Unformatted text preview: 69 11. How many elastic scatters, on the average,are required to slowa IMeV neutron to below I eV in (a) 160 and in (b) 56Fe? Solution:
The averagelogarithmic energy loss per elastic scatter, 1;,is given by Eq. (6.29) where a ==(A 1)2j(A+ 1)2. For 160, a = 0.7785and f. = 0.1199. For 56Fe, a = 0.9311 and f. = 0.03529. The average number of scatters to decreasea neutron's kinetic energy from E1 = 106 eV to E2 = 1 eV is given by Eq. (6.30) 1 = ~ In(lO6). For 160, we find n = 116 and 56Fe,n = 392. 12. How many neutrons per secondare emitted spontaneouslyfrom a 1 mg sample of 252Cf? Solution:
The number of neutrons produced by spontaneousfission per second, Sn, from a mass m of a spontaneouslyfissioning radionuclide is Sn = (N atoms)(>'sf Sl)(V n/fiss.) = From Table 6.3, we obtain Pi = 0.0309 and other neededdata to find Sn =
(103)(6.022 X 1023) 252 In 2 x 0.0309 (2.638 y)(3.15 x 107 sly) 3.73 = 2.30 X 109 fils. Alternative: From Table 6.2 we are told Sn = 2.3 X 1012 /(s g). Thus, 1 mg n of 252Cfhas a source strength Sn = 103Sn = 2.3 X 109 n/s. July 24, 2002 ...
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 Spring '06
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 Energy, Kinetic Energy, Mass, Neutron, source strength Sn

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