1. NUCLEAR CHEMISTRY 2F-1 (of 15) NUCLEONS – The particles found in the nucleus Protons (+) Neutrons (0) ATOMIC NUMBER (Z) – The number of protons in the nucleus, also equal to the charge of the nucleus MASS NUMBER (A) – The number of nucleons in the nucleus, or protons plus neutrons in the nucleus

2. and are isotopes NUCLIDE – An atom with a specific number of protons and neutrons ISOTOPES – A set of nuclides with the same number of protons Hg 80 196 A  Z  Protons:80 Neutrons: 196 - 80 = 116 Hg 80 196 Hg 80 198 2F-2 (of 15)

3. NUCLEAR REACTIONS Reactions that produce new atoms TRANSMUTATION – When an atom of one element is changed into an atom of another element 14 N + 7 O + 17 8 Artificial elements are made by bombarding large nuclei with smaller ones 238 U + 92 1 H 1 238 Np + 93 He → 4 2 1 2 n 0 2 H → 1 2F-3 (of 15) In all nuclear reactions (1) the mass number is conserved and (2) atomic number (or charge) is conserved

4. STABILITY SERIES Nuclides are stable when their nuclei have enough neutrons to minimize proton-proton repulsion (a) For Z < 20 Stable nuclei have n:p ratio of 1:1 (b)For Z > 20 As Z increases, stable nuclei have n:p ratio that increases from 1:1 to eventually 1.5:1 STABLE NUCLIDES (STABLE ISOTOPES) – Atoms with nuclei that last forever RADIOACTIVE NUCLIDES (RADIOISOTOPES) – Atoms with nuclei that eventually break down to more stable nuclei 2F-4 (of 15)

5. Nuclides to the left of the line of stability are unstable because they are neutron poor Nuclides to the right of the line of stability are unstable because they are neutron rich Nuclides beyond the line of stability (with Z > 83) are unstable because they have too many total protons Stable 16 O atom: 8 n, 8 p (1:1 ratio) Stable 200 Hg atom: 120 n, 80 p 1.5:1 ratio 2F-5 (of 15)

6. Most of the stable nuclides have even numbers of protons and neutrons Neutrons Protons Even Odd Even Odd 166 57 53 8 2F-6 (of 15) There are 284 known stable nuclides

7. NUCLEAR DECAY – The process in which a radioactive nuclide turns into a more stable nuclide The type of decay depends on whether the radioactive nuclide has too many total protons, if it is neutron rich, or if it is neutron poor 2F-7 (of 15)

8. α’s are emitted from radioisotopes beyond the line of stability, those with too many total protons (Z > 83 or A > 200) (1) ALPHA DECAY (α) – The release of a helium-4 nucleus ( 4 He 2+ ) from a radioactive nucleus to become more stable U → 92 238 Th 90 234 4 α + 2 A and Z are always conserved in nuclear changes Alpha particles can be stopped by the outermost layer of skin 2F-8 (of 15)

9. β - ’s are emitted from radioisotopes that are to the right of the line of stability, those that are neutron rich Essentially a neutron decays into a proton and an electron (2) BETA MINUS DECAY (β - ) – The release of an electron from a radioactive nucleus to become more stable C → 6 14 N 7 0 β - + Beta particles penetrate about 1 cm into the body 2F-9 (of 15)

10. EC occurs in radioisotopes to the left of the line of stability, those that are neutron poor Essentially an electron and a proton turn into a neutron (3) ELECTRON CAPTURE (EC) – An electron is captured by the nucleus to become more stable Be + 4 7 Li 3 7 0 e - → 2F-10 (of 15)

11. β + ’s are just like electrons, but with a positive charge An electron is matter, but a β + is ANTIMATTER When a β + and β - meet, they are ANNIHILATED, meaning all of their mass is converted into energy A β + /β - annihilation forms 2 equal energy EM radiation photons (4) POSITRON DECAY (β + ) – The release of an electron with a positive charge from a nucleus to become more stable 2F-11 (of 15)

12. β + ’s are emitted from radioisotopes to the left of the line of stability, those that are proton rich Essentially a proton decays into a neutron and an antimatter electron C → 6 11 B 5 0 β + + 1 (4) POSITRON DECAY (β + ) – The release of an electron with a positive charge from a nucleus to become more stable 2F-12 (of 15)

