QuestionMay 4, 2025

The half-life of a certain substance is 19 years.How long will it take for a sample of this substance to decay to 82% of its original amount? The exponential decay model is: A=A_(0)e^kt

The half-life of a certain substance is 19 years.How long will it take for a sample of this substance to decay to 82% of its original amount? The exponential decay model is: A=A_(0)e^kt
The half-life of a certain substance is 19 years.How long will it take for a sample of this substance to decay to 82% of its original amount? The exponential decay model is: A=A_(0)e^kt

Solution
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Answer

t \approx 5.12 years Explanation 1. Identify the decay constant Use the half-life formula: t_{1/2} = \frac{\ln(2)}{-k}. Given t_{1/2} = 19 years, solve for k: k = \frac{\ln(2)}{19}. 2. Set up the decay equation Use A = A_0 e^{kt} with A = 0.82A_0. Substitute k from Step 1: 0.82A_0 = A_0 e^{\left(\frac{\ln(2)}{19}\right)t}. 3. Solve for time t Cancel A_0 and take the natural logarithm of both sides: \ln(0.82) = \left(\frac{\ln(2)}{19}\right)t. Solve for t: t = \frac{\ln(0.82)}{\frac{\ln(2)}{19}}.

Explanation

1. Identify the decay constant<br /> Use the half-life formula: $t_{1/2} = \frac{\ln(2)}{-k}$. Given $t_{1/2} = 19$ years, solve for $k$: <br />$$k = \frac{\ln(2)}{19}.$$<br /><br />2. Set up the decay equation<br /> Use $A = A_0 e^{kt}$ with $A = 0.82A_0$. Substitute $k$ from Step 1:<br />$$0.82A_0 = A_0 e^{\left(\frac{\ln(2)}{19}\right)t}.$$<br /><br />3. Solve for time $t$<br /> Cancel $A_0$ and take the natural logarithm of both sides:<br />$$\ln(0.82) = \left(\frac{\ln(2)}{19}\right)t.$$<br /> Solve for $t$:<br />$$t = \frac{\ln(0.82)}{\frac{\ln(2)}{19}}.$$
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