Explanation
Given that the initial sample has a mass of 22 grams and that its k-value is 0.138, We can use the formula below to find its half-life.
[tex]N=N_0e^{-kt}[/tex]At half life, the substance decays to half of itself, therefore;
[tex]\begin{gathered} \frac{22}{2}=22e^{-0.138t} \\ 11=22e^{-0.138t} \\ divide\text{ both sides by 22} \\ \frac{11}{22}=e^{-0.138t} \\ 0.5=e^{\left\{-0.138t\right\}} \\ ln0.5=ln(e^{-0.138t}) \\ \ln \left(0.5\right)=-0.138t \\ t=-\frac{\ln \left(0.5\right)}{0.138} \\ t=5.02280\approx5.0days \end{gathered}[/tex]Answer: 5.0 days
