To solve this problem, we can use the formula for exponential decay related to half-life:

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

Where:

• $N(t)$ is the amount of substance remaining after time $t$.
• $N_0$ is the initial amount of substance.
• $T_{1/2}$ is the half-life of the substance.
• $t$ is the elapsed time.

Given:

• $N_0 = 3$ grams (initial amount)
• $T_{1/2} = 24.3$ days (half-life of phosphorus-32)
• $t = 9$ days (elapsed time)

Plugging the values into the formula:

$N(9) = 3 \times \left(\frac{1}{2}\right)^{\frac{9}{24.3}}$

First, calculate the exponent:

$\frac{9}{24.3} \approx 0.370$

Now calculate the remaining amount:

$N(9) = 3 \times \left(\frac{1}{2}\right)^{0.370}$

$N(9) = 3 \times 0.766 \approx 2.298$

So, approximately 2.298 grams of the phosphorus-32 would remain after 9 days.

Would you like more details on this calculation or have any other questions?

Here are some related questions to consider:

1. How would the remaining amount change if the half-life was longer or shorter?
2. How much phosphorus-32 would remain after one full half-life period (24.3 days)?
3. What would be the remaining amount after 50 days?
4. How is half-life used to date ancient objects using carbon dating?
5. What are some practical applications of knowing the half-life of a substance?

Tip: Remember, the concept of half-life applies not only to radioactive decay but also to many other processes, such as drug metabolism and certain chemical reactions.