Physics Term 3 STPM Chapter 25 Nuclear Physics 345 25 23. (a) Radioactive decay is a spontaneous and random process. (i) Explain the terms in italics. (ii) State the quantities that are conserved in a radioactive decay. (b) Technetium-99m (99 43Tc) has a half-life of 6.0 hours and decays by emitting a beta-particle and gamma-ray. A sample of Tc-99m of initial activity 80 µ Ci is used as a radioactive tracer in medical diagnosis. (i) Write an equation to represent the decay of Tc-99m. (ii) Calculate the initial mass of Tc99m in the sample used. (iii) What is the activity of the sample after 24 hours? (iv) What is the mass of Tc-99m that remains in the sample after 24 hours? (v) State two precautions that should be taken by persons handling Tc99m. [1 Ci = 3.70 × 1010 Bq] 24. (a) Define mass defect and nuclear binding energy. (b) Sketch a graph of binding energy per nucleon against nucleon number. Using the graph you have drawn, explain a fusion reaction. (c) In a typical nuclear reaction, a tritium nucleus is bombarded by a deuterium nucleus to produce a helium nucleus, a neutron and release an energy Q. (i) Write a complete equation of the nuclear reaction. (ii) Determine the value of Q in MeV. (iii) Calculate the binding energy for each of the deuterium, tritium and helium nuclei. (iv) State, with a reason, which of the three nuclei involved in the reaction have the highest and lowest stability. [Mass of proton = 1.00728 u, mass of neutron = 1.00866 u, mass of electron = 0.00055 u; atomic mass of deuterium = 2.01410 u, atomic mass of tritium = 3.01605 u, atomic mass of helium = 4.002603 u; 1 u = 931.5 MeV.] 25. When a slow neutron is captured by a stationary 10 5 B nucleus, a lithium nucleus and a helium-4 nucleus are produced. The speed of the lithium nucleus is 5.19 × 106 m s–1. (a) Write an equation to represent the nuclear reaction. (b) What is the speed of the helium nucleus? (c) Calculate the Q value of the reaction. (mn = 1.0087 u, mB = 10.0130 u, mLi = 7.0160 u, mHe = 4.0026 u] 26. (a) What is meant by the half-life and decay constant of a radioisotope?Write an equation relating the two quantities. (b) Radium-226, 226 88Ra decays to radon (Rn) by emitting an alpha particle. (i) Write an equation to represent the decay. (ii) Calculate the reaction energy Q in MeV. Is the reaction exothermic or endothermic? [Mass of radium = 226.025402 u, mass of radon = 222.017570 u, mass of alpha-particle = 4.002603 u] (c) What is meant by a radioactive tracer in medical diagnosis? A radioisotope is used as a tracer to monitor the function of an internal organ of a patient. (i) Name the radiation emitted by the radioisotope for it to be detected by a detector outside the patient’s body. (ii) Discuss whether the radioactive tracer should have a half-life of several hours, several years or thousands years. (d) Iodine-131 has a half-life of 8.0 days. The mass of a sample of iodine-131 is 1.8 µg. (i) How many iodine-131 atoms are in the sample? (ii) What is the initial activity of the sample? 27. (a) When beryllium is bombarded with alpha-particle, carbon-12 and a particle X are produced. a + 9 4 Be → C + X
Physics Term 3 STPM Chapter 25 Nuclear Physics 346 25 (i) Complete the above equation. Give the proton number and nucleon number of all the particles. (ii) Identify the particle X. (iii) State two reasons why the particle X was diffi cult to identify experimentally. (b) The particle X is used for the fi ssion of U-235 as represented by the equation X + 235 92U → 141 56Ba + 95 a Kr + b X (i) Complete the above equation, and determine the values of a, and b. (ii) Explain why fi ssion of U-235 is not achievable using fast protons, but fi ssion of U-235 is possible using particle X. 1 1. C 2. B 3. B 4. D: (Δm) = pZ + n(A – Z) – M, binding energy = (Δm)c2 5. B: Products of a nuclear reaction must be higher on the curve. 6. D: Binding energy = (238)(7.4) MeV = 1761 MeV 7. B: EB N = [8(mp + mn) – mO]c2 16 = [8(1.0078 + 1.0087) – 15.995] (1.66 × 10–27)(3.00 × 108 )2 16 = 1.28 × 10-12 J 8. A: E = (Δm)c2 , (Δm) = 1 (3.00 × 108 )2 kg = 1.1 × 10–17 kg 9. D 10. C: m1 < 6mp + 6mn, m2 < 6mp + 7mn m2 – m1 < mn 11. B 12. D 13. C: 1 u = 934 MeV, E = (0.025)(934 MeV) = 23.35 MeV 14. B 15. D: EB = (156)(1.3) pJ = 203pJ 16. (a) 1.1 × 10–17 kg (b) 1.66 × 10–27 kg 17. 3.59 × 10–13 J 18. (a) Refer to page 294 (b) (i) In the nucleus, nucleons are in lower energy state. Energy equivalent of mass defect equals the binding energy. (ii) 1.37 × 10–12 J or 8.54 MeV 19. (a) E = mc2 (b) Binding energy, E = [Zmp + (A–Z)mn – mX] c2 20. (a) (i) (ii) (iii) Refer to page 294 (b) Refer to page 294 21. (a) (b) Refer to page 295 (c) Products of decay have higher binding energy per nucleon. Nucleon at lower energy state. Energy released. 22. (a) 0.0402 m q m = E rB2 (b) 0.0804 m r ∝ m q 23. v = 2.19 × 105 m s–1 24. (a) Refer to page 294 - 295 (b) (i) 2.58 × 10–11 J (ii) 1.29 × 10–12 J (c) (iii) 2 1 H + 2 1 H = 3 2 He + 1 0 n 25. (a) (i) u = 1 12 (mass of C-12 atom) (ii) Show u = 1.66 × 10–27 kg (b) (i) 12 protons, 13 neutrons. A = 25 (ii) Proton number : 12 (iii) 25.20654u (iv) 3.297 × 10–11 J Δm = 0.2207u 26. (a) Nuclides have the same proton number but diff erent mass numbers. (b) (i) 6 protons, 8 neutrons (ii) 12 6C (iii) Same chemical properties. Diff erent masses. ANSWERS
