Preparation problems

44-2) In the fission reaction of Example 44-1, the masses are 235.043915 u, 140.0130 u, and 91.8973 u for U-235, Ba-141, and Kr-92 respectively. Use these values, along wtih the neutron mass from the preceding problem to find the energy released in this reaction.
Ereleased = Dm c2
= [ mass of reactants - mass of products ] c2
= [ m(U-235) + mn - m(Ba-141) - m(Kr-92) - 3mn ] c2
= [ 235.043915 + 1.008665 - 140.0130 - 91.8973 - 3(1.008665) ] (931.494 MeV/c2) c2
= 201 MeV
44-4) A U-235 nucleus undergoes a neutron induced fission, resulting in a Cs-141 nucleus, three free neutrons, and another nucleus. What is that nucleus?
We start with 92 protons in the uranium. Since cesium has 55 protons, the other nucleus must have (92-55) = 37 protons.
37 protons means it must be a rubidium nucleus.
We start with 236 nucleons, and 144 of them are accounted for in the other products, so the rubidium must have the other 92.
so the symbol for the other product must be 9237Rb.
44-7) Assuming 200 MeV per fission, determine the number of fission events occurring each second in a reactor whose thermal power output is 3200 MW.
1 MeV = 1.6 x 10-13 J, so
1 fission = 200 MeV = 3.2 x 10-11 J or
1 fission/sec = 3.2 x 10-11 W
1.0 x 1020 fissions/sec = 3.2 x 109 W.
44-8) How much U-235 would be needed to fule the reactor of the preceding problem for one year?
We need 1.0 x 1020 fissions/sec = 3.154 x 1027 fissions/year.
Each fission is one nucleus, so we need 3.154 x 1027 nucleii.
Each nucleus has a mass of 235 u = (235 u) x (1.66054 x 10-27 kg/u) = 3.9 x 10-25 kg.
(3.154 x 1027 nucleii) x (3.9 x 10-25 kg/nucleus) = 1231 kg of Uranium
44-15) The effective multiplication factor in a typical nuclear weapon is about 1.5, and the genteration time is about 10 ns. Under these conditions, (a) how many generations would it take to fission a 10-kg mass of U-235? (b) How lond would the process take?
(a) Each U-235 nucleus has a mass of 235 u = 3.9 x 10-25 kg, so in in 10 kg there are
10 kg / (3.9 x 10-25 kg/nucleus) = 25.6 x 1024 nucleii.

The number of nucleii consumed in the chain reaction is the same as the number of fissions, or
N = ( k n+1 - 1 ) / ( k - 1 )
N (k-1) = k n+1
ln[ N(k-1) ] = (n+1) ln[ k ]
n = [ ln(N(k-1)) / ln(k) ] - 1
= [ ln( 25.6 x 1024 * 0.5 ) / ln( 1.5 ) ] - 1
= 142 generations

(b) Since each generation takes 10ns, this process takes 1420 ns or 1.42 msec

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