γ 1 1 H 2 H 3 He 2 1 H 1 H e+ 1 1 H 2 H 3 He 4 He 2 3 Figure 25.7 11. A complete proton-proton cycle starts with four protons to form an alpha-particle and two positrons. The cycle is exothermic with the release of 25 MeV of energy. Example 25 (a) The ionisation energy of an hydrogen atom is 13.6 eV. Calculate the possible change in the mass when a hydrogen ion and a free electron become a hydrogen atom. H+ + e– = H Does the mass increases or decreases. (b) Lithium deutride, LiD, under certain conditions becomes helium. (i) Write a nucleus equation to represent to reaction. Using nuclides given below, calculate (ii) the loss of mass by each gram of lithium deutride that undergoes the reaction. (iii) the energy released from 10.0 kg of lithium deutride. Nuclide Relative nuclear mass 6 3 Li 6.01700 D (= 2 1 H) 2.01474 3 2 He 4.00387 Solution: (a) When an electron and hydrogen ion (or proton) combine to become a atom, the energy released is 13.6 eV H+ + e– = H + 13.6 eV Hence, the mass of H atom is less than the total mass of H+ and e– Loss of mass Δm = E c 2 = (13.6 × 1.6 × 10–19) (3 × 108 )2 = 2.42 × 10–35 kg
Physics Term 3 STPM Chapter 25 Nuclear Physics 336 25 (b) (i) LiD = 6 3 Li + 2 1 D → 2 4 2 He + energy 6.01700u + 2.01474u 2 (4.00387)u + energy 8.00774u + energy 8.03174u (ii) Loss of mass, Δm = (8.03174 – 8.00774)u = 0.02400u for 8.03174u of LiD. Fractional of mass lost of LiD = 0.02400 8.03174 = 0.002988 Loss of mass per gram of LiD = 0.002988 g (iii) Loss of mass from 10.0 kg of LiD = 0.002988 × 10.0 kg = 0.02988 kg Energy released = 0.02988 × (3.0 × 108 )2 = 2.69 × 1015 J Example 26 Energy from the Sun is produced from the following thermonuclear reaction. 4 1 1 H → 4 2 He + 2 0 +1 e The mass of 1 1H and 4 2He are 1.00813u and 4.00386u respectively. The mean distance of the Earth from the Sun is 1.5 × 1011 m and the solar energy per second incident normally on a unit area of the Earth’s surface is 1.35 kW m–2. At what rate hydrogen atoms change into helium atoms in the Sun? (Neglect the mass and energy of the two positron 2 0 +1 e) Solution: If E = energy radiated per second by the Sun. This energy E falls on the surface of a sphere of radius R, 1.5 × 1011 m. Energy per second incident normally on a unit area = E 4πR2 E 4πR2 = 1.35 × 103 W m–2 Hence, E = 4π(1.5 × 1011)2 × (1.35 × 103 ) W = 3.82 × 1026 W 4 1 1 H → 4 2 He + 2 0 +1 e Mass of 4 1 1 H = 4 (1.00813)u = 4.03252u Mass of 4 2 He = 4.00386u Loss of mass = (4.03252 – 4.00386)u = 0.02866 × (1.66 × 10–27) kg Energy released by 4 hydrogen atoms = (Δm) c2 = (0.02866 × (1.66 × 10–27)(3.0 × 108 )2 = 4.28 × 10–12 J Rate hydrogen changes into helium = E 1 4 × 4.28 × 10–12 = 4 (3.82 × 1026) 4.28 × 10–12 = 3.57 × 1038 atom s–1 Sun Earth 1.5 × 1011 m
Physics Term 3 STPM Chapter 25 Nuclear Physics 337 25 Example 27 (a) Identity the numbers and symbols represented by the alphabets q, r, s, t, u, and v in the nuclear equation 3 1 q + 2 r H = 4 s He + t v u + 17.5 MeV (i) Express 17.5 MeV in joules. (ii) The product particles each have a greater mass than when at rest. Account for this and calculate the overall difference. (b) If 200 kg of the mixed materials (denoted by q and H) were used each year to fuel a fusion power station with an overall conversion effi ciency of 10%, estimate the electrical power output and the waste heat produced. Solution: (a) 3 1 q : q is H as proton number of H is 1 2 r H : r = 1, proton number of H 4 s He : s = 2, proton number of He t v u : equating nucleon