Question

4. PROBLEM A planar source at the center of an infinite slab of graphite 2 meters thick emits 108 thermal neutrons per cm/sec. Given that the same system is at room temperature, calculate a) total number of thermal neutrons b) number of thermal neutrons absorbed per cm2/sec of the slab c) neutron current as function of position in the slab. d) Total number of neutrons leaking per cm2/sec from the two surfaces of the slab; e) probability that a source neutron does not leak from the slab.

Answer 1

SOLUTION -1 = -2AL (a) Pt - A sinh [catd - 1x1/6)] SA sinh (Card - 1x1/L)] da = -2Al cosh (a td x)) F2A1 [cosh ( + ) - cosh cate] [cosh(atd) – cos(+)] = 2 (2:275X60°m?s)(59.1606cm) [cash (say) - con (lama) As we know n = 9 5.068 X 10"/cmis - 2303636-364/cm2 (b) Za PT = (2-4x164cm') (5.068 Xlo"/cm?s) = 1.216 x 10*]cm?s = 2AL V 2200 x102cm/s dn 2D 2 (c) J=-D de dx 08, J(x) = -D dd SL sinn [(a+d-X)/4] s cosh [d/L] cosh [catd)/L] 2 Cosh [(a+b)/2] (d) 2J (a) = 25 cosh [d/1] =(18/cms) Cosh (1.7892/59-1608] Cosh (catal] cosh [C100 +1-4892)/59-1608] 08, 23(a) - 3.4698 x 107 neutrons /cm? (e) Probability - 1-2 J(a) | - 3.4698 x 107/cm? S 108 /cm² s Conclusion : (a) Total number of thermal neutrons is 2303636-364/cm2 Cb) Number of thermal neutrons absorbed peo cm?/s is 1.216 x 108/cm's CC) Neutron Current is s cosh (2/6) (d) Total number of neutrons leaking Pex cm/s is 3.4698x107 ce) Probability that a source neutron does not leak is 0-653 = 0.653 2 cosh (ata)