Question

module 6 quiz (24-28)
100 points possible answered: 12/20
question 14
an initial sample of 283 g of a radioactive substance decays according to the function $a(t) = 283e^{-0.032t}$ where $t$ is given in years.
(a) how many grams of the substance will remain after 113 years? round your answer to one decimal place.

(b) what is the half-life of the substance? round your answer to one decimal place.

(c) in what year will the sample decay to 3 grams? round your answer to one decimal place

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Explanation:

Part (a)

Step1: Substitute \( t = 113 \) into the formula

We have the decay function \( A(t)=283e^{-0.032t} \). Substitute \( t = 113 \) into it: \( A(113)=283e^{-0.032\times113} \)

Step2: Calculate the exponent

First, calculate \( - 0.032\times113=-3.616 \)

Step3: Calculate the exponential term

Then, calculate \( e^{-3.616}\approx0.0269 \)

Step4: Calculate the final amount

Multiply by 283: \( 283\times0.0269\approx7.6127 \), round to one decimal place is \( 7.6 \)

Part (b)

Step1: Recall the half - life definition

For a decay function \( A(t)=A_0e^{kt} \), at half - life \( t = T_{1/2} \), \( A(T_{1/2})=\frac{A_0}{2} \). Here \( A_0 = 283 \), \( k=- 0.032 \), so we set \( 283e^{-0.032T_{1/2}}=\frac{283}{2} \)

Step2: Simplify the equation

Divide both sides by 283: \( e^{-0.032T_{1/2}}=\frac{1}{2} \)

Step3: Take the natural logarithm of both sides

\( \ln(e^{-0.032T_{1/2}})=\ln(\frac{1}{2}) \), which simplifies to \( - 0.032T_{1/2}=-\ln(2) \)

Step4: Solve for \( T_{1/2} \)

\( T_{1/2}=\frac{\ln(2)}{0.032}\approx\frac{0.6931}{0.032}\approx21.66 \), round to one decimal place is \( 21.7 \)

Part (c)

Step1: Set up the equation

We want to find \( t \) when \( A(t) = 3 \). So we set \( 283e^{-0.032t}=3 \)

Step2: Divide both sides by 283

\( e^{-0.032t}=\frac{3}{283}\approx0.0106 \)

Step3: Take the natural logarithm of both sides

\( \ln(e^{-0.032t})=\ln(0.0106) \), so \( - 0.032t=\ln(0.0106) \)

Step4: Solve for \( t \)

\( t=\frac{\ln(0.0106)}{- 0.032}\approx\frac{-4.543}{-0.032}\approx142.0 \) (rounded to one decimal place)

Part (a) Answer: \( 7.6 \)

Part (b) Answer: \( 21.7 \)

Part (c) Answer: \( 142.0 \)

Answer:

Step1: Set up the equation

We want to find \( t \) when \( A(t) = 3 \). So we set \( 283e^{-0.032t}=3 \)

Step2: Divide both sides by 283

\( e^{-0.032t}=\frac{3}{283}\approx0.0106 \)

Step3: Take the natural logarithm of both sides

\( \ln(e^{-0.032t})=\ln(0.0106) \), so \( - 0.032t=\ln(0.0106) \)

Step4: Solve for \( t \)

\( t=\frac{\ln(0.0106)}{- 0.032}\approx\frac{-4.543}{-0.032}\approx142.0 \) (rounded to one decimal place)

Part (a) Answer: \( 7.6 \)

Part (b) Answer: \( 21.7 \)

Part (c) Answer: \( 142.0 \)