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HW14-13-14a - 1-8 Radiation Interactions Chap 7(a 27-keV...

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1-8 Radiation Interactions Chap. 7 (a) 27-keV neutrons: The total interaction coefficient or macroscopic cross section is Et(27 keY) = NFea[e(27 keY) = 0.03397 cm-l. Then from Eq. (P7.3), the fraction of 27-keVneutrons transmitted through the slab is Fraction transmitted = exp[-Etx] = 0.712. (b) 28-ke V neutrons: The total interaction coefficient or macroscopiccross section 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 is Fraction transmitted = exp[-Etx] = 6.34 X 10-34. 13. When an electron moving through air has 5 MeV of energy, what is the ratio of the rates of energy loss by bremsstrahlung to that by collision? What is this ratio 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 = me 2 (-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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