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1 Topic 14 – Nuclear Chemistry RADIOACTIVITY A. The first three types of radiation discovered 1. Alpha radiation a. Positive charge b. Helium-4 nucleus 2. Beta radiation a. Negative charge b. High speed electron 3. Gamma radiation a. No charge b. High energy quantum of electromagnetic radiation B. Nuclear Equations 1. Nuclide symbol a. “A” is the mass number, the number of protons plus the number of neutrons in the nucleus. b. “Z” is the atomic number, the number of protons in the nucleus. c. The nuclide symbol has the form A ZX 2. Isotope name a. Gives the value of the mass number of that isotope b. Takes the form “name-A” 3. Reactants and products a. Nuclides b. Other particles (1) Alpha particle or helium nucleus 4 4 2 or 2 He (2) Proton 1 1 1 p or 1 H Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 2 (3) Neutron 1 0n (4) Beta particle or electron 0 or 10 e 1 (5) Positron 0 1 or 0 1 e (6) Gamma photon 0 0 4. Conservation rules for nuclear reactions a. Total charge is conserved conservation of atomic number (1) The sum of the charges of the products must equal the sum of the charges of the reactants. (2) The sum of the subscripts of the products must equal the sum of the subscripts of the reactants. b. Total number of nucleons is conserved conservation of mass number (1) The sum of the nucleons of the products must equal the sum of the nucleons of the reactants. (2) The sum of the superscripts of the products must equal the sum of the superscripts of the reactants. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 3 5. Comparison of chemical reactions and nuclear reactions Chemical Reactions Nuclear Reactions Atoms are rearranged by the breaking and forming of chemical bonds. Only electrons in atomic orbitals are involved in the breaking and forming of bonds Reactions are accompanied by absorption or release of relatively small amounts of energy. Rates of reaction are influenced by temperature, pressure, concentration, and catalysts. Elements or isotopes of the of the same elements are converted from one to another. Protons, neutrons, electrons, and other elementary particles may be involved. Reactions are accompanied by absorption or release of tremendous amounts of energy. Rates of reaction normally are not affected by temperature, pressure, and catalysts. 6. Writing nuclear equations a. Procedure (1) Using a periodic table, determine the value of “Z” (atomic number), or the symbol. (2) Determine the missing value of “A” (mass number). (3) Determine the missing value of “Z”. (4) Determine the identity of the missing nuclide or particle. (5) Write the complete reaction. b. Examples (1) Write the equation for the following reaction, oxygen-20 emits a beta particle producing fluorine-20. Using a periodic table, determine the value of “Z” (atomic number), or the symbol. Oxygen is O. Fluorine is F. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 4 Determine the missing value of “A” (mass number). The mass numbers are given in the name of the isotope. Determine the missing value of “Z”. The atomic numbers are determined from the periodic table. Oxygen is 8. Fluorine is 9. Write the complete reaction. 20 8 O 0 1 + 20 9 F (2) Write the equation for the following reaction, radium-226 emits an alpha particle producing radon-222. Using a periodic table, determine the value of “Z” (atomic number), or the symbol. Radium is Ra. Radon is Rn. Determine the missing value of “A” (mass number). The mass numbers are given in the name of the isotope. Determine the missing value of “Z”. The atomic numbers are determined from the periodic table. Radium is 88. Radon is 86. Write the complete reaction. 226 88 Ra 42 + Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 222 86 Rn 5 (3) 212 84 Po 208 82 Pb + X Determine the missing value of “A”. 212 = 208 + A A=4 Determine the missing value of “Z”. 84 = 82 + Z Z=2 Determine the identity of the missing nuclide or particle. A = 4 and Z = 2, therefore this is an alpha particle. Write the complete reaction. 