Question

Find the half-life of a 25g sample that decayed to 3g after 17 527.5 years.

Answer 1

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 \( T_{1/2} \):

1. Start with the equation:
\[
3 = 25 \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}}
\]

2. Divide both sides by 25:
\[
\frac{3}{25} = \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}}
\]

3. Take the logarithm of both sides. We'll use the natural logarithm (ln):
\[
\ln \left( \frac{3}{25} \right) = \frac{17527.5}{T_{1/2}} \ln \left( \frac{1}{2} \right)
\]

4. Substitute \( \ln \left( \frac{1}{2} \right) = -\ln(2) \):
\[
\ln \left( \frac{3}{25} \right) = -\frac{17527.5}{T_{1/2}} \ln(2)
\]

5. 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**.