Small traces of radioactive substances, mainly from the food we eat, are present in the human body. The most abundant radioisotope in the body is potassium-40. A person weighing 70 kg contains about 150 g of potassium, and about 0.01% of this amount is potassium-40. The half-life of potassium-40 is 1.25 x10⁹ years. Estimate the rate at which radiation is produced by the decay of potassium-40 in the body.

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The rate of decay is given by the equation:$$R = \frac{0.693N}{t_{1/2}}$$where:* R is the rate of decay in decays per second* N is the number of radioactive atoms present* t_1/2 is the half-life of the radioactive isotope in secondsWe are given that the mass of potassium in the body is 150 g, and that 0.01% of this amount is potassium-40. Therefore, the number of potassium-40 atoms in the body is:$$N = (150 g) imes (0.01%) imes (6.022 imes 10^{23} atoms/mol) imes (1 mol/39.1 g) = 2.41 imes 10^{21} atoms$$The half-life of potassium-40 is given as 1.25 x 10⁹ years. Converting this to seconds, we get:$$t_{1/2} = (1.25 imes 10^9 years) imes (3.15 imes 10^7 s/year) = 3.94 imes 10^{16} s$$Substituting these values into the equation for the rate of decay, we get:$$R = \frac{0.693 imes 2.41 imes 10^{21} atoms}{3.94 imes 10^{16} s} = 4.00 imes 10^3 decays/s$$Therefore, the rate at which radiation is produced by the decay of potassium-40 in the body is 4,000 decays/s.

Small Traces of Radioactive Substances, Mainly from the Food We