Math Problem Statement

A breeder reactor converts relatively stable uranium-238 into the isotope plutonium239. After 15 years it is determined that 0.043% of the initial amount A0 of plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration is proportional to the amount remaining.

Solution

To solve this problem, we need to apply the formula for radioactive decay, where the rate of disintegration is proportional to the amount of substance remaining. The decay model is expressed as:

A(t)=A0ektA(t) = A_0 e^{-kt}

Where:

  • A(t)A(t) is the amount of substance remaining at time tt,
  • A0A_0 is the initial amount of the substance,
  • kk is the decay constant,
  • tt is time.

We are given:

  • After 1515 years, 0.043%0.043\% of the initial amount of plutonium has disintegrated. This means that 99.957%99.957\% of the initial amount remains, i.e. A(15)=0.99957A0A(15) = 0.99957A_0.

Step 1: Set up the decay equation

We substitute this information into the decay equation:

A(15)=A0e15kA(15) = A_0 e^{-15k}

Since A(15)=0.99957A0A(15) = 0.99957A_0, we can simplify the equation:

0.99957A0=A0e15k0.99957A_0 = A_0 e^{-15k}

Dividing both sides by A0A_0:

0.99957=e15k0.99957 = e^{-15k}

Step 2: Solve for the decay constant kk

To solve for kk, take the natural logarithm of both sides:

ln(0.99957)=15k\ln(0.99957) = -15k

k=ln(0.99957)15k = -\frac{\ln(0.99957)}{15}

Now we can calculate kk.

kln(1/0.99957)15k \approx \frac{\ln(1/0.99957)}{15}

Let me calculate this value.The decay constant kk is approximately 2.867×1052.867 \times 10^{-5} per year.

Step 3: Find the half-life

The half-life T1/2T_{1/2} is the time it takes for half of the substance to decay. It is related to the decay constant by the formula:

T1/2=ln(2)kT_{1/2} = \frac{\ln(2)}{k}

Now, let's calculate the half-life.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, you can calculate the half-life manually by using the formula:

T1/2=ln(2)kT_{1/2} = \frac{\ln(2)}{k}

where k2.867×105k \approx 2.867 \times 10^{-5} per year. Plugging in the values should give you the result.

Would you like further clarification, or do you have any other questions?

Here are some related questions:

  1. How is the decay constant related to the half-life of an isotope?
  2. What are other common applications of radioactive decay formulas?
  3. How does the exponential decay function model real-life scenarios?
  4. What is the difference between uranium-238 and plutonium-239 in terms of stability?
  5. How does a breeder reactor work in converting uranium into plutonium?

Tip: When solving exponential decay problems, always double-check that the time and rate units match!

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Radioactive Decay
Logarithms

Formulas

A(t) = A_0 e^{-kt}
k = -\frac{\ln(\text{remaining percentage})}{t}
T_{1/2} = \frac{\ln(2)}{k}

Theorems

Exponential Decay Theorem
Properties of Logarithms

Suitable Grade Level

Grades 11-12