Nuclear Radiation solutions

Nuclear Radiation solutions - PHYSICS 6 HOMEWORK 5 SPRING...

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PHYSICS 6 HOMEWORK 5 SPRING 2005 due Thursday Apr.28 _________________________________ | number of fissions | generation | in the generation | -------------------------------- | 1 2 0 = 1 | 2 2 1 = 2 3 2 2 = 2x2 = 4 | 4 2 3 = 4x2 = 8 | 5 2 4 = 8x2 = 16 | . . . . | . . | . . | 79 2 78 = 3.0x10 23 | 80 2 79 = 6.0x10 23 | 81 2 80 = 1.2x10 24 | 82 2 81 = 2.4x10 24 | ________________________________ | I. The fission process n + 235 92 U -> 90 37 Rb + 143 ? Cs + ? n + Energy releases energy that you will calculate below. a. How many neutrons are released? 236 nucleons in -> 90+143+X out so X = 3 neutrons out b. What is Z for Cs (cesium)? 92 in -> 37 + Z out so Z = 55 c. The masses of the nuclei are given below. What is the amount of mass that disappears (i.e. gets converted to other forms of energy) in the reaction? (Don’t forget the neutrons.) M( 235 U) = 3.9184x10 -25 Kg M( 90 Rb) = 1.4925x10 -25 Kg M( 143 Cs)= 2.3892x10 -25 Kg M(n) = 0.0167x10 -25 Kg Mass in = Mass out + missing mass ( m) M(n) + M(U) = M(Rb) + M(Cs) + 3M(n) + m So m = M(U) - M(Rb) - M(Cs) - 2M(n) = 3.3x10 -28 Kg d. The mass that disappears is converted to how much energy? (Express in MeV units.) m c 2 = 3.3x10 -28 Kg (3.0x10 8 m/s) 2 ÷ 1.6x10 -13 J/MeV = 186 MeV For the remainder, suppose 2.35 Kg of 235 U undergo fission in a bomb.
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