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SKILLS Project

SKILLS Project. Basic Nuclear Chemistry. So, what’s different?. Unlike regular chemistry, nuclear chemistry: Does NOT always follow the laws of conservation of mass and energy. Allows for the creation of new or different elements in reactions.

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SKILLS Project

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  1. SKILLS Project Basic Nuclear Chemistry

  2. So, what’s different? • Unlike regular chemistry, nuclear chemistry: • Does NOT always follow the laws of conservation of mass and energy. • Allows for the creation of new or different elements in reactions. • Describes changes in the nucleus of the atom, rather than just chemical bonding.

  3. Radioactivity and Decay • Nuclear reactions produce radiation as a product. In movies and also in real life, this is what causes the eerie “glow.” • There are three main types of radioactive decay: • Alpha • Beta • Gamma

  4. Alpha Decay α 4 4 • Alpha decay occurs when the nucleus of a larger atom ejects a helium nucleus to become more stable. • Due to their large size and lower energy, alpha particles (radiation) can be stopped using a sheet of paper and are not particularly hazardous. He Symbol: or 2 2

  5. Beta Decay 0 0 e • Beta decay occurs when a neutron ejects an electron and turns into a proton. • Beta particles have both high energy and velocity. They can be stopped by thin metal plates, but are able to penetrate skin and cause mutations. β Symbol: or -1 -1

  6. Gamma Decay 0 γ • Gamma decay occurs when the nucleus releases a high-energy photon (particle) of light. • Gamma rays tend to have the highest energy of all forms of light, along with x-rays. As a result, they are incredibly dangerous and may only be stopped using massive amounts of shielding. Symbol: 0

  7. Nuclear Stability • Radioactive decay occurs when the nucleus of the atom becomes unstable. • Neutrons play a key role by shielding protons from one another. • Decay occurs if: • Too many protons, not enough neutrons • Too many neutrons, not enough protons • The atom is too heavy (elements > #82)

  8. Isotopes • Isotopes are atoms of the same element that have different numbers of neutrons and, as a result, different masses. • We write isotopes as follows: - - Uranium Name Mass 238

  9. Writing Nuclear Equations • All reactants and products are written with both the mass and charge shown: 4 Mass He Symbol 2 Charge • We balance these reactions by both mass and charge. The charge typically tells us the element number (# of protons).

  10. Example 1: Alpha Decay • Show the alpha decay of Uranium-235 α 235 U  4 + 231 Th 92 2 90 • First, we need to write the symbol of our reactant- the nucleus of a Uranium atom with 92 protons (atomic #92) and an isotope mass of 235. • This nucleus is undergoing alpha decay, so we’ll draw the symbol for an alpha particle in the products. • Now, we’ll need to find out the final product of this decay reaction. Remember, the top and bottom rows of numbers have to be equal on both sides. • The alpha particle has a mass of 4, so our product will have a mass of 231. (231 + 4 = 235, 235 amu in, 235 amu out) • The alpha particle contains 2 protons, so our product will have the remaining 90. • Finally, we’ll need to write the symbol for this new nucleus. From the periodic table, element #90 is Thorium, Th. • That’s it! The alpha decay of Uranium-235 produces Thorium-231.

  11. Example 2: Alpha Decay • Show the alpha decay of Plutonium-243 α 243 Pu  4 + 239 U 94 2 92 • Once again, start by writing the symbol of your reactant. Hint: Plutonium is element 94 and is written as Pu. • We are undergoing alpha decay, so we’ll use the appropriate symbol and values in the products. • Our alpha particle has a mass of 4, so our product should have a mass of 239 amu to compensate. • Balancing our protons/charge, we find that the product will have 92 protons. • Element #92 is Uranium, U. • The alpha decay of Plutonium-243 produces Uranium-239 as a product.

  12. Example 3: Beta Decay • Show the beta decay of Carbon-14 14 C  0 + 14 N β 6 -1 7 • Element 7 happens to be nitrogen, N. • Our reactant, Carbon-14, is element #6 and has an isotope mass of 14 amu. Note: carbon-14 is the isotope used for “carbon dating.” • This time around, we’ll be ejecting a beta particle from the nucleus. These high-velocity electrons have an insignificant mass and a charge of -1. • Beta decay doesn’t affect mass, so our final nucleus will have the same mass. • However, the ejection of an electron creates a proton and increases our atomic number to 7. Remember: Neutron  Proton + Electron • The beta decay of Carbon-14 produces the (stable) isotope Nitrogen-14.