numbers before and after the reaction 3 + 2 = 4 + t t = 1 Equating proton numbers before and after the reaction, 1 + r = s + v 1 + 1 = 2 + v v = 0 Hence t v u = 1 0 u, u = n (neutron) (i) 17.5 MeV = (17.5 × 106 )(1.6 × 10–19) J = 2.80 × 10–12 J (ii) Each of the product particles moves at a high speed, hence its mass is greater than its rest mass. Using E = (Δm)c 2 Overall mass difference, Δm = E c 2 = 2.80 × 10–12 (3.0 × 108 )2 = 3.11 × 10–29 kg (b) From the nucleon number of 3 1 q + 2 1 H, a mass of (3 +2)u = 5u of the mixture of 3 1 q + 2 1 H produces 2.80 × 10–12 J energy. Hence, energy produced by 200 kg of mixture in one year = 200 5 (1.66 × 10–27) × 2.80 × 10–12J = 6.75 × 1016 J Electrical power output = 10% of rate of energy produced s–1 = 10 100 × 6.75 × 1016 365 × 24 × 60 × 60 = 2.14 × 108 W Rate of production of waste heat = 90 100 × 6.75 × 1016 365 × 24 × 60 × 60 = 1.93 × 109 W
Physics Term 3 STPM Chapter 25 Nuclear Physics 338 25 Example 28 Fusion reaction in the sun produces energy. One of fusion reactions is a multistage process known as the proton-proton cycle shown below. I. Two protons collide to form deuteron 2 1 H 1 1 H + 1 1 H → 2 1 H + 0 +1 e + Q1 II. A deuteron collides with another proton to form 3 2 He nucleus. 2 1 H + 1 1 H → 3 2 He + Q2 III. Two helium-3 nuclei collide to form an α-particle 4 2 He and 2 protons. 3 2 H + 3 2 He → 4 2 He + 2 1 1 H + Q3 where Q1 , Q2 , and Q3 are the energy released. (a) Calculate the energy released Q1 in joules. (b) How many protons are required to form an α-particle 4 2 He ? (c) Calculate the total energy, in joules, that was released when an α-particle 4 2 He is formed. (d) What is the fraction of the total mass of the protons that is converted into energy for a complete proton-proton cycle? (e) The Sun radiates energy at a constant rate of 4.0 × 1026 W. If the total mass of protons in the sun is 2.0 × 1030 kg, estimate the life of the Sun if the energy radiated is from the protonproton cycle. [Mass of proton, 1 1 H = 1.00728u. Mass of deuteron, 2 1 H = 2.01355u. Mass of positron, 0 +1 e = 0.00055u. Mass of α-particle, 4 2 He = 4.00150u] Solution: (a) 1 1 H + 1 1 H → 2 1 H + 0 +1 e + Q1 2(1.00728)u 2.01355u + 0.00055u. Mass difference = 2(1.00728)u – (2.01355 + 0.00055)u = 0.00046u Energy released, Q1 = (Δm)c2 = (0.00046)(1.66 × 10–27)(3.0 × 108 )2 = 6.87 × 10–14 J (b) Each 3 2 He requires 31 1 H. Two 1 1 H are produced when two 3 2 He collide to form 4 2 He (α-particle) Hence net number of protons required to form an α-particle = (6 – 2) = 4
Physics Term 3 STPM Chapter 25 Nuclear Physics 339 25 (c) 3 2 He + 3 2 He → 4 2 He + 2 1 1 H 2 [ 2 1 H + 1 1 H] → 4 2 He + 2 1 1 H 2 2 1 H + 2 1 1 H → 4 2 He + 2 1 1 H 2 (1 1 H + 1 1 H – 0 +1 e) → 4 2 He 4 1 1 H → 4 2 He + 2 0 +1 e 4 (1.00728)u 4.00150u + 2 (0.00055)u Mass difference, Δm = 4 (1.00728)u – (4.00150 + 2 (0.00055))u = 0.02652u Energy released, E = (Δm)c2 = (0.02652)(1.66 × 10–27)(3.0 × 108 )2 = 3.96 × 10–12 J (d) Fraction of proton mass converted into energy = Δm 4mp = 0.02652u 4 (1.00728)u = 6.58 × 10–3 (e) Mass of protons in the Sun = 2.0 × 1030 kg Mass of protons that can be converted into energy = (6.58 × 10–3)(2.0 × 1030) kg = 1.32 × 1028 kg Energy radiated by Sun, E = (Δm)c 2 . Life of Sun = E Energy radiated per second = (1.32 × 1028)(3.0 × 108 )2 4.0 × 1026 = 3 × 1018 s = 9 × 1010 years Quick Check 6 1. After an 238 92 U nucleus captures a neutron, it decays in succession a β-particle followed by another β-particle. The nuclide