problem43_65

problem43_65 - ⋅ ⋅ × = c Equivalent dose rem 10 1.3...

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43.65: a) s decays 10 9.6 Ci) s decays 10 Ci(3.70 10 6 . 2 6 10 4 × = × × = - dt dN so in one second there is an energy delivered of s. J 10 62 . 9 ) eV J 10 (1.60 eV) 10 (1.25 s) 00 . 1 ( ) s 10 6 . 9 ( 2 1 2 1 7 19 6 1 6 - - - × = × × × = = γ E t dt dN E b) Absorbed dose kg 500 . 0 s J 10 6 . 9 7 - × = = m E rad. 10 9 . 1 s kg J rad 100 s kg J 10 9 . 1 4 6 - - × =
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Unformatted text preview: ⋅ ⋅ × = c) Equivalent dose rem. 10 1.3 rad ) 10 9 . 1 ( 7 . 4 4--× = × = d) days. 17 s 10 5 . 1 s rem 10 1.3 rem 200 6 4 = × = ×-...
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This note was uploaded on 05/12/2008 for the course PHYS 262 taught by Professor Dougherty during the Spring '08 term at Lafayette.

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