212 84 (4) 137 55 Cs 137 56 Po 208 82 Pb + 42 Ba + X Determine the missing value of “A”. 137 = 137 + A A=0 Determine the missing value of “Z”. 55 = 56 + Z Z = 1 Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 6 Determine the identity of the missing nuclide or particle. A = 0 and Z = 1, therefore this is a beta particle. Write the complete reaction. 137 55 (5) 78 33 As 0 1 Cs 137 56 Ba + 0 1 +X Determine the missing value of “A”. 78 = 0 + A A = 78 Determine the missing value of “Z”. 33 = 1 + Z Z = 34 Determine the identity of the missing nuclide or particle. A = 78 and Z = 34, since Z = 34, therefore this is Se, and the missing nuclide is 78 34 Se Write the complete reaction. 78 33 As 0 1 78 Se + 34 C. Nuclear Stability 1. The strong nuclear force holds the nucleus together. a. It is very strong compared to the other three fundamental forces. b. It acts over a distance of only 1015 m. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 7 2. Predicting nuclear stability a. “Magic numbers” (1) Nuclei that contain 2, 8, 20, 50, 82, or 114 protons are generally more stable than those that do not possess these numbers. (2) Nuclei that contain 2, 8, 20, 50, 82, or 126 neutrons are generally more stable than those that do not possess these numbers. b. Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of one or the other. c. Nuclei with an even number of either protons or neutrons and an odd number of the other are generally more stable than those with odd numbers of both. d. All nuclei with atomic numbers higher than 83 are unstable and are radioactive. e. All isotopes of technetium (Tc, Z = 43) and promethium (Pm, Z = 61) are unstable and radioactive. 3. There is a “band of stability” in a plot of number of protons versus number of neutrons. a. Up to Z = 20 this ratio of neutrons to protons ranges from 1 up to about 1.1 near Z = 20. b. At increasingly higher Z the band of stability falls in ratios of neutrons to protons which are continually increasing (up to 1.5 at the highest values of Z). 4. Example 38 K or 38 Which is radioactive: 19 18 Ar ? Argon-38 is doubly even, and potassium-38 is doubly odd, so the radioactive nuclide would have to be potassium-38. D. Types of radioactive decay 1. 5 common types a. Alpha emission (1) Description The emission of a 42 He nucleus, or alpha particle Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 8 (2) Symbol (3) Effect on Z (atomic number) 2 (4) Effect on A (mass number) 4 (5) Usual nuclear condition Z > 83 (6) Radiation 4 4 2 or 2 He b. Beta emission (1) Description The emission of a high-speed electron (2) Symbol (3) Effect on Z (atomic number) +1 (4) Effect on A (mass number) 0 (5) Usual nuclear condition Neutron/proton ratio is too high (6) Radiation 0 1 or 0 1 e c. Positron emission (1) Description The emission of a positron (2) Symbol + (3) Effect on Z (atomic number) 1 Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 9 (4) Effect on A (mass number) 0 (5) Usual nuclear condition Neutron/proton ratio is too low (6) Radiation 0 1 or 0 1 e d. Electron capture (1) Description The capture of an electron from the innermost orbital of an atom, made possible by quantum mechanical effects (2) Symbol EC (3) Effect on Z (atomic number) 1 (4) Effect on A (mass number) 0 (5) Usual nuclear condition Neutron/proton ratio is too low (6) Radiation x rays e. Gamma emission (1) Description The emission of a gamma photon from an excited nucleus (2) Symbol (3) Effect on Z (atomic number) 0 (4) Effect on A (mass number) 0 (5) Usual nuclear condition Excited nucleus Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 10 (6) Radiation 0 0 2. Predicting types of radioactive decay a. Procedure (1) Determine whether Z > 83. Alpha emission (2) Determine neutron/proton ratio. (a) Z 20 Neutron/proton ratio less than 1.0 + or EC Neutron/proton ratio greater than 1.1 (b) Z 20 Neutron/proton ratio less than “stable band” + or EC Neutron/proton ratio greater than “stable band” b. Examples (1) Predict the type of decay for Z=7 13 7 N Z 20? YES Neutron/proton ratio = 6/7 = 0.86 Neutron/proton ratio less than 1.1? YES + or EC (2) Predict