604 Applications of Nuclear PhysicsP45.271102radJ kg=−QmcT=∆P∆∆tm=∆∆tmc Tmm==⋅°°⋅=×≈−418650010 10209 102426Jkg CCJkgss days!bgafafej..Note that power is the product of dose rate and mass.P45.28Qm=−absorbed energyunit massradJkgrad100010110 02.The rise in body temperature is calculated from where c=4186 J kg for water and thehuman body∆TQmc⋅°°−10 01239 103..±JkgC(Negligible).P45.29If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed isEE=×FHGIKJ×FHGIKJLNMMOQPP×=−−01402100 10602 10426 104 26 101 60 100 682823121213.....MeVg98.9 g molnuclei1 molMeVMeVJ MeVJafejThus, the dose received isDose =FHGIKJ=−06821101142..J60.0 kgradrad .P45.30The nuclei initially absorbed areN09231289 9670 10×FHGIKJ−....gnuclei molgmol.The number of decays in time tis∆NN NNeNettT=−=−=−−−000212λejln.At the end of 1 year,tT10000344..yr29.1 yrand∆NN Ne
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .