Nuclear Reactions

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

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

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

Artificial Transmutation 27Al + 4He  30P + 1n 15 13 2 Original isotope or target nucleus “Bullet” -what hits isotope

Artificial Transmutation 27Al + 4He  30P + 1n All of these equations have 2 reactants! 13 2 15 14N + 4He  17O + 1H 1 2 8 7 75As + 4He  78Br + 1n 2 35 33 37Cl + 1n  38Cl 17 17

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 +)

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

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

Fission Reactant side: 2 terms Product side: at least 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

Fission 235U + 1n  91Kr + 142Ba + 31n + energy 56 92 36 92 36 235U + 1n  72Zn + 160Sm + 41n + energy 62 92 30 More than 200 different product isotopes identified from fission of U-235 A small amount of mass is converted to energy according to E = mc2

Fission Chain Reaction

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

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

FermiLab 4 miles in circumference!

Balancing Nuclear Equations

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

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

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

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

Writing Equations Write the equation for the decay of Thorium-232 Use Table N to find the decay mode: α Write the initial equation: 232Th  4He + X 90 2 figure out what element it turned into

What’s under the hat? Little cats X, Y, & Z!

Write an equation for the α decay of Am-241 241 Am  4He + YX What’s X? 95 2 Z

so Y = 228 232 = 4 + Y Conservation of Mass Number: 232Th  4He + X Y Z 90 2 Conservation of Mass Number: sum of mass numbers on left side must = sum of mass numbers on right side

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

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

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

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 22688Ra  42 + X so X has to be 22286X X is Rn-222 mass number decreases by 4; atomic number decreases by 2

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

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)

β- 14C  14N + 0e 6 7 -1 β+ 18F  18O + 0e 9 8 +1

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 226Ra  4 + X so X has to be 222X X is Ra-222 mass number decreases by 4 atomic number decreases by 2 88 86 2