To determine how long it will take for strontium-90 (Sr-90) to be reduced to 1/32 of its present amount, we can use the same half-life formula:

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

Here:

• $N(t)$ is the remaining quantity of Sr-90 after time $t$.
• $N_0$ is the initial quantity of Sr-90.
• $t$ is the time elapsed.
• $T_{1/2} = 28$ years is the half-life of Sr-90.

We are asked to find $t$ when $N(t) = \frac{1}{32} N_0$. This means:

$\frac{1}{32} N_0 = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{28}}$

We can simplify by dividing both sides by $N_0$:

$\frac{1}{32} = \left(\frac{1}{2}\right)^{\frac{t}{28}}$

To solve for $t$, we can rewrite $\frac{1}{32}$ as a power of 2:

$\frac{1}{32} = \frac{1}{2^5} = \left(\frac{1}{2}\right)^5$

So, we have:

$\left(\frac{1}{2}\right)^5 = \left(\frac{1}{2}\right)^{\frac{t}{28}}$

Since the bases are the same, we can equate the exponents:

$5 = \frac{t}{28}$

Now, solve for $t$:

$t = 5 \times 28 = 140 \text{ years}$

It will take 140 years for the Sr-90 to be reduced to 1/32 of its present amount.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How does the decay of strontium-90 affect the environment?
2. What are some common uses of strontium-90?
3. How does the decay rate of Sr-90 compare to other radioactive isotopes?
4. What safety measures are taken when handling radioactive materials like Sr-90?
5. Can the half-life of a substance change under different conditions?

Tip: When dealing with half-life problems, remember that each half-life reduces the remaining quantity by half, making logarithmic calculations useful.