the type of decay for Z = 11 Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 26 11 Na 11 Z 20? YES Neutron/proton ratio =15/11 = 1.36 Neutron/proton ratio greater than 1.1? YES NUCLEAR BOMBARDMENT REACTIONS A. Transmutation Is the changing of one element into another by bombarding the target nucleus with nuclear particles or nuclei B. Particle accelerator Is the device used to accelerate electrons, protons, alpha particles, or other ions to the very high speeds needed to cause transmutation reactions through nuclear bombardment C. Transuranium elements Are elements with atomic numbers higher than uranium (Z = 92) All of these are man-made through various transmutation reactions D. Bombardment reactions 1. Rules for writing the abbreviated notation for bombardment reactions a. Write the nuclide symbol for the target nucleus. b. Write an open parenthesis, then the symbol for the projectile particle and a comma. c. After the comma, write the symbol for the ejected particle and a closed parenthesis. d. After the closed parenthesis, write the nuclide symbol for the product nucleus. 2. Examples of writing the abbreviated notation for bombardment reactions a. 147 N + 11 p 116 C + 42 He 14 7 N (p, ) 11 6 C Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 12 b. 63 29 Cu + 42 He 63 29 Cu (, n) 66 31 Ga + 01 n 66 33 Ga 3. Examples of writing the nuclear equations from abbreviated notation for bombardment reactions 42 a. 45 21Sc (n, ) 19 K 45 21 Sc + 01 n b. 2963 Cu (p, n) 63 29 63 30 42 19 K + 42 He Zn Cu + 11 p 63 30 Zn + 01 n BIOLOGICAL EFFECTS OF RADIATION A. Two types of biological damage from radiation 1. Somatic a. Somatic injuries affect the organism during its lifetime. b. Examples of somatic injuries: sunburn skin rash cancer cataracts 2. Genetic Genetic injuries cause inheritable gene changes (gene mutations) B. Chemical basis for radiation damage 1. Due to the ionizing ability of these types of radiation 2. Alpha and beta particles, as well as gamma photons, can cause ionization, and are therefore called “ionizing radiation”. 3. These can directly ionize biological and organic molecules. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 13 4. In addition, these can and do form radicals (also called “free radicals”). a. Definition A radical is a molecular fragment having one or more unpaired electrons b. Radicals are highly reactive and are usually short-lived. 5. Formation of radicals H2O radiation H2O+ + e a. H2O+ + H2O H3O+ + OH (hydroxyl radical) b. e + O2 O2 (superoxide ion, which is also a radical) C. Factors that affect the amount of damage caused by radiation 1. The intensity of the radiation – the number of particles or quanta absorbed 2. The energy of the radiation absorbed 3. The type of radiation absorbed D. Units of radiation 1. Curie – disintegrations a. Symbol – Ci b. Is exactly 3.70 x 1010 nuclear disintegrations per second. c. This is the same rate of decay as that of 1 gram of pure radium. 2. Roentgen – output a. symbol – R b. The amount of x-rays or gamma rays that produce ions carrying 2.1 x 109 units of electrical charge in 1 cm3 of dry air at 0 C and 1 atm pressure c. The amount of ionization of body tissue is proportional to the radiation exposure measured in roentgens. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 14 3. rad – absorbed a. The name is an acronym: radiation absorbed dose b. Is the dosage of radiation that deposits 1 x 102 J of energy per kilogram of tissue c. Is a way of measuring the intensity and energy of radiation absorbed d. With soft body tissue the “rad” and the “roentgen” are equivalent. 4. rem – damage a. The name is an acronym: roentgen equivalent for man b. 1 rem = 1 rad x 1 RBE c. RBE is from Relative Biological Effectiveness. d. The RBE factor depends on how destructive to biological tissues a type of radiation happens to be for the same amount of energy delivered to the tissue e. RBEs for selected radiation (1) X-rays: RBE = 0.7 (2) beta: RBE = 1 (3) gamma: RBE = 1 (4) neutron: RBE = 5 (5) alpha: RBE = 10 E. Duration of exposure is critical for determining the effects 1. The body can repair damage from ionizing radiation at only a certain rate. 2. The shorter the period of time over which the same dose of radiation is received, the more harmful it is. