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Published byJob Harrington Modified over 4 years ago

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So far we’ve studied chemical reactions where only electrons have changed. Chemical properties are determined by electrons! › Nucleus was not primarily important in these reactions, as it did not undergo any changes. › Identities remained the same in chemical reactions because protons remained the same. This is no longer true in nuclear reactions!

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Nucleus is extremely small, dense, and contains a huge amount of energy. › Millions of times more E than chemical reaction. Nucleus = neutrons + protons › Made of even smaller parts, such as quarks. A Z X where A = mass # & Z = charge/# of protons. › Isotopes = atoms of the same element with different #’s of neutrons (protons stay the same).

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Radioactive Decay & Nuclear Stability

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Already discussed these in the packet you completed. Additional note: decay types can be broken into two categories: those that change mass # and those that don’t. Changes mass #: alpha emission. › Giving off a He atom decreases mass. Example: 238 92 U 4 2 He + 234 90 Th › This is a type of spontaneous fission – splitting a heavy nuclide into 2 lighter nuclides.

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Do not change mass #: particle emitted/used has no mass (mass # = 0). Beta emission: 131 53 I 0 -1 e + 131 54 Xe Gamma ray emission: 238 92 U 4 2 He + 234 90 Th + 0 0 Positron emission: 22 11 Na 0 1 e + 22 10 Ne Electron capture: 201 80 Hg + 0 -1 e 201 79 Au + 0 0

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Radiation emitted has different levels of penetration. The more penetrating the emission, the more dangerous. The order is as follows: alpha < beta/positron < gamma ray Therefore, alpha particles are least penetrating and gamma rays are by far most penetrating.

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Which of the following statements is true about beta particles? a) They are electrons with a mass number of 0 and a charge of -1. b) They have a mass number of 0, a charge of -1, and are less penetrating than α particles. c) They are electrons with a charge of +1 and are less penetrating than α particles. d) They have a mass number of 0 and a charge of +1.

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When 226 88 Ra decays, it emits 2 α particles, followed by a β particle, followed by an α particle. The resulting nucleus is: a) 212 83 Bic) 214 82 Pb b) 222 86 Rnd) 214 83 Bi Total of 3 α particles, so subtract 12 from mass # and 6 from atomic #: 214 82 Pb β particle means get rid of a neutron and add a proton (and an electron): 214 83 Bi

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The formation of 230 90 Th from 234 92 U occurs by: a) Electron capture. b) α decay. c) βdecay. d) Positron decay.

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An atom of 238 92 U undergoes radioactive decay by α emission. What is the product nuclide? a) 230 90 Th. b) 234 90 Th. c) 230 92 U. d) 230 91 Pa.

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If a nucleus is unstable it will undergo radioactive decay to become stable. Can be tricky to determine if a nuclide is stable and how it will decay, but several generalizations have been. › Note: nuclide = a specific nucleus of an isotope or atom.

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Of 2000 known nuclides, only 279 are stable. › Tin has the greatest number of stable isotopes at 10. Ratio of neutrons: protons determine stability.

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What is the stable ratio of n:p + at the lower end of the belt? What is the stable ratio of n:p + at the upper end of the belt? Based on this information, what can you conclude about the ratio of n:p + in stable nuclides? This can be found on pg. 842 in your textbook.

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Developed by plotting # of neutrons vs. # of protons of known, stable isotopes. Low end of the belt shows a stable ratio of about 1n:1p +. As the belt gets higher (more protons), the stable ratio begins to increase to about 1.5n:1p +. For isotopes with less than 84 p +, the ratio of n:p + is a good way to predict stability.

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For light nuclides (<20 p + ), 1:1 ratio of n:p + are stable. For heavier nuclides (20 to 83 p + ) ratio increases (to ~1.5:1). › Why? › More neutrons needed to stabilize repulsive force of more protons. All nuclides with 84 or more p + are unstable (because they’re so big). › Alpha decay occurs- giving off a He atom lessens both mass and atomic #. Even #’s of n & p + are more stable than odd #’s.

