The formula given in the image for the radioactive decay is:

$A = A_0 e^{-0.025t}$

Where:

• $A_0$ is the initial amount of the isotope.
• $A$ is the remaining amount after time $t$ years.
• The decay constant is $-0.025$.

You are asked to find the half-life, which is the time when the remaining amount $A$ is 50% of the initial amount $A_0$. So, $A = \frac{A_0}{2}$.

We can solve for the half-life $t_{1/2}$ by setting $A = \frac{A_0}{2}$ and solving the equation:

$\frac{A_0}{2} = A_0 e^{-0.025t}$

Dividing both sides by $A_0$ (assuming $A_0 \neq 0$):

$\frac{1}{2} = e^{-0.025t}$

Now, take the natural logarithm (ln) of both sides to solve for $t$:

$\ln\left(\frac{1}{2}\right) = -0.025t$

We know that $\ln\left(\frac{1}{2}\right)$ is $-\ln(2)$, so the equation becomes:

$-\ln(2) = -0.025t$

Solve for $t$:

$t = \frac{\ln(2)}{0.025}$

Using the value of $\ln(2) \approx 0.693$:

$t = \frac{0.693}{0.025} \approx 27.72 \, \text{years}$

Thus, the half-life of the isotope is approximately 27.720 years, rounded to three decimal places.

Would you like further details or have any questions? Here are some related questions:

1. What is the significance of half-life in radioactive decay?
2. How would the formula change if the decay constant were different?
3. Can you explain why natural logarithms are used in this type of problem?
4. How does changing the decay constant affect the half-life of an isotope?
5. How would you find the time for 75% of the isotope to decay?

Tip: Remember that the half-life of an isotope is independent of the initial amount; it only depends on the decay constant.