Physics Term 3 STPM Chapter 25 Nuclear Physics 347 25 (c) (i) 1.4 × 1044 m–3, 4 πR3 3 A (ii) 2.4 × 1017 kg m–3 2 1. A: – dN dt = λN and λ = In2 T1/2 2. C: dN dt = –λN, N = – dN dt λ m = ( N NA )(32 g) = 5.2 × 10-9 g 3. A: dN dt = –λ N, λ dt = – dN N 4. C 5. B: 1 16 = 1 24 , 1.0 min = 60 s = 4t 1 2 , t 1 2 = 15 s, λ = ln 2 t 1 2 = ln 2 15 = 4.6 × 10–2 s–1 6. A: A = A0 e−λt , ln A = ln A0 – λt. Graph is straight line with negative gradient. Both are sections parallel. 7. A: t 1 2 = ln 2 λ = ln 2 2.4 × 10–8 = 2.9 × 107 s 8. C: 7 8 decayed, remains 1 8 = 1 23 , 12 days = 3t1 2 , 24 days = 6t1 2 , 1 26 = 1 64 remains 9. B: At time T, NX = NY . At time = 0, N0 = NX + NY = 2NX, NX = 1 2 N0 . T = half-life, t 10. D: After the same time t, 1 4 = 1 22 of X remains, and 1 8 = 1 23 of Y remains. Hence t = 2(t 1 2 )X = 3 (t 1 2 )Y . (t 1 2 )X/(t1 2 )Y = 3 2 11. C: 10 min = 2t 1 2 , mass = 1 22 (32 mg) = 8 mg. Total mass = (8 + 4) mg = 12 mg t = 15 min or 5 min (t 1 2 ) later, mass = 1 2 (12 mg) = 6 mg. 12. D: 48 days = 6t1 2 . Fraction remains = 1 26 = 1 64 fraction decayed = 1 – 1 64 = 63 64 13. (a) Refer to page 302 (b) (i) A0 = dN dt = –λN = ln 2 t 1/2 (40.5 × 10–6)(6.02 × 1023) 99 = 7.90 × 1012 Bq (ii) A = A0 e–λt = 7.04 × 1012 Bq 14. 3.8 × 10–12 kg – dN dt = λN Mass = N NA (32 × 10–3) kg 15. (a) – dN dt = λN [λ] = T–1 (b) 7.1 × 105 , time t1 2 = 0.5, N = (0.5)0.5 N0 16. 24 minutes N NA = 1 2 x , x = t t 1 2 17. (a) Use GM-tube and ratemeter. (b) R = C – background count rate (c) (ii) λ = 1.17 × 10–2 s–1 t 1 2 = 59 s (d) The radioactive gas is constantly being formed by the decay of another radioactive nuclide. 3 1. C: For 5 g sample, A0 = 100 min–1, A = 50 2 = 25 min-1 = 1 4 = 1 22 , t = 2t 1 2 = 11 200 yr 2. A: 25% = 1 4 = 1 22 , t = 2t 1 2 = 11 400 yr 3. B: A0 = (160 – 40) min-1, A = (70 – 40) min–1 = 1 4, t = 2t 1 2 = 11 000 yr (estimate) 4. B Ra-226 is a member of a radioactive series. 5. C: g- rays penetrate pipe and soil. Short half-life so that activity is negligible after procedure. 6. D: After 2 yr = t 1 2 , activity is 1 2. Time required = 2(10 min) = 20 min 7. D: N = N0 e−λt , ln N = ln N0 – λt. λ = 3 6 = 0.5, N0 = e3 = 20. Hence N = 20 e-0.5t 8. B: 4 h = 1 2t 1 2 = 0.5 t 1 2 . Fraction remains = 1 20.5, fraction decayed = 1 - 1 20.5 = 0.29 9. C: X: N after 3 t1 2 (60 min), 6 t 1 2 (120 min): 4.0 × 109 , 5.0 × 108 Y: N after 2 t 1 2 (60 min), 4 t 1 2 (120 min): 2.0 × 109 :, 5.0 × 108 10. A: – dN dt = λN = λ(nNA) 11. D: A0 = (200 – 20) = 180 s-1, 8 h = 2t 1 2 , A = 1 4(180) = 45 s-1. Count rate = (45 + 20) s-1 = 65 s-1 12. 9 000 years 12 = 1 2 x × 36 t = xT
Physics Term 3 STPM Chapter 25 Nuclear Physics 348 25 13. (a) 9.5 per minute A = 1 2 x A0 (b) 10.5 per minute 14. (a) Decay constant, λ = In 2 T (b) (i) λ = In 2 T = 1.56 × 10–18 s–1 (ii) Use: – dN dt = λN = λ m M NA Mass, m = 2.13 × 10–27 kg (iii) Use A = 0.90 A0 = A0 2x x = 0.152 Time t =(0.152 T ) = 2.14 × 109 years (c) Health hazards – respiration – alpha-particle, strong ionisation power: damage tissues – long half-life: – daughter products may be more harmful 15. T = 75 minutes. Plot graph of N against t 16. T = 29 minutes, λ = 6 250 minute–1, T = ln 2 λ 17. (a) λ = 4.18 × 10–9 s–1 λ = In 2 T (b) – dN dt = 4.19 × 1013 Bq – dN dt = λN 18. 1.55 × 104 s A = 1 2 x A0 t = xT 19. (a) 4 × 109 years U Pb = 1.17, U U + Pb = 1.17 2.17 = N N0 (b) 4 200 years A = 1 2 x A0 t = xT 20. (a) 5 200 ml x = t T , A = 1 2 x A0 A ∝ V, 6 ∝ 12 ml (b) Background count rate negligible. Measurement background count rate. 4 1. A: 226 88Ra → 222 86X + 4 2 a + g 2. B: 4 2 a + 2( 0 –1e): no change in proton number 3. C: A ZX → A – 4 ZY + 4 2 a+ 2( 0 –1e) 4. B: (Refer to answers for question 2 and 3) 5. D: A ZX → A Z – 4 – 1Y + 4 2 a+ 0 –1e + g 6. D: 226 88X → 206 82Y + 5 4 2 a+ 0 –1e 7. A: 238 92U → 234 90X + 4 2 a, 234 90X → 234 91Pa + 0 –1e, 234 91Pa → 234 92U + 0 –1e, 234 92U → 230 90Th + 4 2 a, 230 90Th → 230 91Pa + 0 –1e 8. A: 111 47X → 111 48Cd + 0 –1e 9. D: Two b + one a 10. C: 220 90Th → 216 88X + 4 2 a, Momentum conserved: 4v + 216V = 0, V = – 4 216v. Ea/EX = ( 1 2(4u)v2 )/( 1 2(216u)V2 ) = ( 4 216)/( 4 216)2 , EX = Ea 54 11. D: From 10 above, ETh = 4 234Ea= 0.017 Ea. Hence Ea is slightly less than T. 12. (b) Stable nuclides: Z < 20, N = Z, N/Z increases to 1.5 for large values of Z. 5 1. D: E = mc2 = (10-11)(3.00 × 108 )2 J = 9.0 × 105 J 2. D: E = (Δm)c2 = [mRa – (mRn + ma)] c2 = hv + K.E. of Rn and a-particle 3. C: Neutron is not charged – no repulsive force 4. A: 14 7 N + 1 0 n → 14 6 C + 1 1 H 5. A: +1 proton, -1 neutron 6. B: 238 92U → 234 92W + 4 2 a + 2 0 –1e, N/Z = 234 – 92 92 = 1.54 7. A 8: D: 14 7 N + 4 2 a → 17 8 O+ 1 1 H + g 9. 