Chapter 16 What is radioactivity? Radioactivity is the spontaneous disintegration of nuclei. The first radioactive elements discovered were the heavy atoms thorium and uranium. These heavy atoms and others near to them in the Periodic Table emit alpha particles, beta particles and gamma rays. It is now clear that other radioactive nuclei are found in nature. Some of these are the result of the impact of high energy radiation on the nucleus of a non-radioactive atom. Radioactive carbon-14, used in radiocarbon dating, is formed in constantly replenished trace amounts by the interaction of high energy cosmic rays with atoms in the upper atmosphere. Others, such as radon, a radioactive gas associated with granite; result from the radioactive decay of natural uranium ores. Large numbers of other artificial radioactive elements have now been prepared. These lighter elements tend to emit electrons and positrons rather than alpha particles. How does carbon dating work? To date organic materials less than about 50,000 years old, the radioactive decay of carbon-14 is preferred. Radiocarbon dating rests upon two main facts. The first is that the production of radioactive carbon 4 6C in the upper atmosphere due to the interaction of cosmic ray neutrons with nitrogen is constant and has been over past ages. The production reaction is: 14 7 N + 1 0 n 14 6 C + 1 1 H The second is that the rate of decay of this radioactive carbon is constant and has not changed over time via the reaction: 146 C 14 7 N + 0-1 e; t ½ = 5730 years.
The carbon-14 is distributed throughout the atmosphere in the form of carbon dioxide, CO 2, molecules and due to atmospheric diffusion a fairly constant proportion of all CO 2 is radioactive due to the presence of carbon-14. Living plants absorb CO 2 and so incorporate carbon-14 into their tissues. The relative quantity of carbon-14 in an organism remains constant until it dies. At this point the carbon-14 begins to decay at its normal rate. A measurement of the radioactivity of the once-living samples can then be used to determine the age of the sample itself. What produces energy in a nuclear power station? Energy is produced in a nuclear power station by nuclear fission of U-235. Nuclear fission is the breaking apart of atomic nuclei into two or more pieces. The energy produced is derived from the difference in masses of the reactants and the products. This mass difference is liberated as heat. The amount of energy per gram of U-235 is calculated to be 7.5 x 10 10 J. In contrast to this, one gram of coal burnt in air produces about 3 x 10 4 J. That is, uranium-235 fission produces about 1 million times more energy than burning fossil fuels. The fuel used in almost all nuclear power reactors is uranium dioxide. Natural uranium ores consist of approximately 99 % of the U-238 isotope, about 0.71% U-235 and small amounts of U-234. For power production, the amount of U- 235 in the UO 2 is increased to 2 4 %, the resulting material being enriched uranium dioxide. Quick quiz 1. a; 2. b; 3. c; 4. b; 5. a; 6. c; 7. a; 8. a; 9. b; 10. c; 11. b; 12. c; 13. c; 14. b; 15. a; 16. b; 17. c; 18. b; 19. b; 20. a; 21. b.
Calculations and questions 1. (a) 166 75Re, 75 protons, 91 neutrons, 166 nucleons; (b) 140 56Ba, 56 protons, 84 neutrons, 140 nucleons; (c) 18 8O, 8 protons, 10 neutrons, 18 nucleons; (d) 14 5B, 5 protons, 9 neutrons, 14 nucleons. 2. (a) 181 80Hg, 80 protons, 101 neutrons, 181 nucleons; (b) 169 77Ir, 77 protons, 92 neutrons, 169 nucleons; (c) 117 53I, 53 protons, 64 neutrons, 117 nucleons; (d) 98 40Zr, 40 protons, 58 neutrons, 98 nucleons. 3. (a) 27 14Si 27 13 Al + 0 1e; (b) 28 13Al 28 14 Si + 0-1e; (c) 24 11Na 24 12 Mg + 0-1e; (d) 17 9F 17 8O + 0 1e. 4. (a) 24 8O 24 9 F + 0-1e; (b) 47 23V 47 22 Ti + 0 1e; (c) 32 15P 32 16 S + 0-1e; (d) 39 20Ca 39 19 K + 0 1e. 5. (a) 243 100Fm 239 98 Cf + 4 2He; (b) 241 95Am 237 93 Np + 4 2He; (c) 241 94Pu 241 94 Am + 0-1e; (d) 237 92U 237 93 Np + 0-1e. 6. 5.21 min. 7. 49.8 h. 8. 3.7 x 10 10 disintegrations s -1 g -1. 9. 3.83 x 10 12 disintegrations s -1 g -1. 10. 8.25 x 10 17 disintegrations s -1 g -1. 11. 4.20 h. 12. 13.96 x 10 3 y. 13. (a) 92.5 nuclei min -1 ; (b) 17.6 d. 14. (a) 9.5 x 10 4 atoms dm -3 ; (b) 12.6 d. 15. (a) 3.11 x 10-10 g m -2 ; (b) 794.3 Bq. 16. 1.13 x 10-12 J.
