Question

A sample of ${}^{18}F$ is used internally as a medical diagnostic tool to look for the effects of the positron decay $(T_{1/2}=110min)$. How long it take for 99% of the ${}^{18}F$ to decay?

Answer 1

From law of radio activity$,N=N_0{e}^{-\lambda t}$

Here decay constant, $\lambda =\frac{0.693}{t_{1/2}}=0.693/110=0.0063mi{n}^{-1}$

Let, $N_0=100$ so rest amount after decay at t $N=(100-99)=1$

So, $1=100e^{-\left(0.0063\right)t}$

or $e^{0.0063t}=100$ or $0.0063t=ln100=ln{10}^2=2ln10$

or $t=730$ min