214 84Po → 210 82Pb + 4 2 [(mPo – (mPb + ma)]c2 = K.E. of lead nucleus and a 10. (a) 40 19K → 20Ca + 0 –1e (beta-particle) (b) E = [mK – mCa]c2 = [39.96400 – 39.96258] (934 MeV) = 1.33 MeV or 2.12 × 10–13J 11. (a) 232 Th → 4 He + 228 Ra 090 2 088 (b) 14 N + 4 He → 17 O + 1 H 07 2 08 1 (c) 9 Be + 4 He → 12 C + 1 n 4 2 06 0 (d) 7 Li + 1 H → 2 4 He 3 1 2 (e) 235 U + 1 n → 140 Ba + 93 Kr + 3 1 n 092 0 056 36 0 12. Unsuccessful. Δm = –0.003224u = 3.01 MeV > 2.50 MeV 13. 4.002611u 14. (a) Refer to page 323 (b) (i) 0.578 MeV 15. (a) – Protons are bounded by strong nuclear force. – Protons are in lower energy state.
Physics Term 3 STPM Chapter 25 Nuclear Physics 349 25 (b) (i) –0.0012u = –1.99 × 10–30 kg (ii) 1.79 × 10–13 J Possible Eα = 1 2 mv2 = 2.99 × 10–12 J > 1.79 × 10–13 J (iii) 3.51 × 106 m s–1 6 1. C: 238 92U + 1 0 n → 239 94X + 2 0 –1e 2. A: Momentum is conserved 3. D: 2 1 H → 1 1 H + 1 0 n (1876 MeV) + 3MeV → (939 + 940) MeV 4. D: N(200 MeV) = 1.0 MW N = 1.0 × 106 (200 × 106 )(1.69 × 10–16) = 3.1 × 1016 5. B: Proton-proton cycle 6. C 7. A 8. C: Energy released = 3(2.54) – 2(1.09 × 2) MeV = 3.26 MeV 9. (a) Mass of U-235 = 3.2%(1 kg). E = (0.09%)(Δm)c2 E = (0.09%)(3.2% × 1 kg)(3.00 × 108 )2 J = 2.59 × 1012 J (b) Mass of coal = 2.59 × 1012 (4.0 × 106 )(3600) ton = 180 ton 10. (a) (i) + 21 0 n (ii) Fission Q = (144 + 90)(8.5) – (235)(7.5) MeV = 226.5 MeV Energy released in the form of heat, and kinetic energy of nuclei of Ba, Kr and the two neutrons. 11. (a) (ii) 3 1 n 0 Cause fission in other U-235 nuclei. Continue the chain reaction. (b) (i) 2.68 × 10–11 J (ii) 22.6 kg 12. (a) Refer to page 301-302 (b) (i) Δm = 0.00460 u E = (Δm)c2 = 6.87 × 10–13 J (ii) mav + mThV = mU(0) 1 2 mav2 + 1 2 mThV2 = 6.87 × 10–13 J v = 1.44 × 107 m s-1 (c) (i) N0 = (1.25 × 10–12)(8.50) ( 6.02 × 10 23 14 ) = 4.57 × 1011 (ii) dN dt = – λN = – (ln 2 T1 2 ) N N = 1.83 × 1011 (iii) N = N0 2x where x = ( t T1 2 ) t = 7600 years 13. (a) Refer to page 332 (b) (i) Rate of decay of Xe-139 equals the rate at which is formed. (ii) 4.64 × 10–15 kg (c) (i) 4 1 0 n, x = 4, C = 1 0 n (iii) Reaction (i) because energy released bigger. STPM Practice 25 1. D: Value of binding energy per nucleon is an indication of stability. 2. A 3. D: Daughter nucleus of decay has higher binding energy per nucleon. 4. B: At t = 2t 1 2 , NX = N0 4 , NY = 3N0 4 , NX NY = 1 3 5. C 6. A 7. D 8. B: [112.9044 + 1.0087] = 113.9034 u + hf hf = 0.0097 u = (0.0097)(1.66 × 10-27)c2 f = (0.0097)(1.66 × 10–27)(3.00 × 108 )2 6.63 × 10–34 Hz = 2.19 × 1021 Hz 9. A: Dm = [(2.014102 + 13.003355) – (14.003074 + 1.008665)]u = 0.005718 u 10. B: 232 90Th → 208 82Pb + x( 4 2 a) + y ( 0 –1 β) 4x = (232 – 208) hence x = 6 2x – y = (90 –82), hence y = 4 11. B: Mass of constituents of O-17 nucleus = 8mp + 9mn 12. C 13. B: A ZX → A Z+1Y + 0 –1e 14. C: Number of nuclei in 5.0 g of Th-234, N = ( 5.0 234)(6.02 × 1023) Activity, – dN dt = λN = ( ln 2 24(24 × 3600))( 5.0 234) (6.02 × 1023) Bq = 4.30 × 1015 Bq 15. A: mNa + mHe = (22.9898 + 4.0026) u = 26.9924 > 22.9898 u (mass of 23 11Na) 16. (a) Decay constant, λ = ln 2 t 1 2 = ln 2 3.8(24 × 3600) s–1 = 2.11 × 10–6 s–1 (b) – dN dt = λN, mass, m = ( N NA )(0.222 kg) m = (– dN dt /λ) 0.222 NA = (8.5 × 106 )(0.222) (2.11 × 10–6)(6.02 × 1023) kg = 1.49 × 10-12 kg
Physics Term 3 STPM Chapter 25 Nuclear Physics 350 25 17. Energy required per day = Pt Number of U-235 required per day, N = Pt E Mass of U-235 per day = N NA (0.235 kg) = (2000 × 106 )(24 × 3600)(0.235) (6.02 × 1023)(200 × 106 )(1.60 × 10–19) kg = 2.11 kg 18. (a) Activity, A0 = λN = (ln 2 t 1 2 )(nNA) = ( ln 2 (3.8)(24 × 3600))(0.045) (6.02 × 1023) Bq = 5.72 × 1016 Bq (b) A = A0 2x , 2x = 5.72 × 1016 3.70 × 107 = 1.546 × 109 Time t = xt1 2 = 116 days 19. (a) 3 1H + 2 1H → 4 2He + 1 0n (b) (3.016 049 + 2.014 102)u = mY + 17.6 931 u + 1.008 665 u mY = 4.002 582 20. (a) Strong nuclear force (b) (i) Neutrons only experience attractive nuclear force. Protons are charge positive and experience repulsive coulomb forces. The repulsive coulomb forces are compensated by more