CHAPTER FIVE23 Nuclear Chemistry Chapter 5 Nuclear Chemistry

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CHAPTER FIVE(23) Nuclear Chemistry

CHAPTER FIVE(23) Nuclear Chemistry

Chapter 5 / Nuclear Chemistry Chapter Five outline Chapter Five Contains: 5. 1 The

Chapter 5 / Nuclear Chemistry Chapter Five outline Chapter Five Contains: 5. 1 The Nature of Nuclear Reactions 5. 2 Nuclear Stability 5. 3 Natural Radioactivity 5. 4 Nuclear Transmutation 5. 5 Nuclear Fission 5. 6 Nuclear Fusion 2

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions §A stable

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions §A stable isotope is one that does not spontaneously decompose into another nuclide. §An radioactive nuclide is one that spontaneously decomposes into another nuclide §Some nuclei are unstable; they emit particles and/or electromagnetic radiation spontaneously. The name for this phenomenon is radioactivity. For example, the isotope of polonium, polonium-210 ( 210 Po ), decays spontaneously to 206 Pd by 84 82 emitting an α particle. §Another type of radioactivity, known as nuclear transmutation, results from the bombardment of nuclei by neutrons, protons, or other nuclei. An example of a nuclear transmutation is the conversion of atmospheric 147 N to 146 C and 11 H , which results when the nitrogen isotope captures a neutron (from the sun). In some cases, heavier elements are synthesized from lighter elements. This type of transmutation occurs naturally in outer space, but it can also be achieved artificially 3

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions Atomic number

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Mass Number A Atomic Number Z proton 1 p 1 or neutron 1 H 1 1 n 0 X Element Symbol electron 0 e -1 or a particle positron 0 b -1 0 e +1 or 0 b +1 4 He 2 or A 1 1 0 0 4 Z 1 0 -1 +1 2 4 4 a 2

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions Balancing Nuclear

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions Balancing Nuclear Equations 1. Conserve mass number (A). The sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants. 235 U + 1 n 92 0 138 Cs + 96 Rb 55 37 +2 1 n 0 235 + 1 = 138 + 96 + 2 x 1 2. Conserve atomic number (Z) or nuclear charge. The sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants. 235 U + 1 n 92 0 138 Cs + 96 Rb 55 37 92 + 0 = 55 + 37 + 2 x 0 +2 1 n 0 5

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions 212 Po

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions 212 Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212 Po. 4 He 2 alpha particle 212 Po 84 4 He 2 or + ZAX 212 = 4 + A A = 208 84 = 2 + Z Z = 82 212 Po 84 4 He 2 4 a 2 + 208 Pb 82 6

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions 7

Chapter 5 / Nuclear Chemistry 5. 1 The Nature of Nuclear Reactions 7

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability • Certain numbers of neutrons

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability • Certain numbers of neutrons and protons are extra stable • n or p = 2, 8, 20, 50, 82 and 126 • Like extra stable numbers of electrons in noble gases (e- = 2, 10, 18, 36, 54 and 86) • Nuclei with even numbers of both protons and neutrons are more stable than those with odd numbers of neutron and protons • All isotopes of the elements with atomic numbers higher than 83 are radioactive • All isotopes of Tc (technetium) and Pm (promethium)are radioactive 8

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability §The principal factor that determines

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability §The principal factor that determines whether a nucleus is stable is the neutron-to-proton ratio (n /p). §For stable atoms of elements having low atomic number, the n/p value is close to 1. As the atomic number increases, the neutron-to-proton ratios of the stable nuclei become greater than 1. §This deviation at higher atomic numbers arises because a larger number of neutrons is needed to counteract the strong repulsion among the protons and stabilize the nucleus. 9

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability §The stable nuclei are located

