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The text below is verbatim of what my question is.

The half life of $\textrm{Ra-}226$ is approximately $1599$ years. If the amount left after $1000$ years is $1.7$ grams, what is the amount left after $2000$ years? Give an approximation to three decimal places. If you use a formula you must derive it first.

So, I already know the decay formula which is $$ y= Ce^{kt}$$

but where I get stuck is when trying to solve for one of the variables, because when I use the current amount of $y=1.7$ grams at $t=1000$ like so $$1.7=Ce^{1000k}$$ I still have unknown variables and I cannot solve for $k$ or $C$ without knowing the other.

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3 Answers 3

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You have not used the fact that the half life is $1599$ years. That allows you to derive the value of $k$, because at $t=1599$ the exponential factor is $\frac 12$

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  • $\begingroup$ So doing what you're saying would be $y=\frac{1}{2}e^{1599k}$ but I still have Y to solve for. $\endgroup$
    – Eric Brown
    Sep 25, 2019 at 4:08
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    $\begingroup$ No, you need to derive $k$ from $1599$. You are told that if you start with $1$, after $1599$ years you have $\frac 12$. That is the definition of half-life $\endgroup$ Sep 25, 2019 at 4:14
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The half-life $t_{1/2}$ represents the moment at which your initial quantity reaches half.

Thus, $y(1599)=\frac{1}{2}C$.

This means, $$1/2=e^{1599k} \Longrightarrow k=-\frac{\ln 2}{1599}$$

Now it is easy to solve for $C$.

Having $C$ and $k$ you can get your desired output.

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If you want a more intuitive solution you can think of it like this. In every period of a single half-life the amount of radioactive material decreases to half its original value. Hence the amount left at time $t$ is given by: $A(t) = A_0(\frac 12)^N$, where $N$ is the number of half-lives elapsed.

Here, a thousand years have elapsed since the "initial" $1.7$ gram, so $N=\frac{1000}{1599}$, and you can immediately compute $A(t) = 1.7(\frac 12)^{\frac{1000}{1599}}=1.102$ gram.

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