In considering the effects of radiation on the human body, it is necessary to de
ID: 1998566 • Letter: I
Question
In considering the effects of radiation on the human body, it is necessary to define units for the amount of radiation absorbed. One of these is the rad (radiation absorbed does), 1 rad indicates an average of 0.01 joule of absorbed energy per kg of body tissue, regard less of which part of the body actually was exposed. A 75 kg worker at a hospital radiology lab inadvertently swallows a capsule containing 5 mg of of^88 Ra^226 (half-life = 1600 years). This isotope of radium undergoes alpha-decay, each alpha particle carrying an energy of 4.87 MeV. If 90% of these particles are stopped inside the man's body, what radiation dose does he receive in 12 hours?Explanation / Answer
Radio active decay follows 1st order reaction =>N =No*exp(-k*t) and k=0.693/t1.2
=>Isotope consumed N-No = No*(1-exp(-k*t))
k =0.693/1600 =0.000433yr^-1 =>k=0.693/(1600*365*24) =4.944*10^-8 hr^-1
=> N-No =5mg*(1-exp(-4.944*10^-8*12) =2.967*10^-6 mg
=> No. of moles of Radium =2.967*10^-6 /226 =1.312*10^-8 moles
=> No. of atoms of Radium = no. of alpha particles = Avagadro No. *No.of moles =1.312*10^-8*NA =7.905*10^15 alpha particles
=> no. of particles stoppped inside =0.9*7.905*10^15 =7.1145*10^15 Alpha particles
=>Energy possessed by them =7.1145*10^15*4.87MeV =3.465*10^15 MeV =5543.63 J
=>dose = 5543.63 J/(75*100) =0.7391 rad
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