CHAPTER 43 - 5

CHAPTER 43 - 5 - H 1 2 + H 1 3 → He 2 4 + n H 1 2 + H 1 2...

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Unformatted text preview: H 1 2 + H 1 3 → He 2 4 + n H 1 2 + H 1 2 → He 2 3 + n H 1 1 + H 1 1 → H 1 2 + e +1 + ν 26. The number of fissions in one second is n = t /? t = (1.0 s)/(1.0 × 10 –3 s) = 1.0 × 10 3 . For each fission the number of neutrons is 1.0004 times the number from the previous fission. Thus the reaction rate will increase by (1.0004) n = (1.0004) 1000 = 1.49 . 27. When 236 U decays by α emission, the resulting nucleus is 232 Th. Thus the radii of the two particles are r α = (1.2 fm)(4) 1/3 = 1.90 fm; r Th = (1.2 fm)(232) 1/3 = 7.37 fm. At the instant of separation, the two particles are in contact, so the Coulomb energy is U α = Z α Z Th e 2 /4p Å ( r α + r Th ). If we assume fission into equal parts, the resulting nucleus has A = 118. Thus each radius is r f = (1.2 fm)(118) 1/3 = 5.89 fm....
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at Berkeley.

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CHAPTER 43 - 5 - H 1 2 + H 1 3 → He 2 4 + n H 1 2 + H 1 2...

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