The question provides the radioactive decay function of Strontium-90:

$y = y_0 e^{-0.024t}$

where $y_0$ is the initial amount and $y$ is the amount remaining after $t$ years. We are asked to find the half-life of Strontium-90, which is the time required for the substance to reduce to half of its initial amount.

Steps to Solve

1. Set up the half-life equation: At the half-life $t = T_{1/2}$, the amount $y$ will be half of $y_0$: $\frac{y_0}{2} = y_0 e^{-0.024 T_{1/2}}$

2. Divide by $y_0$ (assuming $y_0 \neq 0$): $\frac{1}{2} = e^{-0.024 T_{1/2}}$

3. Take the natural logarithm of both sides: $\ln\left(\frac{1}{2}\right) = -0.024 T_{1/2}$

   Since $\ln\left(\frac{1}{2}\right) = -\ln(2)$, we have: $-\ln(2) = -0.024 T_{1/2}$

4. Solve for $T_{1/2}$: $T_{1/2} = \frac{\ln(2)}{0.024}$

5. Calculate $T_{1/2}$: Using $\ln(2) \approx 0.693$, $T_{1/2} \approx \frac{0.693}{0.024} \approx 28.9 \text{ years}$

Answer

The half-life of Strontium-90 is approximately 28.9 years.

Would you like a further breakdown of the calculation steps or have any questions?

Here are some related questions for practice:

1. What is the general formula for the half-life of a radioactive substance given an exponential decay model?
2. How would you calculate the remaining amount of a substance after a certain number of years?
3. If the decay rate was different, say -0.03 instead of -0.024, how would the half-life change?
4. How does the half-life formula change if the exponential base is different from $e$?
5. Can this decay model apply to other processes besides radioactive decay?

Tip: When working with exponential decay problems, remember that the half-life is independent of the initial amount and only depends on the decay constant.