Question

At any one time, the dominant naturally occurring radioisotopes in the human body are 14.0 mg of Potassium-40 (half-life 1.25 X 109 years) and 16.5 ng of Carbon-14 (half-life   5730 years). Calculate the body’s natural radioactivity in becquerels (Bq).

Answer 1

SOLUTION:

FOR Potassium-40:

Half life period:

For a first order reaction, we have:

Therefore, rate constant is given by:

Now, Given that, Mass of Potassium-40 = 14 mg = 0.014 g

No.of particles in 14 mg potassium-40 is given by:

Therefore, activity is given by:

FOR Carbon-14:

Half life period:

For a first order reaction, we have:

Therefore, rate constant is given by:

Now, Given that, Mass of Carbon-14 = 16.5 ng = 16.5 x 10^-9 g

No.of particles in 16.5 mg carbon is given by:

Therefore, activity is given by:

Therefore, Total body’s natural radioactivity is given by:

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