`._(19)K^(40)` consists of 0.012% potassium in nature. The human body contains 0.35% potassium by weight. Calculate the total radioactivity resulting from `._(19)K^(40)` decay in a 75 Kg human body. Half life of `._(19)K^(40)` is `1.3 xx 10^(9)` years

Explanation

Weight of radioactive potassium `= (75000 xx 0.012)/(100) xx (0.35)/(100)`
= 0.0315 g
Activity `= (0.693)/(t_(1//2)) xx ("Weight")/("Atomic weight") xx "Avogadro's number"`
Activity `= (0.693)/(1.3 xx 10^(9) xx 365 xx 24 xx 60) xx (0.0315)/(40) xx 6.023 xx 10^(23)`
`= 4.81 xx 10^(5)` dpm