13. γ ’s are emitted along with other forms of decay, or when an excited nucleus releases energy (5) GAMMA DECAY ( γ ) – The release of any high energy photon of electromagnetic radiation Ho → 67 163m Ho 67 163 γ + 0 0 Gamma rays are deeply penetrating 2F-13 (of 15) K → 19 40 Ca 20 0 β - + + γ 0 0

14. 2F-14 (of 15)

15. Several neutrons, and lots of energy are released when nuclei fission (6) SPONTANEOUS FISSION – When a large nucleus (Z > 80) breaks into two, approximately equal halves U → 92 239 Cd + 48 120 Ru + 44 116 3 n 0 1 U → 92 239 Ag + 47 119 Rh + 45 118 2 n 0 1 Daughter Products usually very radioactive, and always different 2F-15 (of 15)

17. THE RATE OF NUCLEAR DECAY Each radioisotope undergoes nuclear decay at its own unique rate HALF-LIFE (t 1/2 ) – The time required for half of the radioisotopes in a sample to decay The shorter the half-life, the more unstable the radioisotope Half-life for 125 I = 60 days At 0days: At60days: At120days: At180days: At240days: 16 125 I atoms 8 125 I atoms 4 125 I atoms 2 125 I atoms 1 125 I atom 2G-1 (of 17)

18. Half-lives range from 1 x 10 -21 seconds for 18 Na 5 x 10 15 years for 142 Ce Common half-lives 5,730 years for 14 C 4.5 x 10 9 years for 238 U 2G-2 (of 17)

19. THE DECAY EQUATION n = n 0 e -kt n 0 = at time 0, number of atoms of a radioisotope (or g or disintegrations/time) k= decay constant of a radioisotope (disintegrations atom -1 time -1 ) t= time of decay n= at time t, number of atoms of a radioisotope (or g or disintegrations/time) 2G-3 (of 17)

20. Half-life (t 1/2 ) is the time needed so that ½ of n 0 disintegrates n 0 = n 0 e -kt 1/2 ___ 2 ln (1/2) = -kt 1/2 ln 2= t 1/2 _____ k 1 = e -kt 1/2 ___ 2 ln 2 = kt 1/2 ln 2= k _____ t 1/2 or n = n 0 e -kt 2G-4 (of 17)

21. THE DECAY EQUATION n = n 0 e -kt n = n 0 e - (ln2/t 1/2 )t 2G-5 (of 17)

22. Calculate the mass of 110 Ag remaining after 2.00 minutes if you start with 1.00 g 110 Ag and its half-life is 24 seconds. = (1.00 g)e -(ln2/24 s)(120. s) = 0.031 g n = n 0 e -(ln2/t 1/2 )t n and n o can be anything proportional to the number of the radioactive atoms: (1) grams, (2) moles, (3) disintegrations per time, (4) percentages, or of course (5) atoms 2G-6 (of 17)

23. Starting with 2.00 g of a radioisotope, after 1.00 hour only 0.63 g remain. Calculate the half-life. n = n 0 e -(ln2/t 1/2 )t n = e -(ln2/t 1/2 )t ___ n 0 ln (n/n 0 ) = -(ln2/t 1/2 )t t 1/2 = -(ln2)t __________ ln (n/n 0 ) = 0.60 h= (ln2)t __________ ln (n 0 /n) = (ln2)(1.00 h) _____________________ ln (2.00 g/0.63 g) 2G-7 (of 17)

24. CARBON DATING In the atmosphere 14 N + 7 C + 14 6 n → 1 0 H 1 1 C + O 2 → CO 2 14 6 6 The carbon in all living organisms has the same percentage of 14 C that the atmosphere has 15.3 dist. min -1 g -1 of carbon When an organism dies, it stops taking in 14 C, so the percentage starts dropping 2G-8 (of 17)

25. An axe with an elk antler sleve produces 4.8 cpm g -1 of carbon. How old is the axe? n = n 0 e -(ln2/t 1/2 )t n = e -(ln2/t 1/2 )t ___ n 0 ln (n/n 0 ) = -(ln2/t 1/2 )t = 9,600 y t 1/2 ln (n 0 /n) = t ______________ ln 2 = (5,730 y) ln (15.3 cpm g -1 /4.8 cpm g -1 ) _______________________________________________ ln 2 2G-9 (of 17)

