To determine the half-life of argon-39, we can use the information about the decay of the sample over time. The original mass of the argon-39 sample is 1578 grams, and after 538 years, only 394.5 grams remain. We can use the following formula related to half-life:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life,
- \( t \) is the elapsed time.
First, we need to express the remaining quantity in terms of the original quantity:
\[ N(t) = 394.5 \text{ grams} \] \[ N_0 = 1578 \text{ grams} \] \[ t = 538 \text{ years} \]
Substituting these values into the half-life formula, we get:
\[ 394.5 = 1578 \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}} \]
Next, we will isolate the half-life term \( T_{1/2} \):
- Divide both sides by 1578:
\[ \frac{394.5}{1578} = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}} \]
Calculating the left side:
\[ \frac{394.5}{1578} \approx 0.25 \]
So we have:
\[ 0.25 = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}} \]
- Recognize that \( 0.25 = \left( \frac{1}{2} \right)^2 \):
\[ \left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}} \]
- Since the bases are the same, we can equate the exponents:
\[ 2 = \frac{538}{T_{1/2}} \]
- Now, solve for \( T_{1/2} \):
\[ T_{1/2} = \frac{538}{2} = 269 \text{ years} \]
Thus, the half-life of argon-39 is 269 years.
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