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- 2. <ul><li>Radioactivity – spontaneous emission of particles and or radiation </li></ul><ul><li>(proposed by Marie Curie) </li></ul><ul><li>Three types of ray are produced by the decay, or breakdown, of radioactive </li></ul><ul><li>substances </li></ul><ul><li>A. Alpha rays ( α ) – consist of positively charged particles called alpha particles and therefore are deflected by the positively charged plate. </li></ul><ul><li>B. Beta rays ( β ) – or beta particles – are electrons and deflected by negatively charged plate </li></ul><ul><li>C. Gamma rays ( γ ) – have no charged and are not affected by an external electric field or magnetic field </li></ul>
- 3. – + Lead block Radioactive substance
- 4. – + Lead block
- 5. – + Lead block Radioactive substance
- 6. – + Lead block Radioactive substance
- 7. – + Lead block Radioactive substance
- 8. – + Lead block Radioactive substance
- 10. 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 1 1 1 0 0 -1 0 +1 4 2 23.1 X A Z Mass Number Atomic Number Element Symbol 1 p 1 1 H 1 or proton 1 n 0 neutron 0 e -1 0 -1 or electron 0 e +1 0 +1 or positron 4 He 2 4 2 or particle
- 11. Balancing Nuclear Equations <ul><li>Conserve mass number (A). </li></ul>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 <ul><li>Conserve atomic number (Z) or nuclear charge. </li></ul>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 1 n 0 U 235 92 + Cs 138 55 Rb 96 37 1 n 0 + + 2 1 n 0 U 235 92 + Cs 138 55 Rb 96 37 1 n 0 + + 2
- 12. 212 Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212 Po. 212 = 4 + A A = 208 84 = 2 + Z Z = 82 23.1 4 He 2 4 2 or alpha particle - 212 Po 4 He + A X 84 2 Z 212 Po 4 He + 208 Pb 84 2 82
- 13. 23.1
- 14. I. Nuclear Stability and Radioactive Decay
- 15. n/p too large beta decay n/p too small positron decay or electron capture 23.2 X Y
- 18. Predicting the mode of decay <ul><li>High n/p ratio (too many neutrons; lie above band of stability) --- undergoes beta decay </li></ul><ul><li>Low n/p ratio (neutron poor; lie below band of stability) --- positron decay or electron capture </li></ul><ul><li>Heavy nuclides ( Z > 83) --- alpha decay </li></ul>
- 19. II. Nuclear Transmutations <ul><li>Nuclear reactions that are induced in a way that nucleus is struck by a neutron or by another nucleus (nuclear bombardment) </li></ul><ul><li>Positive ion bombardment </li></ul><ul><li>- alpha particle is the most commonly used positive ion </li></ul><ul><li>- uses accelerators </li></ul><ul><li>Examples: </li></ul><ul><li>14 7 N + 4 2 He -> 17 8 O + 1 1 H </li></ul><ul><li>9 4 Be + 4 2 He -> 12 6 C + 1 0 n </li></ul><ul><li>Neutron bombardment </li></ul><ul><li>- neutron is bombarded to a nucleus to form a new nucleus </li></ul><ul><li>- most commonly used in the transmutation to form synthetic isotopes because neutron are neutral, they are not repelled by nucleus </li></ul><ul><li>Example: </li></ul><ul><li> 58 26 Fe + 1 0 n -> 59 26 Fe -> 59 27 Co + 0 -1 e </li></ul>
- 20. Transuranium elements <ul><li>Element with atomic numbers above 92 </li></ul><ul><li>Produced using artificial transmutations, either by: </li></ul><ul><ul><li>alpha bombardment </li></ul></ul><ul><ul><li>neutron bombardment </li></ul></ul><ul><ul><li>bombardment from other nuclei </li></ul></ul><ul><ul><li>Examples: </li></ul></ul><ul><ul><li>a. </li></ul></ul><ul><ul><li>b. </li></ul></ul>c. 4 He 2 Pu 239 94 + Cm 242 96 1 n 0 + 1 n 0 U 238 92 + U 239 92 0 e -1 Pu 239 94 + Np 239 93 + 0 -1 e 14 N 7 U 238 92 + Es 247 99 1 n 0 + 5
- 21. 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 2n 0 = 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?