17. 1.27 x 10-12 J. 18. 1.38 x 10-12 J. 19. (a) 4.52 x 10-8 kg Ra; (b) 4.44 x 10-8 kg Rn. 20. 1.614 x 10 10 kj mol -1. 21. (a) 4.10 x 10-12 J; (b) 2.44 x 10 37 fusions s -1. Solutions 1 Write the nuclear symbol, and the number of protons, neutrons and nucleons in: (a) rhenium-166; (b) barium-140; (c) oxygen-18; (d) boron-14. (a) 166 75Re, 75 protons, 91 neutrons, 166 nucleons; (b) 140 56Ba, 56 protons, 84 neutrons, 140 nucleons; (c) 18 8O, 8 protons, 10 neutrons, 18 nucleons; (d) 14 5B, 5 protons, 9 neutrons, 14 nucleons. 2 Write the nuclear symbol, and the number of protons, neutrons and nucleons in: (a) mercury-181; (b) iridium-169; (c) iodine-117; (d) zirconium-98. (a) 181 80Hg, 80 protons, 101 neutrons, 181 nucleons; (b) 169 77Ir, 77 protons, 92 neutrons, 169 nucleons; (c) 117 53I, 53 protons, 64 neutrons, 117 nucleons; (d) 98 40Zr, 40 protons, 58 neutrons, 98 nucleons.
3 Write nuclear equations for the decay of the following radioactive isotopes; the particles emitted in the decay are given in brackets: (a) 27 14Si (positron); (b) 28 13Al (electron); (c) 24 11Na (electron); (d) 17 9F (positron). (a) 27 14Si (b) 28 13Al (c) 24 11Na 27 13 Al + 0 1e; 28 14 Si + 0-1e; 24 12 Mg + 0-1e; (d) 17 9F 17 8 O + 0 1e. 4 Write nuclear equations for the decay of the following radioactive isotopes; the particles emitted in the decay are given in brackets: (a) 24 8O (electron); (b) 47 23V (positron); (c) 32 15P (electron); (d) 39 20Ca (positron). (a) 24 8O 24 9 F + 0-1e; (b) 47 23V 47 22Ti + 0 1e; (c) 32 15P (d) 39 20Ca 32 16 S + 0-1e; 39 19 K + 0 1e. 5 Write nuclear equations for the decay of the following radioactive isotopes; the particles emitted in the decay are given in brackets: (a) 243 100Fm (alpha); (b) 241 95Am (alpha); (c) 241 94Pu (electron); (d) 237 92U (electron). (a) 243 100Fm 239 98Cf + 4 2He; (b) 241 95Am (c) 241 94Pu 237 93 Np + 4 2He; 241 94 Am + 0-1e;
(d) 237 92U 237 93 Np + 0-1e. 6 Determine the half-life of a radioactive isotope from the variation of the number of radioactive disintegrations observed, in counts per minute, over a period of time, given in the Table. Time / min 0 2 4 6 8 10 12 14 Activity / counts 3160 2512 1778 1512 1147 834 603 579 per min Plot ln N versus t, slope = - = -ln 2 / t ½ As activity is proportional to N, plot activity versus t. Time / min 0 2 4 6 8 10 12 14 Activity 3160 2512 1778 1512 1147 834 603 579 Ln N 8.085 7.829 7.483 7.321 7.045 6.726 6.402 6.361 slope (8.058-6) / (0-15.5) t ½ ln 2 x 15.5 / 2.058 = 5.2 min
