Chapter 19 TAVSS

Chapter 19 TAVSS - CHAPTER 19 CHEM 162-2007 POST-EXAM III...

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CHAPTER 19 CHEM 162-2007 POST-EXAM III PRACTICE PROBLEMS CHAPTER 19 - THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS RADIOACTIVE DECAY KINETICS NUCLEAR BINDING ENERGY MISCELLANEOUS E. Tavss, PhD 1
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CHAPTER 19 - THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS 32 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 19 – THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS An isotope with a high neutron to proton (N/Z) ratio will tend to decay through A. β-decay B. α-decay C. γ-decay D. electron capture E. positron emission An isotope with a high neutron to proton ratio will tend to form a stable nuclide by decreasing the neutrons or increasing the protons. Losing an electron (β-decay) will do this. 10n → 11H + 0 -1 e Capturing a positron will also do this. 10n + 0 +1 e → 11H 9 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 19 – THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS Complete the following nuclear reactions. U 238 92 + X → Np 239 93 + e 0 1 - Np 239 93 → Y + e 0 1 - N 13 7 → Z + e 0 1 X Y Z A n 1 0 Pu 239 94 C 13 6 B n 1 0 Pu 240 94 N 13 6 C e 0 1 - Pu 239 94 C 13 6 D e 0 1 Pu 239 94 C 13 6 E n 1 0 U 239 93 C 13 6 238 92 U + X → 239 93 Np + 0 -1 e 238 92 U + 1 0 n → 239 93 Np + 0 -1 e 239 93 Np → Y + 0 -1 e 239 93 Np → 239 94 Pu + 0 -1 e 2
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13 7 N → Z + 0 1 e 13 7 N → 13 6 C + 0 1 e X Y Z A n 1 0 Pu 239 94 C 13 6 B n 1 0 Pu 240 94 N 13 6 C e 0 1 - Pu 239 94 C 13 6 D e 0 1 Pu 239 94 C 13 6 E n 1 0 U 239 93 C 13 6 51. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 19 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Balancing the following nuclear reaction would require the use of one of the following particles 234 Th --> 234 Pa + [ ] 234 90 Th --> 234 91 Pa + A Z ? Using algebra: 234 = 234 + A A = 0 90 = 91 + Z Z = -1 Particle = 0 -1 e A. 4 2 He B. 0 1 e C . 0 1 - e D. 0 0 γ E. 0 0 e 3
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52. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 19 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Which one of the following nuclear emissions has the largest mass number? A . alpha emission 4 2 He B. beta emission 0 -1 e C. positron emission 0 +1 e D. gamma emission 0 0 γ E. emission from electron capture 0 0 X-ray 1. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 19 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS When Po decays, the emission consists consecutively of an α particle, then two β - particles, and finally another α particle. The resulting stable nucleus is 214 84 Po ----> 4 2 He + 0 -1 e + 0 -1 e + 4 2 He + Y X ? 84 = 2 -1 -1 + 2 + X X = 82 = Pb 214 = 4 + 0 + 0 + 4 + Y Y = 206 206 82 Pb A. Bi B. Bi C . Pb D . Pb E. Tl 20 CHEM-2001 FINAL EXAM + ANSWERS 4
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. CHAPTER 19 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS When the nuclide 22 44 Ti undergoes electron capture, the product nuclide is 44 22 Ti + 0 -1 e ----> 44 21 Sc A .
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Chapter 19 TAVSS - CHAPTER 19 CHEM 162-2007 POST-EXAM III...

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