Question

${}_{19}{K}^{40}$ consist 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 $1.3\times {10}^9$ years.

Answer 1

Given Half life of ${}_{19}{K}^{40}=1.3\times {10}^9$ yrs

Since radioactive decay reactions follow first order kinetics so, for first order reactions we have,

$K=\frac{0.693}{t_{1/2}}$

$K=\frac{0.693}{1.3\times {10}^9}=0.53\times {10}^9$ yrs${}^{-1}$

Now, $75$kg human body has $0.35$% of potassium by weight in which % of ${}_{19}{K}^{40}$ is further $0.012$%

So, Amount of ${}_{19}{K}^{40}$ in $75$ kg human body

$=\frac{0.35}{100}\times \frac{0.012}{100}\times 75\times 1000g$

$=0.0315g$

No. of molecules of ${}_{19}{K}^{40}$ in $0.0315g$ of it

$=\frac{0.0315}{40}\times 6.022\times {10}^{23}$

$=0.47\times {10}^{21}$

So, Total radioactivity=$0.47\times {10}^{21}\times 0.53\times {10}^{-9}$

$=0.25\times {10}^{12}$ atoms/year

$=0.25\times {10}^{12}$ decay/year