Question

# 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?

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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 $\textcolor{#c34632}{k = \dfrac{\text{ln} \bigg ( \dfrac{A}{A_0} \bigg)}{t}}$.

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