HW16-1 - K(Gyjh) = KE(J) of beta particles released per h...

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PROBLEMS 1. In an infinite homogeneous medium containing a uniformly distributed radionu- clide source emitting radiation energy at a rate of E MeV cm-3 S-1, energy must be absorbed uniformly by the medium at the same rate. Consider an infinite air medium with a density of 0.0012 gjcm3 containing tritium (half-life 12.33 y and emitting beta particles with an average energy of 5.37 keV jdecay) at a concentration of 2.3 pCijL. What is the air-kerma rate (Gyjh)? Solution: The kerma rate, by definition, is the rate at which kinetic energy of secondary charged particles is released per unit mass of the medium. Here we have
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Unformatted text preview: K(Gyjh) = KE(J) of beta particles released per h per kg of air = [decays per hour per L][av. beta KE(J) per decay]/[kg of air/L] ,= {(2.3 X 10-12 CijL)(3.7 X 1010 decays per sjCi)(3600 sib)] x [(5.37 keY jdecay) (1.602 x 10-16 JjkeV)]j[0.0012 kgjL] = 2.196 x 10-10 J kg-1 h-1 = 2.196 X 10-10 Gyjh 22.0 nGy Ih. Note that for this infinite homogeneous medium, the rate of energy emitted by all radioactive sources in a unit mass must equal the radiation energy absorbed in a unit mass. Thus the kerma rate equals the absorbed dose rate. July 24, 2002...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).

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