Radioactive Decay Phosphorus-32 (P-32) has a half-life of 14.2 days. If 350 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.)
Q(t) = 350(1/2)^(t/14.2)

Correct: Your answer is correct.
What amount (in grams) will be left after 5.3 days? (Round your answer to three decimal places.)

Incorrect: Your answer is incorrect.
g
How fast (in grams per day) is the P-32 decaying when t = 5.3? (Round your answer to three decimal places.)

Incorrect: Your answer is incorrect.
g/day

We are working with the radioactive decay model for Phosphorus-32 (P-32). Let's break it down step by step.

### Step 1: Decay Formula
The amount \( Q(t) \) of a radioactive substance after \( t \) days is modeled by the equation:

\[
Q(t) = Q_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}
\]

Where:
- \( Q_0 = 350 \) g is the initial amount.
- \( T_{1/2} = 14.2 \) days is the half-life.
- \( t \) is the time in days.
  
We are given the formula:

\[
Q(t) = 350 \left(\frac{1}{2}\right)^{\frac{t}{14.2}}
\]

### Step 2: Calculate the Amount Left After 5.3 Days
We want to find \( Q(5.3) \), i.e., the amount left after 5.3 days. Substituting \( t = 5.3 \) into the equation:

\[
Q(5.3) = 350 \left(\frac{1}{2}\right)^{\frac{5.3}{14.2}}
\]

Let's calculate:

\[
Q(5.3) = 350 \times \left(\frac{1}{2}\right)^{\frac{5.3}{14.2}} \approx 350 \times 0.7679
\]
\[
Q(5.3) \approx 268.763
\]

So, the amount left after 5.3 days is approximately **268.763 grams**.

### Step 3: Calculate the Rate of Decay at \( t = 5.3 \)
The rate of decay is given by the derivative of \( Q(t) \), i.e., \( Q'(t) \).

The decay model can be rewritten as:

\[
Q(t) = 350 \times e^{kt}
\]

Where \( k \) is the decay constant. The relationship between \( k \) and the half-life is:

\[
k = \frac{\ln(1/2)}{14.2} \approx \frac{-0.6931}{14.2} \approx -0.0488
\]

Thus, \( Q(t) = 350 e^{-0.0488 t} \).

Now, take the derivative to find the rate of decay:

\[
Q'(t) = 350 \times (-0.0488) \times e^{-0.0488 t}
\]

Substituting \( t = 5.3 \):

\[
Q'(5.3) = 350 \times (-0.0488) \times e^{-0.0488 \times 5.3}
\]

We already calculated \( e^{-0.0488 \times 5.3} \approx 0.7679 \), so:

\[
Q'(5.3) \approx 350 \times (-0.0488) \times 0.7679 \approx -13.110
\]

So, the rate of decay at \( t = 5.3 \) is approximately **13.110 grams per day** (in absolute value, it's decaying at this rate).

---

Would you like further clarifications or details on any of the steps?

Here are 5 related questions to further explore this topic:
1. What is the amount left after 10 days?
2. How would the decay constant change if the half-life were shorter?
3. Can you calculate the time at which 100 grams will remain?
4. What would the rate of decay be after 10 days?
5. How does the decay formula change for a different isotope with a longer half-life?

**Tip:** The decay rate at any given time is directly proportional to the amount of substance present at that time, making it an example of an exponential decay process.

Radioactive Decay Phosphorus-32 (P-32) has a half-life of 14.2 days. If 350 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.) Q(t) = 350(1/2)^(t/14.2) What amount (in grams) will be left after 5.3 days? (Round your answer to three decimal places.) How fast (in grams per day) is the P-32 decaying when t = 5.3? (Round your answer to three decimal places.)

Explore the radioactive decay of Phosphorus-32 with a half-life of 14.2 days. This detailed problem demonstrates how to calculate the remaining substance after 5.3 days and the rate of decay. Ideal for calculus students studying exponential decay and differentiation.

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