26. Much older objects can be dated with radioisotopes of longer half-lives 238 U decays to 206 Pb, so a material containing uranium can be dated by measuring the amount of 206 Pb compared to 238 U 2G-10 (of 17)

27. A rock weighing 4.267 g contains 1.023 g 238 U and 0.112 g 206 Pb. Calculate the age of the rock. = 7.7 x 10 8 y t 1/2 ln (n 0 /n) = t ______________ ln 2 = (4.5 x 10 9 y) ln (1.152 g/1.023 g) _______________________________________ ln 2 t 1/2 = 4.5 x 10 9 y n= 1.023 g n 0 = the original mass of 238 U 0.112 g 206 Pb x mol 206 Pb ______________ 206 g 206 Pb x 1 mol 238 U ______________ 1 mol 206 Pb x 238 g 238 U _____________ mol 238 U = 0.129 g + 1.023 g = 1.152 g 2G-11 (of 17)

28. STABILITY OF NUCLEI Mass of proton + electron: Mass of neutron: 1.007825 amu 1.008665 amu Calculate the mass of a 23 Na atom. 11 p + + e - 12 n =11.086075 amu =12.103980 amu 11 x 1.007825 amu 12 x 1.008665 amu =23.190055 amu Mass spectrometer data Mass 23 Na :22.989773 amu 2G-12 (of 17)

29. 23.190055 amu – 22.989773 amu = 0.200282 amu BINDING ENERGY – The mass of an atom that has been converted into energy to hold the nucleus together Through E = mc 2 mass units can be converted into energy units Mass loss of a 23 Na atom: 1.000 amu = 1.492 x 10 -10 J = 9.315 x 10 8 eV = 931.5 MeV (Joule) (Electron Volt) (Million Electron Volt) 2G-13 (of 17)

30. 0.200282 amu x 931.5 MeV ______________ 1.000 amu = 186.6 MeV This is the BINDING ENERGY of the 23 Na nucleus The stability of a nucleus is measured by its BINDING ENERGY PER NUCLEON 186.6 MeV ________________ 23 nucleons = 8.113 MeV/nucleon 2G-14 (of 17)

31. Calculate the binding energy per nucleon for 56 Fe if it has a mass of 55.934930 amu. 26 p + + e - 30 n =26.203450 amu =30.259950 amu 26 x 1.007825 amu 30 x 1.008665 amu =56.463400 amu 56.463400 amu – 55.934930 amu = 0.528470 amu x 1 _______________ 56 nucleons 0.528470 amu x 931.5 MeV ______________ 1.000 amu = 8.791 MeV/nucleon 2G-15 (of 17)

32. 56 Fe is the most stable atom When large atoms break down they release energy When small atoms combine they release energy 2G-16 (of 17)

33. FUSION – The combining of small nuclei to produce large nuclei Fusion occurs in stars 4 H → 1 1 He 2 4 Very high temperatures or pressure are needed to overcome the repulsion of the positive hydrogen nuclei Fusion releases much more energy than fission Stars can fuse atoms to create even atomic numbered elements all the way up to 56 Fe 2G-17 (of 17)

35. NUCLEAR REACTORS 235 U is used as a fuel U + 92 235 U 92 236 n → 0 1 236 U decays by spontaneous fission CHAIN REACTION – When at least one neutron per fission produces a new 236 U Nuclear reactions release over 100 times more energy than chemical reactions 2G’-1 (of 4)

36. Not enough neutrons are captured for a chain reaction CRITICAL MASS – The minimum amount of 235 U needed to support a chain reaction Enough neutrons are captured to just maintain a chain reaction So many neutrons are captured the chain reaction is an explosion 2G’-2 (of 4)

37. Water – Acts as a MODERATOR to slow down the neutrons, as a COOLANT to keep the reactor core from overheating, and as PROTECTION because it absorbs radiation Cd or B Control Rods – Absorb neutrons to control the rate of the chain reaction Fuel Elements – Metal casings containing 235 U Reactor Core 2G’-3 (of 4)

38. San Onofre Nuclear Generating Station Heat from the nuclear fission boils water, and steam turns a turbine, which produces electricity Used up full elements contain radioactive daughter products, which must be disposed of safely 2G’-4 (of 4)