Monte Carlo Particle Transport Methods: Neutron and Photon - gnssn

LFT?and A and B follow from the boundary conditionsM 1(O) - Mj(O) == 1(0) - l'(0) (5.63)andM 1(X) + MUX) = I(X) + I'(X) (5.(4)Simple algebra shows that the final solutions areMf(X) = k[(l + v)e v * - (1 - v)e-""J (5.65)for transmission andM 1 P(X) = k[(l + v)c HX -*> - (I -- V)Sfor reflection, wherek = 2/[(l + V) 2 C* - (1 - v) 2 e~* x ] = [(1 + y 2 )shvX. + 2vchvX]1(5.67;Now the probability of transmission through the slab is the expected number of particlesleaving the slab due to a starter at x = 0. Thus, from Equation (5.65)t(X) = 2vk ' (5.68)while the reflection probability is the expected number of particles hitting x = 0 from apositive direction. This means that MWf(O) in Equation (5.66) is to be decreased by one,as the starter also crosses the surface x = 0, to obtain the reflection probability asr(X) = k[(l + v)e" x - ( 1 - V)C-"*]-' = kc • sbvX (5.69)For the estimation of the absorption rate, the contribution, function f(x) = 1 - c is to beinserted into the source term of Equation (5.61) to yieldI(x) = (1 - c)[2 - e"" - e- < x - x ) lNotice that if M(x) is the solution of Equation (5.61) with I(x) =absorption rate due to a starter at x is1, then, the expectedM 2 (x) = (1 - c)[2M(x) - M ( , r) (x) - Mf(x)| (5,70)(This follows from the fact that Equation (5.61) is linear in M 1and has a unique solution.)Now, from Appendix 5A again, the solution of the Equation (5.61) with l(x) =1 isM(x) = Ae" x + Be " + 1/(1 - c) (5,7!)where A and B are determined from the boundary conditions in Equations (5.63) and (5.64)The calculations will not be detailed here, but it is easy to see that after solving Equations(5.63) and (5.64) for the constants A and B in Equation (5,71) and inserting M(x) intoEquation (5.70), we obtain the expected absorption rate asMf(X) = 1 - Mf(x) - Mf(x) (5.72)