1-8Radiation InteractionsChap. 7(a) 27-keV neutrons: The total interaction coefficient or macroscopic crosssection is Et(27 keY) = NFea[e(27 keY) = 0.03397 cm-l. Then fromEq. (P7.3), the fraction of 27-keVneutrons transmitted through the slabisFraction transmitted = exp[-Etx] = 0.712.(b) 28-ke V neutrons: The total interaction coefficient or macroscopiccrosssection is Et(27 keY) = NFea[e(28 keY) = 7.644 cm-l. From Eq. (P7.3),the fraction of 28-keV neutrons transmitted through the slab isFraction transmitted = exp[-Etx] = 6.34 X 10-34.13. When an electron moving through air has 5 MeV of energy, what is the ratioof the rates of energy loss by bremsstrahlung to that by collision? What is thisratio for lead?Solution:(a) The Z number for air is zair ~ O.8ZN +O.2Z0 = 7.2. Then from Eq. (7.43)we find for M = me2(-dE/dS)rad ~ ~(~)2 = i~(~)= 0.05.(-dE/ds)coll 700 M 700 1(b) The Z number for lead is 82. Hence, from Eq. (7.43) we find
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