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 15 KINETICS OF RADIOACTIVE DECAY A. All radioactive decays obey first-order kinetics. B. The mathematics of radioactive decay 1. The rate law for radioactive decay is rate = Nt a. is the nuclear physics equivalent of “k” in chemical rate expressions b. Nt is the number of radioactive nuclei present at time “ t ” 2. The integrated rate law is ln N0 = t Nt 3. The expression for the half-life is 0.693147 t = ½ 4. The expression for the rate constant is 0.693147 = t1 2 C. Calculations involving radioactive decay 1. Calculating the decay constant from a measure of the activity of a substance A 2.8 x 106 g sample of plutonium-238 decays at a rate of 1.8 x 106 disintegrations per second. Calculate the decay constant of plutonium-238 in s1. mass of 238 94 GIVEN Pu = 2.8 x 106 g disintegrations rate = 1.8 x 10 s FIND Nt = ? 6 MM 238 94 2.8 x 106 g Nt = =? Pu = 238 g/mol 238 94 Pu 6.022 x 1023 nuclei 1 mol 238 g 238 94 Pu Nt = 7.08 x 1015 nuclei Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 1 mol 16 rate = 1.8 x 106 nuclei s (the number of nuclei that disintegrate per second) rate = Nt = rate Nt nuclei s = 15 7.08 x 10 nuclei 1.8 x 106 = 2.54 x 1010 s1 = 2.5 x 1010 s1 2. Calculating the half-life from the decay constant Neptunium-237 has a decay constant of 1.03 x 1014 s1. Calculate its half-life in years. GIVEN FIND t =? ½ = 1.03 x 1014 s1 t1 = 2 t ½ = 0.693147 0.693147 1.03 x 1014 s1 1 hr 3600 s 1 da 24 hr 1 yr 365.24 da = 2.132 x 106 yr = 2.13 x 106 yr 3. Calculating the half-life from activity measurements A 0.01746 g sample of potassium-40 K-40 decays at a rate of 4.5 x 103 disintegrations per second. Calculate the halflife of potassium-40 in years. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 17 mass of GIVEN K = 0.01746 g 40 19 rate = 4.5 x 103 MM 40 19 0.01746 g Nt = FIND Nt = ? disintegra tions s t =? ½ K = 40.00 g/mol 40 19 K =? 6.022 x 1023 nuclei 1 mol 40.00 g 40 19 K 1 mol Nt = 2.6286 x 1020 = rate Nt nuclei s = 20 2.6286 x 10 nuclei 4.5 x 103 = 1.71 x 1017 s1 t = ½ t ½ = 0.693147 0.693147 1.71 x 1017 s1 1 hr 3600 s 1 da 24 hr 1 yr 365.24 da t = 1.3 x 109 yr ½ 4. Calculating the decay constant from the half-life Promethium-147 has a half-life of 2.5 yr. Calculate its decay constant in s1. = 0.693147 t1 2 = 0.693147 2.5 yr 1 yr 365.24 da Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 1 da 24 hr 1 hr 3600 s 18 = 8.786 x 109 s1 = 8.8 x 109 s1 5. Using radioisotopes for dating samples a. Deriving the radiodating equation N ln 0 = t Nt = 0.693147 t1 2 0.693147 N0 ln = (t) Nt t1 2 ln N0 = 0.693147 Nt t t 12 b. Approach to radiodating (1) Determine the decay constant from a measure of the activity of a substance. (2) Calculate the half-life from the decay constant. (3) For rocks, determine the amounts of the daughter products and the amounts of the parent isotopes in the sample. (a) This uses the radioactive decay series of various radioisotopes found in the minerals of which igneous rocks are formed. (b) A radioactive decay series is a sequence of steps in which one radioactive nucleus decays to a second, and then to a third, and so on, until a stable nucleus is formed. (4) For carbon containing substances, convert all of the carbon to CO2, trap it, and then measure the level of activity per gram of total carbon. Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby 19 c. Example Carbon from the Dead Sea scrolls gave 12.1 disintegrations of carbon-14 per minute per gram of carbon. Carbon from today’s living material gives 15.3 disintegrations of carbon-14 per minute per gram of carbon. Calculate the age of the scrolls. GIVEN dis N0 15.3 min g Nt 12.1 FIND dis min g t=? t = 5730 yr ½ ln t N0 = 0.693147 Nt t1 2 ln t 15.3 = 0.693147 12.1 5730 yr t 0.2346 = 0.693147 5730 yr t 0.3385 = 5730 yr t = 1.94 x 103 yr Topic 14 – Nuclear Chemistry © 2006 Lloyd Crosby