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Chapter 18

Chapter 18. Nuclear Chemistry. Introduction. 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.

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Chapter 18

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  1. Chapter 18 Nuclear Chemistry

  2. Introduction • 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!

  3. Introduction • 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. • AZX where A = mass # & Z = charge/# of protons. • Isotopes = atoms of the same element with different #’s of neutrons (protons stay the same).

  4. 18.1 Radioactive Decay & Nuclear Stability

  5. Types of decay • 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: 23892U  42He + 23490Th • This is a type of spontaneous fission – splitting a heavy nuclide into 2 lighter nuclides.

  6. Types of Decay Do not change mass #: particle emitted/used has no mass (mass # = 0). • Beta emission: 13153I 0-1e + 13154Xe • Gamma ray emission: 23892U 42He + 23490Th + 00 • Positron emission: 2211Na 01e + 2210Ne • Electron capture: 20180Hg + 0-1e  20179Au + 00

  7. Penetration Ability • 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.

  8. AP Practice Question Which of the following statements is true about beta particles? • They are electrons with a mass number of 0 and a charge of -1. • They have a mass number of 0, a charge of -1, and are less penetrating than α particles. • They are electrons with a charge of +1 and are less penetrating than α particles. • They have a mass number of 0 and a charge of +1.

  9. AP Practice Question When 22688Ra decays, it emits 2 α particles, followed by a β particle, followed by an α particle. The resulting nucleus is: a) 21283Bi c) 21482Pb b) 22286Rn d) 21483Bi • Total of 3 α particles, so subtract 12 from mass # and 6 from atomic #: 21482Pb • β particle means get rid of a neutron and add a proton (and an electron): 21483Bi

  10. AP Practice Question The formation of 23090Th from 23492U occurs by: • Electron capture. • α decay. • βdecay. • Positron decay.

  11. AP Practice Question An atom of 23892U undergoes radioactive decay by α emission. What is the product nuclide? • 23090Th. • 23490Th. • 23092U. • 23091Pa.

  12. Nuclear Stability • 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.

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

  14. 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.

  15. Belt/Zone of Stability • 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.

  16. Ways to Predict Stability • 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.

  17. Pg. 843 in textbook

  18. Ways to Predict Stability Cont. • 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)

  19. Examples: Predicting Stability 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 126C 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

  20. Section 1 Homework Pg. 869 #3, 4, 12, 13, 20

  21. 18.2 The Kinetics of Radioactive Decay

  22. Rate of Decay • 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

  23. Rate of Decay • Two formulas are used to solve calculations involving decay and half-life: • ln[A]t – ln[A]0 = -kt • t1/2 = 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

  24. Half Life- ‘Easy’ Example • 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.

  25. Half Life- ‘Easy’ Example • 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!

  26. Half Life- ‘Harder’ Example • 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.

  27. Half Life- ‘Harder’ Example • Steps: • 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. • Then use t1/2 = ln2 & value of k from above to solve for the half life (units will vary depending on what is given in the problem). k

  28. Half Life- ‘Harder’ Example • 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 • t1/2 = ln2/0.00702min-1  t1/2 = 98.7min

  29. Another Way to Use Half-Life • 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 t1/2 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!

  30. Carbon-Dating Example • 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

  31. Mass-Energy Relationships • 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!

  32. Mass-Energy Relationships • Einstein’s equation is used to perform calculations involving the mass-energy change in nuclear reactions: E = mc2. • 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 108 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!

  33. Sample Calculation • When one mole of uranium-238 decays into thorium-234, 5 x 10-6kg of matter is changed into energy. How much energy is released during this reaction? E = (5 x 10-6kg)(3.00 x 108m/s)2 E = 5 x 1011 J

  34. Section 2 Homework • Pg. 870 # 21, 24, 35

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