The half-life of plutonium-239 is 24,300 years

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  1. The half-life of plutonium-239 is 24,300 years. If a nuclear bomb released 8 kg of this isotope, how many years would pass before the amount is reduced to 1 kg ?
  2. The half-life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days?

The Correct Answer and Explanation is :

Let’s address both questions step by step, using the half-life formula:

Formula for Half-Life:

The general formula for the amount of a substance remaining after a certain period is: A=A0×(12)tTA = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}}

Where:

  • AA is the amount remaining after time tt,
  • A0A_0 is the initial amount,
  • TT is the half-life of the substance,
  • tt is the time elapsed.

1. Plutonium-239 Question:

Given:

  • Initial mass (A0A_0) = 8 kg,
  • Final mass (AA) = 1 kg,
  • Half-life (TT) = 24,300 years.

We need to solve for tt, the time it will take for the mass to reduce from 8 kg to 1 kg.

Using the formula: 1=8×(12)t24,3001 = 8 \times \left( \frac{1}{2} \right)^{\frac{t}{24,300}}

Solving for tt: 18=(12)t24,300\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t}{24,300}}

Taking the natural logarithm of both sides: ln⁡(18)=t24,300×ln⁡(12)\ln \left( \frac{1}{8} \right) = \frac{t}{24,300} \times \ln \left( \frac{1}{2} \right)

Since ln⁡(12)=−ln⁡(2)\ln \left( \frac{1}{2} \right) = -\ln(2), we have: ln⁡(1)−ln⁡(8)=t24,300×(−ln⁡(2))\ln(1) – \ln(8) = \frac{t}{24,300} \times (-\ln(2)) ln⁡(8)≈2.0794,ln⁡(2)≈0.6931\ln(8) \approx 2.0794, \quad \ln(2) \approx 0.6931

Thus: −2.0794=t24,300×(−0.6931)-2.0794 = \frac{t}{24,300} \times (-0.6931)

Now, solve for tt: t=2.0794×24,3000.6931≈72,500 yearst = \frac{2.0794 \times 24,300}{0.6931} \approx 72,500 \text{ years}

Thus, it will take approximately 72,500 years for the plutonium-239 to reduce from 8 kg to 1 kg.

2. Radon-222 Question:

Given:

  • Initial mass (A0A_0) = 100 grams,
  • Half-life (TT) = 3.8 days,
  • Time elapsed (tt) = 15.2 days.

We need to determine how much of the original 100-gram sample remains after 15.2 days.

Using the same half-life formula: A=100×(12)15.23.8A = 100 \times \left( \frac{1}{2} \right)^{\frac{15.2}{3.8}}

First, calculate 15.23.8\frac{15.2}{3.8}: 15.23.8=4\frac{15.2}{3.8} = 4

Now: A=100×(12)4A = 100 \times \left( \frac{1}{2} \right)^4 A=100×116=6.25 gramsA = 100 \times \frac{1}{16} = 6.25 \text{ grams}

Thus, after 15.2 days, 6.25 grams of the original 100 grams of radon-222 will remain.

Explanation:

  1. Plutonium-239:
    • The half-life of plutonium-239 is 24,300 years, meaning after each 24,300-year period, half of the initial quantity decays. To find how long it will take for 8 kg to reduce to 1 kg, we apply the half-life formula. The calculation shows that it takes approximately 72,500 years for the mass to decrease to 1 kg.
  2. Radon-222:
    • Radon-222 has a much shorter half-life of 3.8 days. After 4 half-lives (which equals 15.2 days), the amount of radon reduces by a factor of 116\frac{1}{16}, or 6.25 grams remaining from the original 100 grams.

These calculations illustrate how substances decay exponentially over time, with each “half-life” reducing the remaining amount by half, which can be modeled using the formula above.

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