HW8-11-12

# HW8-11-12 - 6-9 11 How many elastic scatters on the...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6-9 11. How many elastic scatters, on the average,are required to slowa I-MeV 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 S-l)(V n/fiss.) = From Table 6.3, we obtain Pi = 0.0309 and other neededdata to find Sn = (10-3)(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 = 10-3Sn = 2.3 X 109 n/s. July 24, 2002 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online