Pregunta
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The radioactive substance uranium- 240 has a half-life of 14 hours. The amount \( A(t) \) of a sample of uranium- 240 remaining (in grams) after \( t \) hours is giv the following exponential function. \[ A(t)=2800\left(\frac{1}{2}\right)^{\frac{t}{14}} \] Find the amount of the sample remaining after 9 hours and after 60 hours. Round your answers to the nearest gram as necessary. Amount after 9 hours: grams

Ask by Blake Bowen. in the United States
Jan 23,2025

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After 9 hours, there are approximately 1793 grams remaining. After 60 hours, there are approximately 144 grams remaining.

Solución

To determine the amount of uranium-240 remaining after a given time, we use the exponential decay formula: \[ A(t) = 2800 \left(\frac{1}{2}\right)^{\frac{t}{14}} \] Where: - \( A(t) \) is the amount remaining after \( t \) hours. - 2800 grams is the initial amount. - The half-life of uranium-240 is 14 hours. ### 1. Amount After 9 Hours Let's calculate \( A(9) \): \[ A(9) = 2800 \left(\frac{1}{2}\right)^{\frac{9}{14}} \] To simplify: \[ \left(\frac{1}{2}\right)^{\frac{9}{14}} \approx 0.6403 \] Now, multiply by the initial amount: \[ A(9) = 2800 \times 0.6403 \approx 1793 \text{ grams} \] **Amount after 9 hours:** **1793 grams** ### 2. Amount After 60 Hours Now, let's calculate \( A(60) \): \[ A(60) = 2800 \left(\frac{1}{2}\right)^{\frac{60}{14}} = 2800 \left(\frac{1}{2}\right)^{4.2857} \] Breaking it down: \[ \left(\frac{1}{2}\right)^{4} = \frac{1}{16} \approx 0.0625 \] \[ \left(\frac{1}{2}\right)^{0.2857} \approx 0.820 \] \[ A(60) = 2800 \times 0.0625 \times 0.820 \approx 144 \text{ grams} \] **Amount after 60 hours:** **144 grams**

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To find the amount of uranium-240 remaining after 9 hours, we can substitute \( t = 9 \) into the function: \[ A(9) = 2800\left(\frac{1}{2}\right)^{\frac{9}{14}} \] Calculating this gives: \[ A(9) \approx 2800\left(0.593\right) \approx 1660.29 \] After rounding to the nearest gram, the amount remaining after 9 hours is approximately **1660 grams**. Now, let's calculate the amount remaining after 60 hours by substituting \( t = 60 \): \[ A(60) = 2800\left(\frac{1}{2}\right)^{\frac{60}{14}} \] \[ A(60) \approx 2800\left(0.088\right) \approx 246.40 \] After rounding to the nearest gram, the amount remaining after 60 hours is approximately **246 grams**.

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