QuestionApril 30, 2025

20. A certain material decays at a rate of 1.9% per year. The sample is 260 grams. How much will be left in 11 years? How long will it take to have only a 100g sample left?

20. A certain material decays at a rate of 1.9% per year. The sample is 260 grams. How much will be left in 11 years? How long will it take to have only a 100g sample left?
20. A certain material decays at a rate of 1.9% per year. The sample is 260 grams. How much will be left in 11 years? How long will it take to have only a 100g sample left?

Solution
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Answer

After 11 years, approximately 210.86 grams will remain. It will take about 50.32 years for the sample to decay to 100 grams. Explanation 1. Write the decay formula The exponential decay formula is A = A_0 e^{-kt}, where A is the remaining amount, A_0 is the initial amount, k is the decay constant, and t is time. 2. Calculate the decay constant k Given the decay rate is 1.9\%, k = 0.019. 3. Find the remaining amount after 11 years Substitute A_0 = 260, k = 0.019, and t = 11 into the formula: A = 260 e^{-0.019 \cdot 11}. A = 260 e^{-0.209} \approx 260 \cdot 0.811 \approx 210.86 grams. 4. Solve for time when A = 100 Use the same formula, 100 = 260 e^{-0.019t}. Divide both sides by 260: \frac{100}{260} = e^{-0.019t}. Take the natural logarithm: \ln\left(\frac{100}{260}\right) = -0.019t. Solve for t: t = \frac{\ln(100/260)}{-0.019} \approx \frac{\ln(0.3846)}{-0.019} \approx \frac{-0.956}{-0.019} \approx 50.32 years.

Explanation

1. Write the decay formula<br /> The exponential decay formula is $A = A_0 e^{-kt}$, where $A$ is the remaining amount, $A_0$ is the initial amount, $k$ is the decay constant, and $t$ is time.<br /><br />2. Calculate the decay constant $k$<br /> Given the decay rate is $1.9\%$, $k = 0.019$.<br /><br />3. Find the remaining amount after 11 years<br /> Substitute $A_0 = 260$, $k = 0.019$, and $t = 11$ into the formula:<br />$A = 260 e^{-0.019 \cdot 11}$. <br />$A = 260 e^{-0.209} \approx 260 \cdot 0.811 \approx 210.86$ grams.<br /><br />4. Solve for time when $A = 100$<br /> Use the same formula, $100 = 260 e^{-0.019t}$. <br />Divide both sides by 260: <br />$\frac{100}{260} = e^{-0.019t}$. <br />Take the natural logarithm: <br />$\ln\left(\frac{100}{260}\right) = -0.019t$. <br />Solve for $t$: <br />$t = \frac{\ln(100/260)}{-0.019} \approx \frac{\ln(0.3846)}{-0.019} \approx \frac{-0.956}{-0.019} \approx 50.32$ years.
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