71. (a) Adapting Eq. 43-20, we find (2.5 × 10-3 g)(6.02 × 1023/mol) N0 = =6.3 × 1018 . 239 g/mol (b) From Eq. 43-14 and Eq. 43-17,
|"N| = N0 1 - e-t ln 2/T1/2
= (6.3 × 1018) 1 - e-(12 h) ln 2/(24,100 y)(8760 h/y) = 2.5 × 1011 . (c) The energy absorbed by the body is
(0.95)EÄ…|"N| =(0.95) (5.2MeV) 2.5 × 1011 1.6 × 10-13 J/MeV =0.20 J . (d) On a per unit mass basis, the previous result becomes (according to Eq. 43-31) 0.20 mJ =2.3 × 10-3 J/kg = 2.3 mGy . 85 kg (e) Using Eq. 43-32, (2.3mGy)(13) = 30mSv.