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1 Nuclear Chemistry Chapter 19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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1 1 Nuclear Chemistry Chapter 19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 2 Chemical Processes vs. Nuclear Processes Chemical reactions involve changes in the electronic structure of the atom –atoms gain, lose, or share electrons –no change in the nuclei occurs Nuclear reactions involve changes in the structure of the nucleus –when the number of protons in the nucleus changes, the atom becomes a different element 2Tro: Chemistry: A Molecular Approach

3 3 X A Z Mass Number Atomic Number Element Symbol Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons A Z 1p1p 1 1H1H 1 or proton 1n1n 0 neutron 0e0e 00 or electron 0e0e +1 00 or positron 4 He 2 44 2 or  particle 1 1 1 0 0 0 +1 4 2 Review

4 4 Important Atomic Symbols 4Tro: Chemistry: A Molecular Approach

5 5 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. 1n1n 0 U 235 92 + Cs 138 55 Rb 96 37 1n1n 0 ++ 2 235 + 1 = 138 + 96 + 2x1 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. 1n1n 0 U 235 92 + Cs 138 55 Rb 96 37 1n1n 0 ++ 2 92 + 0 = 55 + 37 + 2x0

6 6 212 Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212 Po. 4 He 2 44 2 or alpha particle - 212 Po 4 He + A X 84 2Z 212 = 4 + AA = 208 84 = 2 + ZZ = 82 212 Po 4 He + 208 Pb 84 282

7 7

8 8 Nuclear Stability and Radioactive Decay Beta decay 14 C 14 N + 0  6 7 40 K 40 Ca + 0  19 20 1 n 1 p + 0  0 1 Decrease # of neutrons by 1 Increase # of protons by 1 Positron decay 11 C 11 B + 0  6 5 +1 38 K 38 Ar + 0  19 18 +1 1 p 1 n + 0  1 0 +1 Increase # of neutrons by 1 Decrease # of protons by 1

9 9 Electron capture decay Increase number of neutrons by 1 Decrease number of protons by 1 Nuclear Stability and Radioactive Decay 37 Ar + 0 e 37 Cl 18 17 55 Fe + 0 e 55 Mn 26 25 1 p + 0 e 1 n 1 0 Alpha decay Decrease number of neutrons by 2 Decrease number of protons by 2 212 Po 4 He + 208 Pb 84 282 Spontaneous fission 252 Cf 2 125 In + 2 1 n 98 490

10 10 Particle Changes 10Tro: Chemistry: A Molecular Approach

11 11 Chapter 23: The Nucleus: A Chemist’s View 1.Balance the following nuclear reactions: 1. 211 Po  207 Pb + ____ 2. 61 Ni (d, 2n) _____ 3. 27 Al + α  30 P + _____ 4. _____  187 Os + β 5. 235 U + n  135 Xe + _____ + 2 1 n 6. 75 As (α, ___) 78 Br 7. 20 Na  _____ + +1 β

12 12 n/p too large beta decay X n/p too small positron decay or electron capture Y

13 13 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 and Pm are radioactive

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15 15 Nuclear binding energy is the energy required to break up a nucleus into its component protons and neutrons. Nuclear binding energy + 19 F 9 1 p + 10 1 n 9 1 0  m = 9 x (p mass) + 10 x (n mass) – 19 F mass  E = (  m)c 2  m= 9 x 1.007825 + 10 x 1.008665 – 18.9984  m = 0.1587 amu  E = 2.37 x 10 -11 J  E = 0.1587 amu x (3.00 x 10 8 m/s) 2 = -1.43 x 10 16 amu m 2 /s 2 Using conversion factors: 1 kg = 6.022 x 10 26 amu1 J = kg m 2 /s 2

16 16 = 2.37 x 10 -11 J 19 nucleons = 1.25 x 10 -12 J/nucleon binding energy per nucleon = binding energy number of nucleons  E = (2.37 x 10 -11 J) x (6.022 x 10 23 /mol)  E = -1.43 x 10 13 J/mol  E = -1.43 x 10 10 kJ/mol Nuclear binding energy = 1.43 x 10 10 kJ/mol