7 A sample of radioactive sodium-24 with a half-life of 15.0 h is used to measure the diffusion coefficient of Na in NaCl. How long will it take for the activity to drop to 0.1 of its original activity? ln (N 0 / N) = t = (ln 2 / t ½ )t ln (1 / 0.1) = (ln 2 / 15) x t t = 49.8 hr 8 What is the specific activity of radium-226, which has a half-life of 1600 y? specific activity = ln 2 x N A / (molar mass x t ½ ) = 4.174 x 10 23 / (225 x 1600 x 365 x 24 x 60 x 60) = 3.68 x 10 10 disintegrations s -1 g -1 9 What is the specific activity of plutonium-241, which has a half-life of 14.35 y? specific activity = ln 2 x N A / (molar mass x t ½ ) = 4.174 x 10 23 / (241 x 14.35 x 365 x 24 x 60 x 60) = 3.827 x 10 12 disintegrations s -1 g -1 10 What is the specific activity of neptunium-233, which has a half-life of 36.2 min? specific activity = ln 2 x N A / (molar mass x t ½ ) = 4.174 x 10 23 / (233 x 36.2 x 60) = 8.248 x 10 17 disintegrations s -1 g -1
11 A purified sample of a radioactive compound is found to have an activity of 1365 counts per minute at 10am but only 832 counts per minute at 1 pm. What is the half-life of the sample? ln (N 0 / N) = t = (ln 2 / t ½ )t ln (1365 / 832) = (ln 2 / t ½ ) x 3 t ½ = 4.20 hr 12 A 250 mg piece of carbon from an ancient hearth showed 1530 carbon-14 disintegrations in 36 h. 250 mg of fresh carbon from charcoal gave 8280 disintegrations in the same time. What is the date of the carbon sample? The half-life of carbon-14 is 5.73 x 10 3 y. ln (N 0 / N) = t = (ln 2 / t ½ )t ln (8280 / 1530) = (ln 2 / 5.73 x 10 3 ) t t = 13.96x 10 3 years 13 A cellar of dimensions 2 x 3 x 2.5 m is found to show an activity of 0.37 Bq dm -3, due to the presence of radon. (a) How many nuclei decay per minute in the cellar? (b) How long will it take for the activity to fall to 0.015 Bq, if no more radon leaks into it? The half-life of radon is 3.8 d. (a) An activity of 0.37 Bq dm -3 means that there are 0.37 disintegrations per second in each dm 3 of the cellar, so the total number of disintegrations is:
(0.37 x 20 x30 x25) / 60 = 92.5 disintegrations per minute = number of nuclei decaying (b) ln (N 0 / N) = t = (ln 2 / t ½ )t ln (0.37 / 0.015) = (ln 2 / 3.8) t t = 17.6 days 14 The suggested upper limit for radon concentration in a building is equivalent to an activity of 200 Bq m -3. (a) How many radon atoms are needed per dm 3 of air to give this figure? (b) How long will it take for the activity to fall to one tenth of this value in a sealed room? The half-life of radon is 3.8 d. (a) -dn / dt = N = (ln 2 / t ½ ) N activity = 200 Bq = 200 disintegratins per sec = dn / dt hence: 200 = (ln 2 / 3.8 x 24 x 60 x 60) N N = 9.47 x 10 7 atoms m -3 = 9.47 x 10 4 atoms dm -3 (b) ln (N 0 / N) = t = (ln 2 / t ½ )t ln (200 / 20) = (ln 2 / 3.8) t t = 12.6 days 15 In 1986 a nuclear reactor at Chernobyl exploded, depositing caesium-137 over large areas of northern Europe. An initial activity of 1000 Bq m -2 of vegetation was found over parts of Scotland. (a) How many grams of caesium-137 were deposited per square metre of vegetation? (b) What was the activity after 10 years? The half-life of caesium-137 is 30.1 y.
