1. CHAPTER FIVE(23) Nuclear Chemistry

2. Chapter 5 / Nuclear Chemistry 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 Chapter Five outline 2

3. 5.1 The Nature of Nuclear Reactions Chapter 5 / Nuclear Chemistry 3  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 ( ), decays spontaneously to by 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 to and, 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 210 Po 84 206 Pd 82 14 N 7 14 C 6 1H1H 1

4. 5.1 The Nature of Nuclear Reactions Chapter 5 / Nuclear Chemistry 4 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

5. 5.1 The Nature of Nuclear Reactions Chapter 5 / Nuclear Chemistry 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. 5.1 The Nature of Nuclear Reactions Chapter 5 / Nuclear Chemistry 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. 5.1 The Nature of Nuclear Reactions Chapter 5 / Nuclear Chemistry 7

8. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 8 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

9. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 9  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.

10. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 10 n/p too large beta decay X n/p too small positron decay or electron capture Y  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.

11. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 11 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 Without changing the mass number 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 Without changing the mass number

12. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 12 Electron capture decay Increase # of neutrons by 1 Decrease # of protons by 1 37 Ar + 0 e 37 Cl 18 17 55 Fe + 0 e 55 Mn 26 25 1 p + 0 e 1 n 1 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.

13. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 13 Alpha decay Decrease # of neutrons by 2 Decrease # of protons by 2 212 Po 4 He + 208 Pb 84 282 It is electromagnetic radiation of very short wavelength. Dose not cause change in the mass number or in the atomic number of the nucleus. Gama radiation 240 Pu → 236 U * + 4 He 236 U * → 236 U + γ 94 92 2

14. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 14 Nuclear binding energy (BE) is the energy required to break up a nucleus into its component protons and neutrons. BE + 19 F 9 1 p + 10 1 n 9 1 0 BE = 9 x (p mass) + 10 x (n mass) – 19 F mass E = mc 2 BE (amu) = 9 x 1.007825 + 10 x 1.008665 – 18.9984 BE = 0.1587 amu 1 amu = 1.49 x 10 -10 J BE = 2.37 x 10 -11 J binding energy per nucleon = binding energy number of nucleons = 2.37 x 10 -11 J 19 nucleons = 1.25 x 10 -12 J

15. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 15 Nuclear binding energy per nucleon vs Mass number nuclear stability nuclear binding energy nucleon

16. 5.2 Nuclear Stability Chapter 5 / Nuclear Chemistry 16 Example: calculate the energy released from the following decay : If you know : Atomic mass of Pt = 191.9614 amu, Os= 187.956, He= 4.0026 amu 192 Pt 4 He + 188 Os 72 276 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

17. 5.3 Natural Radioactivity Chapter 5 / Nuclear Chemistry 17 Kinetics of Radioactive Decay N daughter NN tt = N Rate =- ln(N/ N 0) = - t N = the number of atoms at time t N 0 = the number of atoms at time t = 0 λ is the decay constant ln2 = t½t½ Radioactive decays obey a first order rate law ln(N/ N 0) = - t = -ln2 t/t 1/2 Since the disintegration rate R is proportional to the number of radioactive atoms R/ R 0 = N/N 0 ln(R/ R 0) =-ln2 t/t 1/2

18. ln [N] 5.3 Natural Radioactivity Chapter 5 / Nuclear Chemistry 18 Kinetics of Radioactive Decay [N] = [N] 0 exp(- t) ln[N] = ln[N] 0 - t [N]

19. 5.3 Natural Radioactivity Chapter 5 / Nuclear Chemistry 19 Example: The decay rate of a sample containing 131 I is 561 disintegrations per minute. Exactly 7.00 days later the same sample decays at 307 disintegrations / min. Calculate the half-life of 131 I. ln(R/ R 0 )=-ln2 t/t 1/2 ln(307/ 561)=-ln2 x7/t 1/2 t 1/2 = -4.85 / lin (0.547) = 8.04 days

20. 5.3 Natural Radioactivity Chapter 5 / Nuclear Chemistry 20 Radiocarbon Dating 14 N + 1 n 14 C + 1 H 716 0 14 C 14 N + 0  6 7 t ½ = 5730 years 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. 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) = -ln2 t/t 1/2 ln(11.5/15.3)= - ((ln2 )/ 5730) x t t= 2361 year

21. 5.3 Natural Radioactivity Chapter 5 / Nuclear Chemistry 21 The activity of source Activity = - dN / dt = KN Activity units = curi, (ci) Ci = 3.7 x 10 10 dismtegration / s µci = 3.7 x 10 4 disintegration / s

22. 5.4 Nuclear Transmutation Chapter 5 / Nuclear Chemistry 22 Cyclotron Particle Accelerator 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

23. 5.4 Nuclear Transmutation Chapter 5 / Nuclear Chemistry 23

24. 5.5 Nuclear Fission Chapter 5 / Nuclear Chemistry 24 235 U + 1 n 90 Sr + 143 Xe + 3 1 n + Energy 92 54 3800 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.

25. 5.5 Nuclear Fission Chapter 5 / Nuclear Chemistry 25 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

26. 5.5 Nuclear Fission Chapter 5 / Nuclear Chemistry 26 Schematic Diagram of a Nuclear Reactor

27. 5.6 Nuclear Fusion Chapter 5 / Nuclear Chemistry 27 Nuclear fusion is the combination of two light nuclides to form a larger one. 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

28. 28 Chapter 5 / Nuclear Chemistry