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.

Answer 1

The annual radiation dose due to potassium-40 can be calculated using the activity, energy deposited per decay, and a conversion factor. The result is 0.00359 mSv/year. This contributes approximately 0.92% to the average annual dose of 0.39 mSv due to internal radioisotopes.

Step-by-step explanation:

To calculate the annual radiation dose due to potassium-40, we can use the information given in the question. The average source activity of potassium-40 in the body is 4630 Bq, and each radioactive decay deposits 0.39 MeV of energy. We can calculate the number of decays per year by multiplying the activity by the time (1 year) and then multiply it by the energy deposited per decay. The result will give us the total energy deposited per year. To convert this to the dose in millisieverts (mSv), we need to use the conversion factor that states that 1 mSv is equivalent to 5.02 x 10^8 MeV. Dividing the total energy deposited per year by this conversion factor will give us the annual radiation dose due to potassium-40.

Let's calculate:

1. Number of decays per year = 4630 Bq * 1 year = 4630 decays/year
2. Total energy deposited per year = 4630 decays/year * 0.39 MeV/decay = 1803.7 MeV/year
3. Annual radiation dose in mSv = 1803.7 MeV/year / (5.02 x 10^8 MeV/mSv) = 0.00359 mSv/year

Therefore, the annual radiation dose due to potassium-40 is 0.00359 mSv/year.

To find out how much this contributes to the average annual dose of 0.39 mSv due to internal radioisotopes, we can calculate the percentage by dividing the dose due to potassium-40 by the total dose and multiplying by 100.

Let's calculate:

1. Contribution of potassium-40 to the total dose = (0.00359 mSv/year / 0.39 mSv/year) * 100 = 0.92%

Therefore, potassium-40 contributes approximately 0.92% to the average annual dose of 0.39 mSv due to internal radioisotopes.