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problem43_71

# problem43_71 - 10 8.22 5 3 10-× = × So the absorbed dose...

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43.71: Mass of K 40 atoms in 1.00 kg is kg. 10 2.52 kg ) 10 2 . 1 ( ) 10 1 . 2 ( 7 4 3 - - - × = × × Number of atoms . 10 793 . 3 ) u kg 10 u(1.661 40 kg 10 52 . 2 18 27 7 × = × × = - - N . y decays 10 054 . 2 10 28 . 1 ) 10 793 . 3 ( ) 693 . 0 ( λ 9 9 18 × = × × = = y N dt dN So in 50 years the energy absorbed is: = × = y) decay 10 (2.054 ) y 50 ( decay) MeV 50 . 0 ( 9 E
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Unformatted text preview: 10 8.22 MeV 10 14 . 5 3 10-× = × So the absorbed dose is = ×-rad) J 100 ( J) 10 22 . 8 ( 3 82 . rad and since the RBE = 1.0, the equivalent dose is 0.82 rem....
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