neutrons in nucleus. Hence it prevents the nucleus from breaking up. (ii) Nuclide P has more neutrons. To reduce the number of neutrons, P decays by emitting a beta-particle. Nuclide Q has more protons. To reduce the number of protons, Q decays by an emitting alpha-particle. (c) (i) Δm = (1.007275 + 1.008665) u – 2.013552 u = 0.002388 u Binding energy = (0.002388)(931 MeV) = 2.22 MeV (ii) Energy required = binding energy per nucleon = 1.11 MeV (iii) By nuclear fusion: Possible reactions: 2 1H + 2 1H → 3 2He + 1 0n 2 1H + 2 1H → 3 1H + 1 1H 21. (a) Refer to page 294 - 295 (b) Dm = 7mp + 8mn – mN = [7(1.007276) + 8(1.008665) - 15.000108] u = 0.120144 u Binding energy per nucleon = (0.120144)(931.5) 15 MeV = 7.46 MeV (c) (i) Nucleons are in a lower energy state. (ii) 238 92U → 234 90Th + 4 2 He The daughter nucleus 234 90Th has higher binding energy per nucleon. The nucleons in Th-234 are in a lower energy state. The difference in energy is released. (iii) Energy released = (binding energy of 4 2 He) – (binding energy of two 1 2 H) = [4(7) – 2(2 × 1)] MeV = 24 MeV 22. (a) Nuclear fusion: two nuclei combine to form a larger nucleus. Nuclei are positively charged. For fusion to occur the two nuclei must have sufficient kinetic energy to overcome the repulsive coulomb force. Large nuclei have more charge, hence greater repulsive force. (b) (i) 2 1 H + 2 1 H → 3 2 He + 1 0 n + 3.27 MeV (ii) 2(2.014102) u = mX + 1.008665 u + 3.27 931.5 u mX = 3.016029 u (c) (i) Electric potential energy, U = qq 4πe0 r = (1.60 × 10–19)2 4π(8.85 × 10–12)(1.0 × 10–14) (1.60 × 10–19)106 MeV = 0.144 MeV (ii) Minimum kinetic energy of a deuteron = 1 2 (0.144 MeV) = 0.072 MeV (iii) 3 2 kT = 0.072 MeV T = 2(0.072 × 106 )(1.60 × 10–19) 3(1. 38 × 10–23) = 5.57 × 108 K 23. (a) (i) Refer to page 301 (ii) Charge and mass energy are conserved. (b) (i) 99 43Tc → 99 44X + 0 –1e + g (ii) – dN dt = λN, N = (– dN dt )/λ Initial mass, m0 = ( N NA )M = ( (80 × 10–6)(3.7 × 1010) (In 2/6.0 × 3600)(6.02 × 1023))(0.099)kg = 1.52 × 10-14 kg (iii) 24h = 4t 1 2 , activity A = ( 80 24 ) µCi = 5.0 µCi
Physics Term 3 STPM Chapter 25 Nuclear Physics 351 25 (iv) Mass = ( 1 24 )(1.52 × 10-14) kg = 9.50 × 10-16 kg (v) Use lead plated as shielding against gamma-rays Dispose of syringe or contain used to store Tc-99m according to safety procedure. 24. (a) Refer to text. (b) Binding energy per nucleon / MeV 9 0 H-2 He-3 U-238 240 A 2 1 H + 2 1 H → 3 2 He + 1 0 n In the fusion of H-2 to form He-3, – He-3 has greater binding energy per nucleon. – Nucleons in He-3 are in a lower energy state than nucleons of H-2. – Difference in energy of the nucleons are released. (c) (i) 2 1 H + 3 1 H → 4 2 He + 1 0 n (ii) Q = [mD + mT – mHe – mn] (931.5 MeV) = [ 2.01410 + 3.01605 – 4.002603 – 1.00866] (931.5 MeV) = 17.6 MeV (iii) Binding energy of deuterium = [1.00728 + 1.00866 + 0.00055 – 2.01410](931.5) MeV = 1.71 MeV Binding energy of tritium = [1.00728 + 2(1.00866) + 0.00055 – 3.01605](931.5) MeV = 8.48 MeV Binding energy of helium nucleus = [2(1.00728) + 2(1.00866) + 2(0.00055) – 4.002603](931.5) MeV = 28.3 MeV (iv) Binding energy per nucleon of H-2, H-3, and He-4 are respecively 0.855 Mev, 2.83 MeV and 7.08 MeV Least stable: deuterium (H-2) Most stable: helium (He-4) 25. (a) 1 0 n + 10 5 B → 7 3 Li + 4 2 He (b) Momentum is conserved (7.0160 u)(5.19 × 106 ) + (4.0026 u)v = 0 Velocity, v = - 9.10 × 106m s-1 Speed of He-4 = 9.10 × 106m s-1 (c) (Dm) = [(1.0087 + 10.0130) – (7.0160 + 4.0026)] u = 0.0031 u (Dm)c2 = K.E. of Li + K.E. of He + Q Q = (0.0031)(1.66 × 10-27)(3.00 × 108 ) 2 – 1 2 (7.0160)( 1.66 ×10-27)(5.19 × 106 ) 2 – 1 2 (4.0026)(1.66 × 10-27)(9.10 × 106 ) 2 J = 3.12 × 10-14 J 26. (a) Refer to text. (b) 226 88Ra → 222 86Rn + 4 2 He Dm = [226.025402 – (222.017570 + 4.002603)] u = 0.005229 u Reaction energy, Q = (0.005229)(934 MeV) = 4.88 MeV Since Q is positive, the reaction is exothermic. (c) Radioactive tracer: a radioisotope that is introduced into the body by injection or orally. The passage of the radioisotope is traced by an external detector. (i) Radiation is gamma-ray. (ii) Half-life of several hours. Reason: after several hours the activity of the sample introduced into the body will be negligible. No harm to the patient. (d) Number of atoms, N = ( 1.8 × 10–6 131 )(6.02 × 1023) = 8.27 × 1015 dN dt = –λN = – ln 2 t 1 2 N = – ( ln 2 (8.0(24 × 3600))(8.27 × 1015) Bq Activity = 8.29 × 109 Bq 27. (a) (i) 4 2 a + 9 4 Be → 12 6 C + 1 0 X (ii) X in a neutron (1 0 n) (iii) – Neutron is not charged. – Has high penetrating power – Does not cause ionization. – is not deflected by electric field or magnetic field (any two) (b) (i) 1 0 X + 235 92U → 141 56Ba + 92 36Kr + 31 0 X a = 36, b = 3 (ii) Proton is positively charged. Strong repulsive force between proton and nucleus. Kinetic energy of accelerated proton is insufficient for it to reach the nucleus. Neutron is not charged. It does not encounter any repulsion
