Math Problem Statement

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

Solution

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)Q(t) of a radioactive substance after tt days is modeled by the equation:

Q(t)=Q0(12)tT1/2Q(t) = Q_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}

Where:

  • Q0=350Q_0 = 350 g is the initial amount.
  • T1/2=14.2T_{1/2} = 14.2 days is the half-life.
  • tt is the time in days.

We are given the formula:

Q(t)=350(12)t14.2Q(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)Q(5.3), i.e., the amount left after 5.3 days. Substituting t=5.3t = 5.3 into the equation:

Q(5.3)=350(12)5.314.2Q(5.3) = 350 \left(\frac{1}{2}\right)^{\frac{5.3}{14.2}}

Let's calculate:

Q(5.3)=350×(12)5.314.2350×0.7679Q(5.3) = 350 \times \left(\frac{1}{2}\right)^{\frac{5.3}{14.2}} \approx 350 \times 0.7679 Q(5.3)268.763Q(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=5.3t = 5.3

The rate of decay is given by the derivative of Q(t)Q(t), i.e., Q(t)Q'(t).

The decay model can be rewritten as:

Q(t)=350×ektQ(t) = 350 \times e^{kt}

Where kk is the decay constant. The relationship between kk and the half-life is:

k=ln(1/2)14.20.693114.20.0488k = \frac{\ln(1/2)}{14.2} \approx \frac{-0.6931}{14.2} \approx -0.0488

Thus, Q(t)=350e0.0488tQ(t) = 350 e^{-0.0488 t}.

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

Q(t)=350×(0.0488)×e0.0488tQ'(t) = 350 \times (-0.0488) \times e^{-0.0488 t}

Substituting t=5.3t = 5.3:

Q(5.3)=350×(0.0488)×e0.0488×5.3Q'(5.3) = 350 \times (-0.0488) \times e^{-0.0488 \times 5.3}

We already calculated e0.0488×5.30.7679e^{-0.0488 \times 5.3} \approx 0.7679, so:

Q(5.3)350×(0.0488)×0.767913.110Q'(5.3) \approx 350 \times (-0.0488) \times 0.7679 \approx -13.110

So, the rate of decay at t=5.3t = 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Exponential Decay
Differentiation
Half-Life Calculation

Formulas

Q(t) = Q_0 (1/2)^(t/T_1/2)
k = ln(1/2) / T_1/2
Q'(t) = Q_0 * k * e^(kt)

Theorems

Exponential Decay Theorem

Suitable Grade Level

College-level (Calculus)