Answer:
[tex]\large \boxed{\text{1.38 g}}[/tex]
Explanation:
Two important formulas in radioactive decay are
[tex](1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt[/tex]
1. Calculate the decay constant k
[tex]\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{12.32 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{12.32 yr}}\\\\& = & \text{0.056 26 yr}^{-1}\\\end{array}[/tex]
2. Calculate the mass remaining
[tex]\begin{array}{rcl}\ln \left(\dfrac{A_{0}}{A}\right) &= &kt \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &\text{0.056 26 yr}^{-1}\times \text{15 yr} \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &0.8439 \\\\\dfrac{\text{3.20 g}}{A} &= &e^{0.8439} \\\\\dfrac{\text{3.20 g}}{A}&= &2.325 \\\\A &= &\dfrac{\text{3.20 g}}{2.325}\\\\&= & \textbf{1.38 g}\\\end{array}\\\text{The final mass of the sample will be $\large \boxed{\textbf{1.38 g}}$}[/tex]
