Homework 1 Solutions Due: Wednesday 1/27 1. Read the article: Isaac Asimov, “The Future of Energy, Climate, Education and Humanity,” in the course’s portal. In no more than one page, answer the following: a. What predictions by Asimov were realized? (2) b. What predictions by Asimov have not been realized? (2) c. Discuss what you agree with, and disagree with Asimov about the future of humanity. (3) Realized predictions: i. Moon landings and resistance to space travel ii. Energy prices will continue to increase iii. Lower birth rates iv. Women’s lib – end of sexism (getting there) v. Continuing education Unrealized predictions: i. We will run out of fossil fuels ii. Climate will continue to get colder iii. Seven billion people by 2000 with stagnant food supplies leading to global famines iv. Multiple mothers rare v. End of ageism, racism, war vi. Education for all vii. Manned based on the Moon 2. Calculate the speed in meters per second of neutrons possessing the following energies: a. Fast neutron from fission at 2 MeV, (1) b. Intermediate energy neutron at 10 keV, (1) c. Thermal energy neutrons at 0.025 eV. (1) Fast neutron energies € E =12mv2v =2Em=2 ⋅ 2 ×106eV( )1.61×10−12ergeV( )1.675 ×10−24g= 1.96 ×107ms Intermediate neutron energies € v =2Em=2 ⋅ 10 ×103eV( )1.61×10−12ergeV( )1.675 ×10−24g= 1.39 ×106ms Thermal neutron energies € v =2Em=2 ⋅ 0.025eV( )1.61 ×10−12ergeV( )1.675 ×10−24g= 2.192 ×103ms≈ 2200msHomework 2 Solutions Due: Friday 1/29 Access the Table of the Nuclides and data mine for the following information about the naturally occurring isotopes of the given elements of interest in nuclear power generation: a) Natural abundances in atomic percent (a/o), b) Atomic mass in atomic mass units (amu). Uranium, Thorium, Lithium, Carbon, Hydrogen, Lead, Beryllium, Boron, Cadmium, Sodium. Isotope Natural abundance (a/o) Atomic mass (amu) U-234 0.0055 % 234.0409456 € ± 0.0000021 U-235 0.72 % 235.0439231 € ± 0.0000021 U-238 99.2745 % 238.0507826 € ± 0.0000021 Th-232 100 % 232.0380504 € ± 0.0000022 Li-6 7.5 % 6.0151223 € ± 0.0000005 Li-7 92.5 % 7.0160040 € ± 0.0000005 C-12 98.89 % 12.000000 € ± 0.0000000 C-13 1.11 % 12.0033548 € ± 0.0000000 H-1 99.985 % 1.0078750 € ± 0.0000000 H-2 0.015 % 2.0141018 € ± 0.0000000 Pb-204 1.4 % 203.9730288 € ± 0.0000031 Pb-206 24.1 % 205.9744490 € ± 0.0000031 Pb-207 22.1 % 206.9758806 € ± 0.0000031 Pb-208 52.4 % 207.9766359 € ± 0.0000031 Be-9 100 % 9.0121821 € ± 0.0000004 B-10 19.9 % 10.0129370 € ± 0.0000004 B-11 80.1 % 11.0093055 € ± 0.0000005 Cd-106 1.25 % 105.9064580 € ± 0.0000065 Cd-108 0.89 % 107.9041834 € ± 0.0000061 Cd-110 12.49 % 109.9030056 € ± 0.0000032 Cd-111 12.80 % 110.9041816 € ± 0.0000032 Cd-112 24.13 % 111.9027572 € ± 0.0000030 Cd-113 12.22 % 112.9044009 € ± 0.0000030 Cd-114 28.73 % 113.9033581 € ± 0.0000030 Cd-116 7.49 % 115.9047554 € ± 0.0000035 Na-23 100 % 22.9897697 € ± 0.0000002Homework 3 Solutions Due: Monday 2/1 If a single fission reaction produces about 180 MeV of energy, excluding the unrecoverable energy of antineutrinos, use Avogadro’s law to calculate the number of grams of the fissile elements that would release 1 kT of TNT equivalent energy that is equal to 1012 calories. 