Phosphorus-32 (P-32) has a half-life of 14.2 days. If 500 g of this substance are present initially, find the amount Q present after t days. (Round your growth constant to four decimal places.)

What amount will be left after 13.3 days? (Round your answer to three decimal places.)

The decay of a radioactive substance like Phosphorus-32 (P-32) can be modeled using the formula:

\[
Q(t) = Q_0 e^{-kt}
\]

Where:
- \( Q(t) \) is the amount of substance remaining after time \( t \) days.
- \( Q_0 \) is the initial amount of substance.
- \( k \) is the decay constant (to be determined).
- \( t \) is the time that has passed.

### Step 1: Calculate the decay constant \( k \)

The decay constant \( k \) is related to the half-life \( T_{1/2} \) by the formula:

\[
k = \frac{\ln 2}{T_{1/2}}
\]

Given:
- \( T_{1/2} = 14.2 \) days

Thus:

\[
k = \frac{\ln 2}{14.2} \approx \frac{0.6931}{14.2} \approx 0.0488
\]

### Step 2: Apply the formula to find the amount after \( t = 13.3 \) days

Now, we use the formula \( Q(t) = Q_0 e^{-kt} \) to find the amount left after 13.3 days.

Given:
- \( Q_0 = 500 \) grams
- \( k = 0.0488 \)
- \( t = 13.3 \) days

\[
Q(13.3) = 500 e^{-0.0488 \times 13.3}
\]

First, calculate the exponent:

\[
0.0488 \times 13.3 \approx 0.648
\]

Now, calculate \( e^{-0.648} \):

\[
e^{-0.648} \approx 0.5233
\]

Finally, calculate \( Q(13.3) \):

\[
Q(13.3) = 500 \times 0.5233 \approx 261.65
\]

### Answer:

After 13.3 days, approximately **261.650 grams** of Phosphorus-32 will be left.

---

Would you like more details or explanations on any part of this process?

Here are five related questions for further exploration:
1. What would the amount of P-32 be after 20 days?
2. How does the decay constant \( k \) affect the rate of decay?
3. How long will it take for the P-32 to decay to 100 grams?
4. Can the same formula be used for substances with different half-lives?
5. How does the decay curve of P-32 look over time?

**Tip**: Always check units and significant digits in calculations involving scientific constants to ensure precision!

Phosphorus-32 (P-32) has a half-life of 14.2 days. If 500 g of this substance are present initially, find the amount Q present after t days. (Round your growth constant to four decimal places.) What amount will be left after 13.3 days? (Round your answer to three decimal places.)

Solve a step-by-step exponential decay problem involving Phosphorus-32 with a half-life of 14.2 days. Learn how to calculate the decay constant and determine the remaining amount after 13.3 days, starting with 500 grams. Suitable for Grades 10-12 students interested in radioactive decay processes.

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