71. (a) Adapting Eq. 43-20, we fndN0=(2.5×10−3g)(6.02×1023/mol)239 g/mol=6.3×1018.(b) From Eq. 43-14 and Eq. 43-17,|∆N|=N0h1−e−tln 2/T1/2i=(6.3×1018)h1−e−(12 h) ln 2/(24,100 y)(8760 h/y)i=2.5×1011.(c) The energy absorbed by the body is(0.95)Eα|∆N|=(0.95) (5.2MeV)(2.5×1011)(1.6×10−13J/MeV
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.