problem43_39

problem43_39 - 43.39: a) Etotal NE Nhc (6.50 1010 ) (6.63...

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43.39: a) m 10 2.00 ) s m 10 (3.00 s) J 10 (6.63 ) 10 (6.50 11 8 34 10 total × × × × = λ = = - γ Nhc NE E J. 10 6.46 4 total - × = E b) The absorbed dose is the energy divided by tissue mass: ` rad. 0.108 kg J rad 100 ) kg J 10 (1.08
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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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