Question

How many half-lives must elapse until (a) $90\%$ and (b) $99\%$ of a radioactive sample of atoms has decayed?

Short answer

(a) Half-lives must elapse until $90\%$radioactive sample of atoms has decayed is, $t=3.33t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

(b) Half-lives must elapse until $99\%$radioactive sample of atoms has decayed is, $t=6.67t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Step-by-step solution

Part (a) Step 1: Given information

We need to find out that,

Time elapse with half lives when $90\%$sample of atoms is decayed.

Part (a) Step 2: Simplify

Let final amount of sample after decay is, $N$

And initial amount of sample is,$N_0$

And half life of sample is, $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

And time elapse with half life is, $t$

We know that,

The no. of radioactive atoms decreases exponentially with time.

And,

$$\begin{gathered} N=0.10N_0=N_0{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$}\right.} \\ 0.10={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$}\right.} \\ \frac{t}{t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\ln \left(\frac{1}{2}\right)=\ln \left(0.10\right) \\ t=\frac{\ln \left(0.10\right)}{\ln \left(0.50\right)}{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ t=\frac{-2.30}{-0.69}{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ t=3.33t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Part (b) Step1: Given information

We need to find out that,

Time elapse with half lives when$99\%$sample of atoms is decayed.

Part (b) Step 2: Simplify

Let final amount of sample after decay is,$N$

And initial amount of sample is,$N_0$

And half life of sample is,$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

And time elapse with half life is,$t$

We know that,

$$\begin{gathered} N=0.010N_0=N_0{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$}\right.} \\ {\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$}\right.}=0.010 \\ \frac{t}{t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\ln \left(\frac{1}{2}\right)=\ln \left(0.010\right) \\ t=\frac{\ln \left(0.010\right)}{\ln \left(0.50\right)}{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ t=\frac{-4.60}{-0.69}{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ t=6.67t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$