Natural Radioactivity – Unstable Nuclei Emit Radiation

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Presentation transcript:

Natural Radioactivity – Unstable Nuclei Emit Radiation Spontaneous nuclear change to attain good n/p ratio. Form a new kind of atom. Each isotope or nuclide decays in a certain manner to get a better n/p ratio. The decay mode is named for the particle emitted. See Table N.

Balancing Nuclear Equations

Nuclear Equations Describe the decay process. reactant or starting side (left)  product or ending side (right).  separates two sides

Nuclear Equations - tasks Have to identify type (out of 4) Have to balance to find 1 unknown term.

Natural Transmutation – I.D. 1 term on the reactant side – starting isotope. 2 terms on the product side – ending isotope and emitted particle. Type of particle emitted is characteristic of the isotope – look up particle in Table N.

Nuclear Equations Use conservation of atomic number & conservation of mass number to balance them. Mass number = left superscript. Atomic Number = left subscript.

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

Alpha decay, Th-232 232Th  4He + YX 90 2 Z

Alpha decay, Th-232 232 = 4 + Y so Y = 228 232Th  4He + X Y Z 90 2 Conservation of Mass Number: The sum of the mass numbers on the left side must equal the sum of the mass numbers on the right side.

Alpha decay, Th-232 so Z = 88 90 = 2 + Z 232Th  4He + 228X 90 2 Z 90 = 2 + Z so Z = 88 Conservation of Atomic Number: The sum of the atomic numbers on the left side must equal the sum of the atomic numbers on the right side.

Alpha decay, Th-232 232Th  4He + 228X 90 2 88 Use the P.T. to find X: X = Ra 232Th  4He + 228Ra 90 2 88

Nuclear Equations If there is only 1 unknown term you can figure out what it is. Doesn’t matter which one isn’t known. Don’t forget – you can look up the decay mode in Table N. Decay mode means what particle is emitted.

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

Write equations for α decay 218Rn + 4He 222Ra  208Po  256Lr  88 86 2 204Pb + 4He 84 82 2 252Md + 4He 103 101 2

Writing equations for α decay 231Pa  225Ac  211Fr  185Au  227Ac + 4He 91 89 2 221Fr + 4He 89 87 2 207At + 4He 87 85 2 181Ir + 4He 79 77 2

α decay 229Th + 4He 233U  149Gd  232Th  175Pt  237Np  90 2 92 145Sm + 4He 2 62 64 228Ra + 4He 88 2 90 171Os + 4He 76 78 2 233Pa + 4He 91 93 2

α decay 234Th  144Nd  146Sm  151Ho  192Pt  4He + 230Ra 90 88 2 4He + 140Ce 60 2 58 4He + 142Nd 62 2 60 4He + 147Tb 67 2 65 4He + 188Os 78 2 76

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).

Decay series for U-238.

Decay series for Thorium-232

Decay series for U-235

Alpha Decay 238 = 4 + 234 238 = 4 + 234 92 = 2 + 90

Positron Emission 1p  1n + 0e 1 +1

Beta minus emission 1n  1p + 0e 1 -1

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

How does the mass number or atomic number change in  (or  or ) decay? Just go to Table N, find an isotope that decays by alpha decay, write the equation, and see how the mass number (or atomic number) changes. 226Ra  4 + X so X has to be 222X X is Ra-222. The mass number decreases by 4 and the atomic number decreases by 2. 2 86 88