1. U235 2. Pu239 3. U233 4. Np237 From Avogadro’s law: € N =mMAv and so € m =NMAv.""We"know"that"the"energy"release"needed"is"1kT"of"TNT,"so"the"equivalent"number"of"nuclei"required"can"be"calculated"using"a"simple"unit"conversion.""€ N =1kT ×1012cal1kT×2.6 ×1025MeV1012cal×1 fission180MeV= 1.44 ×1023nuclei Now we just need to calculate the mass in grams of the following fuels for a given value of N: 1. U235 € m =1.44 ×1023nuc( )235gmol( )6.02 ×1023nucmol= 56.21g 2. Pu239 € m =1.44 ×1023nuc( )239gmol( )6.02 ×1023nucmol= 57.17g 3. U233 € m =1.44 ×1023nuc( )233gmol( )6.02 ×1023nucmol= 55.73g 4. Np237 € m =1.44 ×1023nuc( )237gmol( )6.02 ×1023nucmol= 56.69gHomework 4 Solutions Due: Monday 2/8 The yield from the Hiroshima gun-barrel device was 12.5 kT of TNT equivalent, and the yield from the Nagasaki implosion device was 22 kT of TNT equivalent. Assuming that one critical mass of lead reflected U235 Oralloy at about 30 kg, and one critical mass of Pu239 at about 10 kg were used to generate these yields, compare the energy release efficiencies of the two devices as the fraction or percentage of the fissile material converted into energy in the case of the gun barrel versus the implosion process. We are using the same equations that we used in HW 3 to convert energy released into mass consumed. For the Hiroshima gun-barrel device, € N =12.5kT ×1012cal1kT×2.6 ×1025MeV1012cal×1 fission180MeV= 1.8 ×1024nuclei € m =1.8 ×1024nuc( )235gmol( )6.02 ×1023nucmol= 702.66g So the efficiency can be calculated as € mconvertedmtotal=702.66g3 ×104g= 0.0234 = 2.34% Similarly, for the Nagasaki implosion device, € N = 22kT ×1012cal1kT×2.6 ×1025MeV1012cal×1 fission180MeV= 3.18 ×1024nuclei € m =3.18 ×1024nuc( )239gmol( )6.02 ×1023nucmol= 1261.61g So the efficiency can be calculated as € mconvertedmtotal=1.262 ×103g1×104g= 0.1261 = 12.61% So, in the case of these two devices, the implosion device dropped on Nagasaki was more efficient than the gun-barrel device dropped on Hiroshima.Homework 5 Solutions Due Monday 2/8 Calculate the Q values or energy releases in MeV from the following nuclear reactions: In order to calculate Q values, first we need to balance the equation using conservation of charge and mass. Then we use the mass-energy relationship given by the following equation. Note that all masses are recorded to at least the fourth decimal place. Masses are taken from KAERI. € Q = m a + b( )− m c + d( )[ ]amu × 931.481MeVamu 1. € 1D2+1T3→0n1+2He4 (DT fusion reaction) € Q = m1D2+1T3( )− m0n1+2He4( )[ ]amu × 931.481MeVamuQ = 2.0141018 + 3.0160493( )− 1.0086649 + 4.0026032( )[ ]amu × 931.481MeVamuQ = 17.59MeV 2. € 1D2+1D2→1H1+1T3 (Proton branch of the DD fusion reaction) € Q = m1D2+1D2( )− m1H1+1T3( )[ ]amu × 931.481MeVamuQ = 2.0141018 + 2.0141018( )− 1.0078250 + 3.0160493( )[ ]amu × 931.481MeVamuQ = 4.03MeV 3. € 1D2+1D2→0n1+2He3 (Neutron branch of the DD fusion reaction) € Q = m1D2+1D2( )− m0n1+2He3( )[ ]amu × 931.481MeVamuQ =