To determine what fraction of the radioactive strontium-90 absorbed in 1964 remained in people's bones in 1996, we can use the concept of radioactive decay, governed by the formula:

$N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{\text{half}}}}$

Where:

• $N(t)$ is the amount of substance remaining after time $t$.
• $N_0$ is the initial amount of substance.
• $T_{\text{half}}$ is the half-life of the substance.
• $t$ is the elapsed time.

Given:

• The half-life ($T_{\text{half}}$) of strontium-90 is 28 years.
• The time elapsed from 1964 to 1996 is $t = 1996 - 1964 = 32$ years.

Now, we calculate the fraction of strontium-90 remaining after 32 years:

$\text{Fraction remaining} = \left(\frac{1}{2}\right)^{\frac{32}{28}} = \left(\frac{1}{2}\right)^{1.14}$

We'll now calculate this value.The fraction of strontium-90 remaining in people's bones in 1996, from what was absorbed in 1964, is approximately 0.453, or 45.3%.

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:

1. What is the general formula for radioactive decay?
2. How does the half-life of a substance influence its long-term presence in the environment?
3. What are the health impacts of long-term exposure to radioactive substances like strontium-90?
4. How is the half-life of a substance measured in a laboratory setting?
5. How do scientists use half-life data to track the environmental impact of nuclear tests?

Tip: When working with exponential decay, always ensure you understand how the base (like $\frac{1}{2}$) affects the rate at which the quantity decreases over time.