problem43_34

problem43_34 - mJ 5.0 J 10 5.0 kg J(0.20 kg(0.025 3 = ×...

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43.34: (a) rem = rad × RBE 200 = x (10) x = 20 rad (b) 1 rad deposits 0.010 kg J , so 20 rad deposit kg J 0.20 . This radiation affects 25 g (0.025 kg) of tissue, so the total energy is
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Unformatted text preview: mJ 5.0 J 10 5.0 ) kg J (0.20 kg) (0.025 3 = × =-(c) Since RBE = 1 for β-rays, so rem = rad. Therefore 20 rad = 20 rem...
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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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