produced is A 240 93 Np C 239 94 Pu B 240 94 Pu D 239 90 Th 2. The fi ssion of a heavy nucleus gives, in general, two smaller nuclei, two or three neutrons, some β-particles and some γ-radiation. It is always true that the nuclei produced A travel in exactly opposite directions B have identical neutron-to-proton ratios C have a total rest-mass greater than that of the original nucleus D have large kinetic energies that carry off the greater part of the energy release 3. The rest mass of the deuteron 2 1 H is equivalent to an energy of 1 876 MeV the rest mass of a proton is equivalent to 939 MeV and that of a neutron to 940 MeV. A deuteron may disintegrate to a proton and a neutron if it A emits a γ-ray photon of energy 2 MeV B captures a γ-ray photon of energy 2 MeV C emits a γ-ray photon of energy 3 MeV D captures a γ-ray photon of energy 3 MeV
Physics Term 3 STPM Chapter 25 Nuclear Physics 340 25 4. A nuclear reactor uses U-235 as fuel. 200 MeV of energy is released from the fission of an U-235 nucleus. The output power of a nuclear power station is 1.0 MW. How many U-235 nucleus undergo fission per second? A 5.0 × 103 C 1.3 × 1014 B 1.2 × 106 D 3.1 × 1016 5. Energy from the Sun is produced by A fission of uranium atoms by neutron B fusion of hydrogen nuclei to form helium nucleus C the release of energy when unstable nuclei decay D fission of nuclei that are bombarded by protons 6. Moderator is used in a nuclear reactor for A cooling the reactor B absorbing some of the neutrons C decelerating the neutrons D preventing the neutrons from disintegrating into protons 7. What is the function of control rods in a nuclear reactor? A To control the chain reaction by absorbing excess neutrons. B To slow down the neutrons in the reactor. C To remove heat produced in the chain reaction. D To produce secondary neutrons. 8. Two deuterium nuclei fuse together to form a helium-3 nucleus, with the release of a neutron. The reaction is represented by 2 1 H + 2 1 H → 3 2 He + 1 0 n + energy The binding energies per nucleon are: 2 1 H : 1.09 MeV 3 2 He : 2.54 MeV How much energy is released in this reaction? A 0.36 MeV C 3.26 MeV B 1.45 MeV D 5.44 MeV 9. A nuclear reactor uses enriched uranium as fuel. The mass of enriched uranium consists of 3.2% U-235 which undergoes fission. In the fission of U-235, 0.09% of the mass of U-235 is converted into energy. (a) Calculate the energy released by 1 kg of enriched uranium. (b) If one ton of coal produces 4.0 megawatt-hour of thermal energy, calculate the mass of coal in tonnes which would produce the same thermal energy produced by 1 kg of enriched uranium. 10. Uranium nuclei when bombarded by neutrons may undergo nuclear reactions. One such reaction is 235 92 U + 1 0 n → 144 56 Ba + 90 36 Kr + (a) (i) Complete the equation for this nuclear reaction. (ii) Name this type of nuclear reaction. (b) The binding energy per nucleon of U-235 is 7.5 MeV and that of barium-144 and kryton-90 is 8.5 MeV. (i) Estimates the energy change in this nuclear reaction. (ii) Suggest two forms of energy into which the energy in (i) is transformed during a reaction of this type. 11. (a) (i) Name the quantities that are conserved in a nuclear reaction. (ii) What is the third product, including its number, in the fission reaction represented by 235 92 U + 1 0 n → 148 57 La + 85 35 Br + What is the role of this product in the operation of a nuclear reactor? (b) A nuclear reactor’s efficiency is 10% and the maximum power produced is 1.00 MW. The nuclear fuel used is 235 U, and the nuclear reaction is represented by the equation given in (a) (ii) above. Calculate (i) the energy released by fission of an atom of 235 U. (ii) the minimum mass of 235U required if the reactor operates at maximum rate for 180 days
