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Unformatted text preview: 1-8 Radiation Interactions Chap. 7 (a) 27-keV neutrons: The total interaction coefficient or macroscopiccross section is Et(27 keY) = NFea[e(27 keY) = 0.03397cm-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 sectionis Et(27 keY) = NFea[e(28 keY) = 7.644cm-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. When an electron moving through air has 5 MeV of energy, what is the ratio of the rates of energyloss by bremsstrahlungto 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 (-dE/dS)rad ~~
(-dE/ds)coll 700 (~ ) 2 = i~
M 700 (~) 1 2 = 0.05. (b) The Z number for lead is 82. Hence, from Eq. (7.43) we find ~ (~ ) 2= ~ 700 (!1) 700 M 2 =0.59. 14. About what thickness of aluminum is neededto stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha particles? Hint: For parts (a) and (b), use Table 7.2 and compare your values to ranges shown in Fig. 7.16. For part (c) use the range interpolation rules on page 196. Solution: (a) The range in aluminum of a 2.5-MeV electron is obtained with the empirical formula of Eq. (7.47) and the data in Table 7.2. With x = loglO2.5 = 0.3979
pRe(2.5 MeV) = lO-O.27957+1.2492x+O.18247x2 = 1.764gfcm2, Since p = 2.7 gfcm3 for aluminum, Re(2.5 MeV) = 0.653 cw. This value agreeswith data of Fig. 7.16. (b) The range in aluminum of a 2.5-MeV proton is obtained with the empirical formula of Eq. (7.47) and the data in Table 7.2. With x = loglO2.5 = 0.3979
pRp(2.5 MeV) = lO-2.3829+1,3494x+O.1967x2 0.01532gfcm2, =
July 24, 2002 13. ...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).
- Spring '06