17 17 Nuclear binding energy per nucleon vs mass number nuclear stability nuclear binding energy nucleon

18 18 Kinetics of Radioactive Decay N daughter rate = k N N = the number of atoms at time t N 0 = the number of atoms at time t = 0 k is the decay rate constant NtNt ln N0N0 ktkt = = t½t½ k 0.693

19 19 Half-Lives of Various Nuclides NuclideHalf-LifeType of Decay Th–2321.4 x 10 10 yralpha U–2384.5 x 10 9 yralpha C–145730 yrbeta Rn–22055.6 secalpha Th–2191.05 x 10 –6 secalpha 19Tro: Chemistry: A Molecular Approach

20 20 2.The half-life of Co-60 is 5.30 years. How many grams of a 4.567 g sample of Co-60 is left after 16.7 years? 3.Given the following atomic mass units: n = 1.008665 and p = 1.007838. Calculate a)The mass defect per C-12 atom b)The binding energy per C-12 atom c)The binding energy per nucleon for C-12 atom d)The amount of energy released per mole of C-12 4.How much energy is lost or gained when one mole of Polonium- 211 decays by alpha emission? The atomic masses are: Po-211:210.98665 Pb-207:206.97590  : 4.0026036

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22 22 Radiocarbon Dating 14 N + 1 n 14 C + 1 H 716 0 14 C 14 N + 0  + 6 7 t ½ = 5730 years Uranium-238 Dating 238 U 206 Pb + 8 4  + 6 0  92822 t ½ = 4.51 x 10 9 years

23 23 Nuclear Transmutation 14 N + 4  17 O + 1 p 7 2 8 1 27 Al + 4  30 P + 1 n 13 2 15 0 14 N + 1 p 11 C + 4  7 1 6 2

24 24 Nuclear Transmutation

25 25 Nuclear Fission 235 U + 1 n 90 Sr + 143 Xe + 3 1 n + Energy 92 54 3800 Energy = [mass 235 U + mass n – (mass 90 Sr + mass 143 Xe + 3 x mass n )] x c 2 Energy = 3.3 x 10 -11 J per 235 U = 2.0 x 10 13 J per mole 235 U Combustion of 1 ton of coal = 5 x 10 7 J

26 26 Nuclear Fission 235 U + 1 n 90 Sr + 143 Xe + 3 1 n + Energy 92 54 3800 Representative fission reaction

27 27 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.

28 28

29 29 Schematic Diagram of a Nuclear Reactor U3O8U3O8 refueling

30 30 Chemistry In Action: Nature’s Own Fission Reactor Natural Uranium 0.7202 % U-235 99.2798% U-238 Measured at Oklo 0.7171 % U-235

31 31 Nuclear Fusion 2 H + 2 H 3 H + 1 H 1 1 1 1 Fusion ReactionEnergy Released 2 H + 3 H 4 He + 1 n 1 1 2 0 6 Li + 2 H 2 4 He 3 1 2 6.3 x 10 -13 J 2.8 x 10 -12 J 3.6 x 10 -12 J Tokamak magnetic plasma confinement solar fusion

32 32 Thyroid images with 125 I-labeled compound normalenlarged

33 33 Radioisotopes in Medicine 98 Mo + 1 n 99 Mo 42 0 235 U + 1 n 99 Mo + other fission products 92042 99m Tc 99 Tc +  -ray 43 99 Mo 99m Tc + 0  42 43 Research production of 99 Mo Commercial production of 99 Mo t ½ = 66 hours t ½ = 6 hours Bone Scan with 99m Tc

34 34 Geiger-Müller Counter

35 35 Biological Effects of Radiation Radiation absorbed dose (rad) 1 rad = 1 x 10 -5 J/g of material Roentgen equivalent for man (rem) 1 rem = 1 rad x QQuality Factor  -ray = 1  = 1  = 20

36 36 Chemistry In Action: Food Irradiation


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