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

$31.45mg$of the sample will be left from $50$mg in $40$days.

Step-by-step solution

Step 1. Given information   

It is given that the half-life of the element is $60$days.

Step 2. Finding the value of k.

For an original amount, $A_0$, that grows or decays at a rate, $k$, for a certain time, $t$, the final amount,

$A=A_0{e}^{kt}$

Then substituting the given values in the formula we get:

$$\begin{gathered} A=A_0{e}^{kt} \\ 25=50e^{60k} \\ \frac{1}{2}=e^{60k} \end{gathered}$$

Step 3. Use logarithmic properties   

Taking logarithmic properties on both sides, we get:

$$\begin{gathered} \frac{1}{2}=e^{60k} \\ \ln \frac{1}{2}=\ln e^{60k} \\ -0.693=60k\left[\ln e^a=a\right] \\ k=\frac{-0.693}{60} \\ k=-0.0116 \end{gathered}$$

Step 4. Finding the number of decreased bacteria  

Substituting the obtained value of k and the given time in the given formula, we get:

$$\begin{gathered} A=A_0{e}^{kt} \\ A=50e^{-0.0116·40} \\ A=50e^{-0.464} \\ A=50·0.629 \\ A\approx 31.45 \end{gathered}$$

$31.45mg$ of the sample will be left in$40$ days.