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability §The stable nuclei are located in an area of the graph known as the belt of stability. Most radioactive nuclei lie outside this belt. §Above the stability belt, the nuclei have higher neutron-to-proton ratios than those within the belt (for the same number of protons). To lower this ratio (and hence move down toward the belt of stability), these nuclei undergo the process, called beta-particle emission. §Below the stability belt the nuclei have lower neutron-to-proton ratios than those in the belt (for the same number of protons). To increase this ratio (and hence move up toward the belt of stability), these nuclei either emit a positron or undergo electron capture. n/p too large beta decay X Y n/p too small positron decay or electron capture 10

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Beta decay + 0 b

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Beta decay + 0 b 14 C 6 14 N 7 40 K 19 40 Ca 20 Decrease # of neutrons by 1 Increase # of protons by 1 Without changing the mass number -1 + 0 b -1 1 n 0 1 p 1 + 0 b -1 Positron decay 11 C 6 11 B 5 38 K 19 38 Ar 18 + 0 b Increase # of neutrons by 1 Decrease # of protons by 1 Without changing the mass number +1 + 0 b +1 1 p 1 1 n 0 + 0 b +1 11

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Electron capture decay 37 Ar

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Electron capture decay 37 Ar 18 + 0 e 55 Fe 26 + 0 e -1 -1 37 Cl 17 Increase # of neutrons by 1 55 Mn 25 Decrease # of protons by 1 1 p 1 + 0 e -1 1 n 0 In electron capture an electron in a low energy orbital of the atom is capture by the nucleus and converts a proton to a neutron. X rays ( not rays ) accompany electron capture , because the atom produced is in an excited electronic state. 12

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Alpha decay Decrease # of

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Alpha decay Decrease # of neutrons by 2 212 Po 84 4 He 2 + 208 Pb 82 Decrease # of protons by 2 Gama radiation It is electromagnetic radiation of very short wavelength. Dose not cause change in the mass number or in the atomic number of the nucleus. 240 Pu 94 236 U* 92 → → 236 U* 92 236 U 92 + 24 He + γ 13

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Nuclear binding energy (BE) is

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Nuclear binding energy (BE) is the energy required to break up a nucleus into its component protons and neutrons. BE + 19 F 91 p + 1001 n 9 1 E = mc 2 BE = 9 x (p mass) + 10 x (n mass) – 19 F mass BE (amu) = 9 x 1. 007825 + 10 x 1. 008665 – 18. 9984 BE = 0. 1587 amu BE = 2. 37 x 10 -11 J binding energy per nucleon = 1 amu = 1. 49 x 10 -10 J binding energy number of nucleons = 2. 37 x 10 -11 J 19 nucleons = 1. 25 x 10 -12 J 14

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Nuclear binding energy per nucleon

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Nuclear binding energy per nucleon vs Mass number nuclear binding energy nucleon nuclear stability 15

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Example: calculate the energy released

Chapter 5 / Nuclear Chemistry 5. 2 Nuclear Stability Example: calculate the energy released from the following decay : 192 Pt 72 4 He 2 + 188 Os 76 If you know : Atomic mass of Pt = 191. 9614 amu , Os= 187. 956, He= 4. 0026 amu Δm= 191. 9614 – ( 187. 965 +4. 0026) = 2. 8 x 10 -3 amu 1 amu= 931. 5 Mev 2. 8 x 10 -3 amu = ? Mev = 2. 8 x 10 -3 x 931. 5 = 2. 608 Mev 16

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Kinetics of Radioactive Decay Radioactive

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Kinetics of Radioactive Decay Radioactive decays obey a first order rate law N daughter DN Rate == l. N Dt ln(N/ N 0)= - lt l = ln(N/ N 0)= - lt = -ln 2 t/t 1/2 ln 2 t½ N = the number of atoms at time t N 0 = the number of atoms at time t = 0 λ is the decay constant Since the disintegration rate R is proportional to the number of radioactive atoms R/ R 0 = N/N 0 ln(R/ R 0)=-ln 2 t/t 1/2 17

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Kinetics of Radioactive Decay ln[N]

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Kinetics of Radioactive Decay ln[N] = ln[N]0 - lt ln [N] [N] = [N]0 exp(-lt) 18

