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(b) The fission reaction taken to its stable endpoint fission products is An+ 2~~U---+ ~gZr+ l~~Ce+ 4(An). The photons produced in the fission event and in the decay of the fission products are not shown. (c) The prompt energyreleaseis the Q-value of the initial fission reaction, i.e.,
Ej = Q = [M(2~~U)+1nn-M(~gKr)-Mr~~Ba)-4mn]C2 = 169.5 MeV. (d) The energy released over the time required for the fission fragments to decay to their stable endpoints is Ejnd = Q = [M(2~~U)+mn-M(~gZr)-Mr~~Ce)-4mn]c2 = 190.0 MeV. 15. A 10 g sample of 235Uis placed in a nuclear reactor where it generates 100 W of thermal fission energy. (a) What is the fissionrate (fission/s) in the sample? (b) After one year in the core, estimate the number of atoms of ~~Tc in the sample produced through the decay chain shown in Eq. (6.37). Notice that all fission products above ~~Tc in the decay chain have half-lives much shorter than 1 year; hence all of these fissionproducts can be assumedto decayto ~~Tc immediately. Solution:
( a) From page 148 of the text, we find that a fission power of 1 W corresponds to 3.1 x 1010fission/second. Thus, in our sample generating 100 W the fission rate is Rf = (100 W)(3.1 x 1010fiss/(s W)) = 3.1 X 1012 fiss/s. (b) After one year, the number of fissions that have occurred in the sample is N f = (Rf fiss/s)(3.16 x 107 sly) = 9.78 X 1019fissions/year. From the Chart of the Nuclides (or Fig. 6.6), we seethe fission chain yield for A = 99 is y(99) = 6.1%. This is the yield per fission of one of the nuclides in the fission product chain of Eq. (6.37). If we assumethere is negligible direct production of 99Ru as a fission product, and because of all the half-lives of the chain members above 99Tc are much less than one year, we can assumethat for every fission in the sample, about 0.061 atoms of 99Tc are produced. Hence, the number of 99Tc atoms in the sample after one year is estimated as N(99Tc) = 0.061 x N f = 6.0 X 1018 atoms. July 24.2002 ...
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- Spring '06