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Pg. 843 in textbook

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Magic Numbers: certain #’s of n or p + give especially stable nuclides. 2,8,20,28,50,82,126 › A nuclide with this number of n or p + would be very stable. › If there is a magic number of n & p +, this is called a double magic number (usually seen in heavier nuclides were extra stability is needed). The stability of magic numbers is similar to atoms being stable with certain #s of e - › 2(He), 8(Ne), 18(Ar), 36(Kr), 54(Xe), 86(Rn)

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1) Is the isotope Ne-18 stable? › 10p + and 8n 0.8:1 › Ratio is <1:1, so it is unstable! › Undergoes decay to either increase #n & decrease #p + 2) Is the isotope 12 6 C stable? › Yes! Ratio = 1:1 (also even # n & p + ) 3) Is the isotope sodium-25 stable? › Ratio of n:p + = 14:11 1.3:1 › Not stable! How will it become stable? › Decrease #n beta emission

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Pg. 869 #3, 4, 12, 13, 20

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The Kinetics of Radioactive Decay

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Rate of decay = - change in #of nuclides › Negative sign = number decreasing › Tough to predict when a certain atom will decay, but if a large sample is examined trends can be seen. Trends indicate that radioactive decay follows first-order kinetics. Thus first-order formulas are used! change in time

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Two formulas are used to solve calculations involving decay and half-life: ln[A] t – ln[A] 0 = -kt t 1/2 = ln2 Usually takes 10 half lives for a radioactive sample to be ‘safe’. Half-lives can be seconds or years! Formulas above are used when multiples of half lives are not considered. k

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Example #1: I-131 is used to treat thyroid cancer. It has a half life of 8 days. How long would it take for a sample to decay to 25% of the initial amount? Each half-life cycle decreases the initial amount by half: after 8 days, one half- life, 50% remains. After another 8 days, 25% remains. Thus it would take 16 days.

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Every half-life cycle will follow the following order of percentages of radioactive isotopes that remain : 100%, 50%, 25%, 12.5%, 6.25%, 3.12%, etc. If a question asks about half-life and involves one of these ‘easy’ percentages, you can simply count how many half-life cycles have occurred. However, not all are this simple!

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Example #2: What is the half-life of a radioactive isotope (radioisotope) that takes 15min to decay to 90% of its original activity? 90% is not a multiple of half-life cycles, so we need to use the previously mentioned equations to calculate this.

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Steps: (1) Use ln[A] t – ln[A] 0 = -kt to solve for the rate constant, k. Often times specific amounts/concentrations will not be given! Usually given as percentages of original sample left over. Just assume 100 as the original ([A] 0 ) and use the percent asked about as [A] t. (2) Then use t 1/2 = ln2 & value of k from above to solve for the half life (units will vary depending on what is given in the problem). k

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Example #2: What is the half-life of a radioactive isotope (radioisotope) that takes 15min to decay to 90% of its original activity? ln(90) – ln(100) = -k(15min) k = 0.00702/min t 1/2 = ln2/0.00702min -1 t 1/2 = 98.7min

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If both the half-life of a radioactive isotope is given and the amount of radioactive substance remaining, the amount of time it took for this substance to decay to this point can be calculated. First, use t 1/2 formula to find k. Then, use other formula to solve for t. This is how carbon dating (uses radioactive C-14 isotope) is used to determine ages of objects!

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If a wooden tool is discovered, and its C- 14 activity has decreased to 65% of its original amount, how old is the tool? The half-life of C-14 is 5,730 years. 5,730yr = ln2/k k = 1.21 x 10 -4 /year ln65 – ln100 = -(1.21 x 10 -4 /year)t t = 3,600 years

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During nuclear reactions and nuclear decay, energy is given off. › Gamma rays, x-rays, heat, light, and kinetic E Why does E always accompany these reactions? › Small amount of matter is turned into E. › Law of conservation of matter is not followed during nuclear reactions!

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Einstein’s equation is used to perform calculations involving the mass-energy change in nuclear reactions: E = mc 2. › E = energy released › m = mass converted into energy (units need to be in kg in order to get J as unit of E) › c = speed of light = 3.00 x 10 8 m/s Note that the amount of matter turned into E in a nuclear reaction is small, but is amplified by the speed of light to produce a lot of E!

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When one mole of uranium-238 decays into thorium-234, 5 x 10 -6 kg of matter is changed into energy. How much energy is released during this reaction? E = (5 x 10 -6 kg)(3.00 x 10 8 m/s) 2 E = 5 x 10 11 J

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Pg. 870 # 21, 24, 35

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