Physics Term 3 STPM Chapter 25 Nuclear Physics 341 25 per year for 10 years. [mass of 235 92 U = 235.0439u, [mass of 148 57 La = 147.9314u, [mass of 85 35 Br = 84.9156u, [mass of 1 0 n = 1.0087u] 12. (a) Explain the statement: radioactive decay is a random and spontaneous process. (b) A stationary uranium-238 nucleus decays by alpha-emission according to the equation 238 92U → 234 90Th + 4 2He Calculate (i) the energy released in the reaction. (ii) the velocity of the alpha-particle. [Mass of 238 92U = 238.050 786 u. mass of 234 90Th = 234.043 583 u, mass of 4 2He = 4.002 603 u] (c) A sample of carbon of mass 8.50 g was prepared from a piece of ancient wood. The activity of the sample was 0.70 Bq. (i) If the ratio of carbon-14 to carbon-12 in the atmosphere is 1.25 × 10–12, calculate the number of carbon-14 atoms in 8.50 g of living wood. (ii) Determine the number of carbon-14 in 8.50 g of the ancient wood. (iii) Hence estimate the archeological age of the ancient wood. [ 1 year = 3.154 × 107 s] 13. (a) Explain what is meant by nuclear fission. (b) Xenon-139 has a half-life of 41 s and is produced at a constant rate during the fission of a particular sample of uranium-235. No xenon-139 escapes from the sample and it is found that eventually, the number of xenon-139 nuclei present becomes constant. (i) Suggest why the number of xenon-139 nuclei in the sample becomes constant. (ii) When the number of xenon-139 nuclei has become constant, the activity of the xenon-139 is 3.4 ×108 Bq. Calculate the mass of xenon-139 present in the sample. (c) When uranium-235 nuclei are fissioned by slow-moving neutrons, two possible reactions are 1 : 235 92 U + 1 0 n → 139 54 Xe + 95 38 Sr + 2 1 0 n + energy 2 : 235 92 U + 1 0 n → 2 116 46 Pd + xC + energy (i) For reaction 2, identify the particle C and state the number x of such particles produced in the reaction. The binding energy per nucleon E for a number of nuclides is given below. Nuclide E/MeV 95 38 Sr 8.74 139 54 Xe 8.39 235 92 U 7.60 (ii) Show that the energy released in reaction 1 is 211 MeV. (iii) The energy released in reaction 2 is 163 MeV. Suggest, with a reason, which one of the two reactions is more likely to occur. Important Formulae 1. Binding energy, E = (Dm)c2 = [Smp + Smn – mN]c2 2. Rate of decay or activity A = dN dt = −λN 3. Decay constant, λ = ln 2 t 1 2 4. Number of radioactive nuclei N, and activity A decay exponentially. N = N0 e–λt A = A0 e–λt Also if t t 1 2 = n, then N = N0 2n A = A0 2n 5. In the nuclear reaction: a + X → Y + b Mass difference, Dm = [(ma + mX) – (mY + mb )] Reaction energy, Q= (Dm)c2 = [(ma + mX) – (mY + mb )] c2
Physics Term 3 STPM Chapter 25 Nuclear Physics 342 25 STPM PRACTICE 25 1. The reason why the iron-56 is more stable than carbon-14 is A iron-56 has larger nucleus B iron-56 has more protons C iron-56 has lower binding energy D iron-56 has higher binding energy per nucleon 2. Which of the following is emitted when beryllium is bombarded with alphaparticles? A Neutron B Proton C Beta-particle D Gamma-ray 3. The variation of nuclear binding energy per nucleon with nucleon number of nuclei P, Q, R and S is shown by the graph below. Nuclear binding energy per nucleon Nucleon number P Q R S