Basic Nuclear Chemistry 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. Describes changes in the nucleus of the atom, rather than just chemical bonding.

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

α He Alpha Decay 4 4 Symbol: or 2 2 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.

e β Beta Decay Symbol: or -1 -1 e β Symbol: or -1 -1 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.

Gamma Decay γ Symbol: 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.

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)

- - Uranium Name 238 Mass 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 238 Mass

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).

α U  + Th Example 1: Alpha Decay 235 4 231 92 2 90 Show the alpha decay of Uranium-235 α 235 U  4 + 231 Th 92 2 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. The alpha particle contains 2 protons, so our product will have the remaining 90. 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) 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. 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.

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

C  + N β Example 3: Beta Decay 14 14 6 -1 7 Show the beta decay of Carbon-14 14 C  + 14 N β 6 -1 7 Element 7 happens to be nitrogen, N. The beta decay of Carbon-14 produces the (stable) isotope Nitrogen-14. However, the ejection of an electron creates a proton and increases our atomic number to 7. Remember: Neutron  Proton + Electron Beta decay doesn’t affect mass, so our final nucleus will have the same mass. 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. 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.”

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

Ni  + Ni γ Example 5: Gamma Decay 60 60 28 28 Show the gamma decay of Nickel-60 60 Ni  γ + 60 Ni 28 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. The gamma particle that is produced has no mass or charge. This powerful photon is pure energy and has no effect on either value. Nickel-60 is element 28 with an isotope mass of 60 amu.

Decay Sequences γ U 3 α 2 β 2 Ra 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 γ 226 U 3 α 2 β 2 Ra  + + + 92 2 -1 88

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

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.

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.

Example 1: Half-Life 100g 99Tc x ½ x ½ x ½ = ? 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 = ? 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. 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. 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.

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 As a result, we can estimate that the bone is 11,460 years old. Note: this is an actual method of dating biological materials. 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. To begin this problem, use the percentage or ratio of 14C remaining to determine the amount of half-lives that have passed.

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

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

How many days will pass before a 24g sample of Sodium-24, t1/2 = 14 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