Question

37. Plutonium-239 The half-life of the plutonium isotope is 24,360 years. If 10 g of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for 80% of the isotope to decay?

Accepted Answer

Identify the decay model: The amount of plutonium remaining after time $$t$$ can be modeled by the exponential decay formula $$A(t) = A_0$ \cdot e$$^{-kt}$$, where A_0$ is the initial amount, k$ is the decay constant, and t$ is time in years. Use the half-life to find the decay constant k$: Since the half-life T$$_{1/2}$$ is 24,360 years, use the relation k$$ = \frac{\ln(2)}{T_{1/2}} = \frac{\ln(2)}{24360}$$ to calculate k$. Set up the equation for 80% decay: If 80% of the plutonium decays, 20% remains. So, A(t) = 0.2$$ 	imes $$A_0. $$Substitute into the decay formula to get $$0.2 A_0$ = A_0$ \cdot e$$^{-kt}$$. Simplify the equation: Cancel A_0$ from both sides to get 0.2$$ = $$e^{-kt}. $$Then take the natural logarithm of both sides to solve for $$t$$: $$\ln(0.2) = -kt. $$Solve for $$t$$: Rearrange to find $$t = -\frac{\ln(0.2)}{k}. $$Substitute the value of $$k$$ from step 2 to express $$t in $$terms of known quantities.

37. Plutonium-239 The half-life of the plutonium isotope is 24,360 years. If 10 g of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for 80% of the isotope to decay?

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Key Concepts

Exponential Decay

Half-Life

Decay Constant and Time Calculation