Which process is most likely to represent a radioactive decay? A P → Q C R → S B Q → P D S → R 4. A radioisotope X which has a half-life of t 1 2 decays to a stable isotope Y. A sample initially contains only X. After a time of 2t 1 2 , there NX atoms of X, and NY atoms of Y. What is the ratio of NX/ NY? A 1 4 C 1 2 B 1 3 D 2 5. A detector is used to measure the activity of a radioactive sample which has a short halflife. After a period, the activity of the sample is negligible compared to the background count. Which graph shows the variation with time t of the count rate R recorded by the detector? A 0 t R C 0 t R B 0 t R D 0 t R 6. A nuclide decays by emitting a positron. Which of the following correctly describes the changes, if any, to the proton number and nucleon number? Proton number Nucleon number A –1 Remains the same B +1 Remains the same C Remains the same –1 D Remains the same +1 7. The decay constant of a radioisotope is A the time taken for rate of decay to be halved B the time taken for half the number of nuclei in a sample to decay C the number of nuclei that decays in one half-life D the probability of a nucleus decaying per second 8. When a slow neutron is captured by a stationary 113 48Cd nucleus, the result is a 114 48Cd nucleus and a gamma photon. What is the frequency of the gamma photon? [Mass of 113 48Cd = 112.9044 u, mass of 114 48Cd = 113.9034 u, mass of neutron = 1.0087 u] A 2.43 × 1018 Hz C 3.49 × 1024 Hz B 2.19 × 1021 Hz D 1.46 × 1031 Hz
Physics Term 3 STPM Chapter 25 Nuclear Physics 343 25 9. A nuclear reaction which involves carbon-13, 13 6 C, and deuterium, 2 1 H is represented by the equation 13 6 C + 2 1 H→ 14 7 N + 1 0 n If the kinetic energy of the incident 2 1H nuclei is 36.8 MeV, what is the total kinetic energy of the products of the reaction? [Mass of 2 1 H = 2.014102 u, mass of 13 6 C = 13.003355 u, mass of 14 7 N = 14.003074 u, mass of 1 0 n = 1.008665 u; 1u = 931 MeV] A 5.3 MeV B 31.5 MeV C 36.8 MeV D 42.1 MeV 10. Nuclide 232 90Th decays to 208 82Pb through a series of a and β decay processes. How many a-particle and β-particle are produced when 232 90Th decays to 208 82Pb ? a-particle β-particle A 4 6 B 6 4 C 6 8 D 8 6 11. The masses of proton, neutron, electron and the 17 8 O nucleus are mp , mn, me and mO respectively. What is the mass defect, ∆m of the 17 8 O nucleus? A ∆m = 8 mp + 8mn – mO B ∆m = 8 mp + 9mn – mO C ∆m = 8 mp + 8mn + 8me – mO D ∆m = 8 mp + 9mn – mO – 8me 12. Which statement about nuclear fusion is not true? A Energy is released. B High temperature is required. C The reaction occurs spontaneously. D The binding energy per nucleon increases. 13. A nuclide decays by beta-emission. Which of the following is true? A The number of neutrons remains unchanged B The nucleon number remains unchanged C The nucleon number decreases by one D The nucleon number increases by one 14. Thorium-234 has a half-life of 24 days. What is the activity of a sample of 5.0 g of thorium-234? A 4.30 × 1015 Bq B 1.03 × 1017 Bq C 8.94 × 1019 Bq D 1.29 × 1022 Bq 15. The equation below suggests the decay of aluminium-27 by emitting an alphaparticle. 