352 STPM Model Paper (960/3) Paper 3 Kertas 3 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fi fteen questions in this section. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam bahagian ini. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. What is the condition for a simple pendulum to be in simple harmonic motion? Apakah syarat bagi sebuah bandul ringkas melakukan gerakan harmonik ringkas? A The pendulum bob must be heavy. Ladung bandul perlu berat. B The length of the string must be long. Tali bandul perlu panjang. C The amplitude of oscillation must be small. Amplitud ayunan perlu kecil. D The period of oscillation must be short. Tempoh ayunan perlu pendek. 2. A particle is in simple harmonic motion of amplitude A. At time t = 0, the particle is at the mean position O and moving in the negative x direction as shown in the diagram below. Suatu zarah adalah dalam gerakan harmonik ringkas dengan amplitud A. Pada masa t = 0, zarah di titik keseimbangan dan bergerak menuju arah negatif x seperti ditunjukkan dalam rajah di bawah. –A O +A t = 0 Which graph correctly shows the variation of displacement, x with time t? Graf yang manakah menunjukkan dengan betul ubahan sesaran x dengan masa t? A 0 t x C 0 x t B 0 x t D 0 x t
Physics Term 3 STPM Model Paper Term 3 353 3. Five simple pendulums W, A, B, C and D hang from a wire stretched between two points X and Y. Pendulum W has a heavy bob, while the bobs of P, Q, R and S are light. When pendulum P oscillates, it drives the other pendulums into oscillation. Which of the pendulum oscillates with the largest amplitude? Lima bandul ringkas W, A, B, C dan D digantung dari suatu dawai yang diregang antara titik X dan Y. Ladung W adalah berat, manakala ladung P, Q, R dan S adalah ringan. Apabila bandul W berayun, bandul yang lain dipaksa berayun. Bandul yang manakah berayun dengan amplitud yang terbesar? X Y W A B C D 4. A progressive wave of amplitude 0.05 m, wavelength 0.20 m travels in the positive x-direction with a speed of 1.6 m s-1. The progressive wave equation is Suatu gelombang maju yang amplitudnya 0.05 m, dan panjang gelombangnya 0.20 m bergerak dalam arah positif x dengan laju 1.6 m s-1. Persamaan gelombang maju ialah A y = 0.05 sin 2π(8t – 5x) B y = 0.05 sin 2π(8t + 5x) C y = 0.05 sin 4π(4t – 0.06x) D y = 0.05 sin 4π(4t + 0.06x) 5. Which equation correctly relates the speed c of electromagnetic wave in free space to the permittivity ε0 and permeability µ0 free space? Persamaan yang manakah mengaitkan dengan betul laju c gelombang elektromagnet dalam ruang bebas dengan ketelusan ε0 dan ketelapan µ0 ruang bebas? A c = ε0 µ0 B c = ε0 µ0 C = 1 ε0 µ0 D c = 1 ε0 µ0 6. The intensity level of a sound at a distance x from a point source is 16 dB. What is the intensity level of sound at a distance of 2x from the source? Aras keamatan bunyi pada jarak x dari suatu sumber titik ialah 16 dB. Berapakah aras keamatan bunyi pada jarak 2x dari sumber? A 4 dB B 8 dB C 10 dB D 12 dB 7. A tube which is open at both ends is of length 30.0 cm. What are the two lowest resonance frequencies? (Speed of sound in air = 330 m s-1) Panjang sebatang tiub yang terbuka pada kedua-dua hujungnya ialah 30.0 cm. Berapakah dua frekuensi resonans terendah tiub itu? A 550 Hz, 1100 Hz B 550 Hz, 1650 Hz C 1100 Hz, 2200 Hz D 1100 Hz, 3300 Hz 8. A biconvex lens is made of glass of refractive index 1.50. The convex surfaces have the same radius of curvature of 15.0 cm. What is the focal length of the lens? Sebuah kanta dwicembung diperbuat daripada kaca yang indeks biasannya 1.50. Jejari kelengkungan kedua-dua permukaan cembung kanta ialah 15.0 cm. Berapakah panjang fokus kanta? A 5.0 cm B 10.0 cm C 15.0 cm D 20.0 cm
Physics Term 3 STPM 354 9. A spherical convex mirror is of focal length f. An object is at a distance x = 1.5f in front of the mirror. The image formed by the mirror is Panjang fokus sebuah cermin cembung ialah f. Suatu objek adalah pada jarak x = 1.5f dari cermin. Imej yang dibentuk oleh cermin cembung itu adalah A larger than the object, inverted and in front of the mirror. lebih besar daripada objek, songsang dan di hadapan cermin. B larger than the object, upright and behind the mirror. lebih besar daripada objek, tegak dan di belakang cermin. C smaller than the object, inverted and in front of the mirror. lebih kecil daripada objek, songsang dan di hadapan cermin. D smaller than the object, upright and behind the mirror. lebih kecil daripada objek, tegak dan di belakang cermin. 