  13. Example 4: Beta Decay • What isotope undergoes beta decay to produce Lead-209? 209 Tl  0 + 209 Pb β 81 -1 82 • Element 81 is Thallium, Tl. • This time around, we’re solving for our reactant, so we’ll work backwards. According to the problem, we’re ending with element 82 and a mass of 209. • This ending isotope was produced along with a beta particle- both are products of the decay reaction. • Beta decay has no significant impact on mass, so our initial nucleus will have the same mass as our product, 209 amu. • Totaling up the products, the reactant nucleus should have 81 protons. Remember, beta decay increases the number of protons. • So, Thallium-209 undergoes beta decay to produce Lead-209.

  14. Example 5: Gamma Decay • Show the gamma decay of Nickel-60 60 Ni  0 γ + 60 Ni 28 0 28 • As a result, the nucleus that is produced by this form of radiation will be the same as the reactant in all but energy, a value not shown in the equation. • The gamma decay of Nickel-60 produces a Nickel-60 nucleus with less energy. • Nickel-60 is element 28 with an isotope mass of 60 amu. • The gamma particle that is produced has no mass or charge. This powerful photon is pure energy and has no effect on either value.

  15. Decay Sequences • Many isotopes undergo multiple forms of decay in a particular order known as a decay sequence to reach stability. • Ex. Uranium-238 undergoes α,β,β,α,α,γ,γ • You may write all of the symbols into the equation at the same time, multiplying when needed. 238 4 0 0 γ 226 U 3 α 2 β 2 Ra  + + + 92 2 -1 0 88

  16. Practice on Your Own: • For each of the following isotopes, write a decay equation and find the nuclide produced: • Plutonium-239: α • Strontium-90: β • Thorium-232: α,α • Uranium-238: α,β,α,α • Iodine-131: β,γ,γ,β,α,α Uranium-235 Yttrium-90 Radon-224 Francium-226 Antimony-123

  17. Half-Life • A radioactive half-life is the amount of time required for exactly ½ of the atoms of a particular isotope to decay. • The rate of decay is NOT affected by the amount of atoms present. Half-lives are constant. • In other words 10 atoms will decay to 5 in the same time as 100 will decay into 50.

  18. Working with Half-Lives • Every element has a unique half-life value (t1/2), ranging from seconds to billions of years. • Every half-life, each atom of a radioactive isotope has a 50% chance of decaying. • So, you will be left with ½ of the original amount of a substance each half-life. • 100g U  50g U  25g U  12.5g U, etc.

  19. Example 1: Half-Life • Technetium-99 has a t1/2 of 6.0 days. If you have a 100g sample of Tc-99, how much will be left after 18 days? 100g 99Tc x ½ x ½ x ½ = ? • According to the problem, we’re starting with 100 grams of Technetium-99. (You can abbreviate isotopes like this: 99Tc) • Over the first 6 days, ½ of the original amount will have decayed, leaving us with 50 g 99Tc. However, we still have 12 more days. • At day 12, we will have undergone a second half-life. This will leave ¼ (½ x ½) of the original amount or 25g of undecayed Techetium-99. • Finally, we will have undergone our 3rd half-life at day 18, leaving us with ½ x ½ x ½ or 1/8th of the original Techetium-99. • After 18 days, 12.5 g of Techetium-99 will remain.

  20. Example 2: Original Mass • Using radiocarbon dating, scientists find that a sample of bone contains ¼ of the original amount of Carbon-14. If the t1/2 of 14C is 5730 years, how old is the bone? ¼ = ½ x ½ 5730 years x 2 • To begin this problem, use the percentage or ratio of 14C remaining to determine the amount of half-lives that have passed. • We know that each half-life that passes will reduce the amount of carbon-14 to ½ of the original value. So ¼ is the result of two half-lives. • If two half-lives have passed, we can determine the age of the bone by multiplying the half-life accordingly. • As a result, we can estimate that the bone is 11,460 years old. Note: this is an actual method of dating biological materials.

  21. Practice on Your Own : • Given 256g of 32P (t1/2 = 14.3 days) , how much will remain after: • 14.3 days • 42.9 days • 28.6 days • 114.4 days • 85.8 days 128g 32P 32g 32P 64g 32P 1.0g 32P 4.0g 32P

  22. How old are each of these fossils, if they contain the following percentages of 14C, t½ = 5730 years? • 100% • 50% • 12.5% • 25% • 6.25 % 0 years, 0 half-lives 5730 years, 1 half-life 17190 years, 3 half-lives 11460 years, 2 half-lives 22920 years, 4 half-lives

  23. How many days will pass before a 24g sample of Sodium-24, t1/2 = 14.8 hours, has become safe to handle (less than 0.50g radioactive material remaining)? • 90Sr has a half-life of 28.8 days. How much of a 40g sample will be left after 7 half-lives? How long will this take? 3.7 days or 88.8 hours, 6 half-lives 0.3125g will remain after 201.6 days

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