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Nuclear Reactions. Natural Transmutation 1 term on reactant side Original isotope 2 terms on product side Emitted Particle New Isotope Happens all by.

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Presentation on theme: "Nuclear Reactions. Natural Transmutation 1 term on reactant side Original isotope 2 terms on product side Emitted Particle New Isotope Happens all by."— Presentation transcript:

1 Nuclear Reactions

2 Natural Transmutation 1 term on reactant side Original isotope 2 terms on product side Emitted Particle New Isotope Happens all by itself (spontaneous) Not affected by anything in environment

3 Natural Transmutation 16 N  0 e + 16 O 7 8 1 term on reactant side 2 terms on product side

4 Artificial Transmutation Cause it to happen by smashing particles into one another 2 terms on reactant side Original Isotope Particle that hits it – neutron, proton, or  -particle Product side: usually 2 terms

5 Artificial Transmutation 27 Al + 4 He  30 P + 1 n 132 15 0 Original isotope or target nucleus “Bullet” -what hits isotope

6 Artificial Transmutation 27 Al + 4 He  30 P + 1 n13 2150 14 N + 4 He  17 O + 1 H 7 28 1 75 As + 4 He  78 Br + 1 n 33 2350 37 Cl + 1 n  38 Cl 17 0 17 All of these equations have 2 reactants!

7 Bombarding with Protons or  Protons and  -particles have positive charge and mass do some damage when hit target nucleus must be accelerated to high speeds to overcome repulsive forces between nucleus & particle (both are +)

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9 What is an accelerator? vacuum chamber (usually a long pipe) – surrounded by vacuum pumps, magnets, radio- frequency cavities, high voltage instruments and electronic circuits inside the pipe particles are accelerated to very high speeds then smashed into each other

10 Fission Reaction S plitting heavy nucleus into 2 lighter nuclei S plitting heavy nucleus into 2 lighter nuclei  Requires a critical mass of fissionable isotope Controlled – nuclear reactor Uncontrolled – bomb

11 Fission  Reactant side: 2 terms 1 heavy isotope (examples: U-235 or Pu-239) Bombarding particle – usually a neutron Product side: at least 2 terms 2 medium-weight isotopes 1 or more neutrons Huge amount of energy is released Fission = Division

12 Fission 235 U + 1 n  91 Kr + 142 Ba + 3 1 n + energy92036 560 235 U + 1 n  72 Zn + 160 Sm + 4 1 n + energy 92030062 More than 200 different product isotopes identified from fission of U-235 A small amount of mass is converted to energy according to E = mc 2

13 Fission Chain Reaction

14 Fusion Reactant side has 2 small nuclei: – H + H; H + He; He + He Product side: – 1 nucleus (still small) and maybe a particle Source of sun’s energy 2 nuclei unite 2 H + 3 H  4 He + 1 n + energy 112 0

15 CERN Particles travel just below speed of light In 10 hrs: particles make 400 million revolutions of the ring 27 kilometer ring

16 FermiLab 4 miles in circumference!

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18 Balancing Nuclear Equations

19 Nuclear Equations - tasks Identify type (4 types) Balance to find 1 unknown term

20 Natural Transmutation – ID 1 term on reactant side – starting isotope 2 terms on product side – ending isotope and emitted particle Type of particle emitted characteristic of isotope – Table N

21 Nuclear Equations To balance: use conservation of both atomic number & mass number Mass number = left superscript Atomic Number = left subscript

22 Balancing Nuclear Equations 16 N  0 e + 16 O 7 8 Conservation of mass number: 16 = 0 + 16 Conservation of atomic number: 7 = -1 + 8

23 Writing Equations Write the equation for the decay of Thorium-232 Use Table N to find the decay mode: α Write the initial equation: 232 Th  4 He + X figure out what element it turned into figure out what element it turned into 902

24 Write an equation for the α decay of Am-241 241 Am  4 He + Y X What’s X? 952Z

25 232 Th  4 He + X 902 Conservation of Mass Number: sum of mass numbers on left side must = sum of mass numbers on right side Y Z 232 = 4 + Y so Y = 228

26 232 Th  4 He + 228 X902 Conservation of Atomic Number: sum of atomic numbers on left side must = sum of atomic numbers on right side Z 90 = 2 + Z so Z = 88

27 232 Th  4 He + 228 X 90 2 88 Use the PT to find X: X = Ra 232 Th  4 He + 228 Ra 90 2 88

28 Alpha (α) decay: 233 U  229 Th + 4 He 92 90 2 232 Th  228 Ra + 4 He 90 88 2 175 Pt  171 Os + 4 He 78 76 2

29 How does the mass number or atomic number change in α,β or γ decay? go to Table N: – find isotope that decays by alpha or β decay – write the equation – see how the mass number (or atomic number) changes 226 88 Ra  4 2  + X so X has to be 222 86 X X is Rn-222 – mass number decreases by 4; atomic number decreases by 2

30 Write an equation for the  decay of Am-241 241 Am  4 He + Y X 95 2 Z 241 = 4 + Y 95 = 2 + Z so Y = 237 so Z = 93 What’s X? X = Np

31 Radioactive Decay Series Sometimes 1 transmutation isn’t enough to achieve stability Some radioisotopes go through several changes before they achieve stability (and are no longer radioactive)

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33 β- 14 C  14 N + 0 e β+ 18 F  18 O + 0 e 67 8 +19

34 How does the mass number or atomic number change in  or  decay? Go to Table N; find an isotope that decays by α,  or , write the equation; see how the mass number (or atomic number) changes 226 Ra  4  + X so X has to be 222 X X is Ra-222 – mass number decreases by 4 – atomic number decreases by 288286

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