10. A beam of monochromatic light is incident on a narrow slit. The point O is at the centre of the screen. Suatu alur cahaya monokromatik ditujukan pada satu celah yang sempit. Titik O adalah di tengah skrin. O Monochromatic light Cahaya monokromatik Single slit Celah tunggal Screen Skrin The diffraction pattern produced on the screen is Corak belauan yang dibentuk ialah A O C O B O D O 11. Monochromatic light is incident on the cathode of a photo cell. When the potential difference V across the photocell is varied, the current I in the photocell varies as shown in the graph below. Cahaya monokromatik ditujukan pada katod sebuah selfoto. Apabila beza keupayaan V merentasi selfoto diubah, arus I dalam selfoto berubah seperti ditunjukkan oleh graf di bawah. 0 V I
Physics Term 3 STPM Model Paper Term 3 355 The intensity of monochromatic light incident on the photocell is increased. Which graph shows the new I-V graph? Keamatan cahaya monokromatik pada selfoto ditambah. Graf yang manakah menunjukkan graf I-V yang baru? A 0 V I New Baru C 0 V I New Baru B 0 V I New Baru D 0 V I New Baru 12. The four lowest energy levels in the hydrogen atom are shown in the diagram below. Which electron transition produced a photon of infrared radiation? Empat aras tenaga terendah dalam atom hidrogen ditunjukkan pada rajah di bawah. Peralihan elektron yang mana menghasilkan satu foton sinaran infra merah? E4 E3 E2 E1 BA C D 13. A nuclide is represented by 13 8P. Which nuclide has the same number of neutrons as 13 8P? Suatu nuklid diwakili oleh 13 8P. Nuklid yang manakah mempunyai bilangan neutron yang sama dengan 13 8P? A 13 8W C 13 7Y B 12 7X D 12 8Z 14. The 23.8 MeV energy is released in the fusion reaction 2 1H + 2 1H → 4 2He is 23.8 MeV. The mass of 4 2He is 4.0026 u, what is the mass of 2 1H? [1 u is equivalent to 931.5 MeV] Tenaga sebanyak 23.8 MeV dibebaskan dalam tindak balas pelakuran 2 1H + 2 1H → 4 2He. Jisim 4 2He ialah 4.0026 u, berapakah jisim 2 1H? [1 u setara dengan 931.5 MeV] A 1.9885 u C 2.0141 u B 2.0013 u D 2.00282 u 15. The initial activity of a sample of radioactive isotope is 600 Bq. The half life of the isotope is 20 minutes. What is the activity of the sample after 10 minutes? Keaktifan asal suatu sampel isotop radioaktif ialah 600 Bq. Separuh hayat isotop ialah 20 minit. Berapakah keaktifan sampel itu selepas 10 minit? A 300 Bq C 420 Bq B 360 Bq D 450 Bq
Physics Term 3 STPM 356 Section B (15 marks) Bahagian B (15 markah) Answer all the questions in this section. Jawab semua soalan dalam bahagian ini. 16. One end of the helical spring of force constant 3.20 N m-1 is attached to the wall and a mass of 0.050 kg is attached to the other end. The mass is on a smooth horizontal surface. The mass is displaced to a point P which is 4.00 cm from the equilibrium position O and then released and it performs simple harmonic motion. Satu hujung sebuah spring pusar ditetapkan pada dinding dan suatu jisim 0.050 kg dipasangkan pada hujung yang lain. Pemalar daya spring ialah 3.20 N m-1. Jisim itu di atas permukaan ufuk yang licin. Jisim itu disesar ke titik P, 4.00 cm dari O, dan kemudian dilepaskan untuk melakukan gerakan harmonik ringkas. 4.00 cm O P m = 0.050 kg (a) Calculate Hitungkan (i) the period of the simple harmonic motion. tempoh gerakan harmonik ringkas itu. [2 marks / 2 markah] (ii) the time taken for the mass to move from P to a point 2.00 cm to the left of O. masa untuk jisim itu bergerak dari P ke suatu titik 2.00 cm di sebelah kiri O. [3 marks / 3 markah] (b) (i) Sketch the displacement-time graph of the motion. Lakarkan graf sesaran-masa gerakan itu. [2 marks / 2 markah] (ii) If the mass oscillates on a rough surface, on the same axes, sketch the displacementtime graph. Jika jisim berayun di atas permukaan yang kasar, pada paksi-paksi yang sama, lakarkan graf sesaranmasa. [1 mark / 1 markah] 17. A converging lens and a concave spherical mirror both of focal length 10.0 cm are arranged coaxially and close to each other. A point object O is 30.0 cm from the lens as shown in the diagram. Sebuah kanta penumpu dan sebuah cermin cekung yang mempuyai panjang fokus yang sama 10.0 cm disusun sepaksi dan rapat antara satu sama lain. Suatu objek titik O adalah 30.0 cm dari kanta seperti ditunjukkan dalam rajah. 30.0 cm O The mirror is slowly moved away from the lens. For two values of the separation between the lens and mirror, the image formed by the lens-mirror combination is at the same position as the object O. Determine the values of the two separations between the lens and mirror. Illustrate you answers with ray diagrams.