- 22. Mass defect -mass difference due to the release of energy -this mass can be calculated using Einstein’s equation E =mc 2 2 1 1 H + 2 2 0 H -> 4 2 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: 4 2 He + energy -> 2 1 1 H + 2 2 0 n
- 23. 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.
- 24. Nuclear binding energy per nucleon vs Mass number 23.2 nuclear binding energy nucleon nuclear stability
- 25. -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.
- 26. Nuclear binding energy (BE) is the energy required to break up a nucleus into its component protons and neutrons. 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 = 1.25 x 10 -12 J 23.2 BE + 19 F 9 1 p + 10 1 n 9 1 0 binding energy per nucleon = binding energy number of nucleons = 2.37 x 10 -11 J 19 nucleons
- 27. 23.3
- 28. Radiocarbon Dating t ½ = 5730 years Uranium-238 Dating t ½ = 4.51 x 10 9 years 23.3 14 N + 1 n 14 C + 1 H 7 1 6 0 14 C 14 N + 0 + 6 7 -1 238 U 206 Pb + 8 4 + 6 0 92 -1 82 2
- 29. Nuclear Transmutation 23.4 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
- 30. Nuclear Transmutation 23.4
- 31. Nuclear Fission 23.5 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 235 U + 1 n 90 Sr + 143 Xe + 3 1 n + Energy 92 54 38 0 0
- 32. Nuclear Fission 23.5 Representative fission reaction 235 U + 1 n 90 Sr + 143 Xe + 3 1 n + Energy 92 54 38 0 0
- 33. Nuclear Fission 23.5 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
- 34. Nuclear Fission 23.5 Schematic diagram of a nuclear fission reactor
- 35. Annual Waste Production 23.5 Nuclear Fission 35,000 tons SO 2 4.5 x 10 6 tons CO 2 1,000 MW coal-fired power plant 3.5 x 10 6 ft 3 ash 1,000 MW nuclear power plant 70 ft 3 vitrified waste
- 36. 23.5 Nuclear Fission Hazards of the radioactivities in spent fuel compared to uranium ore From “Science, Society and America’s Nuclear Waste,” DOE/RW-0361 TG
- 37. 23.6 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 2 H + 2 H 3 H + 1 H 1 1 1 1 2 H + 3 H 4 He + 1 n 1 1 2 0 6 Li + 2 H 2 4 He 3 1 2
- 38. 23.7 Radioisotopes in Medicine <ul><li>1 out of every 3 hospital patients will undergo a nuclear medicine procedure </li></ul><ul><li>24 Na, t ½ = 14.8 hr, emitter, blood-flow tracer </li></ul><ul><li>131 I, t ½ = 14.8 hr, emitter, thyroid gland activity </li></ul><ul><li>123 I, t ½ = 13.3 hr, ray emitter, brain imaging </li></ul><ul><li>18 F, t ½ = 1.8 hr, emitter, positron emission tomography </li></ul><ul><li>99m Tc, t ½ = 6 hr, ray emitter, imaging agent </li></ul>Brain images with 123 I-labeled compound
- 39. Geiger-M ü ller Counter 23.7
- 40. 23.8 Biological Effects of Radiation R adiation a bsorbed d ose ( rad ) 1 rad = 1 x 10 -5 J/g of material R oentgen e quivalent for m an ( rem ) 1 rem = 1 rad x Q Q uality Factor -ray = 1 = 1 = 20
- 41. Chemistry In Action: Food Irradiation Dosage Effect Up to 100 kilorad Inhibits sprouting of potatoes, onions, garlics. Inactivates trichinae in pork. Kills or prevents insects from reproducing in grains, fruits, and vegetables. 100 – 1000 kilorads Delays spoilage of meat poultry and fish. Reduces salmonella. Extends shelf life of some fruit. 1000 to 10,000 kilorads Sterilizes meat, poultry and fish. Kills insects and microorganisms in spices and seasoning.

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