To destroy a cancerous tumor, a dose of gamma radiation with a total energy of 2
ID: 1391063 • Letter: T
Question
To destroy a cancerous tumor, a dose of gamma radiation with a total energy of 2.11 J is to be delivered in 29.0 days from implanted sealed capsules containing palladium-103. Assume this isotope has a half-life of 17.0 d and emits gamma rays of energy 21.0 keV, which are entirely absorbed within the tumor. (Enter your answer using one of the following formats: 1.2e-3 for 0.0012 and 1.20e+2 for 120.)
(a) Find the initial activity of the set of capsules.
Bq
(b) Find the total mass of radioactive palladium that these "seeds" should contain.
kg
Explanation / Answer
the decay constant lambds = =0.693/T1/2 = 0.693/17 = 0.0408 /d
let No be the initial number of nulclei present
Let N be the number of nuclei present after 29 days
N = No*e^-lambda*t = No*e^-(0.0408*29) = No*e^-1.1832 = 0.3063*No
the number of nuclei decayed durin 29 days is
No - N = No - 0.3063*No = No*(1 - 0.3063 ) = No*0.6937
energy delivered = 2.11 J
energy released by =(No-N)* 21 Kev
= No*0.6937*21*1000*1.6e-19 = No*2.330832e-15
energy released = energy delivered
No*2.330832e-15 = 2.11
No = 9.053*10^14
part a)
initial activity Ao = lambda*No = 9.053e14*0.0408 = 3.7*10^14 / day
1 day = 24*60*60 = 8.64e+4 s
Ao = 3.7e14/(8.64e+4) = 4.282e9 Bq <-------answer
part (b)
1u = 1.6e-27 kg
mass = 103u = 103*1.67e-27 = 1.7201e-25 kg
initial mass = No*m = 9.053e14*1.7201e-25 = 1.56e-10 kg <-------answer
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