1263_Physics ProblemsTechnical Physics

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

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604 Applications of Nuclear Physics P45.27 11 0 2 rad J kg = Qm cT =∆ P ∆∆ tm = t mc T m m == ⋅° ° 4186 500 10 10 209 10 24 2 6 Jkg C C Jkgs s d a y s ! bg a f af ej . . Note that power is the product of dose rate and mass. P45.28 Q m = absorbed energy unit mass rad Jkg rad 1000 10 1 10 0 2 . The rise in body temperature is calculated from where c = 4186 J kg for water and the human body T Q mc ⋅° ° 10 0 1 239 10 3 .. ±Jkg C (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 × = 0140 2 100 10 602 10 426 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 j Thus, the dose received is Dose = F H G I K J = 0682 1 10 114 2 . . J 60.0 kg rad rad . P45.30 The nuclei initially absorbed are N 0 9 23 12 89 9 670 10 × F H G I K J . . . . g nuclei mol gmo l . The number of decays in time t is NN NN e N e t tT = =−=− 00 0 2 12 λ e j ln . At the end of 1 year, t T 100 00344 . . yr 29.1 yr and NN N e
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