Chemistry - Answer Key

Name 1. Answer Key Write a balanced equation for the fusion reaction between 2H and 3H (one of the products is a neutron). (3) 2 1 H + 31H 6 42He + 10n Calculate the energy released in this fusion reaction. (5) Q = Δ2H + Δ3H - Δ4He - Δn = 13.136 MeV + 14.950 MeV - 2.424 MeV - 8.071 MeV = 17.591 MeV 17.591 MeV Why are fusion reactions involving very light nuclides generally expected to be exoergic? (4) The fusion products have a higher binding energy/nucleon, and hence are more stable (lower energy). 2. (6) The nuclear shell model suggests that the lowest six nucleon energy levels can accommodate 2,4,2,6,2 and 4 nucleons, respectively. On this basis, predict which isobar, 23Mg or 23Na, is more stable. Explain your answer. 23 Mg     12 p     11 n 23 Na     11 p     12 n (Note that proton levels are slightly higher than corresponding neutron levels because of Coulomb repulsions.) As one goes from 23Na to 23 Mg a neutron gets converted to a proton and placed in a higher energy level. Thus, one expects 23Na to be the more stable isobar. (23Mg actually decays to stable 23Na by positron emission.) Cite two pieces of evidence in support of "magic numbers" in nuclear chemistry. (5) any two: especially high EB/A for N = Z = 2, 8, 20 no more than 5 stable isotones except for N = 50 & 82 most stable isotopes for Z = 50 heaviest stable nuclides have Z = 82 and/or N = 126 weak bonding (low σ) of first neutron for N = 50, 82, 126 Name 3. Answer Key Using the information provided, draw detailed decay schemes (relative energy vs. Z) for 59 Fe and 139mCe/139Ce. Indicate the energy involved in all transitions. (Note: In each box the entry preceded by E refers to the total decay energy.) (10) Fe 59 Ce 139 44.51 d 56.4 s β- .464, .274, ... IT 754.2 γ 1099.2, 1291.6, ε γ 165.9 E 1.565 4. 137.6 d E .28 The β+ decay of 14O to its daughter involves an energy change of 5.143 MeV. If Eβ+max is found to be 1.81 MeV, what happens to the remainder of the decay energy? (5) 1.022 MeV of the remaining energy (3.33 MeV) is used to create the positron-electron pair. The rest is emitted as gamma radiation as the excited daughter de-excites. Write the equation for the decay of 49V by electron capture. (3) 49 V 6 49Ti + νe + x-rays + Auger e- If 49V decays solely to the ground state of its daughter, how can one readily detect the decay of 49V. (3) The x-rays (and Auger e-) emitted during this decay can be detected easily. Write a balanced equation (include all particles involved, superscripts, subscripts, charges, etc.) for the β- decay of 108Ag. (3) 108 47 Ag 6 108 48 Cd + e + 00~ νe 0 -1 Name 5. (4) Answer Key After its emission, what is the ultimate fate of: an alpha particle? a positron? An alpha particle will slow down and acquire 2 electrons and become an atom of 4He. A positron will slow down and annihilate an electron, forming 2 gamma rays each with 511 keV of energy. The primary products formed in the fission of uranium are radioactive. By what mode would you expect them to decay? Explain. (5) Since these products are formed by the break up of a heavy nucleus with a large N/Z ratio, they are likely to be neutron-rich compared to stable nuclides in their mass range. Therefore, β- decay is most likely. (Neutron emission is possible, but much less likely.) Briefly explain why the actual cross section for neutron capture may be decidedly smaller or larger than the geometric cross section. (6) Much smaller cross sections may be found if the target nucleus has a magic number of neutrons. Much larger cross sections may be found, especially for slow neutron capture, because the greater de Broglie wavelength of a slow neutron increases its likelihood of interaction with the nucleus. Indicate three probable fast neutron reactions with an aluminum target. (Note: 27Al is the only stable isotope of aluminum.) (6) 27 Al(n,p)27Mg 27 Al(n,2n)26Al 27 Al(n,α)24Na What is an excitation function. (3) An excitation function is a plot of reaction cross section versus bombarding particle energy. Briefly explain why the half life for α-particle emission has a strong inverse relationship to the decay energy. (5) Alpha decay is hindered by a Coulomb barrier to α particle emission. The probability of penetration of the barrier (tunneling) increases markedly with increasing α decay energy (there is a substantially smaller portion of the barrier to tunnel through at higher energies), so α decay rates will increase with increasing decay energies.