10. A X Mass Number Element Symbol Z Atomic Number a particle proton neutron electron positron or or 1p 1H 0b 0e or 1 1 +1 1n 0e +1 0b 0 -1 -1 4a 4He or 2 2 Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons A 1 1 0 0 4 1 0 -1 +1 2 Z 23.1
11. + + + 2 + + + 2 1n 1n 1n 1n 96 96 0 0 0 0 Rb Rb 37 37 235 138 235 138 U Cs U Cs 92 55 92 55 Balancing Nuclear Equations 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 + 1 = 138 + 96 + 2x1 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. 92 + 0 = 55 + 37 + 2x0 23.1
12. or alpha particle - 212Po 4He + AX 2 Z 84 212Po 4He + 208Pb 2 82 84 4a 4He 2 2 212Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212Po. 212 = 4 + A A = 208 84 = 2 + Z Z = 82 23.1
15. X Y n/p too large beta decay n/p too small positron decay or electron capture 23.2
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18. Predicting the mode of decay High n/p ratio (too many neutrons; lie above band of stability) --- undergoes beta decay Low n/p ratio (neutron poor; lie below band of stability) --- positron decay or electron capture Heavy nuclides ( Z > 83) --- alpha decay
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21. Produced using artificial transmutations, either by:alpha bombardment neutron bombardment bombardment from other nuclei Examples: a. b. c.
22. III. Nuclear Energy Recall: Nucleus is composed of proton and neutron Then, is the mass of an atom equal to the total mass of all the proton plus the total mass of all the neutron? Example for a He atom: Total mass of the subatomic particles = mass of 2 p+ + mass of 2n0 = 2 ( 1.00728 amu ) + 2 (1.00867 amu) = 4.03190 amu And atomic weight of He-4 is 4.00150 Why does the mass differ if the atomic mass = number of protons + number of neutrons?
23. Mass defect -mass difference due to the release of energy -this mass can be calculated using Einstein’s equation E =mc2 2 11 H + 2 20 H -> 42 He + energy Therefore: -energy is released upon the formation of a nucleus from the constituent protons and neutrons -the nucleus is lower in energy than the component parts. -The energy released is a measure of the stability of the nucleus. Taking the reverse of the equation: 42He + energy -> 2 11 H + 2 20 n
24. Therefore, -energy is released to break up the nucleus into its component parts. This is called the nuclear binding energy. -the higher the binding energy, the stable the nuclei. -isotopes with high binding energy and most stable are those in the mass range 50-60.
25. Nuclear binding energy per nucleon vs Mass number nuclear binding energy nucleon nuclear stability 23.2
26. -The plot shows the use of nuclear reactions as source of energy. -energy is released in a process which goes from a higher energy state (less stable, low binding energy) to a low energy state (more stable, high binding energy). -Using the plot, there are two ways in which energy can be released in nuclear reactions: a. Fission – splitting of a heavy nucleus into smaller nuclei b. Fusion – combining of two light nuclei to form a heavier, more stable nucleus.
27. BE + 19F 91p + 101n 0 9 1 binding energy per nucleon = binding energy number of nucleons 2.37 x 10-11 J 19 nucleons = Nuclear binding energy (BE) is the energy required to break up a nucleus into its component protons and neutrons. E = mc2 BE = 9 x (p mass) + 10 x (n mass) – 19F mass 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 10-11J = 1.25 x 10-12 J 23.2
32. 235U + 1n 90Sr + 143Xe + 31n + Energy 38 0 0 54 92 Nuclear Fission Energy = [mass 235U + mass n – (mass 90Sr + mass 143Xe + 3 x mass n )] x c2 Energy = 3.3 x 10-11J per 235U = 2.0 x 1013 J per mole 235U Combustion of 1 ton of coal = 5 x 107 J 23.5
34. Non-critical Critical 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. 23.5
36. 35,000 tons SO2 4.5 x 106 tons CO2 70 ft3 vitrified waste 3.5 x 106 ft3 ash 1,000 MW coal-fired power plant 1,000 MW nuclear power plant Nuclear Fission Annual Waste Production 23.5
37. Nuclear Fission Hazards of the radioactivities in spent fuel compared to uranium ore 23.5 From “Science, Society and America’s Nuclear Waste,” DOE/RW-0361 TG
38. 2H + 2H 3H + 1H 1 1 1 1 2H + 3H 4He + 1n 2 1 0 1 6Li + 2H 2 4He 2 3 1 Nuclear Fusion Fusion Reaction Energy Released 6.3 x 10-13 J 2.8 x 10-12 J 3.6 x 10-12 J Tokamak magnetic plasma confinement 23.6
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40. 24Na, t½ = 14.8 hr, b emitter, blood-flow tracer
46. 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 Q Quality Factor g-ray = 1 b = 1 a = 20 23.8
![Radioactivity – spontaneous emission of particles and or radiation (proposed by Marie Curie) ,[object Object], substances A. Alpha rays ( α ) – consist of positively charged particles called alpha particles and therefore are deflected by the positively charged plate. B. Beta rays ( β ) – or beta particles – are electrons and deflected by negatively charged plate C. Gamma rays ( γ ) – have no charged and are not affected by an external electric field or magnetic field](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)