QuestionJuly 28, 2025

Of an initial amount of 1000 g of lead-210, how much will remain in 160 years?Lead-210 decays at a rate of 3.15% /yr square g (Round to one decimal place as needed.)

Of an initial amount of 1000 g of lead-210, how much will remain in 160 years?Lead-210 decays at a rate of 3.15% /yr square g (Round to one decimal place as needed.)
Of an initial amount of 1000 g of lead-210, how much will remain in 160 years?Lead-210 decays at a rate of 3.15% /yr square g (Round to one decimal place as needed.)

Solution
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Answer

6.5 g Explanation 1. Identify the decay formula Use the exponential decay formula: N(t) = N_0 \cdot e^{-kt}, where N_0 is the initial amount, k is the decay rate, and t is time. 2. Convert percentage to decimal Decay rate k = 3.15\% = 0.0315. 3. Calculate remaining amount Substitute values into the formula: N(160) = 1000 \cdot e^{-0.0315 \times 160}. 4. Compute the exponent Calculate e^{-0.0315 \times 160} \approx e^{-5.04}. 5. Evaluate the expression N(160) \approx 1000 \cdot e^{-5.04} \approx 1000 \cdot 0.0065. 6. Final calculation N(160) \approx 6.5 g.

Explanation

1. Identify the decay formula<br /> Use the exponential decay formula: $N(t) = N_0 \cdot e^{-kt}$, where $N_0$ is the initial amount, $k$ is the decay rate, and $t$ is time.<br /><br />2. Convert percentage to decimal<br /> Decay rate $k = 3.15\% = 0.0315$.<br /><br />3. Calculate remaining amount<br /> Substitute values into the formula: $N(160) = 1000 \cdot e^{-0.0315 \times 160}$.<br /><br />4. Compute the exponent<br /> Calculate $e^{-0.0315 \times 160} \approx e^{-5.04}$.<br /><br />5. Evaluate the expression<br /> $N(160) \approx 1000 \cdot e^{-5.04} \approx 1000 \cdot 0.0065$.<br /><br />6. Final calculation<br /> $N(160) \approx 6.5$ g.
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