Find the initial amount in the sample and the amount remaining after 100 years.

Question image

Understand the Problem

The question is asking to find two things: the initial amount of cesium-137 in the sample and the amount that remains after 100 years, using the provided exponential decay formula.

Answer

Initial amount: 266 grams; Remaining after 100 years: 25 grams.
Answer for screen readers

The initial amount of cesium-137 is 266 grams, and the amount remaining after 100 years is approximately 25 grams.

Steps to Solve

  1. Identify the initial amount The initial amount of cesium-137 can be found by evaluating ( A(t) ) at ( t = 0 ).

$$ A(0) = 266 \left(\frac{1}{2}\right)^{\frac{0}{30}} = 266 \left(\frac{1}{2}\right)^{0} = 266 \times 1 = 266 $$

  1. Calculate the amount after 100 years To find the amount remaining after 100 years, substitute ( t = 100 ) into the formula.

$$ A(100) = 266 \left(\frac{1}{2}\right)^{\frac{100}{30}} $$

  1. Simplify the exponent Calculate the exponent:

$$ \frac{100}{30} = \frac{10}{3} \approx 3.33 $$

So,

$$ A(100) = 266 \left(\frac{1}{2}\right)^{3.33} $$

  1. Evaluate the power of ( \frac{1}{2} ) Now calculate ( \left(\frac{1}{2}\right)^{3.33} ):

Using a calculator, we find:

$$ \left(\frac{1}{2}\right)^{3.33} \approx 0.0948 $$

  1. Final calculation for ( A(100) ) Now multiply by the initial amount:

$$ A(100) \approx 266 \times 0.0948 \approx 25.2 $$

  1. Round the answer Rounding to the nearest gram:

Remaining amount after 100 years is approximately 25 grams.

The initial amount of cesium-137 is 266 grams, and the amount remaining after 100 years is approximately 25 grams.

More Information

This problem demonstrates the concept of radioactive decay, where the amount of substance reduces by half over each half-life period. Understanding the decay formula helps quantify how much of a substance remains after a specific time.

Tips

  • Confusing the half-life periods with the total time when calculating decay.
  • Forgetting to evaluate the power of ( \frac{1}{2} ) correctly can lead to wrong results.

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