1263_Physics ProblemsTechnical Physics

1263_Physics ProblemsTechnical Physics - 604 P45.27...

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604 Applications of Nuclear Physics P45.27 1 10 2 rad J kg = Q mc T = P t mc T = t mc T m m = = ⋅° ° = × P 4 186 50 0 10 10 2 09 10 24 2 6 J kg C C J kg s s days! b ga f a f e j a f . . Note that power is the product of dose rate and mass. P45.28 Q m = = = absorbed energy unit mass rad J kg rad J kg 1 000 10 1 10 0 2 b g . The rise in body temperature is calculated from Q mc T = where c = 4 186 J kg for water and the human body T Q mc = = ⋅° = × ° 10 0 1 4 186 2 39 10 3 . . J kg J kg C C b g (Negligible). P45.29 If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed is E E = × F H G I K J × F H G I K J L N M M O Q P P = × = × × = 0 140 2 1 00 10 6 02 10 4 26 10 4 26 10 1 60 10 0 682 8 23 12 12 13 . . . . . . . MeV g 98.9 g mol nuclei 1 mol MeV MeV J MeV J a f e je j Thus, the dose received is Dose = F H G I K J = 0 682 1 10 1 14 2 . . J 60.0 kg rad J kg rad . P45.30 The nuclei initially absorbed are N 0 9 23 12 1 00 10 6 02 10 89 9 6 70 10 = × × F H G I K J = × . . . . g nuclei mol g mol e j . The number of decays in time t is N N N N e N e t t T = = = 0 0 0 2 1 1 1 2 λ e j e j a f ln . At the end of 1 year,
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