Question

${}_{19}{K}^{40}$ 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 $75kg$ human body. Half life of ${}_{19}{K}^{40}$ is $1.3\times {10}^9$ years.

Answer 1

Weight of radioactive potassium $=\frac{75000\times 0.012}{100}\times \frac{0.35}{100}$ $=0.0315g$
Activity $=\frac{0.693}{t_{1/2}}\times \frac{Weight}{Atomicweight}\times$ Avogadro's number
Activity $=\frac{0.693}{1.3\times {10}^9\times 365\times 24\times 60}\times \frac{0.0315}{40}\times 6.023\times {10}^{23}$
$=4.81\times {10}^{5}dpm$