Consider \"one speed\" neutrons incident perpendicular to a three region slab. T
ID: 1709542 • Letter: C
Question
Consider "one speed" neutrons incident perpendicular to a three region slab. The neutron beam is incident on region 1 which is comprised of an ordinary water layer 0.3 cm thick, followed by a 1.5 cm thick iron layer, and finally by a 0.5 cm thick 5 w/o solution of boric acid (H3BO3). Using the data below, calculate: the uncollided flux at the exit of region 3, the first, interaction probability in each region. the probability of escaping each region without and interaction, and the sum of first interaction probabilities and the probability of escape from region 3. (rho(water and boric acid solution) 1 g/cm3, rhoFe = 7.87 g/cm3, phi0 = 1010 n/(cm2 s)).Explanation / Answer
Given density of water W = 1 g / cm 3 total crossection of water = 'a ( H ) + 's (H ) + 'a ( O) + 's (O ) = 0.333 + 38 +0.00028 + 4.2 = 42.53 number density of water N = A v / M = 1 / 18 ( 6.023*10 23 atoms ) = 0.3346*1023 atoms /cm 3 macro scopic crossection of region 1 is 1 = N = ( 0.3346*1023 )( 42.53 ) = 1.423*1024 cm-1 probability of interaction in region 1 is at x = 0.3 cm = 0.003 m P interaction = e -1x P non-interaction = 1 - e -1x _________________________________________________________________ total crossection of iron = 'a (Fe ) + 's (Fe ) = 2.56 +11 = 13.56 number density of iron N = A v / M = 1 / 55.85 ( 6.023*10 23 atoms ) = 1.07*1022 atoms /cm 3 macroscopic rossection of region 2 (iron) is 2 = N = ( 1.07*1022 ) (13.56) = 1.4623 *10 23 probability of interaction in region 2 is at x = 1.5 cm = 0.015 m P interaction = e -2x P non-interaction = 1 - e -2x ___________________________________________________________ total crossection of boric acid is = 'a ( H ) + 's (H ) + 'a ( O) + 's (O )+'a ( B ) + 's (B ) = 0.333+38+0.00028+4.2+760+4 = 806.53328 b number density N = N v / M = ( 1/ 109 ) ( 6.02 *1023 ) = 5.529*1021 macroscopic crossection of region 3 is ( Boric acid ) is 3 = N = ( 5.529*1021 ) ( 806.53328 ) = 4.459*1024 cm -1 probability of interaction in region 3is at x = 0.5 cm = 0.005 m P interaction = e -2x P non-interaction = 1 - e -2x = 1.07*1022 atoms /cm 3 macroscopic rossection of region 2 (iron) is 2 = N = ( 1.07*1022 ) (13.56) = 1.4623 *10 23 probability of interaction in region 2 is at x = 1.5 cm = 0.015 m P interaction = e -2x P non-interaction = 1 - e -2x ___________________________________________________________ total crossection of boric acid is = 'a ( H ) + 's (H ) + 'a ( O) + 's (O )+'a ( B ) + 's (B ) = 0.333+38+0.00028+4.2+760+4 = 806.53328 b number density N = N v / M = ( 1/ 109 ) ( 6.02 *1023 ) = 5.529*1021 macroscopic crossection of region 3 is ( Boric acid ) is 3 = N = ( 5.529*1021 ) ( 806.53328 ) = 4.459*1024 cm -1 probability of interaction in region 3is at x = 0.5 cm = 0.005 m P interaction = e -2x P non-interaction = 1 - e -2x P interaction = e -2x P non-interaction = 1 - e -2xRelated Questions
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