To determine the half-life of argon-39, we can use the formula related to radioactive decay:
\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]
Where:
- \(N(t)\) is the remaining quantity of the substance after time \(t\),
- \(N_0\) is the initial quantity of the substance,
- \(T_{1/2}\) is the half-life,
- \(t\) is the elapsed time.
We have:
- \(N_0 = 1578\) grams,
- \(N(t) = 394.5\) grams,
- \(t = 538\) years.
First, we can find the proportion of the original quantity that remains:
\[ \frac{N(t)}{N_0} = \frac{394.5}{1578} \]
Calculating this gives:
\[ \frac{394.5}{1578} \approx 0.25 \]
Since \(0.25\) is \( \left(\frac{1}{2}\right)^2\), we can conclude that:
\[ \frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^2 \]
This implies that 2 half-lives have passed in this time \(t = 538\) years.
Now we can use the relationship between the time elapsed and the number of half-lives to find the half-life \(T_{1/2}\):
\[ t = n \cdot T_{1/2} \]
Where \(n\) is the number of half-lives (which we've found to be 2). Replacing \(n\) and \(t\):
\[ 538 = 2 \cdot T_{1/2} \]
Solving for \(T_{1/2}\):
\[ T_{1/2} = \frac{538}{2} = 269 \text{ years} \]
Thus, the half-life of argon-39 is approximately 269 years.
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