To find the half-life of a substance, we can use the formula related to exponential 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 time that has passed.
In this case, we know:
- \( N_0 = 25 \) g (initial quantity),
- \( N(t) = 3 \) g (remaining quantity after time \( t \)),
- \( t = 17,527.5 \) years.
We want to find \( T_{1/2} \).
First, we can rewrite the equation to solve for \( T_{1/2} \):
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Start with the equation: \[ 3 = 25 \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}} \]
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Divide both sides by 25: \[ \frac{3}{25} = \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}} \]
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Take the logarithm of both sides. We'll use the natural logarithm (ln): \[ \ln \left( \frac{3}{25} \right) = \frac{17527.5}{T_{1/2}} \ln \left( \frac{1}{2} \right) \]
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Substitute \( \ln \left( \frac{1}{2} \right) = -\ln(2) \): \[ \ln \left( \frac{3}{25} \right) = -\frac{17527.5}{T_{1/2}} \ln(2) \]
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Solve for \( T_{1/2} \): \[ T_{1/2} = -\frac{17527.5 \ln(2)}{\ln \left( \frac{3}{25} \right)} \]
Now we can calculate \( T_{1/2} \).
First, compute \( \ln(2) \): \[ \ln(2) \approx 0.693147 \]
Next, compute \( \ln \left( \frac{3}{25} \right) \): \[ \frac{3}{25} = 0.12 \] \[ \ln(0.12) \approx -2.12026 \]
Now we can substitute back into the equation for \( T_{1/2} \): \[ T_{1/2} = -\frac{17527.5 \cdot 0.693147}{-2.12026} \] \[ T_{1/2} \approx \frac{17527.5 \cdot 0.693147}{2.12026} \approx \frac{12100.7}{2.12026} \approx 5712.556 \text{ years} \]
Therefore, the half-life of the material is approximately 5712.56 years.
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