Physics Term 3 STPM Model Paper Term 3 357 Cermin itu digerakkan dari kanta dengan perlahan. Untuk dua nilai pemisahan di antara kanta dan cermin, imej yang dibentuk oleh gabungan kanta-cermin adalah pada kedudukan objek O. Tentukan nilai dua pemisahan di antara kanta dan cermin itu. Lukiskan gambarajah-gambarajah sinar untuk jawapan anda. [7 marks / 7 markah] Section C (30 marks) Bahagian C (30 markah) Answer two questions in this section. Jawab mana-mana dua soalan dalam bahagian ini. 18. (a) Explain the meaning of diffraction and interference of light. Jelaskan maksud belauan dan interferens cahaya. [2 marks / 2 markah] A beam of monochromatic light of wavelength 480 nm is at normal incidence on a diffraction grating, as shown in the diagram below. The diffraction grating has 750 lines per millimetre. Satu alur cahaya monokromatik berpanjang gelombang 480 nm ditujukan secara normal pada sekeping parutan belauan seperti ditunjukkan dalam rajah. Terdapat 750 garis setiap milimeter parutan itu. Monochromatic light Cahaya monokromatik Diffraction grating Parutan belauan (b) Use the concepts of diffraction and interference to explain the action of the diffraction grating. Dengan menggunakan konsep belauan dan interferens, jelaskan tindakan sekeping parutan belauan. [3 marks / 3 markah] (c) Calculate Hitungkan (i) the angle between the first order line and the zero order line, sudut di antara garis peringkat pertama dengan garis peringkat sifar. [2 marks / 2 markah] (ii) the maximum order of diffraction produced by the grating, peringkat tertinggi belauan yang dihasilkan. [2 marks / 2 markah] (d) Copy the above diagram and draw the different order diffracted beams. Label the order of each beam. Salin rajah di atas dan lukiskan alur terbelau yang dihasilkan. Label peringkat belauan setiap alur. [2 marks / 2 markah] (e) The monochromatic beam is substituted with a beam of white light. With the aid of a diagram, state the changes to the diffraction produced. Alur monokromatik itu digantikan dengan alur cahaya putih. Dengan bantuan satu gambarajah, nyatakan perubahan dalam belauan yang dihasilkan. [4 marks / 4 markah]
Physics Term 3 STPM 358 19. (a) (i) What is meant by the wave-particle duality? Apakah yang dimaksudkan dengan kedualan gelombang-zarah? [2 marks / 2 markah] (ii) Write the expression representing de Broglie's relation. State the quantity in the expression that represents the particle-property, and the quantity that represents the wave property. Tuliskan ungkapan yang mewakili persamaan de Broglie. Nyatakan kuantiti dalam persamaan itu yang mewakili sifat zarah, dan kuantiti yang mewakili sifat gelombang. [4 marks / 4 markah] (b) The kinetic energy of a particle of mass m is K. Deduce, in terms of m and K, the expression for Tenaga kinetik suatu zarah yang berjisim m ialah K. Deduksikan dalam sebutan m dan K, ungkapan bagi (i) the momentum, p and momentum, p dan [2 marks / 2 markah] (ii) the de Broglie's wavelength, λ of the particle. panjang gelombang de Broglie, λ zarah itu. [2 marks / 2 markah] (c) If the de Broglie's wavelength of the particle is doubled, discuss the change in Jika panjang gelombang de Broglie zarah itu adalah berganda, bincang perubahan dalam (i) the kinetic energy, and tenaga kinetik, dan [2 marks / 2 markah] (ii) the velocity of the particle. halaju zarah. [3 marks / 3 markah] 20. (a) Explain radioactivity and decay constant. Terangkan keradioaktifan dan pemalar reputan. [2 marks / 2 markah] (b) An isotope X decays by emitting a beta-particle to 225 89Ac and another isotope Y decays by emitting an alpha-particle to 220 86Rn. Suatu isotop X mereput kepada 225 89Ac dengan mengeluarkan satu zarah-beta, dan satu isotop yang lain Y mereput kepada 220 86Rn dengan mengeluarkan satu zarah-alfa. (i) Write equations to represent the two decays. Tulis dua persamaan untuk mewakili dua reputan itu. [4 marks / 4 markah] (ii) State one common property of X and Y. Nyatakan satu persamaan sifat X dan Y. [1 mark / 1 markah] (c) 226 88Ra has a half-life of 1600 years. Calculate 226 88Ra mempunyai setengah hayat 1600 tahun. Hitungkan (i) the decay constant of 226 88Ra. pemalar reputan 226 88Ra. [2 marks / 2 markah] (ii) the number of 226 88Ra atoms in a sample of 226 88Ra of mass 1.0 g. bilangan atom dalam suatu sampel 226 88Ra yang berjisim 1.0 g. [2 marks / 2 markah] (iii) the activity of 1.0 g of 226 88Ra. aktiviti 1.0 g 226 88Ra. [2 marks / 2 markah] (iv) the number of atoms of 226 88Ra left in the sample after 400 years. bilangan atom 226 88Ra yang tinggal dalam sampel itu selepas 400 tahun. [2 marks / 2 markah]
Physics Term 3 STPM Model Paper Term 3 359 PAPER 1 Section A 1. C 2. B 3. C: Pendulum C has the same length as that of W. 4. A: f = c λ = 8 Hz y = A sin (2πft – 2π λ x) y = 0.05 sin 2π(8t –5x) 5. D 6. C: 16 dB = 10 log10( I I0 ), I∝ 1 d2 At distance of 2d, I’ = 1 4 I Intensity level = 10 log10(I/4 I0 ) = 10 dB 7. A: 0.30 m = λ 2 = v 2f 0 Lowest frequencies = f 0 , 2f 0 8. C: Use 1 f = (n – 1)(1 r + 1 r ) 9. D 10. D 11. A: Higher saturation current. 12. D: A and B: UV light, C: visible light 13. B: Neutron number of 13 8P is (13 – 8) 14. C: mH = 1 2 (4.0026 + 23.8 931.5 ) u 15. C: 10 min T1/2 = 1 2 = 0.50 Activity, A = A0 20.50 = 420 Bq 16. (a) (i) T =2π m k = 0.785 s (ii) x = 4.00 cos ωt and ω = 2π T - 2.00 = 4.00 cos (8t) t = 0.262 s (b) 0 x t (i) (ii) 17. First separation 1 f = 1 u1 + 1 ν1 gives v1 = 15.0 cm 30.0 cm 15.0 cm O I1 30.0 cm 35.0 cm O I1 I Second separation I 1 at v2 = 2f 2 = 20.0 cm from mirror. Separation between lens and mirror = (15.0 + 20.0) cm = 35.0 cm 18. (a) Diff raction: spreading of light waves as they emerge from a narrow slit. Interference: The eff ect of superposition of light waves from two coherent sources. (b) Light waves are diff racted by the slits in the diff raction grating. The diff racted waves are coherent. Construction interference between waves that emerge from the grating produces maxima or bright lines in certain directions. In other directions, no light is observed because of destruction interference. (c) (i) d sin θ1 = (1) λ and d = 1 N θ1 = 21° (ii) sin θ = n(Nλ) < 1.00 n < 2.8 Maximum order = 2 ANSWERS