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Example: The decay rate of

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Example: The decay rate of a sample containing 131 I is 561 disintegrations per minute. Exactly 7. 00 days later the sample decays at 307 disintegrations / min. Calculate the half-life of 131 I. ln(R/ R 0)=-ln 2 t/t 1/2 ln(307/ 561)=-ln 2 x 7/t 1/2 = -4. 85 / lin (0. 547) = 8. 04 days 19

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Radiocarbon Dating of artifacts with

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity Radiocarbon Dating of artifacts with 14 C is based on a constant activity of 15. 3 disintegrations per minute per gram of C for Living organisms. 14 N 7 + 10 n 14 C 6 14 N 7 + 1 H 1 + 0 b -1 t½ = 5730 years Dating with 14 C is valid for objects between 500 and 50. 000 years old. Example: The age of the Dead sea scrolls were measured using 14 C dating methods. If the sample of the scrolls measured had a 14 C activity of 11. 5 disintegration per minute per gram of carbon , what is the age of the scrolls if the fresh scroll recoreded 15. 3 disintegration per minute? ln(N/ N 0)= -ln 2 t/t 1/2 ln(11. 5/15. 3)= - ((ln 2 )/ 5730) x t t= 2361 year 20

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity The activity of source Activity

Chapter 5 / Nuclear Chemistry 5. 3 Natural Radioactivity The activity of source Activity = - d. N / dt = KN Activity units = curi, (ci) Ci = 3. 7 x 1010 dismtegration / s µci = 3. 7 x 104 disintegration / s 21

Chapter 5 / Nuclear Chemistry 5. 4 Nuclear Transmutation 14 N 7 27 Al

Chapter 5 / Nuclear Chemistry 5. 4 Nuclear Transmutation 14 N 7 27 Al 13 14 N 7 + 4 a 2 + 1 p 1 17 O + 1 p 30 P + 1 n 11 C 8 15 + 4 a 6 1 0 2 Cyclotron Particle Accelerator 22

Chapter 5 / Nuclear Chemistry 5. 4 Nuclear Transmutation 23

Chapter 5 / Nuclear Chemistry 5. 4 Nuclear Transmutation 23

Chapter 5 / Nuclear Chemistry 5. 5 Nuclear Fission Nuclear fission forms two nuclei

Chapter 5 / Nuclear Chemistry 5. 5 Nuclear Fission Nuclear fission forms two nuclei of comparable size from a single heavy nucleus. Fission reactions are very exoergic, and produce several neutrons in addition to two nuclides. 235 U 92 + 1 n 0 90 Sr + 143 Xe + 31 n + Energy 38 54 0 24

Chapter 5 / Nuclear Chemistry 5. 5 Nuclear Fission Nuclear chain reaction is a

Chapter 5 / Nuclear Chemistry 5. 5 Nuclear Fission Nuclear chain reaction is a self-sustaining sequence of nuclear fission reactions. The minimum mass of fissionable material required to generate a self-sustaining nuclear chain reaction is the critical mass. Non-critical Critical 25

Chapter 5 / Nuclear Chemistry 5. 5 Nuclear Fission Schematic Diagram of a Nuclear

Chapter 5 / Nuclear Chemistry 5. 5 Nuclear Fission Schematic Diagram of a Nuclear Reactor 26

Chapter 5 / Nuclear Chemistry 5. 6 Nuclear Fusion Nuclear fusion is the combination

Chapter 5 / Nuclear Chemistry 5. 6 Nuclear Fusion Nuclear fusion is the combination of two light nuclides to form a larger one. Fusion Reaction 2 H 1 + 2 H 1 2 H + 3 H 6 Li + 2 H 1 3 1 1 3 H 1 Energy Released + 1 H 1 4 He 2 + 1 n 2 4 He 0 6. 3 x 10 -13 J 2. 8 x 10 -12 J 3. 6 x 10 -12 J 2 27

Chapter 5 / Nuclear Chemistry Thank you for listening 28

Chapter 5 / Nuclear Chemistry Thank you for listening 28