CHAPTER 43 - 4

# CHAPTER 43 - 4 - 18 ln(R/R)=n t t 0 from 1 0...

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n + U 92 235 Ba 56 141 + Kr 36 92 + 3n 18. We use the result from Problem 17 in the form ln ( R t  / R 0 ) = –  n σ   t . We can find the value of  n  from ln ( R 1 / R 0 ) = ln (0.30) = –  n   t 1  = –  n     (0.010 m), which gives  n  = 1.2 × 10 2  m –1 . We find the thickness to reduce the rate by 1/10 6  from ln ( R 2 / R 0 ) = ln (10 –6 ) = –  n   t 2  = – (1.2 × 10 2  m –1 ) t 2   , which gives  t 2  = 0.12 m =       12 cm . 19. We assume a 1% reaction rate allows us to treat the target as thin.  The density of Cd atoms is n  = [(8650 kg/m 3 )(10 3  g/kg)/(113.9 g/mol)](6.02 × 10 23  atoms/mol) = 4.57 × 10 28  m –3 . ( a ) From Fig. 43–3 we find that the cross section for 0.1-eV neutrons is 3000 bn.  Thus we have R / R 0  =  n x ; 0.01 = (4.57 × 10 28  m –3 )(3000 × 10 –28  m 2 ) x , which gives  x  = 7

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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 - 4 - 18 ln(R/R)=n t t 0 from 1 0...

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