One of the most important sources of internal radiation is the naturally occurri
ID: 2119448 • Letter: O
Question
One of the most important sources of internal radiation is the naturally occurring radioisotope, potassium-40, which makesup 0.012% of the potassium found in the body. These nuclei decay by emitting both beta particles and gamma rays. For the sake of this calculation, assume the radiation dose to the body is mostly due to total absorption of the beta particles, and that, on average, each radioactive decay deposits 0.39 MeV of energy. Assume an average source activity of potassium-40 in the body is 4630 Bq and a total body mass of 70 kg. Compute the annual radiation dose due to potassium-40, and find out how much it contributes to the average annual dose of 0.39 mSv due to internal radioisotopes.
Explanation / Answer
Average activity of potassium 40 = 4630 Bq
This means it emits 4630 radiation in 1 second.
So in one year there are = 365*24*60*60*4630 = 1.46*(10^11) Radiations
Energy of 1 radiation = 0.39 MeV
So energy emitted by potassium-40 in one year = 0.39MeV*(Number of radiation) = 5.694*10^16 eV
1 eV = 1.6*10^-19 J
Energy = 5.694*10^16 eV = 5.694*10^16*1.6*10^-19 J = 9.11*10^-3 J
Body mass = 70 Kg
Dose due to potassium - 40 = 9.11*10^-3 J/70 Kg = 1.3*10^-4 J/kg = 1.3*10^-4 Sv
Total Annual dose = 3.9*10^-4 Sv
Contribution = 1.3*10^-4 Sv/3.9*10^-4 Sv = 0.333
percent = 0.3333*100 = 33.33 %
So it contributes 33.33 percent of annual dose.
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