Physics Term 3 STPM 360 (d) n = 1 n = 2 n = 0 n = 1 n = 2 (e) n = 1 n = 2 n = 0 n = 1 n = 2 White light Cahaya putih R R R V V V V R White Putih Changes n = 0, white line n =1 and n = 2, spectrum of white light with violet (V) diffracted more than red (R). 19. (a) (i) Wave-particle duality: Under certain conditions a particle has wave properties such as having a wave length and show diffraction. Under certain conditions a wave has particle properties such as having a mass and momentum. (ii) Wavelength of a particle of mass m moving with a velocity ν is λ = h mν, where h is Plank's constant. Particle property: mν Wave property: λ (b) (i) K = 1 2 mν2 gives ν = 2K m mν = m 2K m = 2mK (ii) λ = h mν = h 2mK (c) (i) λ' = 2λ = h 2mK1 K1 = 1 4 K (ii) K1 = 1 4 K = 1 2 mν2 1 ν1 = 1 2 ν 20. (a) Radioactivity: Random and spontaneous emission of radiations from unstable nuclei. Half-life: Time taken for the activity (or the number of atoms) in a radioactive sample to decay to half of its initial value (or number). (b) (i) 225 88X → 0 –1e + 225 89Ac 224 88Y → 4 2He + 220 86Rn (ii) Same proton number. (c) (i) λ = 1n2 T1/2 = 4.33 × 10-4 year-1 (ii) N0 = ( 1 226)(6.02 × 1023) = 2.66 × 1021 (iii) A0 = –( dN dt ) 0 = λN0 = 3.65 × 1010 Bq (iv) x = 400 1600 = 0.25 N = N0 2x = 2.24 × 1021
Summary of Key, Quantities and Units QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Amount of matter Amaun jirim n mol Electric current Arus elektrik I A Length Panjang l m Mass Jisim m kg Temperature Suhu T K Time Masa t s Other Quantities Kuantiti Lain Acceleration Pecutan a m s–2 Acceleration of free fall Pecutan jatuh bebas g m s–2 Activity of radioactive source Aktiviti sumber radioaktif A s–1, Bq Amplitude Amplitud A m Angular frequency Frekuensi sudut ω rad s–1 Area Luas A m2 Atomic mass Jisim atom ma kg Atomic number (proton number) Nombor atom (nombor proton) Z Capacitance Kapasitans C F Change of internal energy Perubahan tenaga dalam ∆U J Change carrier density Ketumpatan pembawa cas u Conductivity Kekonduksian σ Ω–1 m–1 Critical angle Sudut genting θc ° Current density Ketumpatan arus J A m–2 Decay constant Pemalar reputan λ s–1 Density Ketumpatan ρ kg m–3 Electric charge Cas elektrik Q, q C Electric fi eld strength Kekuatan medan elektrik E N C–1 Electric fl ux Fluks elektrik fi N C–1 m2 Electric potential Keupayaan elektrik V V Electric potential difference Beza keupayaan elektrik V V Electromotive force Daya gerak elektrik ε, E V Electron mass Jisim elektron me kg, u Elementry charge Cas keunsuran e C Emissivity Keterpancaran e Energy Tenaga E, U J Focal length Panjang fokus f m Force Daya F N Force constant Pemalar daya k N m–1 Frequency Frekuensi f, v Hz Gravitational fi eld strength Kekuatan medan graviti g N kg–2 Gravitational potential Keupayaan graviti V J kg–1 Half-life Setengah hayat t 1/2 s Heat Haba Q J Heat capacity Muatan haba C J K–1 Image distance Jarak imej v m Impedance Impedans Z Ω 361
QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Intensity Keamatan I Q m–2 Internal energy Tenaga dalam U J Magnetic flux Fluks magnet φ Wb Magnetic flux density Ketumpatan fluks magnet B T Magnification power Kuasa pembesaran m Mass number (nucleon number) Nombor jisim(nombor nukleon) A Molar heat capacity Muatan haba molar Cm J K–1 mol–1 Molar mass Jisim molar M kg mol–1 Molecular speed Laju molekul c ms–1 Momentum Momentum p N s Mutual inductance Induktans saling M H Neutron mass Jisim neutron mn kg, u Neutron number Nombor neutron N – Object distance Jarak objek u m Period Kala T s Permeability Ketelapan µ H m–1 Permeability of free space Ketelapan ruang bebas µ0 H m–1 Permittivity Ketelusan ε F m–1 Permittivity of free space Ketelusan ruang bebas ε0 F m–1 Phase difference Beza fasa φ J Potential energy Tenaga keupayaan U J Power Kuasa P W Pressure Tekanan p Pa Principle molar heat Muatan haba molar utama CV.m, Cp.m J K–1 mol–1 Ratio of heat capacities Nisbah muatan haba γ Reactance Reaktans X Ω Refractive index Indeks biasan n – Relative atomic mass Jisim atom relatif Ax – Relative molecular mass Jisim molekul relatif Mr – Relative permeability Ketelapan relatif µr – Relative permittivity Ketelusan relatif εr – Resistance Rintangan R Ω Resistivity Kerintangan ρ Ω m Self inductance Swainduktans L H Specific heat capacity Muatan haba tentu c J K–1 kg–1 Speed of electromagnetic waves Laju gelombang (elektromagnet) c m s–1 Stress Tegasan σ Pa Temperature Suhu T, θ °C Tension Tegangan T N Thermal conductivity Kekonduksian terma k, λ W m–1 K–1 Time constant Pemalar masa τ s Velocity Halaju u, v m s–1 Volume Isi padu V m3 Wavelength Panjang gelombang λ m Wave number Nombor gelombang k m–1 Weight Berat W N Work Kerja W J Work function Fungsi kerja φ, W J Young’s modulus Modulus Young E,Y Pa, N m–2 Physics Term 3 STPM Summary of Key Quantities and Units 362