Physics Solution Manual for 1100 and 2101

# To obtain the energy in mev we will use the fact that

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s defect corresponding to a binding energy difference of 5.03 MeV is ( 5.03 MeV) F 1 u I = 0.005 40 u G MeV J H K 931.5 Since the isotope with the larger binding energy has one more neutron ( m = 1.008 665 u ) than the other isotope, the difference in atomic mass between the two isotopes is 1.008 665 u – 0.005 40 u = 1.003 27 u 1596 NUCLEAR PHYSICS AND RADIOACTIVITY 18. REASONING AND SOLUTION a. The mass defect is ∆m = 13.005 738 u + 1.008 665 u − 14.003 074 u = 0.011 329 u The binding energy of the neutron is then F5 J 0 b.011 329 u g931.1 uMeV I = G H K 10.55 MeV b. The mass defect is ∆m = 13.003 355 u + 1.007 825 u − 14.003 074 u = 0.008 106 u The binding energy of the proton is then F5 J 0 b.008 106 u g931.1 uMeV I = G H K 7 .55 MeV c. The neutron is more tightly bound, since it has the larger binding energy. 19. SSM WWW REASONING AND SOLUTION The general form for β + decay is A Z 2P → Parent nucleus D+ 13 2 A Z –1 Daughter nucleus 0 +1 2e β + particle (positron) a. Therefore, the β + decay process for 18 9 F is 18 9 F→ 18 8 b. Similarly, the β + decay process for 15 8 O is 15 8 O→ 15 7 O+ N+ 0 +1 e. 0 +1 e. 20. REASONING The reaction and the atomic masses are: 191 0 → 191 Ir + –1e 76 Os 77 4 14243 14 244 3 190.960 920 u 190.960 584 u 91 91 When the 176 Os nucleus is converted into an iridium 177 Ir nucleus, the number of orbital electrons remains the same, so the resulting iridium atom is missing one orbital electron. However, the given mass includes all 77 electrons of a neutral iridium atom. In effect, then, 91 the value of 190.960 584 u for 177 Ir already includes the mass of the β − particle. Since Chapter 31 Problems energy is released during the decay, the combined mass of the iridium 191 77 Ir 1597 daughter 91 nucleus and the β − particle is less than the mass of the osmium 176 Os parent nucleus. The difference in mass is equivalent to the energy released. To obtain the energy released in MeV, we will use the fact that 1 u is equivalent to 931.5 MeV. SOLUTION The mass decrease that accompanies the β − decay of osmium 190.960 920 u − 190. 960 584 u = 3.36 × 10 energy released is ( −4 191 76 Os is u. Since 1 u is equivalent to 931.5 MeV, the ) 931.5 MeV Energy released = 3.36 ×10−4 u = 0.313 MeV 1u 21. SSM REASONING AND SOLUTION The general form for β – decay is A Z P→ 2 Parent nucleus Therefore, the β – decay process for 35 16 D+ 13 2 A Z +1 Daughter nucleus S is 35 16 0 –1 2e β particle S→ – (electron) 35 17 Cl + 0 –1 e. 22. REASONING Since the released energy E is shared among the three particles, we know that E = E Daughter + E Beta particle + E Antineutrino or EAntineutrino = E − EDaughter − EBeta particle (1) The value for EBeta particle is known. For a given value of E, EAntineutrino will have its maximum possible value when EDaughter is zero. The energy E released during the decay is related to the mass decrease ∆m by E = (∆m)c (Equation 28.5), where c is the speed of light in a vacuum. A mass decrease of one atomic mass unit (∆m = 1 u) corresponds to a released energy of 931.5 MeV. 2 SOLUTION We begin by calculating the decrease in mass for the decay process, which is shown as follows, along with the given atomic mass values: 1598 NUCLEAR PHYSICS AND RADIOACTIVITY 32 P → 32 S + −0 e 164 1 1 15 3 4 24 1 24 3 31.973 907 u γ { + 31.972 070 u Antineutrino When the 32 P nucleus of a phosphorus atom is converted into a 32 S nucleus, the number of 15 16 orbital electrons remains the same, so the resulting sulfur atom is missing one orbital electron. However, the given atomic mass for sulfur includes all 16 electrons of a neutral atom. In effect, then, the value of 31.972 070 u already includes the mass of the β – particle. The mass decrease for the decay is 31.973 907 u – 31.972 070 u = 0.001 837 u. The released energy E is b E = 0.001 837 u F5 J g931.1 uMeV I = 1.711 MeV G H K Using Equation (1), we can now find the maximum possible energy carried away by the antineutrino E Antineutrino = E − E Daughter − E Beta particle b gb gb g = 1.711 MeV − 0 MeV − 0.90 MeV = 0.81 MeV 23. REASONING AND SOLUTION The mass of the products is m = 222.017 57 u + 4.002 60 u = 226. 020 17 u The mass defect for the decay is ∆m = 226.025 40 u − 226.020 17 u = 0.005 23 u, which corresponds to an energy of F5 J 0 b.005 23 u g931.1 uMeV I = G H K 4 .87 MeV 24. REASONING a. An α decay process always takes the form A A− 4 4 ZP → { + { Z −2 D 2 He { Parent nucleus Daughter nucleus α particle Working back from the identity of the daughter nucleus identity of the parent nucleus. ( 207 Pb ) , we will determine the 82 Chapter 31 Problems 1599 a. The β − decay process looks rather different from that of an α decay: A A 0 ZP → { + { Z +1 D −1 e { Parent nucleus Daughter nucleus Knowing the identity of the daughter nucleus nucleus. Electron ( 207 Pb ) will allow us to identify the parent 82 SOLUTION a. An α decay reduces the atomic mass number A of the parent species b...
View Full Document

## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online