(a) -dn / dt = N = (ln 2 / t ½ ) N activity = 1000 Bq = 200 disintegrations per sec = dn / dt hence: 1000 = (ln 2 / 30.1 x 365 x 24 x 60 x 60) N N = 1.369 x 10 12 atoms m -2 The mass of 1 atom molar mass / N A = 137 / 6.02214 x 10 23 Total mass was = (1.369 x 10 12 x 137 / 6.02214 x 10 23 = 3.11 x 10-10 g (b) ln (N 0 / N) = t = (ln 2 / t ½ )t Taking (N 0 / N) to be proportional to the activity: ln (1000 / activity) = (ln 2 / 30.1) x 10 activity after 10 years = 794.3 Bq 16 Calculate the binding energy per nucleon for 4 2 He. The masses are 1 1H, 1.0078 u; 1 0n, 1.0087 u; 4 2He, 4.0026 u. The mass difference: m = mass A ZX [mass of Z protons + (A Z neutrons)] 2 1 1H + 2 1 0n 4 2He m = 4.0026 u - (2 x 1.0078 u + 2 x 1.0087 u) = -0.0304 u 1 u = 1.6605 x 10-27 kg; m = -0.0304 x 1.6605 x 10-27 kg E = mc 2 = 0.0304 x 1.6605 x 10-27 x (2.9978 x 10 8 ) 2 = 4.537 x 10-12 J Binding energy per nucleon = 1.134 x 10-12 J
17 Calculate the binding energy per nucleon for 23 12Mg. The masses are 23 12Mg, 22.9941 u; 1 1H, 1.0078 u; 1 0n, 1.0087 u. The mass difference: m = mass A ZX [mass of Z protons + (A Z neutrons)] 12 1 1H + 11 1 0n 23 12 Mg m = 22.9941 u - (12 x 1.0078 u + 11 x 1.0087 u) = -0.1952 u 1 u = 1.6605 x 10-27 kg; m = -0.1952 x 1.6605 x 10-27 kg E = mc 2 = 0.1952 x 1.6605 x 10-27 x (2.9978 x 10 8 ) 2 = 2.913 x 10-11 J Binding energy per nucleon = 1.266 x 10-12 J 18 Calculate the binding energy per nucleon for 34 16S. The masses are 34 16S, 33.967865 u, 1 1H, 1.0078 u; 1 0n, 1.0087 u. The mass difference: m = mass A ZX [mass of Z protons + (A Z neutrons)] 16 1 1H + 18 1 0n 34 16 S m = 33.967865 u - (16 x 1.0078 u + 18 x 1.0087 u) = -0.31354 u 1 u = 1.6605 x 10-27 kg; m = -0.31354 x 1.6605 x 10-27 kg
E = mc 2 = 0.31354 x 1.6605 x 10-27 x (2.9978 x 10 8 ) 2 = 4.679 x 10-11 J Binding energy per nucleon = 1.376 x 10-12 J 19 On decay, an atom of radium-226 emits one -particle and is converted into an atom of radon-222. A quantity of radium-226 produced 4.48 x 10-6 dm 3 of helium at 273 K and 1 atm pressure. Determine (a) the mass of radium-226 that decayed, (b) the mass of radon-222 produced, if no radon decays. One mole of helium gas occupies 22.4 dm 3 at 273 K and 1 atm pressure. 226 Ra 222 Rn + 4 He (a) The amount of He produced = 4.48 x 10-6 / 22.4 = 2 x 10-7 mol The amount of Ra is thus 2 x 10-7 mol = 0.22603 x 2 x 10-7 = 4.52 x 10-8 kg (b) The amount of Rn is 2 x 10-7 mol = 0.222 x 2 x 10-7 = 4.44 x 10-8 kg 20 Calculate the energy released per mole in the fission reaction: 235 92 U + 1 0n 102 42Mo + 128 50Sn + 6 1 0n The masses are 235 92U, 235.0439 u; 102 42Mo, 101.91025 u; 128 50Sn, 127.91047 u; 1 0n, 1.0087 u. m = mass of products mass of reactants = mass of [6 x 1 0n + 128 50Sn + 102 42Mo] - mass of [ 235 92U + 1 0n] = [6 x 1.0087 + 127.91047 + 101.91025] - [235.0439 + 1.0087] = -0.17968 u m = -0.17968 x 1.6605 x 10-27 kg
E = mc 2 = 0.17968 x 1.6605 x 10-27 x (2.9978 x 10 8 ) 2 = 2.68147 x 10-11 J per atom of U Energy per mole = 2.68147 x 10-11 x 6.02214 x 10 23 = 1.61482 x 10 13 J mol -1 = 1.614 x 10 10 J mol -1 21 Energy generation in the Sun is by way of the reaction of hydrogen to form helium. One reaction is: 4 1 H 4 He + 2 0 +1 e (a) Calculate the energy liberated per fusion reaction. It is estimated that the sun produces approximately 10 26 J s -1. (b) How many fusion reactions per second are required for this? The masses are 1 1H, 1.0078 u; 4 2He, 4.0026 u; 0 +1e, 5.486 x 10-4 u. (a) m = mass of products mass of reactants = mass of [ 4 He + 2 0 +1e] - mass of [4 1 H ] = [4.0026 + 2 x 5.486 x 10-4 ] - [4 x 1.0078] = -0.02750 u per reaction m = -0.02750 x 1.6605 x 10-27 kg E = mc 2 = 0.02750 x 1.6605 x 10-27 x (2.9978 x 10 8 ) 2 = 4.1040 x 10-12 J (b) If there are N fusion reactions per sec N x 4.1040 x 10-12 = 1 x 10 26 N = 2.437 x 1037 fusion reactions per sec.