QuestionMay 5, 2025

(10 points) You have 125 grams of an isotope that decays at 4.5% per day. a. Write an expression representing the quantity Q(t) at time t in days. b. What is the half-life of the isotope?

(10 points) You have 125 grams of an isotope that decays at 4.5% per day. a. Write an expression representing the quantity Q(t) at time t in days. b. What is the half-life of the isotope?
(10 points) You have 125 grams of an isotope that decays at 4.5% per day. a. Write an expression representing the quantity Q(t) at time t in days. b. What is the half-life of the isotope?

Solution
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Answer

a. Q(t) = 125 \cdot e^{-0.045t}; b. T_{1/2} \approx 15.4 days. Explanation 1. Write the decay expression The quantity Q(t) at time t is given by the exponential decay formula: **Q(t) = Q_0 \cdot e^{-kt}**, where Q_0 is the initial quantity and k is the decay constant. Here, Q_0 = 125 grams and the decay rate is 4.5\% per day, so k = 0.045. Thus, Q(t) = 125 \cdot e^{-0.045t}. 2. Calculate the half-life The half-life T_{1/2} is found using the formula: **T_{1/2} = \frac{\ln(2)}{k}**. Substituting k = 0.045, we get T_{1/2} = \frac{\ln(2)}{0.045}.

Explanation

1. Write the decay expression<br /> The quantity $Q(t)$ at time $t$ is given by the exponential decay formula: **$Q(t) = Q_0 \cdot e^{-kt}$**, where $Q_0$ is the initial quantity and $k$ is the decay constant. Here, $Q_0 = 125$ grams and the decay rate is $4.5\%$ per day, so $k = 0.045$. Thus, $Q(t) = 125 \cdot e^{-0.045t}$.<br /><br />2. Calculate the half-life<br /> The half-life $T_{1/2}$ is found using the formula: **$T_{1/2} = \frac{\ln(2)}{k}$**. Substituting $k = 0.045$, we get $T_{1/2} = \frac{\ln(2)}{0.045}$.
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