solut21 - CHAPTER TWENTY-ONE THE NUCLEUS: A CHEMIST'S VIEW...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
577 CHAPTER TWENTY-ONE THE NUCLEUS: A CHEMIST'S VIEW Radioactive Decay and Nuclear Transformations 1. All nuclear reactions must be charge balanced and mass balanced. To charge balance, balance the sum of the atomic numbers on each side of the reaction and to mass balance, balance the sum of the mass numbers on each side of the reaction. a. V 23 51 _ e 1 - 0 + Cr 24 51 b. Xe 54 131 + e 1 - 0 _ I 53 131 c. S 16 32 + e 1 - 0 _ P 15 32 d. Th 90 231 + He 2 4 _ U 92 235 2. a. e 1 - 0 + Ge 32 73 _ Ga 31 73 b. He 2 4 + Os 76 188 _ Pt 78 192 c. e 1 + 0 + Pb 82 205 _ Bi 83 205 d. Am 95 241 _ e 1 - 0 + Cm 96 241 3. a. Zn 30 68 _ e 1 - 0 + Ga 31 68 b. Ni 28 62 + e 1 + 0 _ Cu 29 62 c. At 85 208 + He 2 4 _ Fr 87 212 d. Te 52 129 + e 1 - 0 _ Sb 51 129 4. a. Np 93 237 + He 2 4 _ Am 95 241
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
578 b. Bi 83 209 + e 1 - 0 4 + He 2 4 8 _ Am 95 241 ; The final product is Bi 83 209 . c. α β + Ra 88 225 _ + Th 90 229 _ + U 92 233 _ + Pa 91 233 _ + Np 93 237 _ Am 95 241 + Ac 89 225 _ + Fr 87 221 _ + At 85 217 _ + Bi 83 213 _ + Po 84 213 + Bi 83 209 _ + Pb 82 209 The intermediate radionuclides are: 82 209 and Po, 84 213 Bi, 83 213 At, 85 217 Fr, 87 221 Ac, 89 225 Ra, 88 225 Th, 90 229 U, 92 233 Pa, 91 233 Np, 93 237 . 5. e 1 - 0 ? + He 2 4 ? + Pb 82 206 _ Cm 96 242 ; The change in mass number (242 - 206 = 36) is due exclusively to the alpha particles. A change in mass number of 36 requires 9 He 2 4 particles to be produced. The atomic number only changes by 96 - 82 = 14. The 9 alpha particles change the atomic number by 18, so 4 e 1 - 0 (4 beta particles) are produced in the decay series of 242 Cm to 206 Pb. 6. 26 53 Fe has too many protons. It will undergo either positron production, electron capture and/or alpha particle production. 26 59 Fe has too many neutrons and will undergo beta particle production. (See Table 21.2 of the text.) The reactions are:
Background image of page 2
CHAPTER 21 THE NUCLEUS: A CHEMIST’S VIEW 579 e 1 - 0 + Co 27 59 _ Fe 26 59 He; 2 4 + Cr 24 49 _ Fe 26 53 Mn; 25 53 _ e 1 - 0 + Fe 26 53 e; 1 + 0 + Mn 25 53 _ Fe 26 53 7. Reference Table 21.2 of the text for potential radioactive decay processes. 17 F and 18 F contain too many protons or too few neutrons. Electron capture or positron production are both possible decay mechanisms that increase the neutron to proton ratio. Alpha particle production also increases the neutron to proton ratio, but it is not likely for these light nuclei. 21 F contains too many neutrons or too few protons. Beta particle production lowers the neutron to proton ratio, so we expect 21 F to be a β -emitter. 8. The most abundant isotope is generally the most stable isotope. The periodic table predicts that the most stable isotopes for Exercises a - d are 39 K, 56 Fe, 23 Na and 204 Tl. (Reference Table 21.2 of the text for potential decay processes.) a. Unstable; 45 K has too many neutrons and will undergo beta particle production. b. Stable c. Unstable; 20 Na has too few neutrons and will most likely undergo electron capture or positron production. Alpha particle production makes too severe of a change to be a likely decay process for the relatively light 20 Na nuclei. Alpha particle production usually occurs for heavy nuclei. d. Unstable; 194 Tl has too few neutrons and will undergo electron capture, positron production and/or alpha particle production.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 18

solut21 - CHAPTER TWENTY-ONE THE NUCLEUS: A CHEMIST'S VIEW...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online