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problem44_54

problem44_54 - = × × ×-J 10 50 1 y 1 y 10 1.0(2 ln 10 7...

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44.54: a) The number of protons in a kilogram is . 10 6.7 molecule) protons 2 ( mol kg 10 0 . 18 mol molecules 10 6.023 ) kg 00 . 1 ( 25 3 23 × = × × - Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay. The energy per decay is , J 10 503 . 1 MeV 3 . 938 10 2 p - × = = c m and so the energy deposited in a year, per kilogram, is
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Unformatted text preview: = × × ×-J) 10 50 . 1 ( ) y 1 ( y 10 1.0 (2) ln ) 10 7 . 6 ( 10 18 25 rad. 70 . Gy 10 . 7 3 = ×-b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem....
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