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4. A material is found to have a tenththickness of 2.3 cm for 1.25MeV gamma rays. (a) What is the linear attenuation coefficient for this material? (b) What is the halfthickness? (c) What is the meanfreepath length for 1.25MeV photons in this material? Solution: x= =
IL 1 0.999 cw. 5. In natural uranium, 0.720% of the atoms are the isotope 235U, 0.0055% are 234U,and the remainder 238U.From the data in Table C.I, what is the total linear interaction coefficient (macroscopiccross section) for a thermal neutron in natural uranium? What is the total macroscopic fission cross section for thermal neutrons? Solution: From the data in Table A.3, the atom density of natural uranium is
puNa Au = (18.95)(6.022 x 1023) = 4.794 X 1022atomsfcm3. (a) The atomic abundancesIi of the three isotopes of uranium are found in Table A.4 and the thermal neutron microscopiccross sections are given in Table C.I. The total macroscopiccross section for natural uranium is
~
LIt '\:""""
L..,  N
O"t i i
= N {f
U 340"t 2 234
+ f 235 f 350"t 2 + 380"t 2 238" i = 4.794 X 1022{(0.000055)(116) + (0.0072)(700) +(0.99275)(12.2)} x 1024 = 0.823 cml, (b) Similarly, the macroscopic fission cross section is ~ L..., £.II '""'
i NiO"I i U  N 235 238 {J2340"1 + J2350"1 + J2380"1 } 234 = 4.794 X 1022{(0.000055)(0.465) + (0.0072)(587)
+ (0.99275)(11.8 x 106)} X 1024 = 0.203 cml July 24, 2002 238.0289 Nu= . ...
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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
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