Question

The half-life of radioactive iodine is 60 days. How much of a 50 mg sample will be left in 40 days?

Short answer

The amount left in 40 days will be$79.37mg.$

Step-by-step solution

Step 1. Given information

Half-life of iodine $=60days$

Initial sample $=50mg$

Time for decay$=40days$

Step 2. Writing the concept.

An amount $A_0$that grows or decays at rate k for certain time t, will be given by:

Step 3. Calculation

Put the values of respective variables in the given formula:

$$\begin{gathered} A=2A_0 \\ t=60 \\ 2A_0=A_0{e}^{60k} \\ {e}^{60k}=\frac{2A_0}{A_0} \\ {e}^{60k}=2 \\ Taking\log arithmbothsides. \\ \ln e^{60k}=\ln 2 \\ 60k.\ln e=\ln 2 \\ k=\frac{\ln 2}{60} \\ Nowsubstitute{A}_0=50,t=40 \\ A=50e^{40.\frac{1}{60}\ln 2} \\ =50e^{\frac{2}{3}\ln 2} \\ =79.37mg \end{gathered}$$