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[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
– +    Lead block Radioactive substance
– + Lead block
– +  Lead block Radioactive substance
– +  Lead block Radioactive substance
– +  Lead block Radioactive substance
– +    Lead block Radioactive substance
 
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
Balancing Nuclear Equations ,[object Object],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 ,[object Object],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
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
23.1
I. Nuclear Stability and Radioactive Decay
n/p too large beta decay n/p too small positron decay or electron capture 23.2 X Y
 
 
Predicting  the mode of decay ,[object Object],[object Object],[object Object]
II. Nuclear Transmutations ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Transuranium elements ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],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
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?
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
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.
Nuclear binding energy per nucleon vs Mass number 23.2 nuclear binding energy nucleon nuclear stability
-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.
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
23.3
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
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
Nuclear Transmutation 23.4
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
Nuclear Fission 23.5 Representative  fission reaction 235 U +  1 n  90 Sr +  143 Xe + 3 1 n + Energy 92 54 38 0 0
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
Nuclear Fission 23.5 Schematic diagram of a nuclear fission reactor
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
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
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
23.7 Radioisotopes in Medicine ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Brain images with  123 I-labeled compound
Geiger-M ü ller Counter 23.7
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
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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Nuclear Chemistry Powerpoint 2003

  • 1.  
  • 2.
  • 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
  • 9.  
  • 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.
  • 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
  • 16.  
  • 17.  
  • 18.
  • 19.
  • 20.
  • 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
  • 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.
  • 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.