A 95 kg patient swallows a 30 Ci beta emitter with a half-life of 5.0 days, and

Question

A 95 kg patient swallows a 30 Ci beta emitter with a half-life of 5.0 days, and the radioactive nuclei are quickly distributed throughout his body. The beta particles are emitted with an average energy of 0.35 MeV, 90% of which is absorbed by the body. What does equivalent does the patient receive in the first week? Submit your answer in mSV

I used 1 Curie = 1 Ci =3.7 e10 dis/s

Energy deposited in 7 days is

30*10^-6* 3.7e10 dis/s * 7*24*3600 s/h *0.9 (absorption fraction) * 0.35Mev * 1.602e-13 J/MeV / 95 kg

= 3.56e-4 J/kg = 3.56e-4 Gy

So, dose-equivalent = 3.56e-4 Sv = 3.56e-1mSv
but that' s not the correct answer :(

Explanation / Answer

given
activity, A = 30 uCi = 30*3.7*10^10 =111*10^10 dis / s
consider
lambda = ln(2)/half life = ln(2)/5 = 0.13862 per day
so, disintegrations in 1 week = No - Noe^(-7*0.13862) = 0.621045828 No
No = lambda*A/24*60*60 = 1780881.944444

hence
N = 1106009.30175775
Energy absorbed = N*0.35*0.9 = 348392.93 MeV

E/m = 5.867*10^-10 J/kg
hence dose Eq = 5.867*10^-7 mSv

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