Nuclear Fission and Electrical Energy Electrical energy currently produced by nu
ID: 966044 • Letter: N
Question
Nuclear Fission and Electrical Energy
Electrical energy currently produced by nuclear reactors is derived from the decay of^235_92U, through a process known as nuclear fission, whereby a slow-moving neutron is captured by the^235_92U nucleus, creating^236_92U. The uranium nucleus^236_92U is very unstable, and it will immediately decay into two lighter elements and release neutrons. Some of the neutrons will go on to destabilize other nuclei, creating a chain reaction. During fission reactions the total mass of the products is less than the total mass of reactants. The difference between these two totals is called the mass defect, delta m. The energy equivalent of this mass, delta E, can be calculated when Einstein's equation is written as delta E = delta mc^2 where c = 2.998x10^8s m/s is the speed of light. When^235_92U captures a neutron and decays, one possible set of fission products is rubidium-90,^90_37Rb, and cesium-144,^144_55Cs. The reaction also produces two neutrons as shown in the reaction^235_92U +^1_0n +^90_37Rb +^144_55Cs +2^1_0n Using the mass data given in the table determine how much energy, in joules, is released when one mole of uranium-235 decays. Express your answer to four significant figures and include the appropriate energy per mol units.Explanation / Answer
235U92 + 1n0 -----> 90Rb37 + 144Cs55 +2(1n0)
dm = m of products - m of reactants
dm = (89.9148027) + (143.9320773) + 2(1.0086649) -((235.0439222)+(1.0086649))
dm = 0.1883773
E = 0.1883773 * 1.660540 * 10^-27 kg * (2.998 * 10^8)^2 m^2/Sec^2
E = 2.8115 * 10^-11 J per nucleon
E = 2.8115 * 10^-11 * 6.023 * 10^23 J /mole
E = 1.6934 * 10^13 J/mole
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