The half-life of the radioactive element krypton-9 10 seconds. If 16 grams of krypton-91 are initially present, how many grams are present after 10 seconds? 20 seconds? 30 seconds? 40 seconds? 50 seconds?

To find the amount of krypton-91 present over time, we will use the exponential decay formula.

$$\textcolor{#c34632}{A = A_0 e^{kt}}$$

First, we need to find the decay rate, $\textcolor{#c34632}{k}$, by deriving a formula from the equation above.

$$\begin{aligned} \dfrac{A}{\textcolor{#c34632}{A_0}} &= \dfrac{\cancel{A_0} e^{kt}}{\textcolor{#c34632}{\cancel{A_0}}} \\ \\ \dfrac{A}{A_0} &= e^{kt} \\ \\ \textcolor{#c34632}{\text{ln}} \bigg ( \dfrac{A}{A_0} \bigg) &= \textcolor{#c34632}{\text{ln}} \ e^{kt} \\ \\ \text{ln} \bigg ( \dfrac{A}{A_0} \bigg) &= \textcolor{#c34632}{kt} \text{ln} \ e \\ \\ \text{Since ln $e = 1$,} \\ \\ \text{ln} \bigg ( \dfrac{A}{A_0} \bigg) &= kt \\ \\ \dfrac{\text{ln} \bigg ( \dfrac{A}{A_0} \bigg)}{\textcolor{#c34632}{t}} &= \dfrac{k\cancel{t}}{\textcolor{#c34632}{\cancel{t}}}\\ \end{aligned}$$

The derived formula to solve for $k$ is $\textcolor{#c34632}{k = \dfrac{\text{ln} \bigg ( \dfrac{A}{A_0} \bigg)}{t}}$.

Initially, there are $16$ grams of krypton-91 that is present and its half-time is $10$ seconds. That only means the $16$ grams will be half of its amount after 10 seconds, that is

$$\dfrac{16}{2}=8$$

Therefore, there will be $8$ grams of krypton-91 that will be present in $10$ seconds.