27 13Al → 23 11Na + 4 2 He The masses of the nuclei are 27 13Al = 26.9815 u, 23 11Na = 22.9898 u, 4 2 He = 4.0026 u. Which statement is correct? A The reaction will not occur B The reaction is endothermic C No energy is released in the reaction D The Q-value of the reaction is positive 16. Radon-222 has a half-life of 3.8 days. The initial activity of a sample of radon-222 is 8.5 × 106 Bq. (a) What is the decay constant of radon-222? (b) What is the mass of radon-222 in the sample? (c) What is the activity of the sample after 10 days? 17. The energy released in the fission of a uranium-235 atom is 200 MeV. This energy released provides thermal power of 2000 MW in a nuclear powered station. What is the mass of uranium-235 consumed per day by the power station? [Mass of U-235 = 235.04392 u] 18. Radon-222 gas is radioactive and has a halflife of 3.8 days. Accidentally, 0.045 mol of the gas leaks into a basement storage room. [Assume that the safety level is equivalent to 1 mCi = 3.70 × 107 decay s–1] (a) Determine the initial activity of the gas. (b) Calculate the time elapsed before the basement storage room is considered safe to be used. 19. The fusion of a tritium 3 1 H and a deuterium 2 1 H produces a nuclide Y and a neutron, and releasing an energy of 17.6 MeV.
Physics Term 3 STPM Chapter 25 Nuclear Physics 344 25 [The atomic mass of tritium and deuterium are 2.016 049 u and 2.014 102 u respectively, the mass of the neutron is 1.008 665 u. 1 u = 931 MeV.] (a) Write a complete equation for the nuclear reaction of the fusion. (b) Determine, in atomic mass unit, the atomic mass of nuclide Y. 20. (a) State the forces which are acting between nucleons in a nucleus. (b) The variation of neutron number N with proton number Z for stable nuclides is shown in the graph below. 20 0 40 60 80 100 120 140 20 40 60 80120140 N N = Z Z P Q (i) Explain why heavier nuclides have more neutrons than protons. (ii) Nuclides P and Q are unstable and can become stable nuclides by certain particles. State and explain the probable emission by nuclides P and Q in the process of converting to stable nuclides. (c) A deuteron has a mass of 2.013 552 u and consists of a proton and a neutron. (i) Calculate the binding energy of the deuteron. (ii) Determine the energy required to remove one nucleon from deuteron. (iii) Suggest how the process in (c)(ii) could be done. [The mass of proton and neutron are 1.007 275 u and 1.008 665 u respectively. 1 u = 831 MeV] 21. (a) Define binding energy and mass defect of a nucleus. Write an equation relating the two quantities, (b) Calculate the binding energy per nucleon of 15 7 N. [Mass of 15 7 N = 15.000108 u, mass of proton = 1.007276 u, mass of neutron = 1.008665 u, 1 u = 931.5 MeV] (c) The graph of binding energy per nucleon against nucleon number is shown below. Binding energy per nucleon / MeV Mass number 9 8 7 6 5 4 3 2 1 0 50 H2 1 He 4 2 C12 6 Ni Th 62 29 U 239 92 100 150 200 250 (i) Explain why nuclides with high binding energy per nucleon are stable. (ii) U-238 decays to thorium (Th) by alpha decay. Write an equation to represent the decay. Use the graph to explain why the reaction is exothermic. (iii) Estimate the energy released in the fusion reaction: 2 1 H + 2 1 H → 4 2 He 22. (a) Explain the term nuclear fusion. Explain why fusion occurs only for small nuclei. (b) The equation below represents a fusion reaction of two deuterons. 2 1 H + 2 1 H → X + 1 0 n + 3.27 MeV (i) What is the nuclide X? (ii) Calculate the mass of X. [Mass of 2 1 H = 2.014102 u, mass of neutron = 1.008665 u, 1 u = 931.5 MeV] (c) For fusion between two deuterons to occur the separation must be less than 1.0 × 10–14 m. Calculate (i) the electric potential energy of the deuterons at a separation of 1.0 × 10–14 m in MeV. (ii) the minimum initial kinetic energy of each deuteron. (iii) the temperature of the deuteron.